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![10 PHYSICS
For example : Force = mass × acceleration
= mass ×
1
v–u [LT ]
[M]
t [T]
-
= = [MLT
T–2]
Hence the dimensions of force are 1 in mass 1 in length and (– 2)
in time.
Dimensional Formula :
Unit of a physical quantity expressed in terms of M, L and T is
called dimensional formula. It shows how and which of the
fundamental quantities represent the dimensions.
For example, the dimensional formula of work is [ML2T–2]
Dimensional Equation :
When we equate the dimensional formula with the physical
quantity, we get the dimensional equation.
For example Work=[ML2T–2]
Classification of Physical Quantities (On the basis of dimensions) :
Physical
Quantity
Dimensional
physical quantity
Dimensional
constant
For e.g. c (velocity of light in vaccum)
For e.g. distance, displacement, force, mass, time etc.
For e.g. 0, 1, 2, ....., e, , sin , cos , tan , etc.
q
p q q
For e.g. plane angle, solid angle, strain,
refractive index, dielectric constant, relative density,
specific gravity, poisson's ratio etc.
R(universal gas constant),
stefan's constant), h (Planck's constant),
k (Boltzmann constant), G (universal gravitational constant) etc.
s(
Dimensionless
constant
Dimensional
variable
Dimensionless
variable
Dimensionless
physical quantity
Dimensional Formula of Some Important Physical Quantities :
S.No. Physical quantity Relation with other quantities Dimensional formula
1. Velocity(v)
Length
Time
[M0LT–1]
2. Acceleration (a)
Velocity
Time
[M0LT–2]
3. Momentum (p) Mass × velocity [MLT–1]
4. Force (F) Mass × acceleration [MLT–2]
5. Work Force × displacement [ML2T–2]
6. Power (P) Work
Time
[ML2T–3]
7. Universal gravitational constant
2
1
2
m
m
Fr
G = [M–1L3T–2]
8. Torque F
r ´
=
t [ML2T–2]
9. Surface tension
F
S =
l
[MT–2]
10. Gravitational potential G
W
V
m
= [M0L2T–2]
11. Coefficient of viscosity
dx
dv
A
F
=
h [ML–1T–1]
12. Impulse Force×time(F×t) [MLT–1]](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-14-320.jpg)
![11
Units and Measurements
13. Strain
Changein length L
Originallength L
D
æ ö
ç ÷
è ø
[M0L0T0]
14. Pressure gradient
Pressure P
Distance
æ ö
ç ÷
è ø
l
[ML–2T–2]
15. Plane angle
Arc s
Radius of circle r
æ ö
ç ÷
è ø
[M0L0T0]
16. Angular velocity
Angle
Time t
q
æ ö
ç ÷
è ø
[M0L0T–1]
17. Radius of gyration
i
Moment of inertia of body I
Total mass of the body m
æ ö
ç ÷
ç ÷
å
è ø
[M0L1T0]
18. Moment of force, moment of couple Force × distance (F × s) [ML2T–2]
19. Angular frequency 2p×frequency(2pn) [M0L0T–1]
20. Pressure
Force
Area
[ML–1T–2]
21. Efficiency
Output work or energy W
Input work or energy Q
æ ö
ç ÷
è ø
[M0L0T0]
22. Angular impulse forque×time (t × t) [ML2T–1]
23. Planck’s constant
Energy E
Frequency
æ ö
ç ÷
n
è ø
[ML2T–1]
24. Heat capacity, Entropy
Heat energy Q
Temperature T
æ ö
ç ÷
è ø
[ML2T–2K–1]
25. Specific heat capacity
Heat energy Q
Mass temperature m T
æ ö
ç ÷
´ ´D
è ø [M0L2T–2K–1]
26. Thermal conductivity, K
Heat energy thickness Q T
KA
Area temperature time t x
´ D
æ ö
= -
ç ÷
´ ´ D
è ø
[MLT–3K–1]
27. Thermal Resistance, R
Length
Thermal conductivity × area
[M–1L–2T3K]
28. Bulk modulus (B) or (compressibility)–1
Volume (changein pressure) P
V
Change in volume V
´ D
æ ö
-
ç ÷
D
è ø
[ML–1T–2]
29. Stefan’s constant (s)
4
4 4
Q AtT
(Energy/area)
Time (temperature) E Q/ A.t T
æ ö
= s
ç ÷
ç ÷
´ = = s
è ø
[ML0T–3K–4]
30. Universal gas constant R
Pressure volume PV
Mole temperature nT
´ æ ö
ç ÷
´ è ø
]
mol
K
T
ML
[ 1
1
2
2 -
-
-
31. Voltage, electric potential (V) or
Work W
Charge q
æ ö
ç ÷
è ø
[ 1
3
2
A
T
ML -
- ]
electromotive force (e)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-15-320.jpg)
![12 PHYSICS
32. Capacitance (C)
Charge q
Potentialdifference V
æ ö
ç ÷
è ø
[ 2
4
2
1
A
T
L
M -
- ]
33. Electricfield ( )
E
r Electric force F
Charge q
æ ö
ç ÷
è ø
[ 1
3
A
MLT -
- ]
34. Magnetic field(B)
r
, magnetic induction,
Force
[F I Bsin ]
Current length
= q
´
l [ML0T–2A–1]
magnetic flux density
35. Magnetic flux (fm) f = BAcosq [ML2T–2A–1]
36. Inductance
m
Magneticflux
Current I
f
æ ö
ç ÷
è ø
[ 2
2
2
A
T
ML -
- ]
coefficient of self inductance (L) or
coefficient of mutual inductance (M)
37. Magnetic field strength or
magnetic moment density(I)
Magneticmoment M
Volume V
æ ö
ç ÷
è ø
[M0L–1T0A]
38. Permittivity constant in free space eo
2 2
2 2
(Charge) q
4π electrostaticforce(distance) 4 F r
æ ö
ç ÷
ç ÷
´ p´ ´
è ø
1 3 4 2
M L T A
- -
é ù
ë û
39. Faraday constant (F), charge Avagadro constant × elementry charge ]
mol
TA
L
M
[ 1
0
0 -
40. Mass defect, (Dm) (Sum of masses of nucleons 0 0
[ML T ]
– mass of nucleus)(MP+MN –Mnucleus)
41. Resonant frequency (fr)
r
1
T ]
T
L
M
[ 1
0
0 -
42. Power of lens
-1 1
(Focal length)
f
æ ö
ç ÷
è ø
0 1 0
M L T
-
é ù
ë û
43. Refractive index
Speedof lightin vacuum
Speedof lightin medium
0 0 0
[M L T ]
44. Wave number
2
Wavelength
p 0 1 0
M L T
-
é ù
ë û
45. Binding energy of nucleus Mass defect × (speed of light in vacuum)2 2 2
ML T-
é ù
ë û
46. Conductance (c)
1
Resistance
1 2 2 3
M L A T
- -
é ù
ë û
47. Fluid flow rate
4
(Pressure) (radius)
8 (Viscosity coefficient) (length)
p ´
æ ö
ç ÷
è ø ´
0 3 –1
M L T
é ù
ë û
48. Inductive reactance (Angular frequency× inductance)
2 3 2
ML T A
- -
é ù
ë û
49. Capacitive reactance (Angular frequency × capacitance)–1 2 3 2
ML T A
- -
é ù
ë û
50. Magnetic dipole moment
Torque
or Current area
Magneticfield
´
0 2 0
M L T A
é ù
ë û](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-16-320.jpg)
![13
Units and Measurements
Short cuts / Time saving techniques
1. To find dimensions of a typical physical quantity which is
involved in a number of formulae, try to use that formula
which is easiest for you. For example if you want to find the
dimensional formula of magnetic induction then you can use
the following formulae
q
=
t
=
m
=
q
p
m
= sin
MB
,
qvB
F
,
nI
B
,
r
sin
Idl
4
dB 0
2
0
Out of these the easiest is probably the third one.
2. Ifyou have tofind the dimensional formula of a combination
of physical quantities, then instead of finding the
dimensional formula ofeach, tryto correlate the combination
ofphysical quantities with a standard formula. For example,
if you have to find the dimension of CV2, then try to use
formula
2
CV
2
1
E = where E is energyofa capacitor.
.
3. =
=
e
m
c
1
0
0
velocityof light in vacuum
• Dimensions ofthe following are same
2
2
2
2
LI
t
R
V
c
q
CV
qV
nRT
PV
Work =
´
=
=
=
=
=
=
[ML2T–2]
• Dimensions ofthe following are same
Force = Impulse / time
= q v B = q E
= Thrust
= weight = energygradient [MLT–2]
• The dimension of RC =
R
L
is same as that of time
• Dimensions ofthe following are same
velocity =
0 0
T 1
f
= = ´l
m m e
[M°LT
T–1]
• Dimensions ofthe following are same
Frequency
R k MB 1
L m I LC
= = = = [M°L°T–1]
• Dimensions ofthe following are same
(E) Modulus of elasticity =Y (Young's modulus)
= B (Bulk modulus)
= h (Modulus of rigidity)
= Stress
= Pressure =
ility
Compressib
1
[ML–1T–2]
• Dimensions ofthe following are same
Acceleration, retardation, centripetal acceleration, centrifugal
acceleration, gravitational intensity/strength. [M°LT–2]
• Dimensions ofthe following are same
Water equivalent, thermal capacity, entropy, Boltzmann's
constant. [ML2T–2K–1]
Keep in Memory
The dimensional formula of
• all trigonometricratio is [M0L0T0]
• x in ex is [M0L0T0]
• ex is [M0L0T0]
• x in log x is [M0L0T0]
• log x is [M0L0T0]
Example1.
Find out the unit and dimensions of permittivity of free
space.
Solution :
According to Coulomb’s law
2 2
2
1 3 4 2
0 2 2 2
A T
[q ]
[M L T A ]
4 [F][r ] [MLT ][L ]
- -
-
é ù
ë û
e = = =
p
Its unit =
2 2
2
coulomb (coulomb) coulomb
joule metre volt metre
newton metre
= =
´ ´
´
CV–1m–1
Example2.
Find out the unit and dimensions of coefficient of self or
mutual inductance of.
Solution :
e = ÷
ø
ö
ç
è
æ
dt
d
L
I
or ÷
ø
ö
ç
è
æ
dt
d
M
I
, where e is induced electromotive
force(e.m.f.)
dt W t
L e
d q
æ ö æ ö
= =
ç ÷ ç ÷
è ø è ø
I I
or [L] = ]
A
T
ML
[
]
A
][
AT
[
]
T
][
T
ML
[ 2
2
2
2
2
-
-
-
=
Its unit is volt × sec/amp or ohm × sec or henry.
Example3.
Find out the unit and dimensions of magnetic field intensity .
Solution :
As B = mH, hence
2
r
sin
d
4
1
B
H
q
p
=
m
=
l
I
;
1
2
[A][L]
H [M L T A]
[L ]
-
= = ° °
Its unit is ampere /metre in SI system. In c.g.s. system, the
unit is oersted.
Example4.
Find out the unit and dimensions of magnetic permeability
of free space or medium.
Solution :
According to Biot-Savart’s law
2
0
r
sin
d
4
B
q
p
m
=
l
I
and F = BI l sinq
or 0
2
d sin
F
B
sin 4 r
m q
= =
q p
l
l
I
I
;](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-17-320.jpg)
![14 PHYSICS
2
0 2 2
Fr
or
m m =
l
I
(dimensionally)
Hence [m] or ]
A
MLT
[
]
L
][
A
[
]
L
][
MLT
[
]
[ 2
2
2
2
2
2
0
-
-
-
=
=
m
Its units are
2
2 2 2 2
N m newton joule/ metre volt coulomb
amp amp metre
amp m amp amp
- ´
= = =
´ ´
-
ohm sec henry tesla metre
metre metre amp.
´ ´
= = =
Example5.
The dimensions of physical quantity X in the equation
Force =
X
Density
is given by
(a) [ML4T–2] (b) [M2L–2T–1]
(c) [M2L–2 T–2] (d) [ML–2T–1]
Solution :
(c) Q Force =
X
denisty
X = Force × density
Hence X has dimensions
[MLT–2] 3
[M]
[L ]
= [M2L–2T–2]
Example6.
If force, acceleration and time are taken as fundamental
quantities, then the dimensions of length will be-
(a) [FT2] (b) [F–1A2T–1]
(c) [FA2T] (d) [AT2]
Solution :
(d) x y z
L F A T
=
0 1 0 2 x 2 y z
M L T [MLT ] [LT ] T
- -
=
= x x y 2x 2y z
M L T
+ - - +
x = 0, x + y= 1, – 2x – 2y+z = 0
x =0, y= 1, z = 2
Hence, 2
L AT
=
DIMENSIONAL ANALYSIS AND ITS APPLICATIONS
Principle of Homogeneity:
Only those physical quantities can be added /subtracted/equated
/compared which have the same dimensions.
Uses of Dimensions :
(1) Conversion of one systemof unit intoanother
Example : Convert a pressure of 106 dyne/cm2 in S.I units.
Sol. We know that 1N = 105 dyne Þ 1 dyne = 10–5 N
Also1m = 100 cm Þ 1cm = 10–2 m
Now, the pressure 106 dyne/cm2 in SI unit is
106 2
5
2
2
–
5
6
m
/
N
10
m
10
m
10
N
10
10
cm
cm
dyne
=
´
´
=
´ -
-
(2) Checking the accuracy of various formulae
Example : Check the correctness of the following
equation dimensionally
dv
F = ηA sinθ
dx
where F=force, h =coefficientofviscosity,
,
A = area,
dv
= velocity
dx
gradient w.r.t distance, q = angle of
contact
Sol. L.H.S = force = [MLT–2]
R.H.S=
1
1 1 1 2 2
dv LT
A (sin ) M L T L MLT
dx L
-
- - -
é ù
h q = ´ =
ë û
The equation is dimensionallycorrect.
CAUTION : Please note that the above equation is not correct
numerically. The conclusion is that an equation, if correct
dimensionally, may or may not be numerically correct. Also
remember that if an equation is dimensionallyincorrect, we can
conclude with surety that the equation is incorrect.
(3) Derivationof formula
Example : The air bubble formed by explosion inside water
performed oscillation with time period T which is directly
proportional to Pa db Ec where P is pressure, d is density and E
is the energy due to explosion. Find the values of a, b and c.
Sol. Let us assume that the required expression for time period is
T = K Pa db Ec
where K is a dimensionless constant.
Writing dimensions on both sides,
c
2
2
b
3
a
2
1
1
0
0
]
T
ML
[
]
ML
[
]
T
ML
[
]
T
L
M
[ -
-
-
-
=
]
T
[
]
T
[
]
L
[
]
M
[ 1
c
2
a
2
c
2
b
3
a
c
b
a
=
= -
-
+
-
-
+
+
Equating the powers,
a + b + c = 0 ....(1)
0
c
2
b
3
a =
+
-
- ....(2)
– 2a – 2c = 1 ....(3)
Solving these equations, we get,
a =
6
5
- , b =
2
1
, c =
3
1
.
Limitationsof DimensionalAnalysis:
(1) Noinformation aboutthedimensionlessconstant isobtained
during dimensional analysis
(2) Formula cannot be found if a physical quantityis dependent
on more than three physical quantities.
(3) Formula containing trigonometrical /exponential function
cannot be found.
(4) Ifan equation is dimensionally correct it mayor may not be
absolutely correct.
Example7.
Find the dimensions of a and b in the Van der waal's
equation
æ ö
ç ÷
è ø
2
a
P+ (V - b) = RT
V
where P is pressure and V is volume of gas.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-18-320.jpg)
![15
Units and Measurements
Solution :
Dimensionally
2
a
P [Byprincipleof homogeneity]
V
=
]
T
[ML
a
V
a
T
ML 2
5
2
2
1 -
-
-
=
Þ
=
Þ
AlsodimensionallyV=b [Byprinciple ofhomogeneity]
b= [L3]
Example8.
The formula
2
2
3mg (1 cos )
v
M (1 sin )
- q
=
+ q
l
is obtained as the
solution of a problem. Use dimensions to find whether
this is a reasonable solution (v is a velocity, m and M are
masses, l is a length and g is gravitational acceleration).
Solution :
Dimensions of L.H.S. = [LT–1]2 = [L2T–2]
Dimensions of R.H.S. =
2
2 2
[M][LT ][L]
[L T ]
[M]
-
-
=
Hence formula is reasonable.
Example9.
In the formula; 2 1
2 1
n n
N D
x x
é ù
-
= - ê ú
-
ë û
,
D = diffusion coefficient, n1 and n2 isnumber of molecules
in unit volume along x1 and x2 which represents distances
where N is number of molecules passing through per unit
area per unit time calculate the dimensions of D.
Solution :
By homogeneity theory of dimension
Dimension of (N)
= Dimension of D ×
2 1
2 1
dimension of (n n )
dimension of (x x )
-
-
2
1
L T
= Dimension of D ×
3
L
L
-
Þ Dimensions of 'D' = 3 2
L
L L T
-
´
=
2
L
T
= [L2T–1]
Example10.
Let us consider an equation 2
1
mv mgh
2
= where m is the
mass of the body, v its velocity, g is the acceleration due to
gravity and h is the height. Check whether this equation is
dimensionally correct.
Solution :
The dimensions of LHS are
[M] [L T–1 ]2 = [M] [ L2 T–2] = [M L2 T–2]
The dimensions of RHS are
[M][L T–2] [L] = [M][L2 T–2] = [M L2 T–2]
The dimensions of LHS and RHS are the same and hence
the equation is dimensionallycorrect.
Example11.
A calorie is a unit of heat or energy and it equals about 4.2
J where 1J = 1 kg m2 s–2. Suppose we employ a system of
units in which the unit of mass equals a kg, the unit of
length equals b m, the unit of time isg s. Show that a calorie
has a magnitude 4.2 a–1 b –2 g2 in terms of the new units.
Solution :
1 cal = 4.2 kg m2 s–2.
SI system New system
n1 = 4.2 n2 = ?
M1 = 1 kg M2 = a kg
L1 =1m L2 = b metre
T1 = 1s T2 = g second
Dimensional formula of energyis [ML2T–2]
Comparing with [MaLbTc], we find that
a = 1, b = 2, c = – 2
Now,
a b c
1 1 1
2 1
2 2 2
M L T
n n
M L T
é ù é ù é ù
= ê ú ê ú ê ú
ë û ë û ë û
=
1 2 2
1kg 1m 1s
4.2
kg m s
-
é ù é ù é ù
ê ú ê ú ê ú
a b g
ë û ë û
ë û
= 4.2 a–1 b –2 g2
SIGNIFICANT FIGURES
The number of digits, which are known reliably in our
measurement, and one digit that is uncertain are termed as
significant figures.
Rules to determine the numbers of significant figures:
1. All non-zerodigits are significant. 235.75has fivesignificant
figures.
2. All zeroes between two non-zero digits are significant.
2016.008 has seven significant figures.
3. All zeroes occurring between the decimal point and the non-
zero digits are not significant. provided there is only a zero
to left of the decimal point. 0.00652 has three significant
figures.
4. All zeroes written to the right of a non-zero digit in a number
written without a decimal point are not significant. This rule
does not work if zero is a result of measurement. 54000 has
two significant figures whereas 54000m has five significant
figures.
5. All zeroes occurring to the right of a non-zero digit in a
number written with a decimal point are significant. 32.2000
has six significant figures.
6. When a number is written in the exponential form, the
exponential term does not contribute towards the significant
figures. 2.465 × 105 has four significant figures.
Keep in Memory
1. The significant figures depend upon the least count of the
instrument.
2. The number of significant figure does not depend on the
units chosen.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-19-320.jpg)
![20 PHYSICS
1. Temperature can be expressed as derived quantityin terms
of
(a) length and mass
(b) mass and time
(c) length, mass and time
(d) None of these
2. What is the unit of “a” in Vander Waal’s gas equation?
(a) Atm litre–2 mol2 (b) Atm litre2 per mol
(c) Atm litre–1 mol2 (d) Atm litre2 mol–2
3. Random error can be eliminated by
(a) careful observation
(b) eliminating the cause
(c) measuring the quantitywith more than one instrument
(d) taking large number of observations and then their
mean.
4. Ifeisthecharge, V thepotentialdifference, T thetemperature,
then the units of
eV
T
are the same as that of
(a) Planck’s constant (b) Stefan’s constant
(c) Boltzmann constant (d) gravitational constant
5. If f = x2, then the relative error in f is
(a)
x
x
D
2
(b)
x
x 2
)
(D
(c)
x
x
D
(d) (Dx)2
6. Two quantities A and B have different dimensions which
mathematical operation given below is physically
meaningful?
(a) A/B (b) A+ B
(c) A – B (d) A= B
7. Which of the following systems of units is not based on
units of mass, length and time alone
(a) SI (b) MKS
(b) CGS (d) FPS
8. Unit of latent heat is
(a) J Kg–1 (b) J mol–1
(c) N Kg–1 (d) N mol–1
9. Dyne-sec is the unit of
(a) momentum (b) force
(c) work (d) angular momentum
10. Illuminance of a surface is measured in
(a) Lumen (b) candela
(c) lux (d) luxm–2
11. SI unit of electric polarisation is
(a) Cm–2 (b) coulomb
(c) ampere (d) volt
12. The SI unit of coefficient of mutual inductance of a coil is
(a) henry (b) volt
(c) farad (c) weber
13. Light year is
(a) light emitted by the sun in one year.
(b) time taken by light to travel from sun to earth.
(c) the distance travelled bylight in free space in one year.
(d) time taken by earth to go once around the sun.
14. The SI unit of pressure is
(a) atmosphere (b) bar
(c) pascal (d) mmofHg
15. Electron volt is a unit of
(a) potential difference (b) charge
(c) energy (d) capacity
16. Dimensions of impulse are
(a) ]
T
ML
[ 1
- (b) ]
MLT
[ 2
(c) ]
MT
[ 2
- (d) ]
T
ML
[ 3
1 -
-
17. The S.I. unit of pole strength is
(a) Am2 (b) Am
(c) A m–1 (d) Am–2
18. Which is dimensionless?
(a) Force/acceleration (b) Velocity/acceleration
(c) Volume/area (d) Energy/work
19. Potential is measured in
(a) joule/coulomb (b) watt/coulomb
(c) newton-second (d) None of these
20. Maxwell is the unit of
(a) magnetic susceptibility
(b) intensity of Magnetisation
(c) magneticFlux
(d) magnetic Permeability
21. The mass of the liquid flowing per second per unit area of
cross-section of the tube is proportional to (pressure
difference across the ends)n and (average velocity of the
liquid)m. Which of the following relations between m and n
is correct?
(a) m= n (b) m = – n
(c) m2 = n (d) m = – n2
22. Which of the following is a derived physical quantity?
(a) Mass (b) Velocity
(c) Length (d) Time
23. N kg–1 is the unit of
(a) velocity (b) force
(c) acceleration (d) None of these
24. Which physical quantities have same dimensions?
(a) Moment of couple and work
(b) Force and power
(c) Latent heat and specific heat
(d) Work and power
25. The expression [ML–1 T–2] does not represent
(a) pressure (b) power
(c) stress (d) Young’s modulus](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-24-320.jpg)
![21
Units and Measurements
1. What are the units of magnetic permeability?
(a) Wb A–1 m–1 (b) Wb–1 Am
(c) Wb A m–1 (d) Wb A–1 m
2. The dimensions of pressure gradient are
(a) [ML–2 T–2] (b) [ML–2 T–1]
(c) [ML–1 T–1] (d) [ML–1 T–2]
3. The dimensions of Rydberg’s constant are
(a) [M0 L–1 T] (b) [MLT–1]
(c) [M0 L–1 T0] (d) [ML0 T2]
4. The dimensions of universal gas constant are
(a) [L2 M1 T–2 K–1] (b) [L1 M2 T–2 K–1]
(c) [L1 M1 T–2 K–1] (d) [L2 M2 T–2 K–1]
5. The dimensions of magnetic moment are
(a) [L2 A1] (b) [L2 A–1]
(c) [L2 / A3] (d) [LA2]
6. The dimensions of Wien’s constant are
(a) [ML0 T K] (b) [M0 LT0 K]
(c) [M0 L0 T K] (d) [MLTK]
7. The unit and dimensions ofimpedance in terms of charge Q
are
(a) mho, [ML2 T–2 Q–2] (b) ohm, [ML2 T–1 Q–2]
(c) ohm, [ML2 T–2 Q–1] (d) ohm, [MLT–1 Q–1]
8. If L denotes the inductance of an inductor through which a
current i is flowing, the dimensions of L i2 are
(a) [ML2 T–2]
(b) [MLT–2]
(c) [M2 L2 T–2]
(d) Not expressiblein M, L, T
9. The dimensional formula ofwave number is
(a) [M0 L0 T–1] (b) [M0 L–1 T0]
(c) [M–1 L–1 T0] (d) [M0 L0 T0]
10. The period of a bodyunder S.H.M. is represented by: T = Pa
Db Sc where P is pressure, D is density and S is surface
tension, then values of a, b and c are
(a) 1
,
2
1
,
2
3
- (b) 3
,
2
,
1 -
-
(c)
2
1
,
2
3
,
2
1
-
- (d) 1, 2 1/3
11. The time of oscillation T of a small drop of liquid depends
on radius r, density r and surface tension S. The relation
between them is given by
(a)
3
r
S
T
r
µ (b)
S
r
T
3
r
µ
(c)
r
µ
3
2
r
S
T (d)
S
r
T
3
r
µ
12. Thepotential energyofaparticlevarieswith distance x from
a fixed origin as
B
x
x
A
V
+
= where A and B are constants.
The dimensions ofAB are
(a) 1 5/2 2
[M L T ]
- (b) 1 2 2
[M L T ]
-
(c) 3/ 2 5/2 2
[M L T ]
- (d) 1 7 /2 2
[M L T ]
-
13. Distance travelled by a particle at any instant ‘t’ can be
represented as S =A(t + B) + Ct2. The dimensions of B are
(a) 0 1 1
[M L T ]
- (b) 0 0 1
[M L T ]
(c) 0 1 2
[M L T ]
- - (d) 0 2 2
[M L T ]
-
14. The deBroglie wavelength associated with a particle of mass
m and energy E is h / 2m E . The dimensional formula of
Planck’s constant h is
(a) ]
T
L
M
[ 2
2
2 - (b) ]
T
L
M
[ 1
2 -
(c) ]
LT
M
[ 2
- (d) ]
T
L
M
[ 2
2 -
15. The velocity of a body which falls under gravity varies as ga
hb, where g is acc. due to gravity and h is the height. The
values of a and b are
(a) a = 1, b = 1/2 (b) a = b = 1
(c) a = 1/2, b= 1 (d) a = 1/2; b= 1/2
16. The velocityvofa particle at time t is given by
b
v a t
t c
= +
+
The dimensions of a, b c are respectively
(a) [LT–2], [L], [T] (b) [L2], [T] and [LT2]
(c) [LT2], [LT] and [L] (d) [L], [LT] and [T2]
17. The dimensions of Hubble’s constant are
(a) ]
T
[ 1
- (b) ]
T
L
M
[ 2
0
0 -
(c) ]
MLT
[ 4 (d) 1
[MT ]
-
18. Error in the measurement of radius ofa sphere is 1%. Then
error in the measurement of volumeis
(a) 1% (b) 5%
(c) 3% (d) 8%
19. Subtract 0.2 J from 7.26 J and express the result with correct
number of significant figures.
(a) 7.1J (b) 7.06J
(c) 7.0J (d) 7 J
20. Multiply107.88 by0.610 and express the result with correct
number of significant figures.
(a) 65.8068 (b) 65.807
(c) 65.81 (d) 65.8](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-25-320.jpg)
![22 PHYSICS
21. When 97.52 is divided by2.54, the correct result is
(a) 38.3937 (b) 38.394
(c) 38.39 (d) 38.4
22. Relative density of a metal may be found with the help of
spring balance. In air the spring balancereads (5.00 ± 0.05)
N and in water it reads (4.00 ± 0.05) N. Relative density
would be
(a) (5.00±0.05)N (b) (5.00±11%)
(c) (5.00±0.10) (d) (5.00± 6%)
23. Area of a square is (100 ± 2) m2. Its side is
(a) (10±1)m (b) (10±0.1)m
(c) m
)
2
10
( ± (d) %
2
10 ±
24. Let Q denote the charge on the plate of a capacitor of
capacitance C. The dimensional formula for
C
Q2
is
(a) ]
T
M
L
[ 2
2 (b) ]
LMT
[ 2
(c) ]
MT
L
[ 2
2 - (d) ]
T
M
L
[ 2
2
2
25. IfL and Rdenote inductance and resistance then dimension
of L/R is
(a) ]
T
L
M
[ 0
0
0 (b) ]
T
L
M
[ 0
0
(c) ]
T
L
M
[ 2
0
2 (d) ]
MLT
[ 2
26. The dimensional formula of current densityis
(a) 0 2 1
[M L T Q]
- -
(b) ]
Q
T
L
M
[ 1
1
2
0 -
(c) ]
Q
MLT
[ 1
-
(d) ]
Q
T
ML
[ 2
1
2 -
-
27. The least count ofa stop watch is 0.2 second. The time of20
oscillations of a pendulum is measured to be 25 second.
The percentage error in the measurement of time will be
(a) 8% (b) 1.8%
(c) 0.8% (d) 0.1%
28. The dimensional formula for relative densityis
(a) [M L–3] (b) [M0 L–3]
(c) [M0 L0 T–1] (d) [M0 L0 T0]
29. The solid angle sustended by the total surface area of a
sphere, at the centre is
(a) 4p (b) 2p
(c) p (d) 3p
30. If C and L denote the capacitance and inductance, the
dimensions of LC are
(a) [M0 L0 T–1] (b) [M0 L–1 T0]
(c) [M–1 L–1 T0] (d) [M0 L0 T2]
31. The dimensions of solar constant is
(a) [M0 L0 T0] (b) [MLT–2]
(c) [ML2 T–2] (d) MT–3
32. The dimensions of
hc
e
1 2
o
Î
are
(a) M–1 L–3 T4 A2 (b) ML3 T–4 A–2
(c) M0 L0 T0 A0 (d) M–1 L–3 T2 A
33. The dimensional formula for entropyis
(a) [MLT–2 K1] (b) [ML2 T–2]
(c) [ML2 T–2 K–1] (d) [ML2 T–2 K]
34. Dimensions of specific heat are
(a) [ML2 T–2 K] (b) [ML2 T–2 K–1]
(c) [ML2 T2 K–1] (d) [L2 T–2 K–1]
35. L, C, Rrepresent physical quantities inductance, capacitance
and resistance respectively. The combinations which have
the dimensions of frequency are
(a) 1/RC (b) R/L
(c) LC
/
1 (d) C/L
36. The physical quantity which has the dimensional formula
[M1T–3] is
(a) surface tension (b) solar constant
(c) density (d) compressibility
37. Which of the following is the most accurate?
(a) 200.0m (b) 20 ×101 m
(c) 2 × 102 m (d) data is inadequate
38. The velocity of water waves (v) may depend on their
wavelength l, the density of water r and the acceleration
due to gravity, g. The method of dimensions gives the
relation between these quantities is
(a) v (b) 2
v g
µ l
(c) 2 2
v g
µ l (d) 2 1 2
v g-
µ l
39. The time dependence of a physical quantity p is given by
p = p0 exp. (– a t2), where a is a constant and t is the time.
The constant a
(a) is dimensionless (b) has dimensions T–2
(c) has dimensions T2 (d) has dimensions of p.
40. In the eqn. 2
a
P (V b) constant,
V
æ ö
+ - =
ç ÷
è ø
the unit of a is
(a) dyne × cm5 (b) dyne × cm4
(c) dyne/cm3 (d) dyne × cm2
41. Dimensions of ‘ohm’ are same as (where h is Planck’s
constant and e is charge)
(a)
e
h
(b)
e
h2
(c) 2
e
h
(d) 2
2
e
h
42. The Richardson equation is given by I = AT2e–B/kT. The
dimensional formula for AB2 is same as that for
(a) I T2 (b) kT
(c) I k2 (d) I k2/T
43. The unit of current in C.G..S. system is
(a) 10A (b) 1/10A
(c) 1/100A (d) 1/1000A](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-26-320.jpg)
![23
Units and Measurements
44. Which of the following do not have the same dimensional
formula as the velocity?
Given that m0 = permeabilityof free space, e0 = permittivity
offree space, n = frequency, l = wavelength, P= pressure, r
= density, w = angular frequency, k = wave number,
(a) 0 o
1 m e (b) nl
(c) P/r (d) wk
45. A cubehas numericallyequal volume and surface area. The
volume of such a cube is
(a) 1000unit (b) 200unit
(c) 216unit (d) 300unit
46. A spherical body ofmass m and radius r is allowed tofall in
a medium of viscosity h. The time in which the velocity of
the body increases from zero to 0.63 times the terminal
velocity(v) is called time constant t. Dimensionallytcan be
represented by
(a)
h
p
6
mr2
(b) ÷
÷
ø
ö
ç
ç
è
æ h
p
2
g
r
m
6
(c)
m
6 r v
ph
(d) None of these
47. A quantity is represented by X = Ma Lb Tc. The % error in
measurement of M, L and T are a%, b% and g%
respectively. The % error in X would be
(a) %
)
c
b
a
( g
+
b
+
a (b) %
)
c
b
a
( g
+
b
-
a
(c) %
100
)
c
b
a
( ´
g
-
b
-
a (d) None of these
48. In a Vernier calliper, N divisions of vernier scale coincide
with (N–1) divisions of main scale (in which one division
represents 1 mm). the least count of the instrument in cm.
should be
(a) N (b) N – 1
(c)
N
10
1
(d)
1
N
1
-
49. What is the fractional error in g calculated from
g
/
2
T l
p
= ? Given fraction errors in T and l are ± x and
± y respectively.
.
(a) x+ y (b) x – y
(c) 2x + y (d) 2x – y
50. Conversion of 1 MW power in a New system of units having
basic units of mass, length and time as 10 kg, 1 dm and 1
minute respectively is
(a) 2.16 × 1010 unit (b) 2 × 104 unit
(c) 2.16 × 1012 unit (d) 1.26 × 1012 unit
51. A resistor of 10 k W having tolerance 10% is connected in
series with another resistor of20 k W having tolerance 20%.
The tolerance of the combination will be
(a) 10% (b) 13%
(c) 30% (d) 20%
52. Using mass (M), length(L), time (T) and electric current (A)
as fundamental quantities the dimensions of permittivity
will be
(a) [MLT–1A–1] (b) [MLT–2A–2]
(c) [M–1L–3T+4A2] (d) [M2L–2T–2A2]
53. Thepercentageerrors in the measurement ofmass and speed
are 2% and 3% respectively. The error, in kinetic energy
obtained by measuring mass and speed, will be
(a) 12% (b) 10%
(c) 8 % (d) 2 %
54. The density of a cube is measured by measuring its mass
and length of its sides. If the maximum error in the
measurement ofmass and length are4% and3% respectively,
themaximum error in themeasurement of densitywill be
(a) 7% (b) 9%
(c) 12% (d) 13%
55. The speed of sound in a gas is given by
RT
v
M
g
=
R = universal gas constant,
T = temperature
M = molar mass of gas
The dimensional formula of g is
(a) ]
T
L
M
[ 0
0
0
(b) ]
LT
M
[ 1
0 -
(c) ]
MLT
[ 2
- (d) ]
T
L
M
[ 1
0
0 -
56. Specific gravity has ............ dimensions in mass, ............
dimensions in length and ............ dimensions in time.
(a) 0,0, 0 (b) 0,1, 0
(c) 1,0, 0 (d) 1,1, 3
57. If I is the moment of inertia and w the angular velocity,
,
what is the dimensional formula ofrotational kinetic energy
2
I
2
1
w ?
(a) ]
T
ML
[ 1
2 -
(b) ]
T
L
M
[ 2
1
2 -
-
(c) ]
T
ML
[ 2
2 - (d) ]
T
L
M
[ 2
1
2 -
-
58. Given that r = m2 sin pt , where t represents time. If the unit
of m is N, then the unit of r is
(a) N (b) 2
N
(c) Ns (d) s
N2
59. The dimensional formula of farad is
(a) ]
TQ
L
M
[ 2
1 -
-
(b) 1 2 2 2
[M L T Q ]
- -
(c) ]
TQ
L
M
[ 2
2
1 -
- (d) ]
Q
T
L
M
[ 2
2
1 -
-
60. If time T, acceleration A and force F are regarded as base
units, then the dimensional formula of work is
(a) [FA] (b) ]
FAT
[
(c) ]
FAT
[ 2 (d) ]
T
FA
[ 2](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-27-320.jpg)
![24 PHYSICS
61. Turpentine oil is flowing through a capillary tube of length
l and radius r. The pressure difference between the two
ends of the tube is p. The viscosity of oil is given by :
2 2
p(r x )
4v
-
h =
l
. Here vis velocityof oil at a distance x from
the axis of the tube. From this relation, the dimensional
formula of h is
(a) ]
T
ML
[ 1
1 -
- (b) ]
MLT
[ 1
-
(c) ]
T
ML
[ 2
2 -
(d) ]
T
L
M
[ 0
0
0
62. The dimensional formula of velocity gradient is
(a) ]
T
L
M
[ 1
0
0 -
(b) ]
MLT
[ 1
(c) 0 1
[ML T ]
- (d) ]
LT
M
[ 2
0 -
63. The thrust developed by a rocket-motor is given by
)
P
P
(
A
mv
F 2
1 -
+
= where m is the mass of the gas ejected
per unit time, v is velocity of the gas, A is area of cross-
section of the nozzle, 1
P and 2
P are the pressures of the
exhaust gas and surrounding atmosphere. The formula is
dimensionally
(a) correct
(b) wrong
(c) sometimes wrong, sometimes correct
(d) Data is not adequate
64. IfE, m, J and G represent energy, mass, angular momentum
and gravitational constant respectively, then the
dimensional formula of 2
5
2
G
m
/
EJ is
(a) angle (b) length
(c) mass (d) time
65. In a vernier callipers, ten smallest divisions of the vernier
scale are equal tonine smallest division on the main scale. If
the smallest division on the main scale is half millimeter,
then the vernier constant is
(a) 0.5mm (b) 0.1mm
(c) 0.05mm (d) 0.005mm
66. Avernier calliper has20 divisions on thevernier scale, which
coincide with 19 on the main scale. The least count of the
instrument is 0.1 mm. The main scale divisions are of
(a) 0.5mm (b) 1mm
(c) 2mm (d)
4
1
mm
67. The pitch of the screw gauge is 0.5 mm. Its circular scale
contains 50 divisions. The least count ofthe screw gauge is
(a) 0.001mm (b) 0.01mm
(c) 0.02mm (d) 0.025mm
68. If x = a – b, then the maximum percentage error in the
measurement ofx will be
(a) %
100
b
b
a
a
´
÷
ø
ö
ç
è
æ D
+
D
(b) %
100
b
b
a
a
´
÷
ø
ö
ç
è
æ D
-
D
(c) %
100
b
a
b
b
a
a
´
÷
ø
ö
ç
è
æ
-
D
+
-
D
(d) %
100
b
a
b
b
a
a
´
÷
ø
ö
ç
è
æ
-
D
-
-
D
69. The heat generated in a circuit is given by 2
I
Q = Rt, where
I is current, R is resistance and t is time. If the percentage
errorsin measuring I,Randt are2%,1%and1%respectively,
then themaximum error in measuring heat will be
(a) 2% (b) 3%
(c) 4% (d) 6%
70. The pressure on a square plate is measured by measuring
the force on the plate and length of the sides of the plate by
using the formula 2
F
P =
l
. If the maximum errors in the
measurement of force and length are 4% and 2%
respectively, then the maximum error in the measurement of
pressure is
(a) 1% (b) 2%
(c) 8% (d) 10%
71. In a simple pendulum experiment for the determination of
acceleration due to gravity, time period is measured with an
accuracy of 0.2% while length was measured with an
accuracy of 0.5%. The percentage accuracy in the value of
g so obtained is
(a) 0.25% (b) 0.7%
(c) 0.9% (d) 1.0%
72. The dimensions of a rectangular block measured with
callipershaving least count of0.01 cm are5 mm × 10mm × 5
mm. The maximum percentageerror in the measurement of
the volume of the block is
(a) 5% (b) 10%
(c) 15% (d) 20%
73. Consider the following pairs of quantities
1. Young's modulus; pressure
2. Torque; energy
3. Linear momentum; work
4. Solar day; light year.
In which cases are the dimensions, within a pair, same?
(a) 1 and 3 (b) 1 and 4
(c) 1 and 2 (d) 2 and 4
74. Which one of the following has the same dimension as that
oftime, ifRisresistance, Linductance and C is capacitance?
(a) RC (b) LC
(c)
R
L (d) All of the above
75. The equation of a wave is given by
x
y asin k
v
æ ö
= w -
ç ÷
è ø
, where w is angular velocity and v is
linear velocity. The dimensions of K will be
(a) [T2] (b) [T–1]
(c) [T] (d) [LT]](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-28-320.jpg)
![25
Units and Measurements
76. In the equation X = 3YZ2, X and Z are dimensions of
capacitance and magnetic induction respectively. In MKSQ
system, the dimensional formula for Y is
(a) [M–3 L–2 T–2 Q–4] (b) [M L–2]
(c) [M–3 L–2 Q4 T8] (d) [M–3 L–2 Q4 T4]
77. A force is given by F = at + bt2, where t is time, the
dimensions of a and b are
(a) [M L T–4] and [M L T–1]
(b) [M L T–1] and [M L T0]
(c) [M L T–3] and [M L T–4]
(d) [M L T–3] and [M L T0]
78. The frequency of vibration of a string is given by f =
L
2
n
T
m
, where T is tension in the string, L is the length, n is
number ofharmonics. The dimensional formula for m is
(a) [M0 L T] (b) [M1 L–1 T–1]
(c) [M1 L–1 T0] (d) [M0 L T–1]
79. The dimensions of voltage in terms of mass (M), length (L)
and time (T) and ampere (A) are
(a) [ML2T–2A–2] (b) [ML2T3A–1]
(c) [ML2T–3A1] (d) [ML2T–3A–1]
80. Suppose the kinetic energy of a body oscillating with
amplitude A and at a distance x is given by
2 2
Bx
K
x A
=
+
The dimensions of B are the same as that of
(a) work/time (b) work × distance
(c) work/distance (d) work× time
81. Thedimensions ofmagneticfield in M, L, T and C (coulomb)
are given as
(a) [MLT–1 C–1] (b) [MT2 C–2]
(c) [MT–1 C–1] (d) [MT–2 C–1]
82. Two full turns of the circular scale of a screw gauge cover a
distance of 1mm on its main scale. The total number of
divisions on the circular scale is 50. Further, it is found that
the screw gauge has a zero error of – 0.03 mm. While
measuring the diameter of a thin wire, a student notes the
main scale reading of 3 mm and the number of circulr scale
divisions in line with the main scale as 35. The diameter of
the wire is
(a) 3.32mm (b) 3.73mm
(c) 3.67mm (d) 3.38mm
83. If the dimensions of a physical quantity are given by
Ma Lb Tc, then the physical quantity will be
(a) velocity if a = 1, b = 0, c = – 1
(b) acceleration if a = 1, b = 1, c = – 2
(c) force if a = 0, b = – 1, c = – 2
(d) pressure if a = 1, b = – 1, c = – 2
84. The density ofa material in CGS system ofunits is 4g/cm3.
In a system ofunits in which unit of length is 10 cm and unit
of mass is 100 g, the value of densityof material will be
(a) 0.4 (b) 40
(c) 400 (d) 0.04
85. In an experiment four quantities a, b, c and d are measured
with percentage error 1%, 2%, 3% and 4% respectively.
Quantity P is calculated as follows
P =
3 2
a b
cd
% error in P is
(a) 10% (b) 7%
(c) 4% (d) 14%
86. What is the fractional error in g calculated from
g
/
2
T l
p
= ? Given fractional errors in T and l are ± x
and ± y respectively.
.
(a) x+ y (b) x – y
(c) 2x + y (d) 2x – y
87. The resistance R ofa wire is given bythe relation R = 2
.
r
r
p
l
Percentage error in the measurement ofr, l and r is1%, 2%
and 3% respectively. Then the percentage error in the
measurement ofR is :
(a) 6 (b) 9
(c) 8 (d) 10
88. What are the dimensions of permeability?
(a) [M1L1T1A–2] (b) [M1L1T–2A–2]
(c) [M2L2T1A0] (d) [M1L2T2A–2]
89. The physical quantity having the dimensions
[M–1L–3T3A2] is
(a) resistance (b) resistivity
(c) electrical conductivity (d) electromotive force
90. The time of reverberation of a room Ais one second. What
will be the time (in seconds) of reverberation of a room,
having all the dimensions double of those of room A?
(a) 2 (b) 4
(c)
1
2
(d) 1
91. Which of the following is the unit of molar gas constant?
(a) JK–1 mol–1 (b) Joule
(c) JK–1 (d) J mol–1
92. Densityof liquid is 16.8 gcm–3. Its valuein the International
System of Units is
(a) 3
16.8 kgm- (b) 3
168 kgm-
(c) 3
1680 kgm- (d) 3
16800 kgm-
93. The dimensional formula of couple is
(a) ]
T
ML
[ 2
2 - (b) ]
MLT
[ 2
(c) 1 3
[ML T ]
- - (d) ]
T
ML
[ 2
2 -
-](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-29-320.jpg)
![27
Units and Measurements
10. The mean length ofan object is 5 cm. Which is the following
measurement is most accurate?
(a) 4.9cm (b) 4.805cm
(c) 5.25cm (d) 5.4cm
11. Young's modulus ofsteel is1.9 × 1011 N/m2.When expressed
in CGS units ofdyne/cm2, it will beequal to(1N = 105 dyne,
1 m2 =104 cm2)
(a) 1.9× 1010 (b) 1.9× 1011
(c) 1.9× 1012 (d) 1.9× 1013
12. If momentum (p), area (A) and time (T) are taken to be
fundamental quantities, then energy has the dimensional
formula
(a) [pA–1T1] (b) [p2AT]
(c) [pA–1/2T] (d) [pA1/2T]
NEET/AIPMT (2013-2017) Questions
13. In an experiment four quantities a, b, c and d are measured
with percentage error 1%, 2%, 3% and 4% respectively.
QuantityP is calculated as follows P =
3 2
a b
cd
% error in P is:
(a) 10% (b) 7% [2013]
(c) 4% (d) 14%
14. The pair of quantities having same dimensions is
(a) Young’s modulus and energy [NEET Kar. 2013]
(b) impulse and surface tension
(c) angular momentum and work
(d) work and torque
15. Ifforce(F),velocity(V)and time(T)aretaken asfundamental
units, then the dimensions of mass are : [2014]
(a) [F V T– 1] (b) [F V T– 2]
(c) [F V– 1 T– 1] (d) [F V– 1 T]
16. If energy (E), velocity (V) and time (T) are chosen as the
fundamental quantities, the dimensional formula of surface
tension will be : [2015]
(a) [EV–1T–2] (b) [EV–2T–2]
(c) [E–2V–1T–3] (d) [EV–2T–1]
17. If dimensions of critical velocity ucof a liquid flowing
through a tube are expressed as x y x
[ r ]
h r , where h, r and
r are the coefficient of viscosity of liquid, density of liquid
and radius of the tube respectively, then the values of x, y
and z are given by : [2015 RS]
(a) –1, –1, 1 (b) –1, –1, –1
(c) 1,1, 1 (d) 1, –1, –1
18. A physical quantityof the dimensions of length that can be
formed out of c, G and
2
0
e
4pe
is [c is velocity of light, G is
universal constant of gravitation and e is charge] [2017]
(a)
1/ 2
2
2
0
e
c G
4
é ù
ê ú
pe
ê ú
ë û
(b)
1/2
2
2
0
1 e
G4
c
é ù
ê ú
pe
ê ú
ë û
(c)
2
0
1 e
G
c 4pe
(d)
1/ 2
2
2
0
1 e
G
4
c
é ù
ê ú
pe
ê ú
ë û](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-31-320.jpg)
![28 PHYSICS
EXERCISE - 1
1. (d) Temperature is one of the basic physical quantities.
2. (b) The vander Waal’s gas equation is
RT
)
b
V
)(
V
a
P
(
2
=
-
+ for one mole.
& =
m
-
÷
÷
ø
ö
ç
ç
è
æ m
+ )
b
V
(
V
a
P
2
mRT for m mole
Dimensionallyin first bracket on L.H.S
m
=
Þ
ú
û
ù
ê
ë
é m
=
2
2
V
]
P
[
]
a
[
V
a
]
P
[
Dimension of
ú
ú
û
ù
ê
ê
ë
é
=
mole
litre
.
atm
]
a
[
2
3. (d)
4. (c) R
T
PV
T
W
T
eV
=
=
=
and
N
R
= Boltzmann constant.
5. (a) 6. (a)
7. (a) SI is based on seven fundamental units.
8. (a)
1
kg
J
kg
J
m
Q
L -
=
=
=
9. (a) As force = change in momentum/time.
force × time= change in momentum
10. (c) Illuminanceis intensityofillumination measured in lux.
11. (a)
2
2
i
Cm
m
C
A
Q
P -
=
=
=
12. (a)
13. (c) 1 light year = speed oflight in vacuum × no. ofseconds
in one year = (3 × 108) × (365 × 24 × 60 × 60) = 9.467 ×
1015 m.
14. (c) 1 pascal = 1 N / m2.
15. (c) Electron volt is a unit of energy &
1eV = 1.6×10–19 joule
16. (a) Impulse = force × time ]
LT
M
[
T
MLT 1
1
2 -
-
=
´
= .
17. (b) Pole strength, .
m
A
m
m
A
2
M
m
2
=
=
=
l
18. (d) Both energy and work have same unit.
energy/work is a pure number.
19. (a) Potential is work done per unit charge.
20. (c) Maxwell is the unit ofmagnetic flux in C.G.S system.
1 Wb(S.I unit) = 108 maxwell
21. (b) Let M = pnvm
m
1
n
2
1
1
2
)
LT
(
)
T
ML
(
T
ML -
-
-
-
-
=
m
n
2
m
n
n
T
L
M -
-
+
-
=
2
m
n
;
1
n -
=
+
-
=
1
1
2
n
2
m -
=
+
-
=
+
-
=
n
m -
=
22. (b)
23. (c) N kg–1 = force/mass = acceleration
24. (a) Moment of couple = force × distance ]
T
L
M
[ 2
2
1 -
=
work = force × distance ]
T
L
M
[ 2
2
1 -
= .
25. (b) ]
T
ML
[
T
T
ML
time
energy
Power 3
2
2
2
-
-
=
=
= .
EXERCISE - 2
1. (a) From Biot Savart’s law
2
0
r
sin
dl
i
4
B
q
p
m
=
2 2 2
1 1
0
4 Br Wb m m
Wb A m
idl sin Am
-
- -
p
m = = =
q
2. (a) Pressure gradient
Pressure difference
distance
= .
[Pressure gradient]
1 2
ML T
L
- -
=
2 2
ML T
- -
é ù
=
ë û
3. (c) From ,
n
1
n
1
R
1
2
2
2
1
÷
÷
ø
ö
ç
ç
è
æ
-
=
l
dimensions of ]
T
L
M
[
L
L
1
R 0
1
0
1 -
-
=
=
=
4. (a)
2 2
PV W ML T
R
T T molK
-
= = =
m m
where m is number of mole of the gas
= [M1L2T–2K–1mol–1]
5. (a) M = current × area =AL2 = [L2 A1]
6. (b) 0 1 0 1
m
b T LK [M L T K ]
= l = =
7. (b)
2 2
1
V W ML T
Impedance
Q Q QT
-
-
= = =
I I
]
Q
T
ML
[ 2
1
2 -
-
= .
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-32-320.jpg)
![29
Units and Measurements
8. (a) Energy stored in an inductor ]
T
ML
[
i
L
2
1 2
2
2 -
=
=
9. (b) Wave number ]
T
L
M
[
L
1
1 0
1
0 -
=
=
l
=
n .
10. (a) T = Pa Db Sc
c
2
b
3
a
2
1
1
0
0
)
MT
(
)
ML
(
)
T
ML
(
T
L
M -
-
-
-
=
c
2
a
2
b
3
a
c
b
a
T
L
M -
-
-
-
+
+
=
Applying principle of homogeneity
a + b + c = 0; – a – 3b = 0; – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
11. (b) T
MT
L
ML
S
r
2
3
3
3
=
=
r
-
-
12. (d) x
V
A
;
Vx
x
A
];
L
[
x
B =
=
=
=
2
2
/
5
2
/
1
2
2
T
L
M
L
T
L
M -
-
=
=
]
T
L
M
[
)
L
(
)
T
ML
(
AB 2
2
/
7
1
2
2
/
5 -
-
=
=
13. (b) In S = A(t + B) + Ct2 ; B is added to time t. Therefore,
dimensions of B are those of time.
14. (b) )
T
ML
(
M
L
E
m
2
h 2
2 -
=
l
= ]
T
L
M
[ 1
2 -
=
15. (d) a b
v g h ;
= a
2
b
a
b
a
2
1
0
T
L
L
)
LT
(
]
LT
M
[ -
+
-
-
=
=
2
/
1
a
1
a
2
;
1
b
a =
-
=
-
=
+
2
/
1
b =
16. (a) As c is added to t, c = [T]
1
2
v LT
a [LT ]
t T
-
-
= = = ,
1
b v(t c) LT T [L]
-
= + = ´ =
17. (a) Hubble’s constant,
]
L
[
]
LT
[
ce
tan
dis
velocity
H
1
-
=
=
]
T
[ 1
-
= .
m
/
N
10
70 3
-
´
=
18. (c) ;
r
3
4
V 3
p
=
%
3
%
1
3
100
r
r
3
100
V
V
=
´
=
´
÷
ø
ö
ç
è
æ D
=
´
D
19. (a) Subtraction is correct upto one place of decimal,
corresponding to the least number of decimal places.
7.26–0.2=7.06=7.1J.
20. (d) Number of significant figures in multiplication is three,
corresponding tothe minimum number
107.88× 0.610=65.8068=65.8
21. (d) 4
.
38
393
.
38
54
.
2
52
.
97
=
= (with least number of
significant figures, 3).
22. (d) Relative density =
Weight of body in air
Loss of weight in water
5.00 5.00
5.00 – 4.00 1.00
= =
0.05 0.05
100 100
5.00 1.00
Dr æ ö
´ = + ´
ç ÷
è ø
r
= (0.01 + 0.05) × 100
= 0.06 × 100 = 6%
Relative density = ±
5.00 6%
23. (a) Area = (Length)2
or length 1 2
(Area)
=
1 2
(100 2)
= ±
1 2 1
(100) 2
2
= ± ´
(10 1)m
= ±
24. (c) We know that
C
2
Q2
is energyofcapacitor so it represent
the dimension of energy ]
T
ML
[ 2
2 -
= .
25. (b)
Volt/amp.
sec/amp.
Volt
L/R
´
= = sec. ]
T
L
M
[ 0
0
=
26. (a)
Current Q
Current density
area area t
= =
´
27. (c) 8
.
0
100
25
2
.
0
=
´
28. (d) 29. (a)
30. (d) From
1
2 LC
n =
p
2 0 0 2
2 1 2
1 1
LC T [M L T ]
(2 ) (T )
-
= = = =
p n
31. (d) Solar constant = energy/sec/area
]
MT
[
TL
T
L
M 3
2
2
2
-
-
=
=
32. (c) From
2
2
o r
e
4
1
F
e
p
=
2
o
2
r
F
4
e
p
=
e
(dimensionally)
2 2 2 2
2 1 1
o
e 4 Fr (MLT )L
hc hc ML T [LT ]
-
- -
p
= =
e
]
A
T
L
M
[ 0
0
0
0
= ,
hc
e
o
2
e
is called fine structure constant & has value
137
1
.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-33-320.jpg)
![30 PHYSICS
33. (c)
K
T
ML
T
Q
Entropy
2
2 -
=
= ]
K
T
ML
[ 1
2
2 -
-
=
34. (d)
2 2
2 2 1
Q ML T
s [L T K ]
m MK
-
- -
= = =
q
35. (c)
2 2 2 1 2 4 2
1 1
LC (ML T A ) (M L T A )
- - - -
=
´
=
1
2
1
T
T
-
=
36. (b) Solar constant = energy/area/time
]
T
M
[
T
L
T
L
M 3
1
2
2
2
-
-
=
= .
37. (a)
38. (b) a b c
v k g
= l r
c
2
b
3
a
1
0
)
LT
(
)
ML
(
L
]
LT
M
[ -
-
-
=
c
2
c
b
3
a
b
T
L
M -
+
-
=
1
c
b
3
a
;
0
b =
+
-
=
1
c
2 -
=
- Þ 2
/
1
c =
2
1
a =
1/ 2 0 1/ 2 2
v g or v g
µ l r µ l
39. (b) In p = p0 exp. (– a t2) is dimensionless
]
T
[
T
1
t
1 2
2
2
-
=
=
=
a
40. (b) As P
V
a
2
=
2
a PV
=
3 2 4
2
dyne
(cm ) dyne cm
cm
= = ´
41. (c)
2
3
2
2
1
2
2
A
T
ML
)
AT
(
T
ML
e
h -
-
-
=
= = Resistance (ohm)
42. (c) I = AT2 e–B/kT
Dimensions ofA = I /T2; Dimensions of B = kT
(Q power of exponential is dimensionless)
2 2 2
2
AB (kT) k
T
= =
I
I
43. (a) The C.G.S unit ofcurrent is called biot (Bi) i.e.,
1C (1/10)e.m.u of charge 1
1A Bi
1sec sec 10
= = =
or 1Bi = 10A
44. (d) ]
T
L
[
L
1
T
1
k 1
1 -
-
=
´
=
w
The dimensions of the quantities in a, b, c are of velocity
[LT–1]
45. (c) Volume (L3) = surfacearea (6L2)
L= 6, volume = 63 = 216
46. (d) None of the expressions has the dimensions of time.
47. (a) ;
T
L
M
X c
b
a
=
X a M b L c T
100 100
X M L T
D D D D
æ ö
´ = + + ´
ç ÷
è ø
)%
c
b
a
( g
+
b
+
a
=
48. (c) NVD = (N – 1) MD
MD
N
1
N
VD
1 ÷
ø
ö
ç
è
æ -
=
L.C. = Least count = 1MD – 1VD
MD
N
1
N
1
.
C
.
L ÷
ø
ö
ç
è
æ -
-
=
cm
N
10
1
cm
N
1
.
0
.
D
.
M
N
1
=
=
=
scale
on vernier
parts
of
number
scale
main
on
part
1
of
value
=
whereV.D. = vernier division, M.D. Main scaledivision.
49. (c) From
2
2
T
4
g
;
g
2
T
l
l
p
=
p
=
)
x
2
y
(
T
T
2
g
g
+
=
D
+
D
=
D
l
l
50. (c) We have, n1u1 = n2u2
n2
1
1
2
æ ö
= ç ÷
è ø
u
n
u
2 3
6 1 1 1
2 2 2
10
-
æ ö æ ö æ ö
= ´ ç ÷ ç ÷ ç ÷
è ø è ø è ø
M L T
M L T
3
2
6
1
1 1 1
10
10 60
1 10
kg m s
kg s
m
-
-
æ ö æ ö
æ ö
= ´ ç ÷
ç ÷ ç ÷
è ø
è ø è ø
´
( ) ( )
2 3
6 1
10 10 60
10
æ ö
= ´ ç ÷
è ø
( )3
7
10 60
= ´ = 2.16 × 1012 units.
51. (c) Effective resistance
S
R (10k 10%) (20k 20%)
= W ± + W ±
Tolerance of the combination = 30 30
W ±
( k %)
52. (c) Force, F
2
2
1
0 r
q
q
4
1
pe
=
2
2
1
0
Fr
4
q
.
q
p
=
e
Þ](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-34-320.jpg)
![31
Units and Measurements
So dimension of e0
]
A
T
L
M
[
]
L
][
MLT
[
]
AT
[ 2
4
3
1
2
2
2
-
-
-
=
=
53. (c) Percentage error in mass %
2
100
m
m
=
÷
ø
ö
ç
è
æ
´
D
and
percentage error in speed %
3
100
v
v
=
÷
ø
ö
ç
è
æ
´
D
.
2
1
E mv
2
=
E m V
100 100 2 100
E m V
D D D
´ = ´ + ´
= 2% + 2 × 3% = 8%.
54. (d) Density =
Mass
Volume
3
L
M
=
r ,
L
L
3
M
M D
+
D
=
r
r
D
% error in density= % error in Mass + 3 (% error
in length)
= 4 + 3(3) = 13%
55. (a) Ratio of specific heat,
p
v
C
C
g =
56. (a) Specific gravity is the ratio of density of substance
and densityofwater at 4°C. The ratioof likequantities
is dimensionless.
57. (c) DimensionallyK.E = Work
58. (b) Trigonometric ratio are a number and hence
demensionless
59. (b) 1 2 2 2
Q Q
[C] [M L T Q ]
V W
2
- -
é ù
é ù
= = =
ê ú
ê ú
ë û ê ú
ë û
60. (c) ]
LT
[
]
A
[ 2
-
= or ]
AT
[
]
L
[ 2
=
[Work] = [Force ×Distance] = ]
FAT
[
]
FL
[ 2
=
61. (a) h is the coefficient of viscosity.
.
62. (a) Velocity gradient is velocity per unit distance.
63. (a) Use principle of homogeneity.
64. (a) ]
T
L
M
[
]
T
L
M
][
M
[
]
T
ML
][
T
ML
[ 0
0
0
2
2
3
1
5
2
1
2
2
2
=
-
-
-
-
= angle.
65. (c) 10VD= 9MD, 1VD =
9
MD
10
Vernier constant = 1 MD – 1VD
=
9 1 1 1
1 MD MD
10 10 10 2
æ ö
- = = ´
ç ÷
è ø
=0.05mm
66. (c)
0.1 19
1 MSD
10 20
æ ö
= -
ç ÷
è ø
Þ
1 1
1MSD
100 20
= ´
Þ 1 MSD =
1
10 2
5
´ =
67. (b) Least count
0.5
0.01mm
50
= =
68. (c) Maximum absolute error is b
a D
+
D . Therefore the
percentage error =
absolute error
100
actual value
´
69. (d) 100
t
t
100
R
R
100
I
I
2
100
Q
Q
´
D
+
´
D
+
´
D
=
´
D
2 2% 1% 1%
= ´ + + = 6%.
70. (c)
P F
100 100 2 100
P F
D D D
´ = ´ + ´ =
l
l
4% + 2 × 2%
=8%
71. (c)
2
T
g
,
g
2
T
l
l
µ
p
=
100
g
g
´
D
= 0.5% + 2 × 0.2% = 0.9%
72. (a) % error = 100
5
.
0
01
.
0
100
0
.
1
01
.
0
100
5
.
0
01
.
0
´
+
´
+
´
= 2 + 1 + 2 = 4 + 1 = 5
73. (c)
74. (b) 2 2 2 1 2 4 2
[ LC] ML T A .M L T A T
- - - -
= =
75. (c) The quantity (
v
wx
– wk) has dimension of angle and
hence wk is dimensionless being angle.
76. (d)
1 2 4 2
3 2 4 4
2 2 4 2
[X] M L T A
[Y] M L Q T
[Z ] M T A
- -
- -
- -
= = =
Q
A
T
æ ö
=
ç ÷
è ø
Q
77. (c) [at] = [F] amd [bt2] = [F]
Þ [a] = MLT–3 and [b] = MLT–4
78. (c) Clearly,
2
2 2
n T
m
4f L
= ;
2
2 2
MLT
[m]
T .L
-
-
=
79. (d)
2 2
2 1 3
W ML T
[V] ML A T
Q AT
-
- -
é ù
= = =
ê ú
ë û
80. (b) From 2 2 2
Bx Bx B
K
x
x A x
= = =
+
B = K × x = K.E. × distance = work × distance](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-35-320.jpg)
![32 PHYSICS
81. (c) We know that F = q v B
2
1 1
1
F MLT
B MT C
qv C LT
-
- -
-
= = =
´
82. (d) Least count of screw gauge =
0.5
mm 0.01mm
50
=
Reading = [Main scale reading
+ circular scale reading × L.C] – (zeroerror)
=[3 +35×0.01]–(–0.03)=3.38mm
83. (d) Pressure = [ ]
–2
–1 –2
2
MLT
ML T
L
=
Þ a = 1, b = – 1, c = – 2.
84. (b) In CGS system, 3
g
4
cm
d =
The unit of mass is 100g and unit of length is 10 cm, so
density = 3
100g
4
100
10
cm
10
æ ö
ç ÷
è ø
æ ö
ç ÷
è ø
= 3 3
4
(100g)
100
(10cm)
1
10
æ ö
ç ÷
è ø
æ ö
ç ÷
è ø
3
3
4 100g
(10) ·
100 (10cm)
= ´ = 40 unit
85. (d) P =
3 2
a b
cd
,
P
P
D
× 100% =
a
3
a
D
× 100% +
b
2
b
D
×
100%+
c
c
D
× 100%+
d
d
D
×100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
86. (c) From
2
2
T
4
g
;
g
2
T
l
l
p
=
p
=
)
x
2
y
(
T
T
2
g
g
+
=
D
+
D
=
D
l
l
87. (b) Given = 2
R ,
r
r
=
p
l
then
R
100
R
D
´
=
r
100 100 2 100
r
Dr D D
´ + ´ + ´
r
l
l
= 1% + 2% + 2 × 3% = 9%
88. (b) The magnetic field at a point near a long straight
conductor is given by
B =
0 2
.
4
I
r
m
p
Þ 0
m =
4
2
rB
I
p
[m0] =
[ ][ ]
[ ]
r B
I
=
-2 -1
L.MT A
I
F
B
qv
æ ö
=
ç ÷
è ø
Q
= MLT–2A–2
89. (c) Resistivity,
m
ne2t
r =
3 –3 –2
[ML T A ]
r =
So, electrical conductivity
1
s =
r
–1 –3 3 2
[M L T A ]
s =
90. (a) Reverberation time,
t =
forward backward
d d
V V
æ ö æ ö
+
ç ÷ ç ÷
è ø è ø
when dimensions double then, d¢ = 2d
d d 2d 2d d d
t 2
V V V V V V
¢ ¢ æ ö
¢ = + = + = +
ç ÷
è ø
Þ t¢ = 2t
d d
t
V V
æ ö
+ =
ç ÷
è ø
Q
91. (a)
1
1
mol
K
J
K
mol
J
nT
PV
R -
-
=
=
= .
92. (d) 16.8 gcm–3 =16800Kgm–3.
93. (a) Dimensionallycouple = Torque =Work
94. (a) The mean value ofrefractiveindex,
1.34 1.38 1.32 1.36
1.35
4
+ + +
m = =
and
| (1.35 1.34) | | (1.35 1.38) | | (1.35 1.32) | | (1.35 1.36) |
4
- + - + - + -
Dm =
= 0.02
Thus 100
Dm
´
m
=
0.02
100 1.48
1.35
´ =
95. (b) Least count =
0.1
10
= 0.01 cm
d1
= 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d2
= 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
d3
= 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
Mean diameter =
0.61 0.57 0.59
3
+ +
= 0.59 cm
96. (d) As,
2
2
g 4
T
= p
l
So,
g T
100 100 2 100
g T
D D D
´ = ´ + ´
l
l
0.1 1
100 2 100
20 90
= ´ + ´ ´ = 2.72 ; 3%
97. (c) 98. (b)
99. (b) 100. (a)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-36-320.jpg)
![33
Units and Measurements
EXERCISE - 3
Exemplar Questions
1. (b) In 0.06900, the twozeroes before six are not siginificant
and two zeroes on right side of9 are significant figures.
Hence, number of significant figures are four (6900).
2. (b) In addition the result will be in least number of places
after decimal andminimum number of significant figure.
The sum of the given numbers can be calculated as
663.821. Thenumber with least decimal places is 227.2
is correct to only one decimal place but in addition of
numbers, the final result should be rounded off to one
decimal place i.e., 664.
3. (c) As we know that in multiplication or division, the final
result should retain as many significant figures as are
there in the original number with the least significant
figures.
Thesignificant figure in given numbers4.237 g and2.5
cm3 are four and two respectivey so, density should
be reported to two significant figures.
3
3
4.237 g
Density 1.6948 1.7 g.cm
2.5 cm
-
= = =
As rounding off the number upto 2 significant
figures, we get density = 1.7.
4. (d) Rounding off2.745 upto 3 significant figures here IVth
digit is 5 and its preceding is even, so no change in 4.
Thus answer would be 2.74. Rounding off 2.735 upto3
significant figures, here IV digit is 5 and its preceding
digit is 3 (odd). So 3 is increased by 1 answer become
would be 2.74.
5. (a) If Dx is error in a physical quantity, then relativeerror is
calculated as
Dx
x
.
Given that,
Length l = (16.2 ± 0.1) cm
Breadth b = (10.1 ± 0.1) cm
Q Dl = 0.1 cm, Db= 0.1 cm
Area (A)= l × b= 16.2×10.1 = 163.62cm2
In significant figure rounding off to three significant
digits, area A = 164 cm2
0.1 0.1
16.2 10.1
D D D
= + = +
A l b
A l b
1.01 1.62 2.63
16.2 10.1 163.62
+
= =
´
So,
2.63 2.63
164
163.62 163.62
D = ´ = ´
A A
= 2.636 cm2
Now rounding off up to one significant figure
DA =3cm2.
So,Area A = A ± DA =(164 ±3) cm2.
6. (c) a. Work = force × distance
= [MLT–2][L] = [ML2T–2]
Torque = F × d = [ML2T–2]
LHS and RHS has same dimensions.
b. Angular momentum (L)
= mvr=[M][LT–1][L]
= [ML2T–1]
Planck's constant h =
n
E
2 2
2 1
1
[ML T ]
[ML T ]
[T ]
-
-
-
= =
1
and
æ ö
= n n =
ç ÷
è ø
E h
T
Q
So, dimensions of h and L are equal.
c. Tension = force = [MLT–2]
Surface tension
2
force [MLT ]
length [L]
-
= =
=[ML0T–2]
d. Impulse= force ×time = [MLT–2][T]
= [MLT–1]
Momentum = mass × velocity
= [M][LT–1] = [MLT–1]
LHS and RHS has same dimensions.
Hence only (c) options of Physical quantities does not
have same dimensions.
7. (a) By applying the Rule of significant figure in
multiplication and addition.
As given that,
A = 2.5ms–1 ± 0.5 ms–1, B = 0.10s ± 0.01 s
x = AB = (2.5)(0.10) = 0.25 m (consider onlysignificant
figure value)
D D D
= +
x A B
x A B
0.5 0.01 0.05 0.025 0.075
2.5 0.10 0.25 0.25
D +
= + = =
x
x
Dx =0.075=0.08m
(rounding off to two significant figures.)
AB =(0.25±0.08)m
8. (d) As given that,
A = 1.0m ±0.2 m, B =2.0m ±0.2m
So, X = (1.0)(2.0)
=
AB = 1.414 m
Rounding off upto two significant digit
X =1.4 m=(r) (Let)
1 1 0.2 0.2
2 2 1.0 2.0
D D D
é ù é ù
= + = +
ê ú ê ú
ë û ë û
x A B
x A B
0.6
2 2.0
=
´
0.6 0.6 1.4
0.212
2 2.0 2 2.0
´
Þ D = = =
´ ´
x
x
Rounding off upto one significant digit
Dx =0.2 m = Dr (Let)
So, correct value of
= + D
AB r r =(1.4 ±0.2)m](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-37-320.jpg)
![34 PHYSICS
9. (a) For the most precise measurement, the unit must be
least and number of digits including zeroes after
decimal must be zero.
Now, take first option,
As here 5.00 mm has the smallest unit and the error in
5.00 mm is least (commonly taken as 0.01 mm if not
specified), hence, 5.00 mm is most precise.
10. (a) Now, checking the errors with each options one by
one,
|Dl1|= |5 –4.9|= 0.1cm
|Dl2|=|5–4.805|= 0.195cm
|Dl3|=|5.25–5|=0.25cm
|Dl4|= |5.4–5|= 0.4cm
Error Dl1 isleast or minimum.
So, 4.9 cm is most precise.
11. (c) It is given that Young's modulus (Y) is,
Y =1.9 ×1011 N/m2
1N = 105 dyne
So, Y = 1.9 × 1011 × 105 dyne/m2
Convert meter tocentimeter
Q 1m=100cm
Y = 1.9 ×1011 × 105 dyne/(100)2 cm2
= 1.9 × 1016 – 4 dyne/cm2
Y = 1.9 × 1012 dyne/cm2
12. (d) Given that fundamental quantities are momentum (p),
area (A) andtime (T).
Let us consider the dimensional formula for
[ ]
a b c
E p A T
µ
]
= a b c
E kp A T
where k is dimensionless constant of proportionality.
Dimensions ofenergy[E] = [ML2T–2] and Dimension
of momentum p= mv = [MLT–1]
Dimension ofArea [A] = [L2]
Dimension ofTime [T] = [T]
Dimension of energy[E] = [K] [p]a[A]b[T]c
Putting all the dimensions, value
ML2T–2 = [MLT–1]a [L2]b [T]c
= MaL2b + aT–a + c
Byprinciple of homogeneity of dimensions,
a = 1, 2b +a = 2 Þ 2b + 1 = 2 Þ
1
2
=
b
2
- + = -
a c
c = –2 + a = –2 + 1 = –1
So, Dimensional formula (ofenergy)
E = [pA1/2T–1]
1/2 1
[ ]
-
=
E pA T
NEET/AIPMT (2013-2017) Questions
13. (d) P =
3 2
a b
cd
,
P
P
D
× 100% =
a
3
a
D
× 100% +
b
2
b
D
×
100%+
c
c
D
× 100% +
d
d
D
×100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
14. (d) Work = Force × displacement
Torque = Force × force arm
= mass × acceleration × length
= [M] × [LT–2] × [L] = [M L2T–2]
15. (d) Force = mass × acceleration
Þ [Mass]
=
force
acceleration
é ù
ê ú
ë û
=
force
velocity / time
é ù
ê ú
ë û
= [F V– 1 T]
16. (b) Let surface tension
s = Ea VbTc
b
–2
2 –2 a C
MLT L
(ML T ) (T)
L T
æ ö
= ç ÷
è ø
Equating the dimension ofLHS and RHS
ML0T–2 = MaL2a + b T–2a – b + c
Þ a = 1, 2a + b = 0, –2a – b + c = –2
Þ a = 1, b = – 2, c = – 2
Hence, the dimensions of surface tension are
[EV–2 T–2]
17. (d) Applying dimensional method :
vc = hxryrz
[M0LT–1]= [ML–1T–1]x [ML–3T0]y [M0LT0]z
Equating powers both sides
x + y= 0; –x = –1 x =1
1 + y = 0 y = –1
–x – 3y + z = 1
–1 – 3(–1) + z = 1
–1 + 3+ z = 1
z = –1
18. (d) Let dimensions of length is related as,
L
z
2
x y
0
e
[c] [G]
4
é ù
= ê ú
pe
ê ú
ë û
2
0
e
4pe
= ML3T–2
L= [LT–1]x [M–1L3T–2]y[ML3T–2]z
[L] = [Lx + 3y + 3z M –y + z T–x – 2y – 2z]
Comparing both sides
–y+ z = 0 Þ y= z ...(i)
x+ 3y+ 3z =1 ...(ii)
–x – 4z = 0 (Q y= z) ...(iii)
From(i),(ii)&(iii)
z = y=
1
,
2
x = –2
Hence, L =
1/2
2
2
0
e
c G
4
-
é ù
×
ê ú
pe
ê ú
ë û](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-38-320.jpg)
![BASIC DEFINITIONS
Mechanics : Branch of physics, which deals with the study of
objects in rest and in motion.
Statics : Studyof objects at rest or in equilibrium.
Kinematics : Studyof motion of objects without considering the
cause of motion.
Dynamics : Study of motion of objects considering the cause of
motion.
Rest : An object is said to be at rest if it does not change its
position with time, with respect to its surrounding (a reference
point which is generallytaken as origin in numerical problems)
Motion : An object is said tobe in motion if it changes its position
with time, with respect to its surroundings.
Rest and motion are relative terms.
Point mass/Point object : An object is said to be a point mass if
during its motion it covers distance much greater than its own
size.
One dimensional motion : An object travels in a straight line. It is
also called rectilinear or linear motion. The position change of
the object with time in one dimension can be described by only
one co-ordinate.
Ex. Astone falling freelyunder gravity.
Two dimensional motion or motion in a plane : For an object
travelling in a plane two coordinates sayX and Y are required to
describe its motion.
Ex. An insect crawling over the floor.
Three dimensional motion : An objecttravels in space.Todescribe
motion of objects in three dimension require all three coordinates
x, yand z.
Ex. Akite flying in the sky.
DISTANCE AND DISPLACEMENT
Distance or Path length : The length of the actual path travelled
by an object during motion in a given interval of time is called
the distance travelled by that object or path length. It is a scalar
quantity.
Displacement : It is the shortest distance between the initial and
final position of an object and is directed from the initial position
to the final position. It is a vector quantity.
Keep in Memory
1. Displacement maybe positive, negative or zero but distance
is always positive.
2. Displacement is not affected by the shift of the coordinate
axes.
3. Displacement of an object is independent of the path
followed by the object but distance depends upon path.
4. Displacement and distance both have same unit as that of
length i.e. metre.
5.
Distance
1
|Displacement|
³
6. For a moving body distance always increases with time
7. For a bodyundergoing one dimensional motion, in the same
direction distance = | displacement |. For all other motion
distance > | displacement |.
SPEED
It is the distance travelled per unit time by an object. It is a scalar
quantity. It cannot be negative.
Uniform speed : An object is said to be moving with a uniform
speed, if it covers equal distances in equal intervals of time,
howsoever small the time intervals maybe.
Non-uniform speed : If an object covers unequal distances in
equal interval of time or equal distances in unequal interval of
time.
Instantaneousspeed : Thespeed ofan object at a particular instant
of time is called the instantaneous speed.
Instantaneous speed, Vinst =
0
lim
D ®
D
=
D
t
s ds
t dt
.
Average speed : It is ratio of the total distance travelled by the
object to the total time taken.
Average speed Vav
x
t
D
=
D
Dimensions : [M0LT-1]; Unit: In SI systems.
VELOCITY
It is the displacement of an object per unit time. It is a vector
quantity. It can be positive negative or zero.
Uniform velocity : An object is said tobemoving with a uniform
velocity, if it covers equal displacements in equal intervals of
time, howsoever small the time intervals maybe.
Non-uniform velocity : If an object covers unequal displacements
in equal interval oftime or equal displacements in unequal interval
of time.
Instantaneous velocity : The velocity of an object at a particular
instant of time is called the instantaneous velocity.
3
Motion in a
Straight Line](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-39-320.jpg)
![36 PHYSICS
Variable or non-uniform acceleration : If the velocity of body
changes in different amounts during same time interval, then the
acceleration of the body is known as variable acceleration.
Acceleration is variable if either its direction or magnitude or
both changes with respect to time. A good example of variable
acceleration is the acceleration in uniform circular motion.
EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
When the motion is uniformlyaccelerated i.e., when acceleration
is constant in magnitude and direction :
(i) v = u + at
(ii) 2
at
2
1
ut
s +
=
(iii) v2 – u2 = 2as
where u = initial velocity; v = final velocity;
a = uniform acceleration and
s = distance travelled in time t,
(iv)
u v
s t
2
+
æ ö
= ç ÷
è ø
(v) Distance travelled in nth second sn = 1)
(2n
2
a
u -
+ ;
sn = distance covered in nth second
Above equations in vector form
,
t
a
2
1
t
u
s
,
t
a
u
v 2
+
=
+
=
)
s
.
a
2
u
.
u
v
.
v
or
(
s
.
a
2
u
v 2
2 r
r
r
r
r
r
r
r
r
r
=
+
=
-
t
)
v
u
(
2
1
s
r
r
r
+
=
When displacement (s) is given as a function of time t [s = f(t)]
then
0
t
s µ Body at rest v = 0 a = 0
1
t
s µ Uniform velocity;
acceleration zero 0
t
v µ a = 0
2
t
s µ Uniform acceleration 1
t
v µ 0
t
a µ
more
or
3
t
s µ Non uniform
acceleration more
or
2
t
v µ more
or
1
t
a µ
We use calculus method (integration and differentiation)
for displacement, velocity, acceleration as a function of time.
We know that ò
=
Þ
= dt
v
s
dt
ds
v ; ò
=
Þ
= dt
a
v
dt
dv
a ;
when a = f(s)
ò
ò =
Þ
= dv
v
ds
a
ds
dv
v
a ,
where s = displacement, v = instantaneous velocity,
a = instantaneous acceleration
Instantaneous velocity,
0
lim
D ®
D
= =
D
r r
r
inst
t
r dr
V
t dt
.
Average Velocity : It is ratio of the total displacement to the total
timetaken.
Average velocity,
D
=
D
r
av
r
v
t
Dimensions : [M0LT–1] ; Unit: In SI system, m/s
Keep in Memory
1. |Average velocity| can be zero but average speed cannot be
zero for a moving object.
2.
|Average velocity|
1
Average speed
£
3. | Instantaneous velocity | = Instantaneous speed.
4. A particle may have constant speed but variable velocity. It
happens when particle travels in curvilinear path.
5. If the body covers first half distance with speed v1 and
next half with speed v2 then
Average speed 1 2
1 2
2
=
+
v v
v
v v
6. If a bodycovers first one-third distance at a speed v1, next
one-third at speed v2 and last one-third at speed v3, then
Average speed 1 2 3
1 2 2 3 1 3
3
=
+ +
v v v
v
v v v v v v
7. If a bodytravels with uniform speed v1 for time t1 and with
uniform speed v2 for time t2, then
Average speed 1 1 2 2
1 2
+
=
+
v t v t
v
t t
ACCELERATION
The rate of change of velocity with respect to time is called
acceleration. It is a vector quantity.
Let velocity changes by v
r
D during some interval of time t
D .
Average acceleration av
a
r
is given by
D
=
D
r
r
av
v
a
t
Instantaneous acceleration a
r
is given by
0
lim
D ®
D
= =
D
r r
r
t
v dv
a
t dt
Its SI unit is meter/sec2 (ms–2).
A body moving with uniform velocity has zero acceleration. It
means that neither its speed nor its direction ofmotion is changing
withtime.
Uniform acceleration : If the velocity of the body changes in
equal amount during same time interval, then the acceleration of
the body is said to be uniform. Acceleration is uniform when
neither its direction nor magnitude changes with respect to time.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-40-320.jpg)
![38 PHYSICS
Solution :
dt
k
x
dx
or
x
k
dt
dx
or
x
k
v =
=
=
Given that when t = 0, x = 0 and when t = t, x = x,
Hence ò
ò =
t
0
x
0
dt
k
x
dx
;
ò
ò =
-
t
0
x
0
½
dt
k
dx
x or [ ]
x
1
1
2 t
0
0
x
k t
1
1
2
- +
é ù
ê ú
=
ê ú
ê ú
- +
ê ú
ë û
or
2
t
k
x
or
t
k
x
2 =
=
Now,
2
t
k
2
t
k
k
v
2
=
ú
û
ù
ê
ë
é
´
=
Thus the velocity varies with time.
Example2.
A particle covers each 1/3 of the total distance with speed
vl, v2 and v3 respectively. Find the average speed of the
particle.
Solution :
Average speed
1 2 3
Total distance travelled s
v
s s s
Total time taken
3v 3v 3v
= =
+ +
Þ
1 2 3
1 2 2 3 3 1
3v v v
v
v v v v v v
=
+ +
Example3.
A cheetah can accelerate from0 to 96 km/h in 2 sec., whereas
a cat requires 6 sec. Compute the average accelerations for
the cheetah and cat.
Solution :
For cheetah | r
a av| = f i
| v v |
t
-
D
r r
=
96km / h 0
2sec
-
=
1000m
96
3600sec
2sec
´
= 15 m/s2
For cat |
r
a av| =
10
96
36
6
´
= 5 m/s2.
Example4.
A particle is moving in east direction with speed 5 m/s.
After 10 sec it starts moving in north direction with same
speed. Find average acceleration.
Solution :
| r
v f| = |
r
v i| = 5 m/s
Acceleration ¹ 0
E
S
N
W
(due to change in direction
of velocityAv. acceleration,
D
r
v =
r
v f –
r
v i =
r
v f + (–
r
v i)
r
a =
f i
v v
t
-
D
r r
=
ˆ ˆ
5j 5i
10
-
Þ
r
a = –
1
2
$
i +
1
2
$
j
|
r
a | =
1
2
1
2
2 2
F
HG I
KJ +
F
HG I
KJ =
1
2
m/s2.
Keep in Memory
1. An object moving under the influence of earth's gravity in
which air resistance and small changes in g are neglected is
called a freely falling body.
2. In the absence of air resistance, the velocity of projection
is equal to the velocity with which the body strikes the
ground.
3. Distance travelled by a freely falling body in 1st second is
always half of the numerical value of g or 4.9 m, irrespective
of height h.
4. For a freely falling body with initial velocityzero
(i) Velocity µ time (v = gt)
(ii) Velocity Distance fallen
µ (v2 = 2gs)
(iii) Distance fallen a (time)2 ÷
ø
ö
ç
è
æ
= 2
gt
2
1
s , where g is the
acceleration due to gravity.
5. If maximum height attained bya body projected vertically
upwards is equal to the magnitude of velocityof projection,
then velocityof projection is 2g ms–1 and time of flight is 4
sec.
6. If maximum height attained bya body projected upward is
equal to magnitude of acceleration due to gravity i.e., 'g',
the time of ascent is 2 sec. and velocity of projection is
2
g .
7. Ratio of maximum heights reached by different bodies
projected with velocities u1, u2, u3 etc. are in the ratio of
3
3
2
2
2
1 u
:
u
:
u etc. and ratio of times of ascent are in ratio of
u1 : u2 : u3 etc.
8. During free fall velocity increases by equal amount every
decend and distance covered during 1st, 2nd, 3rd seconds
offall,are4.9m,14.7m,24.5m.
t = 3
4.9 m
14.7 m
24.5 m
t = 2
t = 1
t = 0
u = 0
9.8 m/s
19.6 m/s
29.4 m/s
9. If a body is projected horizontally from top of a tower, the
time taken by it to reach the ground does not depend on
the velocity of projection, but depends on the height of
tower and is equal to t =
g
h
2
.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-42-320.jpg)
![41
Motionina Straight Line
Example12.
The height of a tower is h metre. A body is thrown from the
top of tower vertically upward with some speed, it takes
t1, second to reach the ground. Another body thrown from
the top of tower with same speed downwards and takes t2
seconds to reach the ground. If third body, released from
same place takes ‘t’ second to reach the ground, then
(a) 1 2
t +t
t =
2
(b)
1
2
t
t =
t
(c)
1 2
2 1 1
= +
t t t
(d) 1 2
t = (t t )
Solution : (d)
Let u be the initial velocity of the body. Then at time t1
2
1
1 t
g
2
1
t
u
h +
-
= ...(1)
and at time t2, h =
2
2
2 t
g
2
1
t
u + ...(2)
From eqs. (1) and (2), we get
2
2
2
2
1
1 t
g
2
1
t
u
t
g
2
1
t
u +
=
+
-
or )
t
t
(
u
)
t
t
(
g
2
1
2
1
2
2
2
1 +
=
-
or )
t
t
(
u
)
t
t
(
)
t
t
(
g
2
1
2
1
2
1
2
1 +
=
-
+
or )
t
t
(
g
2
1
u 2
1 -
= ...(3)
Substituting the value of u in eqn. (2) we get
2
2
2
2
1 t
g
2
1
t
)
t
t
(
g
2
1
h +
-
= or 2
1 t
t
g
2
1
h = ...(4)
For third body,
2
t
g
2
1
h = ...(5)
From eqs. (4) and (5) we get, 1 2
t t t
=
Example13.
A bullet moving with a speed 10 m/s hits the wooden plank
and is stopped when it penetrates the plank 20 cm deep.
Calculate retardation of the bullet.
Solution :
v0 = 10 m/s, v = 0 and s = 20 cm. =
2
0.02m
100
=
Using v2 – v0
2 = 2ax
2 100
0 (10) 2a (0.02) a
2 0.02
-
- = Þ =
´
or a = – 2500 m/s2
Retardation = 2500 m/s2
Example14.
A body covers a distance of 20m in the 7th second and 24m
in the 9th second. How much distance shall it cover in
15th sec?
Solution :
From formula, th
n
S u a(2n –1)
2
1
= +
7
a
S th u (2 7 1)
2
= + ´ - but 7
S th 20m
=
a 13a
20 u 13 20 u
2 2
= + ´ Þ = + ...(1)
also 9
s th 24m
=
17a
24 u
2
= + ...(2)
From eqn. (1)
13a
u 20
2
= - ...(3)
Substitute this value of in eqn. (2)
13a 17a
24 20
2 2
= - +
17a 13a
24 20
2 2
- = -
4a
4
2
= Þ 4 = 2a Þ
2
4
a 2 m /s
2
= =
Substituting this value of a in eqn. (3)
13a
u 20
2
= -
Þ
13 2
u 20 u 20 13
2
´
= - Þ = -
u = 7 m/s
Now, s15th =
a
u (2 15 1)
2
+ ´ -
=
2
7 (29) 7 29 36m
2
+ = + =
Example15.
A train starts from rest and for the first kilometer moves
with constant acceleration, for the next 3 kilometers it has
constant velocity and for another 2 kilometers it moves
with constant retardation to come to rest after 10 min.
Find the maximum velocity and the three time intervals in
the three types of motion.
Solution :
Let the three time intervals be t1 min, t2 min, and t3 min.
respectively.
Let the maximum velocityattained be vm/min.
A B C D
1000m 3000m 2000m
t1 v–t2 t3
For Ato B 1
0 v
1000 t
2
+
æ ö
= ç ÷
è ø Þ 1
vt
2000 = ...........(1)
For B to C 2
vt
3000 = ...........(2)
[Using in both equations disp. = mean vel. × time]
and for C to D 3
v 0
2000 t
2
+
æ ö
= ç ÷
è ø Þ 3
vt
4000 = ...... (3)
Adding eqs. (1), (2) and (3), we get
)
t
t
t
(
v
9000 3
2
1 +
+
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-45-320.jpg)
![42 PHYSICS
3
9000 900 10
v 900 m / min. km 54 km / hr.
10 1/ 60 hr.
-
´
= = = =
Now, from eqs. (1), (2) and (3) we get
1
2000 20 2
t 2 min.
900 9 9
= = = ,
2
3000 10 1
t 3 min.
900 9 3
= = =
and 3
4000 40 4
t 4 min.
900 9 9
= = =
Example16.
A falling stone takes 0.2 seconds to fall past a window
which is 1m high. From howfar above the top of the window
was the stone dropped ?
Solution :
2
1
h gt
2
= ;
2
1
h 1 g(t 0.2)
2
+ = +
2 2 2
1 1 1 1
gt 1 gt g(0.2) g 2 0.2t
2 2 2 2
+ = + + ´ ´
u=0
1m
h
1
1 0.2gt
5
= + ;
4 2
2t t
5 5
= Þ =
1 4 4
h g m
2 25 5
= =
Example17.
From the top of a multi-storeyed building 40m tall, a boy
projects a stone vertically upwards with an initial velocity
of 10 ms–1 such that it eventually falls to the ground. (i)
After how long will the stone strike the ground ? (ii) After
how long will it pass through the point from where it was
projected ? (iii) What will be its velocity when it strikes the
ground ? Take g = 10 ms–2.
Solution :
Using, 2
1
S ut at
2
r r
r
= +
(i) 2
1
40 10t gt
2
- = - or 2
40 10t 5t
- = -
[Q taking g = 10m/s2]
or 2
5t 10t 40 0
- - =
h
v
u
u=10m/s
40m
or
2
10 10 4 5( 40) 10 100 800
t
2 5 10
+ - ´ - + +
= =
´
10 30
4 sec.
10
+
= =
(ii) It will pass from where it was projected after
2 10
t 2 sec.
g
´
= =
(iii) Velocity with which stone strikes the ground
V =10 + g× 2 =30m/s
VARIOUS GRAPHS RELATED TO MOTION
(a) Displacement-time graph - In this graph time is plotted on
x-axis and displacement on y-axis.
(i) For a stationary body (v = 0) the time-displacement
graph is a straight line parallel to time axis.
Displ
Velocity = 0
Time
Body is at rest
(ii) When the velocity of a body is constant then time-
displacement graph will be an oblique straight line.
Greater the slope ÷
ø
ö
ç
è
æ
q
= tan
dt
dx
of the straight line,
higher will be the velocity.
(iii) If the velocity of a body is not constant then the time-
displacement curve is a zig-zag curve.
(iv) For an accelerated motion the slope of time-
displacement curve increases with time while for
decelerated motion it decreases with time.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-46-320.jpg)
![55
Motionina Straight Line
Exemplar Questions
1. Among the four graph shown in the figure there is onlyone
graph for which average velocityover the time interval (O,
T) can vanish for a suitably chosen T. Which one is it?
(a) t
x
(b)
t
x
(c)
x
t
(d)
t
x
2. A lift is coming from 8th floor and isjust about to reach 4th
floor. Taking ground floor as origin and positive direction
upwards for all quantities, which one of the following is
correct?
(a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0
(c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a < 0
3. In one dimensional motion, instantaneous speed v satisfies
0
0 v v .
£ <
(a) The displacement in time T must always take non-
negative values
(b) The displacement x in time T satisfies 0 0
– < <
v T x v T
(c) The acceleration is always a non-negative number
(d) The motion has no turning points
4. A vehicle travels half the distance l with speed v1 and the
other half with speed v2, then its average speed is
(a) 1 2
2
+
v v
(b)
1 2
1 2
2 +
+
v v
v v
(c)
1 2
1 2
2
+
v v
v v
(d)
1 2
1 2
( )
+
L v v
v v
5. The displacement of a particle is given byx = (t – 2)2 where
x is in metre and t in second. The distance covered by the
particle in first 4 seconds is
(a) 4m (b) 8m
(c) 12m (d) 16m
6. At a metro station, a girl walks up a stationary escalator in
time t1. If she remains stationary on the escalator, then the
escalator take her up in time t2. The time taken by her to
walk up on the moving escalator will be
(a) 1 2
2
+
t t
(b)
1 2
2 1
-
t t
t t
(c)
1 2
2 1
+
t t
t t
(d) t1 – t2
NEET/AIPMT (2013-2017) Questions
7. Astone falls freely under gravity. It covers distances h1, h2
and h3 in the first 5 seconds, the next 5 seconds and the
next 5 seconds respectively. The relation between h1, h2
and h3 is [2013]
(a) h1 = 2
h
3
= 3
h
5
(b) h2 = 3h1 and h3 = 3h2
(c) h1 = h2 = h3 (d) h1 = 2h2 = 3h3
8. The displacement ‘x’ (in meter) ofa particle of mass ‘m’ (in
kg) moving in one dimension under the action of a force, is
related to time ‘t’ (in sec) by t = 3
x + . The displacement
of the particle when its velocityis zero, will be
[NEET Kar. 2013]
(a) 2m (b) 4m
(c) zero (d) 6m
9. A particle of unit mass undergoes one-dimensional motion
such that its velocity varies according to v(x) = bx–2n
where b and n are constants and x is the position of the
particle. The acceleration ofthe particle as d function of x, is
given by: [2015]
(a) –2nb2x–4n–1 (b) –2b2x–2n+1
(c) –2nb2e–4n+1 (d) –2nb2x–2n–1
10. Ifthe velocityof a particle is v=At + Bt2, whereA and B are
constants, then the distance travelled by it between 1s and
2s is : [2016]
(a)
3
A 4B
2
+ (b) 3A+7B
(c)
3 7
A B
2 3
+ (d)
A B
2 3
+
11. Preeti reached the metrostation and found that the escalator
was not working. She walked up the stationaryescalator in
time t1. On other days, if she remains stationary on the
moving escalator, then the escalator takes her up in time t2.
The time taken by her to walk up on the moving escalator
will be: [2017]
(a) 1 2
2 1
t t
t t
-
(b)
1 2
2 1
t t
t t
+
(c) t1 – t2 (d) 1 2
t t
2
+](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-59-320.jpg)
![56 PHYSICS
EXERCISE - 1
1. (b) 2. (b) 3. (c) 4. (c)
5. (d) 6. (b) 7. (c)
8. (b) 2
)
7
t
(
x
or
)
7
t
(
x +
=
+
=
time
velocity
),
7
t
(
2
dt
dx
µ
+
=
9. (b) a = f t, t
f
dt
dv
a =
= at t = 0, velocity = u
v t
u 0
dv f t dt
=
ò ò , v – u =
2
t
f
2
Þ v = u+
2
t
f
2
NOTE : Do not use v = u + at directly because the
acceleration is not constant.
10. (c) 1 2
dx
v a 2a t
dt
= = + 2
dv
a 2a
dt
= =
11. (d) A B
v tan30º and v tan 60º
= =
A
B
v tan30º 1/ 3 1
v tan 60º 3
3
= = =
12. (a) Let for the first halftime t, the person travels a distance
s1. Hence 1
1 1 1
s
v or s v t
t
= =
For second half time, 2
2 2 2
s
v or s v t
t
= =
Now,
1 2
s s
Total displacement
v
Total time 2t
+
= =
1 2 1 2
v t v t v v
2t 2
+ +
= =
13. (b) When a particle cover half of circle of radius r, then
displacement isAB = 2r
& distance = half ofcircumference of circle = pr
A
B
r
r
pr
14. (d) 2
2
gn
2
1
)
1
n
(
g
2
1
y -
+
=
)
1
n
2
(
2
g
]
n
)
1
n
[(
2
g 2
2
+
=
-
+
= ......(i)
Also, )
1
n
2
(
2
g
h -
= ......(ii)
From (i)and(ii)
y = h + g
15. (a) Differentiate two times and put x = 0.
16. (d) The difference in velocities is increasing with time as
both of them have more constant but different
acceleration.
17. (b) Time taken by the stone to reach the water level
g
h
2
t1 =
Time taken bysound to come to the mouth of the well,
v
h
t2 =
Total time
v
h
g
h
2
t
t 2
1 +
=
+
18. (d) Let the total distance be d. Then for first half distance,
time =
0
v
2
d
, next distance. = v1t and last halfdistance
= v2t
1 2
d
v t v t
2
+ = ;
1 2
d
t
2(v v )
=
+
Now average speed
)
v
v
(
2
d
)
v
v
(
2
d
v
2
d
d
t
2
1
2
1
0 +
+
+
+
=
0 1 2
1 2 0
2v (v v )
(v v ) 2v
+
=
+ +
19. (c) a = bt or bt
dt
dv
= . Integrating, we get
c
2
bt
v
2
+
= , where c is a constant of integration.
At t = 0, v = v0. Thus v0 = c.
Now, o
2
v
2
bt
dt
ds
v +
=
= dt
v
2
bt
ds o
2
÷
÷
ø
ö
ç
ç
è
æ
+
=
Integrating we get, t
v
6
bt
s o
3
+
=
20. (a) 3
kv
dt
dv
-
= or dt
k
v
dv
3
-
=
Integrating we get, c
kt
v
2
1
2
+
-
=
- ...(1)
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-60-320.jpg)
![57
Motionina Straight Line
At t = 0, v = v0 2
o
1
c
2v
- =
Putting in (1)
2 2 2 2
0 0
1 1 1 1
kt or kt
2v 2v 2v 2v
- = - - - = -
or
2
2
0 v
2
1
kt
v
2
1
=
ú
ú
û
ù
ê
ê
ë
é
+ or [ ] 2
2
0
2
0
v
v
kt
v
2
1 =
+
or
kt
v
2
1
v
v
or
kt
v
2
1
v
v
2
0
0
2
0
2
0
2
+
=
+
=
21. (d) )
e
1
1
(
b
a
)
e
1
(
b
a
)
e
1
(
b
a
x 1
b
1
b
-
=
-
=
-
= -
´
-
a (e 1)
b e
-
= b
/
a
3
2
~
b
a
637
.
0
718
.
2
)
718
.
1
(
b
a
718
.
2
)
1
718
.
2
(
b
a
-
=
=
-
=
velocity bt
ae
dt
dx
v -
=
= , v0 = a
accleration ab
a
&
abe
dt
dv
a 0
bt
-
=
-
=
= -
At t = 0, 0
)
1
1
(
b
a
x =
-
= and
At
b
1
t = , b
/
a
3
2
)
e
1
1
(
b
a
)
e
1
(
b
a
x 1
=
-
=
-
= -
At
a
t , x
b
= ¥ =
It cannot go beyond this, so point
a
x
b
> is not reached
by the particle.
At t = 0, x = 0, at
a
t , x
b
= ¥ = , therefore the particle
does not come back to its starting point at ¥
=
t .
22. (c) 2
x t 2t 1
= + +
Hence
dx
v 2t 2
dt
= = + . It increases with time.
23. (b) When a body falls through a height h, it acquires a
velocity gh
2 .
24. (c) Initial relative velocity vA – vB, is reduced to 0 in
distance d¢ (<d) with retardation a.
d
a
2
)
v
v
(
0 2
B
A
2
¢
-
=
-
-
2 2
A B A B
(v v ) (v v )
d d
2a 2a
- -
= >
¢
25. (d)
EXERCISE -2
1. (c) Given that x = At2 – B t3
velocity
2
dx
2At 3Bt
dt
= = -
and acceleration Bt
6
A
2
dt
dx
dt
d
-
=
÷
ø
ö
ç
è
æ
=
For acceleration to be zero 2A – 6Bt = 0.
B
3
A
B
6
A
2
t =
=
2. (a) t
sin
s
dt
x
d
a 2
2
2
w
w
-
=
=
On integrating, t
cos
s
t
cos
s
dt
dx 2
w
w
=
w
w
w
=
Again on integrating, we get
t
sin
s
t
sin
s
x w
=
w
w
w
=
3. (c) 3
dy
v b 2ct 4d t
dt
= = + -
3
0
v b 2c(0) 4d(0) b
= + - =
(Q for initial velocity, t = 0)
Now 2
dv
a 2c 12d t
dt
= = -
2
0
a 2c 12d(0) 2c
= - = , (at t = 0)
4. (d) 1 2
1 2
2 1
1 2 1 2
x x
2v v
1
2 2
x x v v
v v
2v 2v 2v v
+
= =
+
æ ö
+
+ ç ÷
è ø
5. (b) Let u be the initial velocity
1 1 2 1 2
v u a t , v u a (t t )
¢ ¢
= + = + +
and 3 1 2 3
v u a (t t t )
¢ = + + +
Now 1 1
1
u v u (u a t )
v
2 2
¢
+ + +
= = 1
t
a
2
1
u +
=
1 2
2 1 2
v v 1
v u a t a t
2 2
¢ ¢
+
= = + +
2 3
3 1 2 3
v v 1
v u a t a t a t
2 2
¢ ¢
+
= = + + +
So, 1 2 1 2
1
v v a (t t )
2
- = - +
and 2 3 2 3
1
v v a (t t )
2
- = - +
1 2 2 3 1 2 2 3
(v v ) :(v v ) (t t ) :(t t )
- - = + +](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-61-320.jpg)
![59
Motionina Straight Line
16. (b) The stone rises up till its vertical velocity is zero and
again reached the top of the tower with a speed u
(downward). The speed of the stone at the base is 3u.
–
+
v, g, h
u
Hence
2
2 2 4u
(3u) ( u) 2gh or h
g
= - + =
17. (c) Sobysecond equation ofmotion, we get
2
1
S ut at
2
= +
here S = l, u = 0, a = g sinq
2
2 2h 1 2h
t
a sin g
gsin
= = =
q
q
l h
sin
æ ö
q =
ç ÷
è ø
Q
l
h
g
cosq
g
sinq
g
q
18. (a) S = ut + ½at2 here a = g
For first body u1 =0 Þ S1=½g × 9
For second body u2=0 Þ S2= ½g × 4
So difference between them after 3 sec. = S1 – S2
= ½ g×5
If g = 10m/sec2 then S1–S2 = 25 m.
19. (b) Relative speed of each train with respect to each other
be, n = 10 + 15 = 25 m/s
Here distance covered by each train = sum of their
lengths= 50 + 50 = 100 m
Required time .
sec
4
25
100
=
=
20. (d) At highest point of the trajectory velocity becomes
zero and all kinetic energychanges to potential energy
so at highest point, K.E. = 0
21. (b) The distance travel in nth second is
Sn = u + ½ (2n–1)a ....(1)
so distance travel in tth & (t+1)th second are
St = u +½ (2t–1)a ....(2)
St+1= u+½ (2t+1)a ....(3)
As per question,
St+St+1 = 100 = 2(u + at) ....(4)
Now from first equation of motion the velocity, of
particle after time t, ifit moves with an accleration a is
v = u + a t ....(5)
where u is initial velocity
Sofrom eq(4) and (5), we get v = 50cm./sec.
22. (b) Since v2 = u2 + 2as
For first case u1=20m/sec, v1= 0, a1= g = 10, s1= ?
So m
20
10
2
400
s1 =
´
=
For second case (at moon) u2 =20m/sec, v2=0,
?
s
,
sec
/
m
7
.
1
6
g
a 2
2
2 =
=
=
6
/
10
2
400
7
.
1
2
400
s2
´
=
´
= so
6
1
s
s
2
1
=
23. (d) All options are correct :
(i) When two bodies A & B move in opposite
directions then relative velocity between A & B
either VAB or VBA both are greater than VA&VB.
(ii) When twobodiesA& Bmove in parallel direction
then A
AB
B
A
AB V
V
V
V
V <
Þ
-
=
B
BA
A
B
BA V
V
V
V
V <
Þ
-
=
24. (b) From third equation of motion
v2 = u2 – 2gh ( a g)
= -
Q
Given, v = 10 m/sec at h/2. But v = 0, when particle
attainedmaximum height h.
Therefore (10)2 = u2 – 2gh/2
or 100 = 2gh –2gh/2 (Q0 = u2 – 2gh)
Þ h = 10m
25. (b) Since in both case the height of building and down
ward acceleration ‘g’ is same. So both stones reach
simultaneouslyi.e.,
2
2
t
10
2
1
10
gt
2
1
S ´
=
Þ
=
or t 2 sec,
= for both stone.
26. (d) Speed 1
5
v 60
18
= ´ m/s
50
3
= m/s
= = ´ =
1 1
5 100
d 20m, v' 120
18 3
m/s
Let dceleration be a
2
1 1
0 v 2ad
= - ....(1)
or 2
1 1
v 2ad
=
2
1 2
(2v ) 2ad
= ...(2)
(2) divided by (1) gives,
m
80
20
4
d
d
d
4 2
1
2
=
´
=
Þ
=
27. (d) Equation of motion is at
u =
we know that ds
u
dt
= at
dt
ds =
Þ atdt
ds
or =
integrating it we get, ò
=
ò
4
0
s
0 tdt
a
ds
a
8
]
t
[
2
a
s 4
0
2
=
=
28. (d) The onlyforce acting on the ball is the force ofgravity.
The ball will ascend until gravity reduces its velocity
to zero and then it will descend. Find the time it takes
for the ball toreach its maximumheight and then double
the time to cover the round trip.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-63-320.jpg)
![61
Motionina Straight Line
45. (c) 1
(8 1 3 1 1) 0
v 5ms
1
-
´ - ´ ´ -
= =
46. (c) Sincethe initial velocityofjump issame on both planets
So 0 = u2– 2gAhA
0 = u2–2gBhB
or m
18
2
1
9
h
h
g
h
g
B
B
B
A
A
=
´
=
Þ
=
´
47. (d)
q
S
R
P Q
mg
q
sin
m
g
q
cos
mg
Let distance (PR) iscovered by the particle in time ‘t’.
Þ PR = q
=
q
+ sin
gt
2
1
t
.
sin
g
2
1
0 2
2
Further PR =
q
cos
PQ
(Given PQ = constant)
Þ PQ = q
qcos
sin
gt
2
1 2
= q
2
sin
gt
4
1 2
Þ t = 2
q
2
sin
g
PQ
q
µ
2
sin
1
t
So as q increases, q
2
sin first increases and then
decreases. Hence ‘t’ first decreases and then increases.
48. (d) Given x = ae–at + bebt
Velocity, v =
dx
dt
= –aae–at + bbebt = - +
a
e
b e
t
t
a
b
a
b
i.e., goon increasing with time.
49. (b) 2
y t ; v- t';a t°
µ µ µ
50. (c) On differentiating, acceleration = 0.2t )
t
(
f
a =
Þ
51. (d) Use v2 – u2 = 2aS. In both the cases, (u positive or
negative) u2 is positive.
52. (b) Area under a-t graph is change in velocity.
Area
1 1 1
(4 4) 6 4 2 4 2 2
2 2 2
= ´ + ´ + ´ ´ - ´ ´
1
36 2 34 ms-
= - =
As initial velocity is zero therefore, the velocity at 14
second is 1
s
m
34 - .
53. (c) At t = 20s, d = 20 m
54. (a) At six points in the graph the tangents have zero slope
i.e. velocity is zero.
55. (d) Time fall is
1
second.
2
2
1 1 10
h g
2 2 8
æ ö
= =
ç ÷
è ø
=1.25m
56. (b)
3
2
1
av
t
t
t
x
3
x
2
x
v
+
+
+
+
=
1
max
2x
t
v
= , 2
max
2x
t
v
= , 3
max
6x
t
v
=
max
av
6x v
v
10x
=
av
max
v 3
v 5
=
57. (c) Let A
v and B
v are the velocities of two bodies.
In first case, A
v + B
v = 6m/s ........(1)
In second case, A
v – B
v = 4m/s .........(2)
From (1) &(2) we get, A
v = 5 m/s and B
v =1 m/s.
58. (d) BB
v
r
= Relative velocityofballw.r.t. balloon = v
u +
gt
)
v
u
(
0 +
+
-
= of
g
v
u
t
+
=
Total time
g
)
v
u
(
2 +
=
59. (c) 2
2
2
1 )
N
n
(
g
2
1
y
,
gn
2
1
y -
=
=
]
)
N
n
(
n
[
g
2
1
y
y 2
2
2
1 -
-
=
-
Þ
g
1 (2n N)N
2
= - ]
1
y
y
[ 2
1 =
-
Q
Þ
2
N
gN
1
n +
=
60. (b) S1 =
2
1
g
2
2
1
÷
ø
ö
ç
è
æ
, S1 + S2 =
1
2
g(1)2
2
1
g
2
2
3
÷
ø
ö
ç
è
æ
= S1 + S2 + S3
By solving we get
S1 : S2 : S3 = 1 : 3 : 5
61. (b) Distance in last two second
=
2
1
×10× 2 =10m.
Total distance =
2
1
× 10 × (6 + 2) = 40 m.
62. (d) In (a), at the same time particlehas twopositionswhich
is not possible. In (b), particle has two velocities at the
same time. In (c), speed is negative which is not
possible.
63. (a) Velocityat timet is tan 45° = 1.Velocityat time(t = 1) is
tan 60 3
° = . Acceleration is change in velocity in
one second = 3 1
- .
64. (c) Here, the particle B moves upwards. Let the upward
velocity of B be v then
v
tan 60
10
= °](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-65-320.jpg)
![63
Motionina Straight Line
75. (b) v u at
= +
r r r
ˆ ˆ ˆ ˆ
(2 3 ) (0.3 0.2 ) 10
v i j i j
= + + + ´ = ˆ ˆ
5 5
i j
+
| |
v
r
= 2 2
5 5
+ ; | | 5 2
v =
r
76. (a) Q h =
1
2
gt2
h1 =
1
2
g(5)2 = 125
h1 + h2 =
1
2
g(10)2 = 500
Þ h2 = 375
h1 + h2 + h3 =
1
2
g(15)2 = 1125
Þ h3 = 625
h2 = 3h1 , h3 = 5h1
or h1 =
2
h
3
=
3
h
5
77. (a) h =
2
gT
2
1
now for t = T/3 second vertical distance moved is given by
h¢
2
1 T
g
2 3
æ ö
= ç ÷
è ø
Þ h¢
2
1 gT h
2 9 9
= ´ =
position of ball from ground
9
h
h -
=
9
h
8
=
78. (d) x = a + bt2 = 15 + 3t2
v =
dx
dt
= 3 × 2t = 6t
Þ t 3s
v = = 6 × 3 = 18cm/s
79. (a) Motion with constant acceleration is represented by a
quadratic equation of t
Y = (p + qt) (r + pt) = pr + qrt + p2t + pqt2
80. (c) Let the maximum height attained bythe ball be h.
At maximum height , velocityof ball, v = 0
Given, initial velocity, u = 19.6 m/s
Using the equation of motion,
v2 = u2 + 2gh
We get 0= (19.6)2 + 2 (–9.8) ×h
Þ h =
2
(19.6)
2 9.8
´
= 19.6m
81. (a) Here, length of train A, LA = 120 m
length of train B, LB = 130 m
velocity of train A, vA = 20 m/s
velocity of train B, vB = 30 m/s
Train B is running in opposite direction to train B,
velocityof train B relative to train A,
vBA = vB + vA
= (30 + 20) m/s
= 50 m/s
Total distance to be covered by train B
= LA +LB=(120+130)m
=250m
Hence, time required by train B to cross train A
t =
250
50
sec = 5sec
82. (a) Retardation,
2 2 2
0 (25/3)
2 2 8
v u
a
s
- -
= =
´
2
1
25
16
3
a
æ ö
= - ´
ç ÷
è ø
For u = 60 km h–1 =
50
3
ms–1
2
2
0 (50/ 3)
2 [ (25/3) 1/16]
s
-
=
´ - ´
=32m
83. (d) Velocitytime curve will be a straight line as shown:
o
t
v
At the highest point v = 0.
84. (b) Let O be the origin, then
O
Q
N
Car
E
Train P
N
W
S
E
passenger in the train at P observes the car at Q along
the direction PQ; i.e. west north direction.
85. (d) Given : u = 0, t = 5 sec, v= 108 km/hr = 30m/s
By eqn of motion
v = u + at
or 2
v 30
a 6 m / s [ u 0]
t 5
= = = =
Q
2
1
1
S at
2
=
2
1
6 5 75 m
2
= ´ ´ =
Distance travelled in first 5 sec is 75m.
Distance travelled with uniform speed of 30 m/s is S2
395 = S1 + S2 + S3
395 = 75+ S2 +45
S2 =395–120= 275m
Timetaken to travel 275 m =
275
9.2sec
30
=
For retarding motion, we have
02 – 302 = 2 (– a) × 45](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-67-320.jpg)
![64 PHYSICS
We get, a = 10 m/s2
Now by, S = ut +
2
1
at
2
45 = 30t +
2
1
(–10)t
2
45 = 30t – 5t2
Pn solving we get, t = 3 sec
Total time taken = 5 + 9.2 +3 = 17.2 sec.
86. (c) When particle comes to rest,
V = 0 =
dx d
dt dt
= (40 + 12t – t3)
Þ 12 – 3t2 = 0
Þ t2 =
12
4
3
= t = 2 sec
Therefore distance travelled byparticle before coming
to rest,
x = 40+ 12t – t3 = 40 + 12 × 2 – (2)3 = 56m
87. (b) We get acceleration by double differentiation of
displacement.
V=
1 2
0
a t a
dx d
a t
dt dt 2 3
2
æ ö
= + -
ç ÷
è ø
= 1
2
a 2
a t
2 3
-
a =
1
2
2
a 2
d a t
dv 2
2 3
a
dt dt 3
æ ö
-
ç ÷
è ø -
= =
88. (b)
1 1 1
(2 4) 2 1 4 3 4 2m
2 2 2
+ ´ + ´ ´ - ´ ´ =
89. (a) Velocity of boat 1
h
km
8
2
8
8 -
=
+
=
Velocityof water 1
h
km
4 -
=
160
h
3
8
4
8
8
4
8
8
t =
=
+
+
-
= minute
90. (b) Using relative terms
s
/
m
0
u .
rel =
?
t
,
m
9
.
4
S
,
s
m
8
.
9
a 2
=
=
= -
2
t
8
.
9
2
1
t
0
9
.
4 ´
´
+
´
=
Þ 9
.
4
t
9
.
4 2
= Þ t = 1 s
91. (b) From first equation of motion v = u + a t
here u = 40, a = g = – 10, t = 2
so v = 40 – 10 × 2 = 20 m/sec
92. (b) v= [144 × 1000/(60 × 60)] m/sec.
v = u + at
or (144 ×1000)/(60 ×60) =0+ a ×20
2
144 1000
a 2m / sec
60 60 20
´
= =
´ ´
Now
2
1
s u t at
2
= + m
400
)
20
(
2
2
1
0 2
=
´
´
+
=
93. (c) Height of tap = 5m and (g) = 10 m/sec2.
For the first drop, 5 = ut
2
1
2
+ gt
2
1
(0 ) 10
2
= ´ + ´
t t = 5t2 or t2 = 1 or t = 1 sec.
It means that the third drop leaves after one second of
the first drop. Or, each drop leaves after every 0.5 sec.
Distance covered by the second drop in 0.5 sec
2
1 1
(0 0.5) 10
2 2
= + = ´ + ´
ut gt ×(0.5)2
=1.25m.
Therefore, distance of the second drop above the
ground = 5 – 1.25 = 3.75 m.
94. (d) At E, the slope of the curve is negative.
95. (d) Velocity, 2
3 –12 3
= = +
ds
v t t
dt
Acceleration, 6 –12;
= =
dv
a t
dt
For a = 0, we have,
0 = 6 t – 12 or t = 2s. Hence, at t = 2 s the velocitywill be
2 –1
3 2 –12 2 3 9 ms
= ´ ´ + = -
v
96. (a) 4
3
0 (2 4 1)
7
2
5
0 (2 3 1)
2
+ ´ -
= =
+ ´ -
a
D
a
D
97. (b) In one dimensional motion, thebodycan have at a time
one velocity but not two values of velocities.
98. (d) Inuniformmotion theobject moveswithuniformvelocity,
the magnitude of its velocity at different instane i.e., at
t = 0, t =1, sec, t =2sec..... will alwaysbeconstant. Thus
velocity-timegraphfor an objectinuniform motion along
a straight path isa straight lineparallelto timeaxis.
99. (b) 100. (d)
EXERCISE -3
Exemplar Questions
1. (b) Ifwe draw a line parallel to time axis from the point (A)
on graph at t = 0 sec. This line can intersect graph at B.
In graph (b) for one value of displacement there are
twodifferent pointsoftime. so, for onetime, the average
velocity is positive and for other time is equivalent
negative.
As there are opposite velocities in the inteval 0 to T
hence average velocity can vanish in (b). This can be
seen in the figure given below.
t
x
O
A B
T
Here, OA = BT (same displacement) for two different
points of time.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-68-320.jpg)
![65
Motionina Straight Line
2. (a) As the lift is moving downward directions so
displacement isnegative (zero).Wehavetoseewhether
the motion is accelerating or retarding.
Dueto downward motion displacement is negative the
lift reaches 4th floor is about to stop hence, motion is
retarding (–a) downward in nature hence, x < 0; a > 0.
8th floor
x < 0
4th floor
Ground floor
x < 0 O
6th floor
As displacement is in negative direction,
x < 0 velocity will also be negative i.e., v < 0 but net
acceleration is +ve a > 0, that can be shown in the
graph.
3. (b) In one dimensional motion, for the maximum and
minimum displacement we must have the magnitude
and direction of maximum velocity.
As maximum velocityin positive direction is v0, hence
maximum velocity in opposite direction is also –v0.
Maximum displacement in one direction = v0T
Maximum displacement in opposite directions = –v0T.
Hence, 0 0
- < <
v T x v T
4. (c) Time taken totravel first half distance,
1
1 1
/ 2
2
= =
l l
t
v v 1
( / 2)
=
QL l
Time taken to travel second half distance,
2
2
2
=
l
t
v 2
( / 2)
=
QL l
So, total time taken to travel full distance
= t1 + t2
1 2 1 2
1 1
2 2 2
é ù
= + = +
ê ú
ë û
l l l
v v v v
Total time
2 1
1 2
2
é ù
+
= ê ú
ë û
v v
l
v v
So, average speed,
vav. =
Total distance
Total time
=
1 2
av.
1 2
1 2
2
1 1
2
Þ = =
+
é ù
+
ê ú
ë û
v v
l
v
v v
l
v v
1 2
av.
1 2
2
=
+
v v
v
v v
5. (b) As given that, x = (t – 2)2
Now, velocity
2
( 2)
= = -
dx d
v t
dt dt
= 2 (t – 2) m/s
Acceleration, [2( 2)]
= = -
dv d
a t
dt dt
= 2[1–0]=2m/s2 =2ms–2
at t = 0; v0 = 2 (0 – 2) = –4 m/s
t = 2 s; v2 = 2 (2 – 2) = 0 m/s
t = 4 s; v4 = 2 (4 – 2) = 4 m/s
B
D
O A
V
t
4
2
C
–4 m/s
4 m/s
(Time)
v-t graph is shown in diagram.
Distance travelled
= area between time axis of the graph
= area OAC + are ABD
1 1
2 2
= ´ + ´
OA OC AD BD
4 2 1
2 4 8 m
2 2
´
= + ´ ´ =
If displacement occurs
1 1
2 2
= ´ ´ + ´ ´
OA OC AD BD
1 1
2( 4) 2 4 0
2 2
= ´ - + ´ ´ =
6. (c) Let us consider, displacement is L, then
velocity of girl with respect to ground,
1
=
g
L
v
t
Velocity of escalator with respect to ground,
2
=
e
L
v
t
Net velocityof the girl on moving escalator with respect
to ground
1 2
= + = +
g e
L L
v v
t t
Þ vge
1 2
1 2
é ù
+
= ê ú
ë û
t t
L
t t](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-69-320.jpg)
![72 PHYSICS
3. . .
A B B A
=
r r
r r
Dot product of two vectors is commutative.
4. 2
.
A A A
=
r r
5. .( ) . .
A B C A B A C
+ = +
r r r r r
r r
Dot product is distributive.
6. x y z x y z
ˆ ˆ ˆ ˆ ˆ ˆ
A.B (A i A j A k).(B i B j B k)
= + + + +
r r
= (Ax Bx + Ay By + Az Bz)
Vector or Cross Product
The vector product of two vectors is defined as a vector having
magnitude equal to the product of two vectors and sine of the
angle between them. Its direction is perpendicular to the plane
containing the two vectors (direction of the vector is given by
right hand screw rule or right hand thumbrule.
C A B
= ´
u
r ur u
r
= (AB sin q) n̂
The direction of (A B)
´
ur u
r
perpendicular to the plane containing
vectors A
ur
and B
u
r
in the sense of advance of a right handed
screw rotated from A
ur
to B
u
r
is through the smaller angle between
them.
( )
A B C
´ =
ur u
r u
r q
B
A
e.g., ; ;
v r r F L r p
= w´ t = ´ = ´
r r
r
r r r r r r
Properties of Vector or Cross Product
1. ˆ ˆ
ˆ ˆ ˆ ˆ 0
i i j j k k
´ = ´ = ´ =
2. ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
; ;
i j k j k i k i j
´ = ´ = ´ =
i
j
k
3. 0
A A
´ =
r r
4. A B B A
´ ¹ ´
r r
r r
(not commutative) [ ]
´ = - ´
r r
r r
Q A B B A
5. ( ) ( ) ( )
A B C A B A C
´ + = ´ + ´
r r r r r
r r
(followsdistributivelaw)
6. ˆ ˆ
ˆ ˆ ˆ ˆ
( ) ( )
x y z x y z
A B A i A j A k B i B j B k
´ = + + ´ + +
r r
x y z
x y z
ˆ ˆ ˆ
i j k
A A A
B B B
=
= (Ay Bz – Az By) î + (Az Bx – A
Ax Bz) ĵ
+ (Ax By – Ay Bx) k̂
7. The cross product of two vectors represents the area of the
parallelogram formed bythem.
k
q
A
B
P Q
R
S
| | ( sin )
A B A B
´ = q
r r
= area ofparallelogram PQRS
= 2 (area of DPQR)
8. A unit vector which is perpendicular to A as well as B is
sin
| |
´ ´
=
q
´
uu
r uu
r r r
uu
r uu
r
A B A B
AB
A B
Keep in Memory
1.
x x y y z z
2 2 2 2 2 2
x y z x y z
A B A B A B A.B
cos
| A || B |
A A A B B B
+ +
q = =
+ + + +
ur u
r
ur u
r
2. tan q =
q
q
=
´
cos
B
A
sin
B
A
B
.
A
|
B
A
|
r
r
r
r
3.
2 2 2 2
.
| A B | | A B| A B
´ + =
r r
r r
4. |
)
A
(
)
B
(
|
2
|
B
–
A
|
|
B
A
|
r
r
r
r
r
r
´
=
´
+
5. If 0
C
B
A =
+
+
r
r
r
, then A
C
C
B
B
A
r
r
r
r
r
r
´
=
´
=
´
6. q
=
´
- 2
cos
B
A
|
B
A
|
|
B
.
A
| 2
2
2
2
r
r
r
r
7. If B
.
A
|
B
A
| =
´ then angle between A and B is
4
p
.
8. If B
||
A then 0
B
A =
´
9. Division by a vector is not defined. Because, it is not
possible to divide by a direction.
10. The sum and product of vectors is independent of
co-ordinate axes system.
CONDITION OF ZERORESULTANT VECTOR AND LAMI'S
THEOREM
If the three vectors acting on a point object at the same time are
represented in magnitude and direction by the three sides of a
triangle taken in order, then their resultant is zero and the three
vectors are said tobe in equilibrium.
i.e. 1 2 3
F F F 0
+ + =
r r r](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-76-320.jpg)
![73
Motion in a Plane
P Q
R
F3 F2
F1
a
b g
Lami's Theorem
It states that if three forces acting at a point are in equilibrium,
then each force is proportional to the sine of the angle between
the other two forces.
3
1 2 F
F F
sin sin sin
= =
a b g
r
r r
F3
F1
F2
a
b
g
or,
3
1 2 F
F F
PQ QR PR
= =
r
r r
Example10.
Calculate the area of a parallelogram formed from the
vectors k̂
3
ĵ
2
î
A +
+
=
r
and k̂
j
3
î
2
B +
-
=
r
, as adjacent
sides.
Solution :
The area of a parallelogram is given by |
B
A
|
r
r
´
Here,
ˆ ˆ ˆ
i j k
A B 1 2 3
2 3 1
´ = +
-
r r
= î [(2 × 1) – (–3 × 3)] + ĵ [(3 × 2) – (1 × 1)]
+ k̂ [(1 × –3)– (2 × 2)]
= 11 î + 5 ĵ – 7 k̂
2 2 2
| A B | (11) (5) ( 7)
´ = + + -
ur u
r
@ 14
Example11.
A particle suffers three displacements 4m in the northward,
2 m in the south-east and 1 m in the south-west directions.
What is the displacement of the particle and the distance
covered by it?
Solution :
Taking a frame of reference with the x-axis in the eastward
and the y-axisin the northward direction
Y
North (N)
X East (E)
West (W)
South (S)
1 2
ˆ ˆ ˆ
s = 4j, s = 2cos 45ºi 2 sin 45º j
-
ur uu
r
3
ˆ ˆ
s = 2 cos 45º i 2 sin 45º j
- -
uu
r
1 2 3
ˆ ˆ ˆ ˆ ˆ
s = s +s +s = 4j+ 2 i 2 j 2 i 2 j
- - -
r ur uu
r uu
r
Displacement ˆ
s (4 2 2)j
= -
r ˆ
= (1.17)j= (1.17)
(northward)
And total distance covered = 4 + 2 + 1 = 7m
Example12.
Prove that vectors ˆ
ˆ ˆ
A= i +2j+3k
r
and ˆ ˆ
B = 2i j
-
r
are
perpendicular to each other.
Solution :
Here, k̂
3
ĵ
2
î
A +
+
=
r
and ĵ
î
2
B -
=
r
Two vectors are perpendicular to each other if, 0
B
.
A =
r
r
Now B
.
A
r
r
= )
k̂
3
ĵ
2
î
( +
+ . )
ĵ
î
2
( - = 0
= 1 × 2+ 2 × (–1) + 3 ×(0) = 2 – 2+ 0 = 0
Since 0
B
.
A =
r
r
, therefore vectors A
r
and B
r
are
perpendicular to each other.
Example13.
Find the angle between the vectors ˆ
ˆ ˆ
A i j 2k
= + -
r
and
ˆ
ˆ ˆ
B i 2 j k
= - + -
r
.
Solution :
Here k̂
2
ĵ
î
A -
+
=
r
, k̂
ĵ
2
î
B -
+
-
=
r
We know that q
= cos
AB
B
.
A
r
r
or
AB
B
.
A
cos
r
r
=
q
Now 2 2 2
2 2 2
A 1 1 ( 2) 6 ,
B ( 1) 2 ( 1) 6
ˆ ˆ ˆ ˆ ˆ ˆ
A.B (i j 2k).( i 2j k) 1 2 3 3
3 3 1
cos or 60
6 2
6 6
= + + - =
= - + + - =
= + - - + - = - + + =
q = = = q = °
´
r r
Example14.
A particleis displaced from a point (3, – 4, 5) toanother point
(–2, 6, – 4) under a force k̂
ĵ
3
î
2 -
+ . Find the work doneby
the force.
Solution :
k̂
ĵ
3
î
2
F -
+
=
r
The displacement of the particle is
s
r = position vector of point (–2, 6, – 4) – position vector of
point (3, – 4, 5)
s
r = k̂
9
ĵ
10
î
5
)
k̂
5
ĵ
4
î
3
(
)
k̂
4
ĵ
6
î
2
( -
+
-
=
+
-
-
-
+
-](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-77-320.jpg)
![77
Motion in a Plane
Solution : (a)
Refer to fig., the boat man should go along OC in order to
cross the river straight (i.e. along OB).
B
C
O A
2km
q
b
n
r
n
º
30
sin
2
1
10
5
OC
CB
sin
b
r
=
=
=
n
n
=
=
q ; q = 30º;
Boat man should go along in a direction inclined at
90º + 30º = 120º to the direction of river flow.
Example18.
A man swims at an angle q = 120º to the direction of water
flow with a speed vmw = 5km/hr relative to water. If the
speed of water vw = 3km/hr, find the speed of the man.
Solution :
r
v mw = r
v m – r
v w
r
v m = r
v mw + r
v w
Þ vm = | r
v mw + r
v w| = v v v v
mw w mw w
2 2
2
+ + . cosq
Þ vm = 5 3 2 5 3 120
2 2
+ + ( )( )cos º
Þ vm = 25 9 15
+ - = 19 m/sec.
Example19.
A gun throws a shell with a muzzle speed of 98 m/sec. When
the elevation is 45º, the range is found to be 900 m. How
much is the range decreased by air resistance?
Solution :
Without air resistance, the expected range
8
.
9
)
98
(
8
.
9
90
sin
)
98
(
g
2
sin
u
R
2
2
2
=
´
=
q
= =980m
Decrease in range = 980 m – 900 m = 80 m
Example20.
A particle is projected with velocity u at an angle q with
the horizontal so that its horizontal range is twice the
greatest height attained. The horizontal range is
(a) u2/g (b) 2 u2/3 g
(c) 4 u2/5 g (d) None of these
Solution : (c)
Greatest height attained
g
2
sin
u
H
2
2
q
= ……(1)
Horizontal range,
g
cos
sin
u
2
g
2
sin
u
R
2
2
q
q
=
q
= ……(2)
[Qsin 2q = 2 sin q cosq]
Given that R = 2 H
g
2
sin
u
2
g
cos
sin
u
2 2
2
2
q
=
q
q
Solving we get tan q = 2 .....(3)
Hence )
5
/
1
(
cos
and
)
5
/
2
(
sin =
q
=
q
From eqn. (2)
g
5
u
4
5
1
5
2
g
u
2
R
2
2
=
´
´
=
Example21.
A body is projected downwards at an angle of 30º to the
horizontal with a velocity of 9.8 m/s from the top of a tower
29.4 m high. How long will it take before striking the
ground?
Solution :
The situation is shown in fig.
30°
u
B
A
–
+
The time taken bythe bodyis equal to the time taken bythe
freelyfalling bodyfrom the height 29.4m with initial velocity
u sin q. This is given by
9.8
usin 4.9 m / s
2
q = =
Applying the formula, s = u t +
2
1
g t2, we have
29.4 = 4.9 t +
2
1
(9.8) t2 or 4.9 t2 + 4.9 t – 29.4 = 0
(because s, u and g are all in downward direction)
t2 + t – 6 = 0or t = 2 or –3
Time taken to reach ground = 2 second
Example22.
Two boys stationed at A and B fire bullets simultaneously
at a bird stationed at C. The bullets are fired from A and B
at angles of 53° and 37° with the vertical. Both the bullets
fire the bird simultaneously. What is the value of vA if
vB = 60 units? (Given : tan 37° = 3/4)
Solution :
The vertical components must be equal.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-81-320.jpg)
![78 PHYSICS
vA cos 53° = vB cos 37°
or vA = vB
cos 37
cos (90 37 )
°
° - °
or vA = 60 cot 37° =
60
tan 37° [QvB = 60 units]
=
60 4
3
´
= 80 units
PROJECTILE ON AN INCLINED PLANE
Let a body is thrown from a plane OA inclined at an angle a with
the horizontal, with a constant velocity u in a direction making an
angle q with the horizontal.
The body returns back on the same plane OA. Hence the net
displacement of the particle in a direction normal to the plane OA
is zero.
a
q
(q-a)
g sin( )
a
g cos( )
a
O
A
B
u a
g
y
uy
x
ux = u cos (q – a ) along the incline, + x-axis)
uy = u sin (q – a ) along the incline, + y-axis)
ax = g sin a along – x-axis, as retardation
ay = g cos a along – y-axis, as retardation
The time of flight of the projectile is given by
2
at
2
1
ut
s +
=
or 2
1
0 u sin( )T g cos T
2
= q - a - a
usin( )
T
gcos( )
2 q - a
=
a
If maximumheight above the inclined plane is H,
2 2
u sin ( )
H
2g
q - a
=
The horizontal range R of the projectile is given by
OB = u cos q t =
2
u sin( )cos
R
gcos
2 q - a q
=
a
The range of the projectile at the inclined plane is given by
OA=
2
2
OB u sin( )cos
R
cos gcos
2 q- a q
= =
a a
Condition for horizontal range R on the inclined plane to be
maximum:
Since
2
2
u
R [2sin( )cos ]
gcos
= q - a q
a
2
2
u
[sin(2 ) sin ]
gcos
= q - a - a
a
{2 sin Acos B = sin(A+B)+sin(A–B)}
R is maximumwhen sin (2q –a) is maximum
i.e., sin (2q – a) =1 or
4 2
p a
q = +
2
max 2
u
R [1 sin ]
gcos
Þ = - a
a
or Rmax(on inclined plane) max
R (on horizontalplane)
1 sin
=
+ a
where Rmax (on horizontal plane)
g
2
u2
= .
Conditionfor time offlightTto bemaximum:
2 u sin( )
T
g cos
q - a
=
a
so T is max when sin (q – a) is maximum
i.e., sin (q –a) = 1 or
2u
T
2 g cos
p
q = + a Þ =
a
Itmeans that ifq1 istheanglefor projectilefor which Tismaximum
and q2 isthe anglefor which Rismaximum, then q1 = 2q2.
Example23.
The slopes of wind screen of two cars are b1 = 30° and
b2 = 15° respectively. At what ratio v1/v2 of the velocities
of the cars will their drivers see the hailstorms bounced by
windscreen of their cars in the vertical direction? Assume
hailstorms falling vertically.
Solution :
From the fig tan
1
v
v
a = and 1
90 2
a = °- b
where v is velocity of hail
1 1
1
v
tan(90 2 ) cot 2
v
° - b = b =
b1
–v1
b1
b1
v
a
Similarly, 2
2
v
cot 2
v
b =
1 2
2 1
v cot 2 cot30
3.
v cot 2 cot 60
b
= = =
b](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-82-320.jpg)
![85
Motion in a Plane
26. The magnitude of the vector product of two vectors is 3
times the scalar product. The angle between vectors is
(a)
4
p
(b)
6
p
(c)
3
p
(d)
2
p
27. IfA= B + C and the magnitudes ofA, B and C are5, 4 and 3
units, the angle between A and C is
(a) cos–1 (3/5) (b) cos–1 (4/5)
(c)
2
p
(d) sin–1 (3/4)
28. A rectangular sheet of material has a width of 3 m and a
length of 4 m. Forces with magnitudes of 3 N and 4N.
respectively, are applied parallel to two edges of the sheet,
as shown in the figure below.
3m
4m
4N
3N
F
q
Athird force F, is applied to the centre of the sheet, along a
line in the plane of the sheet, at an angle q = tan 0.75 with
respect to the horizontal direction. The sheet will be in
translational equilibrium when F has what value?
(a) F = 3 N (b) F= 4N
(c) F = 5 N (d) F= 7N
29. The linear velocityof a rotating body is given by :
v r
= w ´
uu
r uu
r ur
If k̂
3
ĵ
4
r
and
k̂
2
j
ˆ
2
î -
=
+
-
=
w , then the magnitude of
v is
(a) units
29 (b) units
31
(c) units
37 (d) units
41
30. Two forces are such that the sum of their magnitudes is 18 N
and their resultant is 12 N which is perpendicular to the
smaller force. Then the magnitudes of the forces are
(a) 12 N, 6 N (b) 13 N, 5 N
(c) 10 N, 8 N (d) 16N, 2N.
31. For any two vectors A
r
and B
r
, if A . B | A B |
= ´
r r
r r
, the
magnitude of C A B
= +
r r r
is
(a) A+ B (b) AB
2
B
A 2
2
+
+
(c) 2
2
B
A + (d)
2
AB
B
A2
+
+ 2
32. The angles which the vector ˆ ˆ ˆ
A 3i 6j 2k
= + +
r
makes with
the co-ordinate axes are :
(a)
7
1
cos
,
7
4
cos
,
7
3
cos 1
1
1 -
-
-
(b)
7
2
cos
,
7
6
cos
,
7
3
cos 1
1
1 -
-
-
(c)
7
3
cos
,
7
5
cos
,
7
4
cos 1
1
1 -
-
-
(d) None of these
33. The resultant of two forces 3P and 2P is R. If the first force
is doubled then the resultant is also doubled. The angle
between the two forces is
(a) 60º (b) 120º
(c) 70º (d) 180º
34. Let C A B
= +
r r r
(a) | C |
r
is always greater than | A |
r
(b) It is possible to have | C | | A |
<
r r
and | C | | B |
<
r r
(c) C
r
is always equal to A B
+
r r
(d) C
r
is never equal to A B
+
r r
35. At what angle must the two forces (x + y) and (x – y) act so
that the resultant may be 2 2
(x y )
+ ?
(a) 1 2 2 2 2
cos [ (x y )/ 2(x y )]
-
- + -
(b) 1 2 2 2 2
cos [ 2(x y )/(x y )]
-
- - +
(c) 1 2 2 2 2
cos [ (x y ) /(x y )]
-
- + -
(d) 1 2 2 2 2
cos [ (x y )/(x y )]
-
- - +
36. A force of k̂
F
– acts on O, the origin of the coordinate
system. The torque about the point (1, –1) is
X
Y
Z
O
(a) )
j
ˆ
î
(
F - (b) )
ĵ
î
(
F +
-
(c) )
ĵ
î
(
F + (d) )
j
ˆ
î
(
F -
-
37. The component of vector ˆ ˆ
a 2i 3j
= +
r
along the vector
i + j is
(a)
2
5
(b) 2
10
(c) 2
5 (d) 5](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-89-320.jpg)
![89
Motion in a Plane
4. Thecomponent ofa vector ralong X-axiswill havemaximum
value if
(a) r is along positive Y-axis
(b) r is along positive X-axis
(c) r makes an angle of 45° with the X-axis
(d) r is along negative Y-axis
5. The horizontal range of a projectilefired at an angleof 15° is
50 m. Ifit is fired with the same speed at an angle of 45°, its
range will be
(a) 60m (b) 71m
(c) 100m (d) 141m
6. Consider the quantities, pressure, power, energy, impulse,
gravitational potential, electrical charge, temperature, area.
Out of these, the only vector quantities are :
(a) impulse, pressure and area
(b) impulseand area
(c) area and gravitational potential
(d) impulse and pressure
7. In a two dimensional motion, instantaneous speed v0 is a
positive constant. Then, which of the following are
necessarily true?
(a) The average velocity is not zero at any time
(b) Average acceleration must always vanish
(c) Displacements in equal time intervals are equal
(d) Equal path lengths are traversed in equal intervals
8. In a two dimensional motion, instantaneous speed v0 is a
positive constant. Then, which of the following are
necessarily true?
(a) The acceleration ofthe particle is zero
(b) The acceleration of the particle is bounded
(c) The acceleration of the particle is necessarily in the
plane of motion
(d) The particle must be undergoing a uniform circular
motion
NEET/AIPMT (2013-2017) Questions
9. The velocity ofa projectile at the initial pointA is $
( )
2 3
i j
+
$
m/s. It’s velocity (in m/s) at point B is [2013]
(a) $
2 3
i j
- +
$ (b) $
2 3
i j
-
$
(c) $
2 3
i j
+
$ (d) $
2 3
i j
- -
$
10. Vectors ,
A B
r r
and C
r
are such that 0
A B
× =
r r
and 0.
A C
× =
r r
Then the vector parallel to A
r
is [NEET Kar. 2013]
(a) and
B C
r
r
(b) A B
´
r r
(c) B C
+
r
r
(d) B C
´
r
r
11. A particle is moving such that its position coordinate (x, y)
are
(2m,3m) attimet = 0
(6m, 7m) at time t = 2 s and
(13m, 14m) attimet = 5s.
Average velocity vector av
(V )
r
from t = 0 to t = 5s is : [2014]
(a)
1 ˆ ˆ
(13i +14j)
5
(b)
7 ˆ ˆ
(i + j)
3
(c) ˆ ˆ
2(i + j) (d)
11 ˆ ˆ
(i + j)
5
12. A ship A is moving Westwards with a speed of 10 km h–1
and a ship B 100 km South ofA, is moving Northwards with
a speed of 10 km h–1. The time after which the distance
between them becomes shortest, is : [2015]
(a) 5 h (b) 5 2 h
(c) 10 2 h (d) 0 h
13. If vectors ˆ ˆ
A cos ti sin tj
= w + w
r
and
t t
ˆ ˆ
B cos i sin j
2 2
w w
= +
r
are functions of time, then the value of t at which they are
orthogonal to each other is : [2015 RS]
(a) t
2
p
=
w
(b) t
p
=
w
(c) t = 0 (d) t
4
p
=
w
14. The position vector of a particle R
r
as a function of time is
given by:
ˆ ˆ
R 4sin(2 t)i 4cos(2 t)j
= p + p
r
Where R is in meter, t in seconds and î and ĵ denote unit
vectors along x-and y-directions, respectively.
Which one of the following statements is wrong for the
motion of particle? [2015 RS]
(a) Magnitude of acceleration vector is
2
v
R
, where v is
the velocity of particle
(b) Magnitude of thevelocityofparticle is 8 meter/second
(c) path of the particle is a circle ofradius 4 meter.
(d) Acceleration vector is along - R
r
15. A particle moves so that its position vector is given by
ˆ ˆ
r cos tx sin ty
= w + w
r
. Where w is a constant. Which ofthe
following is true ? [2016]
(a) Velocityand acceleration both are perpendicular to r
r
(b) Velocityand acceleration both are parallel to r
r
(c) Velocity is perpendicular to r
r
and acceleration is
directed towards the origin
(d) Velocity is perpendicular to r
r
and acceleration is
directed awayfrom the origin
16. If the magnitude of sum of two vectors is equal to the
magnitude of difference of the two vectors, the angle
between these vectors is : [2016]
(a) 0° (b) 90°
(c) 45° (d) 180°
17. The x and ycoordinates of the particle at anytime are x = 5t
– 2t2 and y = 10t respectively, where x and y are in meters
and t in seconds. The acceleration of the particle at t = 2s is
(a) 5 m/s2 (b) –4 m/s2 [2017]
(c) –8 m/s2 (d) 0](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-93-320.jpg)
![90 PHYSICS
EXERCISE - 1
1. (b) 2.(d)
3. (d) q
=
q
-
q
= 2
2
2
2
2
2
1
tan
g
2
/
)
º
90
(
sin
u
g
2
/
sin
u
H
H
4. (d) 5. (a)
6. (a) In vector addition, the commutative law is obeyed
i.e., A
B
B
A
r
r
r
r
+
=
+
Vector subtraction does not follow commutative law.
7. (d) Since B
.
A
0
C
.
A
r
r
r
r
=
= , itmeans that A
r
isperpendicular
to both B
&
C
r
r
, hence A
r
is parallel to )
C
B
(
r
r
´ or
)
B
C
(
r
r
´ .
8. (a) The unit vector of any vector A
r
is defined as
|
A
|
A
 r
r
=
9. (c) On earth, R = u2 sin 2q/g.
On moon, g' = g/6
R' = u2 sin 2q/g' = 6u2 sin2q/g = 6R.
10. (c)
B
A
2
B
A
R
cos
2
2
2
-
-
=
q 0
B
A
2
R
R 2
2
=
-
=
2
/
p
=
q
11. (c) ]
cos
AB
2
B
A
[
R 2
2
2
q
+
+
=
q
+
+
= cos
R
2
R
R
R 2
2
2
2
2
/
1
cos
or
cos
R
2
R 2
2
-
=
q
q
=
- or 3
/
2p
=
q
12. (c) The dot product should be zero.
13. (c) )
g
/
u
(
2
3
g
º
60
sin
u
R 2
2
º
30 =
=
g
/
u
g
90
sin
u
R 2
2
º
45 =
=
)
2
/
3
(
g
u
g
º
30
cos
u
g
º
120
sin
u
R
2
2
2
º
60 =
=
=
so 30º 60º 45º
R R R
= > or A C B
R R R
= >
14. (a) 2
1
1 mr
F w
= ; 2
2
2 mr
F w
=
since period T is same, so w is same, because
w
p
=
2
T .
Hence ÷
÷
ø
ö
ç
ç
è
æ
=
2
1
2
1
r
r
F
F
15. (a) A parabola
16. (a)
2 2
1
u sin
h
2g
q
=
2 2
2
u sin (90 )
h
2g
-q
= ,
2
u sin 2
R
g
q
=
Range R is same for angle q and (90° – q)
2 2 2 2
1 2
u sin u sin (90 )
h h
2g 2g
q - q
= ´
4 2 2
2
u (sin ) sin (90 )
4g
q ´ -q
= [ sin(90 ) cos ]
- q = q
Q
4 2 2
2
u (sin ) cos
4g
q ´ q
= [ sin 2 2sin cos ]
q = q q
Q
4 2
2
u (sin cos )
4g
q q
=
4 2
2
u (sin 2 )
16g
q
=
2 2 2
2
(u sin 2 ) R
16
16g
q
= =
or, R2 = 16 h1h2 or 1 2
R 4 h h
=
17. (b) Ifthere isnoresistance, bomb will drop ata placeexactly
below the flying aeroplane. But when we take into
account air resistance, bomb will face deceleration in
its velocity. So, it will fall on the earth exactly behind
the aeroplane.
18. (b)
19. (c) From the resultant path of the particle, when it is
projected at angle q with its velocity u is
q
-
q
=
2
2
2
cos
u
gx
2
1
tan
x
y
y
B
D
O
q
A
C
a
n
o
n
Where y denotes the instantaneous height of particle
when it travels an instantaneous horizontal distance x.
herex = D, u = vo
so
2
2 2
0
1 gD
y D tan
2 v cos
= q -
q
20. (d) Max. height =
g
2
)
90
(
sin
v
H
2
2
q
-
= .....(i)
Time of flight,
g
)
90
sin(
v
2
T
q
-
= ...(ii)
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-94-320.jpg)
![91
Motion in a Plane
Horizontal
Vertical
v
q
From (i),
g
H
2
g
cos
v
=
q
,From (ii),
2H 8H
T = 2
g g
=
21. (a) For A: It goes up with velocity u will it reaches its
maximumheight (i.e. velocitybecomeszero) andcomes
back to O and attains velocity u.
Using as
2
u
v 2
2
+
= Þ gh
2
u
v 2
A +
=
O vx
u =
vX
u =
A
v
B
v
C
v
u
u
h
2 2
c x y
v v v
= +
For B, going down with velocity u
Þ gh
2
u
v 2
B +
=
For C, horizontal velocityremains same, i.e. u.Vertical
velocity= gh
2
0 + = gh
2
The resultant C
v = 2
y
2
x v
v + = gh
2
u2
+ .
Hence C
B
A v
v
v =
=
22. (c) Since horizontal component ofthevelocityof thebomb
will be the same as the velocity of the aeroplane,
therefore horizontal displacements remain the same at
anyinstant of time.
23. (a) In uniform circular motion speed is constant. So, no
tangential acceleration.
It hasonlyradial acceleration
R
v
a
2
R = [directedtowards
center]
anditsvelocityisalwaysin tangential direction. Sothese
two areperpendicular toeach other.
24. (d) 2usin ( )
T
gcos
q - a
=
a
25 (d) Velocityand kinetic energyis minimum at the highest
point.
q
= 2
2
cos
v
m
2
1
E
.
K
EXERCISE - 2
1. (b) K.E. isminimum at the highestpoint. So,the horizontal
distanceishalfoftherangeRi.e., 0.5R.
2. (b)
g
30
sin
u
g
2
sin
u
R
2
2
1 =
q
= or 3
g
u
or
g
2
u
5
.
1
2
2
=
=
km
3
g
u
g
90
sin
u
R
2
2
2 =
=
=
3. (b) The bullets are fired at the same initial speed
º
30
sin
º
60
sin
º
30
sin
u
g
2
g
2
º
60
sin
u
H
H
2
2
2
2
2
2
=
´
=
¢
2
2
( 3 / 2) 3
1
(1/ 2)
= =
4. (b)
5. (a) 20
)
k̂
4
j
ˆ
3
).(
k̂
5
î
2
(
F
.
F 2
1 =
+
+
=
r
r
6. (d)
g
2
sin
u
2
g
cos
sin
2
u 2
2
2
q
´
=
q
q
or tan q = 2
7. (a) 2 2
1 3 1
m(u cos ) m u
2 4 2
q = ´
or
4
3
cos 2
=
q or º
30
cos
2
3
cos =
=
q .
8. (c)
9. (d)
g
u
6
or
g
u
R
2
2
=
= ;
q
-
q
= 2
2
2
cos
u
2
x
g
tan
x
y
or ÷
ø
ö
ç
è
æ
-
=
6
1
º
45
cos
2
4
º
45
tan
2
h 2
m
3
4
3
2
2 =
-
=
10. (c) We know that,
g
2
)
sin
u
(
H
y
2
m
q
=
=
g
2
sin
u 2
2
q
=
.
u
u
2
H
H D
=
D
Given %
2
u
u
=
D
H
2 2 4%
H
D
= ´ =
11. (c) As,
2
gt
2
1
t
sin
u
s -
q
=
so 2
t
10
2
1
t
)
2
/
3
(
3
20
40 ´
´
-
´
=
or 5t2 – 30t + 40 = 0 or t2 – 6t + 8 = 0
or t = 2 or 4.
Theminimum timet = 2s.
12. (a) 2
2
5
4
R +
= N
41
=
q
R
5N
4N](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-95-320.jpg)
![93
Motion in a Plane
28. (c) A body is in translational equilibrium when the
components of all external forces cancel.
Forthesheet :Fcos q=4 N, Fsinq =3 N.Themagnitude
of F is found byadding the squares ofthe components:
F2cos2 q+ F2 sin2 q = F2 = 42 + 32 = 25 N2. Therefore F
= 5 N. The F vector points in the proper direction,
since tan q = 0.75 = 3/4.
29. (a)
ˆ ˆ ˆ
i j k
v 1 2 2
0 4 3
= -
-
r
v
r
]
0
4
[
k̂
]
3
0
[
j
ˆ
]
8
6
[
î -
+
+
+
-
=
v
r
k̂
4
j
ˆ
3
î
2 +
+
-
= | v | 29 units
Þ =
uu
r
.
30. (b) Use tan
q
+
q
=
a
cos
P
Q
sin
P
Þ tan 90° ¥
=
q
+
q
=
cos
P
Q
sin
P
Q + P cos q = 0 Þ P cos q = -Q.
R= q
+
+ cos
PQ
2
Q
P 2
2
R = 2
2
2
Q
2
Q
P -
+ or R = 2
2
Q
P - = 12
144 = (P + Q)(P - Q) or P - Q = 144/18 = 8.
P = 13 N and Q = 5 N.
31. (b) | A.B| | A B |
= ´
r r
r r
ABcos ABsin
Þ q = q
1
tanθ =1 cosθ =
2
Þ Þ
Now, 2 2
| A B | A B 2ABcos
+ = + + q
r r
2 2 1
A B 2AB.
2
= + + 2 2
A B 2AB
= + +
32. (b)
ˆ ˆ ˆ
3i bj 2k 3 6 2
ˆ ˆ ˆ
 i j k
7 7 7
9 36 4
+ + + æ ö
= = + +
ç ÷
+ + è ø
. If a, b and g are
angles made by A with coordinate axes, then
7
2
cos
and
7
6
cos
,
7
3
cos =
g
=
b
=
a .
33. (b) Solve two equations : R2 = 9P2 + 4P2 + 12P2 cosq and
4R2 = 36P2 + 4P2 + 24P2 cosq.
34. (b)
35. (a)
2
2 2 2 2
x y (x y) (x y) 2(x y)(x y)cos
æ ö
+ = + + - + + - q
ç ÷
è ø
2 2 2 2 2 2 2 2
x y x y 2xy x y 2xy 2(x y )cos
Þ + = + + + + - + - q
or 2 2 2 2
2(x y ).cos (x y )
- q = - +
2 2
1
2 2
(x y )
cos
2(x y )
-
é ù
- +
Þ q = ê ú
-
ê ú
ë û
36. (c) Torque ˆ ˆ ˆ
r F (i j) ( Fk)
t = ´ = - ´ -
u
r r ur
ˆ ˆ ˆ ˆ
F[ i k j k]
= - ´ + ´ ˆ ˆ ˆ ˆ
F(j i) F(i j)
= + = +
ˆ ˆ ˆ ˆ ˆ ˆ
Since k i j and j k i
é ù
´ = ´ =
ë û
37. (a) Component of a
r
along
a . b
b
| b |
=
r
r
r
r
38. (a) x
u 30cos30º 30 3 /2,
= = , uy = 30 sin 30°,
y
v 30sin30 gt
= °-
0
5
.
1
10
º
30
sin
30
vy =
´
-
=
As vertical velocity = 0,
Angle with horizontal a= 0º
It is a state, when a particle reach to a highest point of
its path.
39. (c) From conservation of linear momentum,
1 2
m m
mv v v
2 2
= +
r r r
q
120 m/s
y
Explode
= 50m/sec
v1
v2
50 m/s
(here we take particle & earth as a system so in this
case external force is zero & linear momentum is
conserved)
Where v
r
is velocity of particle before explosion &
1 2
v ,v
r r
are velocity of its equal pieces.
here ˆ
v 60i
=
r (in x direction),
1
ˆ
v 50j
=
r
(in ydirection)
so 2
ˆ ˆ
v 120i 50 j
= -
r
or 2
| v | 130m/sec
=
r
&
[From conservation of linear momentum]
120
50
tan
-
=
q
40. (b) There is no change in horizontal velocity, hence no
change in momentum in horizontal direction. The
vertical velocity at t = 10sec is
10
)
8
.
9
(
º
60
sin
98
v ´
-
´
= = –13.13 m/sec
sochange in momentum in vertical direction is
= (0.5 98 3/2) [ (0.5 13.13)]
´ ´ - - ´
=42.434+6.56 =48.997»49
41. (b) Maximum possible horizontal range = v2/g
Maximum possible area of the circle
2
2 4
2
v v
g g
æ ö p
= p =
ç ÷
ç ÷
è ø
2
v
Here r
g
é ù
=
ê ú
ê ú
ë û](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-97-320.jpg)
![96 PHYSICS
62. (d) j
ˆ
)
t
sin
a
(
î
)
t
cos
a
(
r w
+
w
=
r
dt
d
dt
)
r
(
d
v =
=
r
}
j
ˆ
)
t
sin
a
(
î
)
t
cos
a
{( w
+
w
ĵ
)
t
cos
a
(
î
)
t
sin
a
( w
w
+
w
w
-
=
]
j
ˆ
)
t
cos
a
(
î
)
t
sin
a
[( w
+
w
-
w
=
r.v 0
=
r r
velocity is perpendicular to the displacement.
63. (b) Since
m
E
2
v
mv
2
1
E 2
=
Þ
=
Now at highest point of flight, the vertical component
of velocity is zero & onlyhorizontal component is non
zero. SoK.E. at highest point is
2
1
E ' m(v cos 45º )
2
=
=E /2
64. (b) Given, u1 = u2 = u, q1 = 60º, q2 = 30º
In Ist case, we know that range
g
)
30
90
sin(
u
g
120
sin
u
g
)
60
(
2
sin
u
R
2
2
2
1
°
+
°
=
°
=
°
=
2 2
u (cos 30 ) 3u
g 2g
°
= =
In IInd case when °
=
q 30
2 , then
g
2
3
u
g
60
sin
u
R
2
2
2 =
°
= Þ R1 = R2
(we get same value of ranges).
65. (a) ˆ ˆ
F = 6ti + 4tj or x y
6t 4t
a ,a
3 3
= =
so 2
t
u = a dt = t
o
x x
ò (u ) = 9m/sec
x t=3
Þ
and
2
2t
t
u = a dt =
y y
o 3
ò sec
/
m
6
)
u
( 3
t
y =
Þ =
(because ux & uy = 0 at t = 0 sec)
66. (b)
2
gt
2
1
s = , s and g are same for both the balls, so time
of fall ‘t’ will also be the same for both of them (s is
vertical height)
67. (c) B
A
sum
vector
P
r
r
r
+
=
=
Q vector differences A B
= = -
r r r
Since P
r
and Q
r
are perpendicular
0
Q
.
P =
r
r
Þ (A B).(A B) 0
+ - =
r r
r r
Þ A
A2 = B2
Þ B
A =
68. (b) Circumferenceofcircleis 2pr = 40m
Total distance travelled in tworevolution is 80m. Initial
velocityu =0, final veloctiyv = 80 m/sec
so from
v2 =u2+2as
Þ (80)2 =02+2×80×a
Þ a = 40m/sec2
69. (b) q
=
´ sin
B
A
|
B
A
|
r
r
q
= cos
B
A
B
.
A
r
r
B
.
A
3
|
B
A
|
r
r
r
r
=
´
Þ AB sin q = Ö3 AB cos q or tan q = Ö3
q = 60º
2 2
| A B| A B 2ABcos60º
+ = + +
r r
AB
B
A 2
2
+
+
=
70. (b) For two vectors to be perpendicular to each other
A B
® ®
× = 0
(2 3 8
i j k
Ù Ù Ù
+ + )· (4 4
j i k
Ù Ù Ù
- +a )= 0
–8 + 12 + 8a = 0 or a = - = -
4
8
1
2
71. (d) 72. (c) 73. (d) 74. (d) 75. (a)
EXERCISE - 3
Exemplar Questions
1. (b) Given, ˆ ˆ
A i j
= +
ˆ ˆ
B i j
= -
As we know that
| || | cos
A B A B
× = q
r r
2 2 2 2
ˆ ˆ ˆ ˆ
( ) ( ) ( 1 1 )( 1 1 )cos
i j i j
+ × - = + + q
( )( ) 2 2 cos
i j i j
+ - = ´ q
where q is the angle between A and B
1 0 0 1
cos 0
2 2
- + -
q = =
q = 90°
2. (d) A scalar quantity does not depend on direction so it
has the same value for observers with different
orientations of the axes.
3. (b) From the diagram, ˆ ˆ
u ai bj
= +
As u is in the first quadrant, so both components a
and b will be positive.
For ˆ ˆ,
v pi qj
= + as it is in positive x-direction and
located downward so x-component p will be positive
and y-component q will be negative.
Hence, a, b and p are positive but q is negative.
4. (b) Let r makes an angle q with positive x-axis component
of r along x-axis.
| | cos
x
r r
= q
maximum maximum
( ) | | (cos )
x
r r
= q
| | cos0 | |
r r
= ° =
( cosq
Q is maximumofq=0°)
q = 0°.
Hence the vector r has maximum value along positive
x-axis.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-100-320.jpg)
![97
Motion in a Plane
5. (c) Consider, projectile is fired at an angle q.
According to question,
q = 15° and R = 50 m
R
X
Y
u
q
Range,
2
sin 2
u
R
g
q
=
2
sin(2 15 )
50m
u
R
g
´ °
= =
2 2 1
50 sin30
2
g u u
´ = ° = ´
50 × g × 2 = u2
2
50 9.8 2 100 9.8 980
u = ´ ´ = ´ =
980
u = = 31.304 m/s 14 5
=
2
( 9.8 m/s )
=
g
Q
Now, 45 ;
q = °
2 2
sin 2 45
u u
R
g g
´ °
= =
2
(14 5) 14 14 5
100 m
9.8
R
g
´ ´
= = =
6. (b) As we know that,
Impulse,
D
æ ö
= D = D = D
ç ÷
è ø
D
p
I F t t p
t
where F is force, Dt is time duration and Dp is change
in momentum.AsDp is a vector quantity, henceimpulse
is also a vector quantity. Sometimes area can also be
treated as vector.
7. (d) As speed is a scalar quantity, hence it will be related
with path length (scalar quantity) only.
Hence, Speed 0
total distance travelled
time taken
v =
So, total distance travelled = Path length
= (speed) × time taken
Hence, path length which is scalar and traversed in
equal intervals.
8. (c) As given that in two dimensional motion the
instanteous speed v0 is positive constant and we know
that acceleration is rate of change of velocity or
instantaneous speed and hence it will also be in the
plane of motion.
NEET/AIPMT (2013-2017) Questions
9. (b) At point B the direction of velocity component of the
projectile along Y - axis reverses.
Hence, B
V
®
= 2i 3j
-
$ $
10. (d) Vector triple product
( ) ( ) ( ) 0
A B C B A C C A B
´ ´ = × - × =
r r r r r r
r r r
Þ || ( )
A B C
´
r r
r
[ 0 and 0]
× = × =
r r r
r
Q A B A C 1.
(a) 2 2
( ) ( )
A B C
+ =
r r
r
2 2
2 .
A B A B C
2
Þ + + =
r r
2 2
3 4 2 . 5
A B
2
Þ + + =
r r
2 . 0
A B
Þ =
r r
or . 0
A B
Þ =
r r
A B
^
r r
Here A2 + B2 = C2. Hence, A B
^
uu
r uu
r
11. (d) av
v
r
=
r (displacement)
t (time taken)
D
D
r
=
ˆ ˆ
(13 2)i (14 3)j
5 0
- + -
-
=
11 ˆ ˆ
(i j)
5
+
12. (a) ( )
A
V 10 –i
=
ur
$
( )
B
V 10 j
=
ur
$
BA
ˆ
V 10 j 10 i 10 2 km / h
= + =
ur
$
DistanceOB = 100 cos 45° = 50 2 km
O
S
100 km
w 10 km/h A
45°
100 km
N(j)
$
BA
V 10 2 km / h
=
B
Time taken to reach the shortest distance between
A and
BA
OB
B
V
= uuuur
50 2
5h
10 2
= =
13. (b) Two vectors are
A
r
= ˆ ˆ
cos ti sin tj
w + w
B
r
=
t t
ˆ ˆ
cos i sin j
2 2
w w
+](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-101-320.jpg)
![101
Lawsof Motion
B
dx
v velocity of block
dt
ù
= = ú
û
Again differentiating we get, 2ap = aB
B
p B
dv
dVp
a and a
dt dt
é ù
= =
ê ú
ë û
ap = acceleration of pulley, aB = acceleration of block
Case 2 Here =
+
+ y
x
h 2
2
constt. On differentiating w.r.t ‘t’
F
x
y
h
q
1 2
[Negative sign with dy/dt shows that with increase in time, y
decreases]
2 2
1 2x dx dy
0
dt dt
2 h x
´
- =
+
Þ cos q (v1 – v2) = 0
2 2
x
cos
h x
é ù
q =
ê ú
ê ú
+
ë û
Q
Case 3 Wedge block system : Thin lines represents thecondition
of wedge block at t = 0 and dotted lines at t = t
ax
c
Ax
Ax A
q
q
B
ay
ax
ay
Ax = acceleration of wedge towards left
ax, ay = acceleration of block as shown
From DABC, y
x x
a
tan
a A
q =
+
Frame of Reference :
Reference frames are co-ordinate systems in which an event is
described.
There are two typesof reference frames
(a) Inertial frame of reference: These are frames of reference
in which Newton’s laws hold good. These frames are at rest
with each other or which are moving with uniform speed
with respect to each other.
All reference frames present on surface of Earth are
supposed to be inertial frame of reference.
(b) Non – inertial frame of reference: Newton’s law do not
hold good in non-inertial reference frame.
All accelerated and rotatory reference frames are non –
inertial frameof reference. Earth isa non-intertial frame.
When the observer is in non-inertial reference framea
pseudo force is applied on the body under observation.
Free Body Diagram (FBD) :
Free body diagram of a mass is a separate diagram of that mass.
All forces acting on the mass are sketched. A FBD is drawn to
visualise the direct forces acting on a body.
Case 1 : Masses M1 and M2 are tied to a string, which goes over
a frictionless pulley
(a) If M2 > M1 and they move with accelerationa
1
M
M2
M g
2
M g
1
T
T
a
a
FBD of M1, FBD of M2
T
M1
1
M g
a
T
M2
M g
2
a
1 1
T M g M a
- = 2 2
M g T M a
- =
where T is the tension in the string. It gives
2 1
1 2
M M
a g
M M
-
=
+
and 1 2
1 2
2M M
T g
M M
=
+
(b) If the pulley begins to move with acceleration f,
downwards
2 1
1 2
( )
M M
a g f
M M
-
= -
+
uu
r uu
r uu
r
and 1 2
1 2
2
( )
M M
T g f
M M
= -
+
ur uu
r uu
r
Case 2 : Three massesM1, M2 and M3 are connected with strings
as shown in the figure and lie on a frictionless surface. They are
pulled with a force F attached to M1.
M3 M2 M1 F
T 1
T1
T2
T2
The forces on M2 and M3 are as follows
2 3
1
1 2 3
M M
T F
M M M
+
=
+ +
and 3
2
1 2 3
M
T F
M M M
=
+ +
;
Acceleration of the system is
1 2 3
F
a
M M M
=
+ +
Case 3 : Two blocks of masses M1 and M2 are suspended
vertically from a rigid support with the help of strings as shown
in the figure. The mass M2 is pulled down with a force F.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-105-320.jpg)
![103
Lawsof Motion
FBD of M2 a
T
M g
2
M gcos
2 b
M
gsin
2
b
M
2
b
Equation of motion for M2
M2g sinb – T = M2a
or T = M2g sin b – M2a ...(2)
Using eqn. (1) and (2) we get,
M1g sin a + M1a = M2g sin b – M2a
Solving we get,
( )
2 1
1 2
sin sin
M M g
a
M M
b - a
=
+
and 1 2
1 2
[sin sin ]
M M g
T
M M
= b + a
+
Case 7 :Aperson/monkey climbing a rope
T
Mg
a
a
(a) A person of mass M climbs up a rope with acceleration a.
The tension in the rope will be M(g+a).
T – Mg = Ma Þ T = M(g + a)
(b) If the person climbs down along the rope with acceleration
a, thetension in the rope will be M(g–a).
a a
Mg
T
Mg – T = Ma Þ T = M(g – a)
(c) When the person climbs up or down with uniform speed,
tension in the string will be Mg.
Case 8 : Abodystarting from rest moves along a smooth inclined
plane of length l, height h and having angle of inclination q.
h
l
q
FBD of body
N=R
mg
mg sinq mg cosq
q
(where N=R is normal reaction applied by plane on the body
of mass m)
For downward motion, along the inclined plane,
q
=
Þ
=
q sin
g
a
ma
sin
mg
Bywork-energytheorem loss in P.E. = gain in K.E.
gh
2
v
mv
2
1
mgh 2
=
Þ
=
Þ
Also, from the figure, h = l sin q. q
=
=
sin
g
2
gh
2
v l
(a) Acceleration down the plane is g sin q.
(b) Its velocity at the bottom of the inclined plane will be
2 2 sin
gh g
= q
l
(c) Time taken to reach the bottom will be
1/ 2
1/ 2
2 1/ 2
2 2 1 1 2
sin sin
sin
sin
2
h h
t
g g
g g
h
æ ö
æ ö
= = = =
ç ÷
ç ÷ ç ÷
q q
q
è ø æ ö
è ø qç ÷
è ø
l
(d) Ifangles ofinclination are q1 and q2 for twoinclined planes
Keeping the length constant then
½
1 2
2 1
sin
sin
t
t
æ ö
q
= ç ÷
q
è ø
Case 9 : Weight of a man in a lift :
(i) When lift is acceleratedupward :In thiscase theman also
movesin upwarddirection with an acceleration a
r
.
N
a a
mg
Then from Newton’ second law
N – mg = ma or N = m(g + a)
or Wapp = m(g + a) (1 / )
o
W a g
= + (as W = mg)
Where Wapp is apparent weight ofthe man in the lift, Wo is
thereal weight, N isthe reaction oflift on theman. It is clear
that N = Wapp
When the lift moves upward and if we measure the weight
of the man byany means (such as spring balance) then we
observe more weight (i.e., Wapp) than the real weight (Wo)
Wapp >Wo
(ii) When lift is accelerated downward : In this case from
Newton’s second law
N
a
mg
mg – N = ma
or N = m(g – a) = Wo(1– a/g)
or W'app= Wo(1– a/g) { }
mg
Wo =
Q
If we measure the weight of man by spring balance, we
observe deficiency because Wapp< Wo.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-107-320.jpg)
![104 PHYSICS
(iii) Whenlift is at rest or movingwithconstantvelocity :From
Newton’s second law N –mg = 0 or N = mg
In this case spring balance gives the true weight of the man.
Case 10 : Three masses M1, M2 and M3 are placed on a smooth
surface in contact with each other as shown in the figure.
A force F pushes them as shown in the figure and the three
masses move with acceleration a,
M3
M2
F2
F1
F2 F1
M1
F
a
M1
F
F1
Þ F – F1 = m1a ...(i)
M2
F2
F1
Þ F1 – F2 = m2a ...(ii)
F2
M3
Þ F2 = M3 a ...(iii)
Adding eqns. (i), (ii) and (iii) we get,
1 2 3
F
a
M M M
=
+ +
Þ 3
2
1 2 3
M F
F
M M M
=
+ +
and 2 3
1
1 2 3
( )
M M F
F
M M M
+
=
+ +
Keep in Memory
1. When a man jumps with load on his head, the apparent
weight of the load and the man is zero.
2. (i) If a person sitting in a train moving with uniform
velocitythrows a coin verticallyup, then coin will fall
back in his hand.
(ii) If the train is uniformlyaccelerated, the coin will fall
behind him.
(iii) Ifthe train is retarded uniformly, then the coin will fall
in front ofhim.
Example1.
A chain of length l is placed on a smooth spherical surface
of radius R with one of its ends fixed at the top of the
sphere. What will be the acceleration a of each element of
the chain when its upper end is released? It is assumed
that the length of chain
R
π
2
æ ö
<ç ÷
è ø
l .
Solution :
Let m be the mass of the chain of length l. Consider an
element of length dlof thechain at an angle q with vertical,
R
q
dl
dq
From figure, dl = R d q ;
Mass of the element,
dm = l
l
d
m
; or dm = q
d
R
.
m
l
Force responsible for acceleration, dF = (dm)g sinq ;
dF = q
q
=
q
÷
ø
ö
ç
è
æ
q d
sin
mgR
)
sin
g
(
d
R
m
l
l
Net force on the chain can be obtained by integrating the
above relation between 0 to a, we have
ò
a a
a
-
=
q
-
=
q
q
=
0 0
]
cos
1
[
R
mg
)
cos
(
R
mg
d
sin
R
mg
F
l
l
l
ú
û
ù
ê
ë
é
-
=
R
cos
1
R
mg l
l
;
Acceleration,
F gR
a 1 cos
m R
æ ö
= = -
ç ÷
è ø
l
l
.
Example2.
A block slides down a smooth inclined plane to the ground
when released at the top, in time t second. Another block
is dropped vertically from the same point, in the absence
of the inclined plane and reaches the ground in t/2 second.
Then find the angle of inclination of the plane with the
vertical.
Solution :
If q is the angle which the inclined plane makes with the
vertical direction, then the acceleration of the block sliding
down the plane of length l will be g cosq.
A
C B
q
l h
Using the formula, 2
at
2
1
ut
s +
= , we have s = l, u = 0, t = t
and a = g cos q.
so 2
2
t
)
cos
g
(
2
1
t
cos
g
2
1
t
0 q
=
q
+
´
=
l ...(i)
Taking vertical downward motion of the block, we get
4
/
gt
2
1
)
2
/
t
(
g
2
1
0
h 2
2
=
+
= ...(ii)
Dividing eqn. (ii) by (i), we get
q
=
cos
4
1
h
l
]
/
h
cos
[ l
Q =
q
or
q
=
q
cos
4
1
cos ; or
4
1
cos2
=
q ; or
2
1
cos =
q
or q = 60º](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-108-320.jpg)
![105
Lawsof Motion
Example3.
A large mass M and a small mass
m hang at the two ends of a string
that passes through a smooth
tube as shown in fig. The mass m
moves around a circular path in
a horizontal plane. The length of
the string from mass m to the top
of the tube is l, and q is the angle
the string makes with the
vertical. What should be the
frequency (n) of rotation of mass
m so that mass
M remains stationary? M
m
l
r
q
T
Solution :
Tension in the string T = Mg.
Centripetal force on the body= mrw2 =mr ( 2pn )2. This is
provided by the component of tension acting horizontally
i.e. T sinq ( = Mg sinq).
mr ( 2pn)2 = Mg sinq = Mgr/l. or
1 Mg
2 m
n =
p l
Example4.
A string of negligible mass going over a clamped pulley of
mass m supports a block of mass M as shown in fig. The
force on the pulley by the clamp is given by
(a) 2 Mg
(b) 2 mg
M
m
(c) 2 2
[ (M m) m ]
+ + g
(d) 2 2
[ (M m) M ]
+ + g
Solution : (c)
Force on the pulley by the clamp = resultant of
T = (M + m)g and mg acting along horizontal and vertical
respectively
F 2 2
[(M m)g] (mg)
= + + 2 2
[ (M m) m ]g
= + +
Example5.
The masses of 10 kg and 20 kg respectively are connected
by a massless spring infig. A force of 200 newton actson the
20 kg mass. At the instant shown, the 10 kg mass has
acceleration 12 m/sec2. What is the acceleration of 20 kg
mass?
10 kg
20 kg
200 newton
Solution :
Force on 10 kg mass = 10 × 12 = 120 N
Themassof10kgwill pullthemassof20kgin thebackward
direction with a force of 120 N.
Net force on mass 20 kg = 200 – 120 = 80 N
Its acceleration
force
a
mass
= 2
s
/
m
4
kg
20
N
80
=
=
Example6.
Two masses each equal to m are lying on X-axis at (–a, 0)
and (+ a, 0) respectively as shownin fig. They are connected
by a light string. A force F is applied at the origin and along
the Y-axis. As a result, the masses move towards each other.
What is the acceleration of each mass? Assume the
instantaneous position of the masses as (– x, 0) and (x, 0)
respectively
m O m
(–a, 0) (a, 0)
–X X
F
Solution :
O
(–x, 0) (x, 0)
F
T T
q
B C
A
From figure F = 2 T cos q or T = F/(2 cos q)
The force responsible for motion of masses on X-axis is T
sin q
q
´
q
=
q
=
sin
cos
2
F
sin
T
a
m
)
x
a
(
x
2
F
OA
OB
2
F
tan
2
F
2
2
-
´
=
´
=
q
=
so,
)
x
a
(
x
m
2
F
a
2
2
-
´
=
Example7.
A block of mass M is pulled along horizontal frictionless
surface by a rope of mass m. Force P is applied at one end
of rope. Find the force which the rope exerts on the block.
Solution :
The situation is shown in fig
M
T O
m
P
Let a be the common acceleration of the system. Here
T = M a for block
P – T = m a for rope
P – M a = m a or P = a (M + m) or
)
m
M
(
P
a
+
=
)
m
M
(
P
M
T
+
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-109-320.jpg)
![107
Lawsof Motion
Example10.
A hammer of mass M strikes a nail of mass m with velocity
of u m/s and drives it ‘s’ meters in to fixed block of wood.
Find the average resistance of wood to the penetration of
nail.
Solution :
Applying the law of conservation of momentum,
m u = (M + m) v0 Þ u
M
m
M
v0 ÷
ø
ö
ç
è
æ
+
=
There acceleration a can be obtained using the formula
(v2 = u2 + 2as).
Here we have 0 – v0
2 = 2as or a = v0
2 /2s
s
2
u
M
m
M
a
2
2
÷
ø
ö
ç
è
æ
+
=
Resistance = (M + m) a
2 2
M u
m M 2 s
æ ö
= ç ÷
ç ÷
+
è ø
Example11.
A ball of mass 0.5 kg is thrown towards a wall so that it
strikes the wall normallywith a speed of 10 ms–1. If the ball
bounces at right angles away from the wall with a speed of
8ms–1, what impulse does the wall exert on the ball ?
Solution :
f
10
8
Approaching wall
u = –10 ms–1
Leaving wall
v = +8 ms–1
Taking the direction of the impulse J as positive and using
J = mv – mu
we have
1 1
J 8 ( 10) 9
2 2
= ´ - - = N-s
Therefore the wall exerts an impulse of9 N-s on the ball.
Example12.
Two particles, each of mass m, collide head on when their
speeds are 2u and u. If they stick together on impact, find
their combined speed in terms of u.
Solution :
2u
Before impact
After impact
m
u
v
2m
m
Using conservation of linear momentum (in the direction of
the velocity 2u) we have
(m) (2u) – mu = 2m × V Þ
1
V u
2
=
The combined mass will travel at speed u/2.
(Note that the momentumofthe second particlebefore impact
is negative because its sense is opposite to that specified
as positive.)
Force vary with time and impulse is area under force versus
time curve
Area=Fext.Dt
t
tf
ti
Fext.
Fav
Fext.
(b)
.
Force constant with time i.e., .
ext
F
r
constant with time (shown
by horizontal line) and it would give same impulse to particle
in time Dt = tf – ti as time varying force described.
It is a vector quantityhaving a magnitude equal to the area under
theforce-timecurve asshown in fig. (a). In thisfigure, it isassumed
that force varieswith time and is non-zeroin time interval Dt = tf–
ti. Fig.(b) shows the time averaged force .
ext
F
r
i.e., it is constant
in time interval Dt, then equation (iii) can be written as
f
i
t
ext. t
I F dt
= ò
r
)
t
t
(
F i
f
.
ext -
=
r
t
F
I .
ext D
=
r
...(iv)
The direction of impulsive vector I is same as the direction of
change in momentum. Impulse I has same dimensions as that of
momentum i.e, [MLT–1]
Rocket propulsion (A case of system of variable mass ) : It is
based on principleofconservation of linearmomentum.
In rocket, the fuel burns and produces gases at high temperature.
These gases are ejected out of the rocket from nozzle at the
backside of rocket and the ejecting gas exerts a forward force on
the rocket which accelerates it.
Let the gas ejects at a rate
dt
dM
r -
= and at constant velocity u
w.r.t. rocket then from the conservation of linear momentum
rt
M
ru
M
ru
dt
dv
0 -
=
= whereM = M0 - rt and M0 is mass of rocket
with fuel and solving this equation, we get ÷
÷
ø
ö
ç
ç
è
æ
-
=
rt
M
M
log
u
v
0
0
e
where v = velocity of rocket w.r.t. ground.
Example9.
Two skaters A and B approach each other at right angles.
Skater A has a mass 30 kg and velocity 1 m/s and skater
B has a mass 20 kg and velocity 2 m/s. They meet and
cling together. Find the final velocity of the couple.
Solution :
Applying principle of conservation of linear momentum,
2
2
2
2
1
1
2
1
2
2
2
1 )
v
m
(
)
v
m
(
v
)
m
m
(
;
p
p
p +
=
+
+
=
( ) ( ) ( )
2 2
30 20 v 30 1 20 2 50
+ = ´ + ´ =
50
v 1 m / s
50
= =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-111-320.jpg)
![110 PHYSICS
Solution :
The situation is shown in fig.
F sinq
F cosq
W cosq
W sinq
F
R
q
W
q
q
f= R
m
Let F be the horizontal force needed to just push the object
up the plane. From figure R = W cos q + F sin q
Now f = mR = m [W cos q + F sin q] ...(1)
Further, F cos q = W sin q + f ...(2)
F cos q = W sin q + m [W cos q + F sin q]
F cos q – m F sin q = W sin q + m W cos q
)
sin
(cos
)
cos
(sin
W
F
q
m
-
q
q
m
+
q
=
CASES OF CIRCULAR MOTIONS
Motion in a Vertical Circle :
Let us consider a particle of mass m attached toa string oflength
R let the particle be rotated about its centre O.
At t = 0 the particle start with velocity u from the point A(lowest
point of vertical circle) and at time t its position is P. Then the
tension at point P is given by
T
P
vP
q
q
O
B
R
A u
mg mg
mg cos q
sin q
R
mv
cos
mg
T
2
P
P =
q
- or
R
mv
cos
mg
T
2
P
P +
q
= ...(1)
So tension at point A(lowest point of vertical circle) is
R
mv
mg
T
2
A
A =
- (Q q = 0º) ...(2)
and tension at point B (highest point of vertical circle) is
R
mv
mg
T
2
B
B =
+ (Q q=180º) ...(3)
Where
2
mv
r
is centripetal force required for the particletomove
in a vertical circle.
Now from law of conservation of energy
mgR
2
mv
2
1
mv
2
1 2
B
2
A =
-
or, 2 2
A B
v v 4gR
- = ...(4)
(change in kinetic energyof particle)
= (change in potential energy of particle)
or
(loss in kinetic energyofthe particle) = (gain in potential energy)
In conservative force system (such as gravity force) the
mechanical energy(i.e., kinetic energy+ potential energy) must
be constant.
Total energy will be constant
Nowfrom eqns.(2) and (3), we get
mg
R
TB
vB B
A
D
mg
R
TA
v = v
A C
B
A
D
)
V
V
(
R
m
mg
2
T
T 2
B
2
A
B
A -
+
=
- )
gR
4
(
R
m
mg
2 +
=
Þ mg
6
T
T B
A =
- ...(5)
or TA =TB + 6mg ...(6)
So it is clear from eqn. (6) that tension in string at lowest point
of vertical circle is greater then the tension at highest point of
vertical circle by 6mg.
Condition to complete a vertical circle :
Ifwe reducethevelocityvA in equation (2), then TA will be reduce
and at some critical velocity vc, TB will be zero, then put TB = 0
and vB = vC in equation (3) and we obtain
C B
v v gR
= = ...(7)
In this condition the necessary centripetal force at point B is
provided by the weight of the particle [see again equation (3)]
then from equation (4), we get
2
4 5
A A
v gR gR v gR
- = Þ = ...(8)
then the tension at the point A will be
mg
6
R
)
gR
5
(
m
mg
TA =
+
= ...(9)
Hence if we rotate a particle in a vertical circle and tension in
string at highest point is zero, then the tension at lowest point of
vertical circle is 6 times of the weight of the particle.
Some Facts of Vertical Motion :
(i) The body will complete the vertical circle if its velocity at
lowest point is equal to or greater then gR
5
(ii) The bodywill oscillate about the lowest point if itsvelocity
at lowest point is less then gR
2 . This will happen when
the velocity at the halfwaymark, i.e.
ú
û
ù
ê
ë
é
=
= mgR
mv
2
1
0
v 2
A
D Q
(iii) The string become slack and fails to describe the circle
when its velocity at lowest point lies between
gR
5
to
gR
2](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-114-320.jpg)
![116 PHYSICS
13. A particle ofmass mmoving eastwardwith a speed vcollides
with another particle of the same mass moving northward
with the same speed v. The two particles coalesce on
collision. The newparticleofmass 2mwill movein thenorth-
external direction with a velocity:
(a) v/2 (b) 2v
(c) 2
/
v (d) None of these
14. A spring is compressed between two toy carts of mass m1
and m2. When the toy carts are released, the springs exert
equal and opposite average forces for the same time on
each toy cart. If v1 and v2 are the velocities of the toy carts
and there is no friction between the toy carts and the ground,
then :
(a) v1/v2 = m1/m2 (b) v1/v2 = m2/m1
(c) v1/v2 = –m2/m1 (d) v1/v2 = –m1/m2
15. Two mass m and 2m are attached with each other by a rope
passing over a frictionless and massless pulley. If the pulley
is accelerated upwards with an acceleration ‘a’, what is the
value of T?
(a)
3
a
g +
(b)
3
a
g -
(c)
3
)
a
g
(
m
4 +
(d)
3
)
a
g
(
m -
16. A rider on a horse back falls forward when the horse
suddenly stops. This is due to
(a) inertia of horse
(b) inertia of rider
(c) large weight of the horse
(d) losing of the balance
17. A ball of mass m is thrown verticallyupwards. What is the
rate at which the momentum of the ball changes?
(a) Zero (b) mg
(c) Infinity (d) Data is not sufficient.
18. A small block is shot into each of the four tracks as shown
below. Each of the tracks rises to the same height. The
speed with which the block enters the track is thesame in all
cases.At the highest point of the track, the normal reaction
ismaximumin
(a)
v
(b)
v
(c)
v
(d) v
19. A weight W rests on a rough horizontal plane. If the angle
of friction be q , the least force that will move the body
along the plane will be
(a) q
cos
W (b) q
cot
W
(c) q
tan
W (d) q
sin
W
20. A particle starts sliding down a frictionless inclined plane.
If n
S is the distance traveled byit from time t = n – 1 sec to
t = n sec, the ratio Sn/Sn+1 is
(a)
1
n
2
1
n
2
+
-
(b)
n
2
1
n
2 +
(c)
1
n
2
n
2
+
(d)
1
n
2
1
n
2
-
+
21. Ablock iskepton a inclinedplaneofinclination qoflength l .
The velocity of particle at the bottom of inclined is (the
coefficient offriction is m )
(a) 2
/
1
)]
sin
cos
(
g
2
[ q
-
q
m
l (b) )
cos
(sin
g
2 q
m
-
q
l
(c) )
cos
(sin
g
2 q
m
+
q
l (d) )
sin
(cos
g
2 q
m
+
q
l
22. A bird is in a wire cage which is hanging from a spring
balance . In the first case, the bird sits in the cage and in the
second case, thebird fliesabout insidethecage. Thereading
in the spring balance is
(a) more in the first case
(b) less in first case
(c) unchanged
(d) zero in second case.
23. In an explosion, a bodybreaks up into twopieces ofunequal
masses. In this
(a) both parts will have numerically equal momentum
(b) lighter part will have more momentum
(c) heavier part will have more momentum
(d) both parts will have equal kinetic energy
24. A block of mass m on a rough horizontal surface is acted
upon by two forces as shown in figure. For equilibrium of
block the coefficient offriction between block and surface is
m
q
F1
F2
(a)
q
+
q
+
cos
F
mg
sin
F
F
2
2
1
(b) q
-
+
q
sin
F
mg
F
cos
F
2
2
1
(c)
q
+
q
+
sin
F
mg
cos
F
F
2
2
1
(d)
1 2
2
F sin F
mg F cos
q -
- q
25. A plate of mass M is placed on a horizontal of frictionless
surface (see figure), and a body of mass m is placed on this
plate. The coefficient of dynamic friction between this body
and the plate is m . If a force 2 m mg is applied to the body
ofmass m along the horizontal, theacceleration of the plate
will be
M
m
2 mg
m
(a) g
M
m
m
(b) g
)
m
M
(
m
+
m
(c) g
M
m
2m
(d) g
)
m
M
(
m
2
+
m
.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-120-320.jpg)
![119
Lawsof Motion
26. Two trolleys of mass m and 3m are connected by a spring.
They were compressed and released at once, they move off
in opposite direction and come to rest after covering a
distance S1, S2 respectively. Assuming the coefficient of
friction to be uniform, ratio of distances S1 : S2 is :
(a) 1: 9 (b) 1: 3
(c) 3: 1 (d) 9: 1
27. A particle of mass 10 kg is moving in a straight line. If its
displacement, x with time t is given by x = (t3 – 2t – 10) m,
then the force acting on it at the end of 4 seconds is
(a) 24N (b) 240N
(c) 300N (d) 1200N
28. When forces F1, F2, F3 are acting on a particle of mass m
such that F2 and F3 are mutually perpendicular, then the
particle remains stationary. If the force F1 is now removed
then the acceleration of the particle is
(a) F1/m (b) F2F3/mF1
(c) (F2 – F3)/m (d) F2/m
29. One end of massless rope, which passes over a massless
and frictionless pulley P is tied to a hook C while the
other end is free. Maximum tension that the rope can
bear is 360 N. With what value of maximum safe
acceleration (in ms–2) can a man of 60 kg moves
downwards on the rope? [Take g = 10 ms–2]
P
C
(a) 16 (b) 6
(c) 4 (d) 8
30. A force k̂
10
ĵ
6
î
8
F -
-
=
r
newton produces an acceleration
of 1 ms–2 in a body. The mass of the body is
(a) 10kg (b) 2
10 kg
(c) 3
10 kg (d) 200kg
31. A uniform chain of length 2 m is kept on a table such that a
length of 60 cm hangs freelyfrom the edge of the table. The
total mass of the chain is 4 kg. What is the work done in
pulling the entire chain on the table ?
(a) 12J (b) 3.6J
(c) 7.2J (d) 1200J
32. A body of mass 1 kg moving with a uniform velocity of
1
ms
1 -
. If the value of g is 2
ms
5 -
, then the force acting on
the frictionless horizontal surface on which the body is
moving is
(a) 5N (b) 1N
(c) 0N (d) 10N
33. A trailer of mass 1000 kg is towed by means of a rope
attached to a car moving at a steady speed along a level
road. The tension in the rope is 400 N. The car starts to
accelerate steadily. Ifthe tension in the rope is now 1650 N,
with what acceleration is the trailer moving ?
(a) 1.75ms–2 (b) 0.75ms–2
(c) 2.5 ms–2 (d) 1.25ms–2
34. Arocket of mass5000 kg is to beprojected verticallyupward.
The gases are exhausted verticallydownwards with velocity
1000 ms–2 with respect tothe rocket. What is the minimum
rate of burning the fuel so as to just lift the rocket upwards
against gravitational attraction ?
(a) 49 kg s–1 (b) 147 kg s–1
(c) 98 kg s–1 (d) 196 kg s–1
35. BlocksAand Bof masses 15 kg and 10 kg, respectively, are
connected bya light cable passing over a frictionless pulley
as shown below. Approximately what is the acceleration
experienced by the system?
(a) 2.0m/s2
(b) 3.3m/s2
(c) 4.9m/s2
A
B
(d) 9.8m/s2
36. A50 kg ice skater, initiallyat rest, throwsa 0.15 kg snowball
with a speed of35 m/s. What is the approximate recoil speed
of the skater?
(a) 0.10m/s (b) 0.20m/s
(c) 0.70m/s (d) 1.4m/s
37. Block A is moving with acceleration A along a frictionless
horizontal surface. When a second block, B is placed on top
of Block Athe acceleration of the combined blocks drops to
1/5 the original value. What is the ratio of the mass ofA to
the mass of B?
(a) 5: 1 (b) 1: 4
(c) 3: 1 (d) 2: 1
38. A force F is used to raise a 4-kg mass M from the ground to
a height of 5 m.
60°
F
M
What is the work done bythe force F? (Note : sin 60° = 0.87;
cos60° = 0.50. Ignore friction andthe weightsofthe pulleys)
(a) 50J (b) 100J
(c) 174J (d) 200J
39. A 5000 kg rocket is set for vertical firing. The exhaust speed
is 800 m/s. Togive an initial upward acceleration of 20 m/s2,
the amount of gas ejected per second to supply the needed
thrust will be (Takeg = 10 m/s2)
(a) 127.5kg/s (b) 137.5kg/s
(c) 155.5kg/s (d) 187.5kg/s](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-123-320.jpg)
![120 PHYSICS
40. A bullet is fired from a gun. The force on the bullet is given
byF = 600 – 2 × 105 t
Where, F is in newtons and t in seconds. The force on the
bullet becomes zero as soon as it leaves the barrel. What is
the average impulse imparted to the bullet?
(a) 1.8 N-s (b) Zero
(c) 9 N-s (d) 0.9 N-s
41. A rifle man, whotogether with his rifle has a mass of100 kg,
stands on a smooth surface and fires 10 shots horizontally.
Each bullet hasa mass10 gand a muzzlevelocityof 800 ms–
1. The velocity which the rifle man attains after firing 10
shots is
(a) 8 1
ms-
(b) 0.8
1
ms-
(c) 0.08 1
ms-
(d) – 0.8 1
ms-
42. A block of mass 4 kg rests on an inclined plane. The
inclination to the plane is gradually increased. It is found
that when the inclination is 3 in 5, the block just begins to
slidedown the plane. The coefficient of friction between the
block and the plane is
(a) 0.4 (b) 0.6
(c) 0.8 (d) 0.75.
43. The minimum velocity(in ms-1)with which a car driver must
traverse a flat curveofradius 150m and coefficient offriction
0.6 to avoid skidding is
(a) 60 (b) 30
(c) 15 (d) 25
44. A body of mass 2 kg is placed on a horizontal surface
having kinetic friction 0.4 and static friction 0.5. If the force
applied on the bodyis 2.5 N, then the frictional force acting
on the body will be [g = 10 ms–2]
(a) 8N (b) 10 N
(c) 20N (d) 2.5N
45. Abag ofsand of mass m is suspended bya rope. A bullet of
mass
20
m
is fired at it with a velocity v and gets embedded
into it. The velocity of the bag finally is
(a)
v
21
20
´ (b)
20v
21
(c)
v
20
(d)
v
21
46. For thearrangement shown in the Figure thetension in the
string is [Given : 1
tan (0.8) 39
-
= ° ]
39°
m = 1 kg
m = 0.8
(a) 6N (b) 6.4N
(c) 0.4N (d) zero.
47. A1kg block anda 0.5 kgblock movetogether on a horizontal
frictionless surface . Each block exerts a force of6 N on the
other. The block move with a uniform acceleration of
a
1 kg 0.5 kg
F
(a) 3 2
ms-
(b) 6 2
ms-
(c) 9 2
ms-
(d) 12 2
ms-
48. A body of mass 32 kg is suspended by a spring balance
from the roof of a vertically operating lift and going
downward from rest. At the instant the lift has covered 20 m
and 50 m, the spring balance showed 30 kg and 36 kg
respectively. Then the velocity of the lift is
(a) decreasing at 20 m, and increasing at 50 m
(b) increasing at 20m and decreasing at 50 m
(c) continuously decreasing at a steady rate throughout
the journey
(d) constantly increasing at constant rate throughout the
journey.
49. An object at rest in space suddenly explodes into three
parts of same mass. The momentum of the two parts are
î
p
2 and ĵ
p . The momentum of the third part
(a) will have a magnitude 3
p
(b) will have a magnitude 5
p
(c) will have a magnitude p
(d) will have a magnitude 2p.
50. A triangular block of mass M with angles 30°, 60°, and 90°
rests with its 30°–90° side on a horizontal table. A cubical
block of mass m rests on the 60°–30° side. The acceleration
which M must have relative to the table to keep m stationary
relativeto the triangularblock assuming frictionless contactis
(a) g (b)
2
g
(c)
3
g (d)
5
g
51. Abodyofmass 1.0 kg is falling with an acceleration of 10 m/
sec2. Its apparent weight will be (g = 10 m/sec2)
(a) 1.0 kg wt (b) 2.0 kg wt
(c) 0.5 kg wt (d) zero
52. In the figure a smooth pulley of negligible weight is
suspended by a spring balance. Weight of 1 kg f and
5 kg f are attached to the opposite ends of a string passing
over the pulley and move with acceleration because of
gravity, During their motion, the spring balance reads a
weight of
(a) 6 kg f
(b) less then 6 kg f
(c) more than 6 kg f
1 kg
5 kg
(d) may be more or less then 6 kg f](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-124-320.jpg)
![123
Lawsof Motion
76. A smooth block is released at rest on a 45° incline and then
slides a distance ‘d’. The time taken to slide is ‘n’ times as
much to slide on rough incline than on a smooth incline.
The coefficient of friction is
(a) k
m = 2
n
1
1- (b) k
m =
2
n
1
1-
(c) s
m =
2
n
1
1- (d) s
m =
2
n
1
1-
77. The upper half of an inclined plane with inclination f is
perfectly smooth while the lower half is rough. A body
starting from rest at the top will again come to rest at the
bottom if the coefficient of friction for the lower half is given
by
(a) 2 cos f (b) 2 sin f
(c) tan f (d) 2 tan f
78. A particle of mass 0.3 kg subject to a force F = – kx with k =
15 N/m . What will be its initial acceleration ifit is released
from a point 20 cm awayfrom the origin ?
(a) 15 m/s2 (b) 3 m/s2
(c) 10 m/s2 (d) 5 m/s2
79. A block is kept on a frictionless inclined surface with angle
ofinclination ‘a ’ . The incline isgiven an acceleration ‘a’ to
keep the block stationary. Then a is equal to
a
(a) g cosec a (b) g / tan a
(c) g tan a (d) g
80. Consider a car moving on a straight road with a speed of100
m/s . The distance at which car can be stopped is [µk = 0.5]
(a) 1000 m (b) 800 m
(c) 400 m (d) 100 m
81. A round uniform bodyof radius R, mass M and moment of
inertia I rolls down (without slipping) an inclined plane
making an angle qwith the horizontal. Then its acceleration
is
(a)
2
gsin
1 MR / I
q
,
(b)
2
gsin
1 I / MR
q
∗
(c)
2
gsin
1 MR / I
q
∗
(d)
2
gsin
1 I/ MR
q
,
82. A block of mass m is placed on a smooth wedge of
inclination q. The whole system is accelerated horizontally
so that the block does not slip on the wedge. The force
exerted by the wedge on the block (g is acceleration due to
gravity) will be
(a) mg/cos q (b) mg cos q
(c) mg sin q (d) mg
83. The coefficient of static friction, ms, between block A of
mass 2 kg and the table as shown in the figure is 0.2. What
would be the maximum mass value of block B so that the
two blocks do not move? The string and the pulley are
assumed to be smooth and massless. (g = 10 m/s2)
A
B
2 kg
(a) 0.4kg (b) 2.0kg
(c) 4.0kg (d) 0.2kg
84. A body under the action of a force
ˆ ˆ ˆ
F = 6 i –8 j+10k,
r
acquires an acceleration of 1 m/s2. The
mass of this body must be
(a) 10kg (b) 20kg
(c) 10 2 kg (d) 2 10 kg
85. Aconveyor belt is moving at a constant speed of 2m/s. Abox
is gently dropped on it. The coefficient of friction between
them is µ= 0.5. Thedistance that theboxwill moverelativeto
belt before coming to rest on it taking g = 10 ms–2, is
(a) 1.2m (b) 0.6m
(c) zero (d) 0.4m
86. A person of mass 60 kg is inside a lift of mass 940 kg and
presses the button on control panel. The lift starts moving
upwards with an acceleration 1.0 m/s2. If g = 10 ms–2, the
tension in the supporting cable is
(a) 8600N (b) 9680N
(c) 11000N (d) 1200N
87. The upper half of an inclined plane of inclination q is
perfectlysmooth while lower half is rough. Ablock
starting from rest at the top of the plane will again come to
rest at the bottom, if the coefficient of friction between the
block and lower half of the plane is given by
(a) m =
2
tan q
(b) m = 2 tan q
(c) m = tan q (d) m =
1
tan q
88. A bridge is in the from of a semi-circle of radius 40m. The
greatest speed with which a motor cycle can cross the bridge
without leaving the ground at the highest point is
(g = 10 m s–2) (frictional force isnegligiblysmall)
(a) 40 m s–1 (b) 20 m s–1
(c) 30 m s–1 (d) 15 m s–1](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-127-320.jpg)
![125
Lawsof Motion
Exemplar Questions
1. A ball is travelling with uniform translatory motion. This
means that
(a) it is at rest
(b) the path can be a straight line or circular and the ball
travels with uniform speed
(c) all parts of the ball have the same velocity(magnitude
and direction) and the velocity is constant
(d) the centre of the ball moves with constant velocityand
the ball spins about its centre uniformly
2. Ametre scale is moving with uniform velocity. This implies
(a) the force acting on the scale is zero, but a torque about
the centre of mass can act on the scale
(b) the force acting on the scale is zero and the torque
acting about centre of mass of the scale is also zero
(c) the total force acting on it need not be zero but the
torque on it is zero
(d) neither the force nor the torque need to be zero
3. A cricket ball of mass 150 g has an initial velocity
1
ˆ ˆ
(3 4 )ms
u i j -
= +
r
and a final velocity 1
ˆ ˆ
(3 4 )ms
v i j -
= - +
r
,
after beinghit. The changein momentum (final momentum-
initial momentum) is (in kgms–1)
(a) zero (b) ˆ ˆ
(0.45 0.6 )
i j
- +
(c) ˆ ˆ
(0.9 1.2 )
j j
- + (d) ˆ ˆ ˆ
5( )
i j i
- +
4. In the previous problem(3), the magnitude ofthemomentum
transferred during the hit is
(a) zero (b) 0.75 kg-m s–1
(c) 1.5 kg-m s–1 (d) 1.4 kg-m s–1
5. Conservation of momentum in a collision between particles
can be understood from
(a) Conservation of energy
(b) Newton's first law only
(c) Newton's second law only
(d) both Newton's second and third law
6. A hockey player is moving northward and suddenly turns
westward with the same speed to avoid an opponent. The
force that acts on the player is
(a) frictional force along westward
(b) muscle force along southward
(c) frictional force along sotuh-west
(d) muscle force along south-west
7. A body of mass 2 kg travels according to the law
2 3
( )
x t pt qt rt
= + + where, q = 4 ms–2, p = 3 ms–1 and r= 5
ms–3. The force acting on the body at t = 2s is
(a) 136N (b) 134N
(c) 158N (d) 68N
8. A body with mass 5 kg is acted upon by a force
ˆ ˆ
( 3 4 ) N
F i j
= - +
r
. If its initial velocity at t = 0 is
1
ˆ ˆ
(6 12 ) ms
v i j -
= - , the time at which it will just have a
velocity along the y-axis is
(a) never (b) 10 s
(c) 2 s (d) 15 s
9. A car of mass m starts from rest and acquires a velocity
along east, ˆ ( 0)
v vi v
= >
r
in two seconds. Assuming the
car moves with uniform acceleration, the force exerted on
the car is
(a)
2
mv
eastward and is exerted by the car engine
(b)
2
mv
eastward and is due to the friction on the tyres
exerted by the road
(c) more than
2
mv
eastward exerted due to the engine and
overcomes the friction of the road
(d)
2
mv
exerted by the engine
NEET/AIPMT (2013-2017) Questions
10. Three blocks with masses m, 2 m and 3 m are connected by
strings as shown in the figure. After an upward force F is
applied on block m, the masses move upward at constant
speed v. What is the net force on the block of mass 2m?
(g is the acceleration due to gravity) [2013]
(a) 2mg
(b) 3mg
(c) 6mg
(d) zero
11. A car is moving in a circular horizontal track of radius 10 m
with a constant speed of 10 m/s. A bob is suspended from
the roof of the car by a light wire of length 1.0 m. The angle
made bythe wire with the vertical is [NEET Kar. 2013]
(a) 0° (b)
3
p
(c)
6
p
(d)
4
p
12. A balloon with mass ‘m’ is descending down with an
acceleration ‘a’ (where a < g). How much mass should be
removed from it so that it starts moving up with an
acceleration ‘a’? [2014]
(a)
2ma
g a
+ (b)
2ma
g a
-
(c)
ma
g a
+
(d)
ma
g a
-](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-129-320.jpg)
![126 PHYSICS
13. Theforce‘F’ acting on a particle ofmass ‘m’ is indicated by
the force-timegraph shown below.Thechange in momentum
of the particle over the time interval from zero to 8 s is :
[2014]
6
3
0
–3
4 6 8
F(N)
t(s)
2
(a) 24 Ns (b) 20 Ns
(c) 12 Ns (d) 6 Ns
14. A system consists ofthree masses m1, m2 and m3 connected
by a string passing over a pulley P. The mass m1 hangs
freely and m2 and m3 are on a rough horizontal table (the
coefficient of friction = m). The pulleyis frictionless and of
negligible mass. The downward acceleration of mass m1 is :
(Assume m1 = m2 = m3 = m) [2014]
(a)
g(1 – g )
g
m
(b)
2g
3
m
P
m1
m2 m3
(c)
g(1 – 2 )
3
m
(d)
g(1 – 2 )
2
m
15. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg
respectively, are in contact on a frictionless surface, as
shown. If a force of 14 N is applied on the 4 kg block then
the contact force between A and B is [2015]
A B C
(a) 6N (b) 8N
(c) 18N (d) 2N
16. A block A of mass m1 rests on a horizontal table. A light
string connected to it passes over a frictionless pulley at
the edge of table and from its other end another block B of
mass m2 is suspended. The coefficient of kinetic friction
between the block and the table is µk. When the block A is
sliding on the table, the tension in the string is [2015]
(a)
2 1
1 2
(m – k m )g
(m m )
m
+ (b)
1 2 k
1 2
m m (1 )g
(m m )
+ m
+
(c)
1 2 k
1 2
m m (1 – )g
(m m )
m
+ (d)
2 k 1
1 2
(m m )g
(m m )
+ m
+
17. Two stones of masses m and 2 m are whirled in horizontal
circles, the heavier one in radius
r
2
and the lighter one in
radius r. The tangential speed of lighter stone is n times that
of the value of heavier stone when they experience same
centripetal forces. The value of n is : [2015 RS]
(a) 3 (b) 4
(c) 1 (d) 2
18. A plank with a box on it at one end is graduallyraised about
the other end.As the angle of inclination with the horizontal
reaches 30º the box starts to slip and slides 4.0 m down the
plank in 4.0s. The coefficients of static and kinetic friction
between the box and the plank will be, respectively :
[2015 RS]
mg
q
(a) 0.6 and 0.5 (b) 0.5 and 0.6
(c) 0.4 and 0.3 (d) 0.6 and 0.6
19. What is the minimum velocitywith which a bodyof mass m
must enter a vertical loop ofradius R sothat it can complete
the loop ? [2016]
(a) gR (b) 2gR
(c) 3gR (d) 5gR
20. One end of string of length l is connected to a particle of
mass 'm' and the other end is connected to a small peg on a
smooth horizontal table. If the particle moves in circle with
speed 'v' the net force on the particle (directed towards
centre) will be (T represents the tension in thestring) : [2017]
(a)
2
mv
T
l
+ (b) T –
2
mv
l
(c) Zero (d) T
21. Two blocks A and B of masses 3 m and m respectively are
connected bya massless and inextensible string. The whole
system is suspended by a massless spring as shown in
figure. The magnitudes of acceleration of A and B
immediatelyafter the string is cut, are respectively: [2017]
(a)
g
,g
3
A
B
3m
m
(b) g, g
(c)
g g
,
3 3
(d)
g
g,
3](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-130-320.jpg)
![127
Lawsof Motion
EXERCISE - 1
1. (c) 2. (b) 3. (d) 4. (a) 5. (c)
6. (a) 7. (a)
8. (d) Let n be the mass per unit length of rope. Therefore,
mass of rope = nL.
Acceleration in the rope due to force F will be
a= F/nL.
Mass of rope of length (L – l) will be n (L – l).
Therefore, tension in the rope oflength (L – l), is equal
to pulling force on it
= n (L – l) a = n (L – l) × F/nL = F (1 – l/L)
9. (a) Impulse = change in momentum = 1
2 v
m
v
m
r
r
-
10. (a) Let a be the acceleration of mass m2 in the downward
direction. Then
T – m2 (g/2) = m1 a ....(i)
and m2 g – T = m2 a ....(ii)
Adding eqs. (1) and (2), we get
(m1 + m2) a = m2g – m2 (g/2) = m2 g/2
)
m
m
(
2
g
m
a
2
1
2
+
=
11. (a) Let acceleration of a
lift = and
let reaction at spring balance = R
R
mg
Applying Newton’s law
R– mg = ma ( )
a
g
m
R +
=
Þ
thus net weight increases,
So reading of spring balance increases.
12. (b) See fig.
ma
mN
N
mg
If a = acceleration ofthe cart, then N = ma
mN = mg or m ma = mg or a = g/m
13. (c) p1 = mv northwards, p2 = mv eastwards
N
E
S
W
m
m
v
v
Let p = momentum after collision. Then,
2
1 p
p
p
r
r
r
+
= or 2
2
)
mv
(
)
mv
(
p +
=
r
2
mv
v
m
2 =
¢ or
2
v
v =
¢ m/sec
14. (c) Applying law of conservation of linear momentum
m1v1 + m2v2 = 0,
1
2
2
1
v
v
m
m
-
= or
1
2
2
1
m
m
v
v
-
=
15. (c) The equations of motion are
2 mg – T = 2ma
T– mg = ma Þ T= 4ma &a = g/3 soT =4mg/3
If pulley is accelerated upwards with an accleration a,
then tension in string is
)
a
g
(
3
m
4
T +
=
16. (b) Inertia is resistance to change.
17. (b) The time rate of change ofmomentum is force.
18. (a) At the highest point of the track,
2
mv '
N mg
r
+ =
mg
N
where r is the radius of curvature at that point and v¢ is
the speed of the block at that point.
Now
2
mv '
N mg
r
= -
N will be maximum when r is minimum (v¢ is thesame
for all cases). Of the given tracks, (a) has the smallest
radius of curvature at the highest point.
19. (c) f W
=m
f W tan
= q [ q
=
m tan ]
20. (a) )
1
n
2
(
2
a
Sn -
=
)
1
n
2
(
2
a
S 1
n +
=
+
1
n
2
1
n
2
S
S
1
n
n
+
-
=
+
21. (b) From the F.B.D.
N = mg cos q
F = ma = mg sin q – N
m
)
cos
(sin
g
a q
m
-
q
=
Þ
x
q
sin
g q
g
N
m
q
N
m m
mg
cos
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-131-320.jpg)
![128 PHYSICS
Now using, as
2
u
v 2
2
=
-
or, l
)
cos
(sin
g
2
v2
q
m
-
q
´
=
( l = length ofincline)
or, v = )
cos
(sin
g
2 q
m
-
q
l
22. (a) Based on Newton’s third law of motion.
23. (a) If m1, m2 are masses and u1, u2 are velocity then by
conservation of momentum m1u1 + m2u2 = 0 or
1 1 2 2
| m u | | m u |
=
24. (a) Here, on resolving force F2 and applying the concept
ofequilibrium
F1
F2sin q
F2
cos q
mg
N
f
m
q
+
= cos
F
mg
N 2 , and f = µN
]
cos
F
mg
[
f 2 q
+
m
= …(i)
Also q
+
= sin
F
F
f 2
1 …(ii)
From (i)and(ii)
]
cos
F
mg
[ 2 q
+
m = F1 + F2 sin q
Þ
q
+
q
+
=
m
cos
F
mg
sin
F
F
2
2
1
25. (a) The frictional force acting on M is µmg
Acceleration =
M
mg
m
EXERCISE - 2
1. (b) Here u = 10 ms–1, v = –2 ms–1,
t = 4 s, a = ?
Using 2
s
/
m
3
4
10
2
t
u
v
a -
=
-
-
=
-
=
Force, F =ma = 10×(–3) = –30 N
2. (b) 2 T cos 60º = mg
or T = mg = 2×10 = 20 N.
3. (c) If k is the spring factor, then P.E. of the spring
compressedbydistance x ÷
ø
ö
ç
è
æ
= 2
kx
2
1
will equal to gain
in P.E. of the dart ( = mgh) i.e. mgh
kx
2
1 2
=
200
g
16
)
4
(
k
2
1 2
´
´
= ....(i)
and h
g
16
)
6
(
k
2
1 2
´
´
= ...(ii)
On solving, (i) and (ii), weget h = 450 cm = 4.5m.
4. (d) Herem = 0.5 kg ; u = – 10 m/s ;
t = 1/50 s ; v = + 15 ms–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
5. (b) As, (1/2)m v2 = Fs
So 4
F
)
30
(
m
2
1 2
´
= and s
F
)
60
(
m
2
1 2
´
=
s/4 = (60)2 / (30)2 = 4 or s = 4 × 4 = 16 m.
6. (c) Acceleration, 2
v u 0 6
a 2 ms
t 3
-
- -
= = = -
Force= m×a =4×2= 8N
7. (a)
taken
time
momentum
in
change
required
Force =
=
3
(50 10 30) 400 (5 0)
10 N
60
-
´ ´ ´ - ´
=
8. (a) Changein momentum = Force × time =Area which the
force-time curve encloses with time axis.
9. (b) 20cm
10cm
32N(F )
2
20N(F )
1
l1 l2
L
F F
It is clear F2 > F1, so rod moves in right direction with
an acceleration a, whereas a is given by
(F2–F1)=mL×a................(i)
where m is mass of rod per unit length.
Now consider the motion of length l1 from first end,
then
F– F1=ml1a..................(ii)
Dividing eq (ii) by(i), we get
L
l
F
F
F
F 1
1
2
1
=
-
-
or 1
1
1
2 F
L
l
)
F
F
(
F +
´
-
=
here l1 = 10 cm., L= 30 cm., F1 = 20 N, F2 = 32N
so F = 24 N
10. (b) Change in momentum = F × t
= 10 × 10 = 100 Ns or 100 kg. m/s
11. (d) See fig.
1 kg
1 kg
T
T
T
From figure, 1 g – T = 1 a ...(i)
and T = 1 a ....(ii)
From eqs. (i) and (ii), we get
1g – 1a = 1a or 2a = g
a = (g/2) = (10/2) = 5 m/s2
So, T =ma = 1 ×5 = 5 N
12. (b) R= mg –ma = 0.5 × 10 – 0.5 × 2= 5 – 1= 4 N
13. (c) Seefig. Let Fbetheforce between theblocksand a their
common acceleration. Then for 2 kgblock,](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-132-320.jpg)
![131
Lawsof Motion
42. (d)
5
3
sin =
q
q
5
3
4
tan
4
3
=
q Þ
3
tan 0.75
4
m = q = =
43. (b) The condition to avoid skidding, v = rg
m
= 10
150
6
.
0 ´
´ = 30 m/s.
44. (d) Limitingfriction = 0.5 2 10 10N
´ ´ =
The applied force is lessthan force offriction, therefore
the force of friction is equal to the applied force.
45. (d) Applying law of conservation of momentum
Momentum of bullet = Momentum of sand-bullet
system
m m 21
v m V mV
20 20 20
æ ö
= + =
ç ÷
è ø
46. (d) Here 8
.
0
tan =
q
where q is angle of repose
°
=
=
q -
39
)
8
.
0
(
tan 1
The given angle of inclination is equal to the angle of
repose. So the 1 kg block has no tendency to move.
=
q
sin
mg force of friction
Þ T = 0
47. (d) For 0.5 kg block, 6 = 0.5 a
48. (b) While moving down, when the lift is accelerating the
weight will beless and when the lift is decelerating the
weight will be more.
49. (b) Total momentum = j
ˆ
p
î
p
2 +
Magnitude of total momentum
= p
5
p
5
p
)
p
2
( 2
2
2
=
=
+
This must be equal to the momentum ofthe third part.
50. (c)
90°
30°
M
mg sin 30°
ma cos 30°
30° 60°
ma (pseudo force)
a
°
=
° 30
sin
mg
30
cos
ma
3
g
a =
51. (d) Apparent weight when mass is falling down is given
by W ' m(g a)
= -
W ' 1 (10 10) 0
= ´ - =
52. (b) Reading of spring balance
2T =
1 2
1 2
4m m 4 5 1 10
m m 6 3
´ ´
= =
+
kgf
53. (a) a = 2v (given)
Þ
dv
v 2v
ds
=
or dv 2ds
=
v
0.1
0
0.1
dv 2[s] 0.2
= =
ò
v 0.1 0.2
- =
Þ 1
v 0.3ms-
=
54. (c) IfW is themaximum weight, then
W = 2T cos 60°
or W= T = 20N
55. (c) The acceleration of both the blocks =
15 5
3x x
=
Force on
5
B 2x 10 N
x
= ´ =
56. (b) The maximum acceleration that can be given is a
a
10
g
10
g
30 +
=
Þ 2
ms
20
g
2
a -
=
=
We know that
2
at
2
1
ut
s +
=
20
10
2
a
s
2
t
´
=
= = 1 s
57. (a) Let the air resistance be F. Then
mg F ma
+ = Þ ]
g
a
[
m
F -
=
Here
2
ms
12
5
.
2
30
a -
=
=
58. (c) N
30
1
.
0
)
20
0
(
15
.
0
t
)
u
v
(
m
F =
-
=
-
=
59. (c) Considering the two masses and the rope a system,
then
Initial net force = [ ]
25 (15 5) g 5g
- + =
Final net force = ( )
25 5 15 g 15 g
+ - =
é ù
ë û
Þ (acceleration)final = 3 (acceleration)initial
60. (d)
mg sin q
mg cos q
F cos q
q
F sin
F
mg
q
N
From figure q
+
q
= sin
F
cos
mg
N](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-135-320.jpg)
![135
Lawsof Motion
Presultant = 2 2
12 16
+
= 144 256
+ = 20
m3v3 = 20 (momentum ofthird part)
or, m3 =
20
4
= 5 kg
90. (b) Mg – T = Ma
T – mg = ma
m
mg mg
T
T
M
On solving, we get
a =
(M m)g
M m
-
+
and T =
2Mmg
M m
+
91. (c) Let velocity of bullet be v. If velocity of gun is V, then
mv + MV = 0
Þ V =
mv
M
-
As spring compresses by d, so
2 2
1 1
kd MV
2 2
=
or
2
2
1 1 mv
kd M
2 2 M
æ ö
= ç ÷
è ø
Þ v =
d
kM
m
92. (b) Here, l2 = 2l1
As for a spring, force constant
1
k
l
µ
1 2
1 2 1 2
1 1 1
, ,
k k k
l l l l
µ µ µ
+
1 1 2
1
k l l
k l
+
= and
2 1 2
2
k l l
k l
+
=
or
2
1
1
1
l
k k
l
é ù
+
= ê ú
ë û
and
1
2
2
1
l
k k
l
é ù
+
= ê ú
ë û
1 [1 2] 3
k k k
= + = [using (i)]
2
3
1
1
2
2
k k k
é ù
= =
+
ê ú
ë û
[using (i)]
93. (a) R = mg cos q –
2
mV
r
V
car
R
O
mg
q mg cos q
Over bridge
q
When q decreases, cosq increases i.e. R increases
94. (c) Impulse experienced by the body
= change in momentum = MV– (–MV) = 2MV.
95. (a) The force of friction on the chain lying on the table
should be equal to the weight of the hanging chain.
Let
r = mass per unit length of the chain
µ = coefficient of friction
l = length of the total chain
x = length of hanging chain
Now, µ(l – x) rg = xrg or µ(l – x) = x
or µl = (µ + 1)x or x = µl/(µ+ 1)
0.25 0.25
0.2
(0.25 1) 1.25
l l
x l
= = =
+
0.2 20%
x
l
= =
96. (a) Masses of the pieces are 1, 1, 3 kg. Hence
2 2 2
(1 21) (1 21) (3 )
V
´ + ´ = ´
That is, 7 2
V = m/s
97. (c)
2
1 1 2
2
2 1 2
K p m
K m p
= ´ [Q p = mv Þ
2
=
2
p
K
m
]
Hence,
1 1
2 2
1 1
4 2
p M
p M
= = =
98. (d) Work done in moving an object against gravitational
force depends only on the initial and final position of
the object, not upon the path taken. But gravitational
force on the bodyalong the inclined plane is not same
as that along the vertical and it varies with angle of
inclination.
99. (b) On a rainy day, the roads are wet. Wetting of roads
lowers the coefficient of friction between the types
and the road. Therefore, grip on a road of car reduces
and thus chances of skidding increases.
100. (a) The force acting on the body of mass M are its weight
Mg acting vertically downward and air resistance F
acting vertically upward.
Acceration of the body ,
M
F
g
a -
=
Now M > m, therefore, the bodywith larger mass will
have great acceleration and it will reach the ground
first.
EXERCISE - 3
Exemplar Questions
1. (c) In a uniform translatory motion if all the parts of the
body moves with (same velocity in same straight line,
so the velocity is constant.
A
v
v
v
The situation is shown in (figure) where a body A is in
unfirom translatory motion.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-139-320.jpg)
![136 PHYSICS
2. (b) According to question we have to apply Newton's
second law of motion, in terms of force and change in
momentum.
We know that
dp
F
dt
=
As given that the meter scale is moving with uniform
velocity, hence
Force (F) = m × 0 = 0
No change in its velocityi.e., acceleration of it zero by
Newton's second law.
Hence, net or resultant force must act on body is zero.
,
r F
t = ´
r
r r so,
As all part ofthe scale is moving with uniform velocity
and total force is zero, hence, torque will also be zero.
3. (c) As given that,
Mass of the ball = 150 g = 0.15 kg
ˆ ˆ
(3 4 ) m/s
u i j
= +
r
ˆ ˆ
(3 4 ) m/s
v i j
= - +
r
(Dp) Change in momentum
= Final momentum – Initial momentum
mv mu
= -
r r
ˆ ˆ ˆ ˆ
( ) (0.15)[ (3 4 ) (3 4 )]
m v u i j i j
= - = - + - +
r r
ˆ ˆ
(0.15)[ 6 8 ]
i j
= - -
ˆ ˆ
[0.15 6 0.15 8 ]
i j
= - ´ + ´
ˆ ˆ
[0.9 1.20 ]
i j
= - +
ˆ ˆ
[0.9 1.2 ]
p i j
D = - +
Hence verifies option (c).
4. (c) From previous solution
ˆ ˆ ˆ ˆ
(0.9 1.2 ) 0.9 1.2
p i j i j
D = - + = - -
Magnitude of
2 2
| | ( 0.9) ( 1.2)
p
D = - + -
0.81 1.44
= +
2.25 1.5
= = kg-m s–1
Verifies the option (c).
5. (d) By Newton's second law :
ext
dp
F
dt
=
r
As ext
F
r
in law of conservation of momentum is zero.
If 0, 0
ext
F dp p
= = Þ = constant
Hence, momentum of a system will remain conserve if
external force on the system is zero.
In case of collision between particle equal and opposite
forces will act on individual particle by Newton's third
law.
So, 12 21
F F
= -
r r
( 0)
ext
F =
r
Q
Total force on the system will be zero.
12 21
dp dp
dt dt
= -
r r
or 12 21
dp dp
= -
r r
12 21
( 0)
dp dp
Þ + =
r r
So prove the law of conservation of momentum and
verifies the option (d).
6. (c) Consider the adjacent diagram.
A
B
B O
E
S
N
P2
P2
P1
–
( )
W
P1
The force on player is due to rate of change of
momentum. The direction offorce acting on player will
be the same as the direction of change in momentum.
Let OA = P1 i.e., Initial momentumof player northward
AB = P2 i.e., Final momentum of player towards west.
Clearly, OB =OA +OB
Changein momentum
= P2 – P1
= AB – OA = AB + (–OA)
= Clearlyresultant AR will be along south-west.
So, it will be also the direction of force on player.
7. (a) As given that, mass = 2 kg
p = 3 m/s, q = 4 m/s2, r = 5 m/s3
As given equation is
2 3
( )
x t pt qt rt
= + +
2
( )
2 3
ds t
v p qt rt
dt
= = + +
2
2
( )
0 2 6
dv d x t
a q rt
dt dt
= = = + +
2
2
( 2)
( )
2 12
t
d x t
q r
dt =
é ù
= +
ê ú
ë û
2 12
q r
= +
=2 ×4 +12 ×5
= 8 + 60 = 68 m/s2
Force acting on body ( ) =
F ma
r
= 2 × 68= 136 N.
8. (b) As given that mass = m = 5 kg
Acting force = ˆ ˆ
( 3 4 ) N
F i j
= - +
r](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-140-320.jpg)
![138 PHYSICS
m1
m2
mk
m g
2
a
m g
1
mk
T
a
a = 2 k 1
1 2
m g – m g
m m
m
+
m2g – T = (m2)
2 k 1
1 2
m g – m g
m m
æ ö
m
ç ÷
+
è ø
solving we get tension in the string
T = 1 2 k
1 2
m m g (1 )g
m m
+ m
+
17. (d) According to question, two stones experience same
centripetal force
i.e. 1 2
C C
F F
=
or,
2 2
1 2
mv 2mv
r (r / 2)
= or, 2 2
1 2
V 4V
=
So, V1 = 2V2 i.e., n = 2
18. (a) Coefficient of static friction,
ms = tan 30°=
1
3
= 0.577@ 0.6
S = ut +
2
1
at
2
4 =
1
2
a(4)2 Þ a =
1
2
= 0.5
[Q s = 4m and t = 4s given]
a = gsinq – mk(g) cosq
Þ mk =
0.9
3
= 0.5
19. (d) To complete the loop a bodymust enter a vertical loop
ofradius R with the minimum velocity v= 5gR .
20. (d) Net force on particle in uniform circular motion is cen-
tripetal force
2
mv
l
æ ö
ç ÷
ç ÷
è ø
which is provided by tension in
string so the net force will be equal to tension i.e., T.
21. (a) m
mg
Before cutting the string
kx=T +3mg ...(i)
T = mg ...(ii)
Þ kx=4mg
After cutting the string T = 0
A
4mg 3mg
a
3m
-
=
4mg
A
g
a
3
=
and B
mg
a g
m
= =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-142-320.jpg)
![WORK
Work Done by a Constant Force
Work done (W) by a force F
ur
in displacing a body through a
displacement x is given by
W = F.x
r r
= Fx cos q
F
x
Body
F cos q
q
Fsinq
Where q istheanglebetween theappliedforce F
r
and displacement
x
r
.
The S.I. unit ofwork isjoule, CGS unit is erg and itsdimensions
are[ML2T–2].
1 joule = 107 erg
(a) When q = 0° then W = Fx
(b) When q is between 0 and p/2 then
W = Fx cos q = positive
(c) When q = p/2 then W = Fx cos 90° = 0 (zero)
Work done bycentripetal force is zero as in this case angle
q =90°
(d) When q is between / 2 and
p p then
W = Fx cos q = negative
Work Done by a Variable Force
When the force is an arbitrary function of position, we need the
techniques of calculus toevaluate the work done byit. The figure
shows Fx as function of the position x. We begin by replacing
the actual variation of the force by a series of small steps.
The area under each segment of the curveis approximatelyequal
tothe area ofa rectangle. The height ofthe rectangle is a constant
valueof force, and its width is a small displacement Dx. Thus, the
step involves an amount of work DWn = Fn Dxn. The total work
done is approximately given by the sum of the areas of the
rectangles.
i.e., W » n
F
å Dxn.
As the size of the steps is reduced, the tops of the rectangle more
closelytrace the actual curve shown in figure. Ifthe limit Dx ®0,
which is equivalent to letting the number ofsteps tend to infinity,
the discrete sum is replaced by a continuous integral.
0
lim
n
n n x
x
W F x F dx
D ®
= D =
å ò
Thus, the work done bya force Fx from an initial point A tofinal
point B is
n
A
x
A B x
x
W F dx
® = ò
The work done bya variable force in displacing a particle from x1
to x2
ò
=
2
1
x
x
Fdx
W = area under force displacement graph
CAUTION : When we find work, we should be cautious about
the question, work done bywhich force? Let us take an example
to understand this point. Suppose you are moving a body up
without acceleration.
Work done by applied force
app app
app
W F x F x
·
= =
uu
r uu
r
Work done by gravitational force x
mg
Fapplied
g
grav
W F x mgx
= = -
uuu
r uu
r uu
r
6
Work, Energy
and Power](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-143-320.jpg)
![142 PHYSICS
Then byrearranging equation (i), (iii) & (iv), we get
÷
÷
ø
ö
ç
ç
è
æ
´
÷
÷
ø
ö
ç
ç
è
æ
=
=
1
2
2
1
2
2
1
1
2
1
u
u
x
x
a
/
u
a
/
u
t
t
÷
÷
ø
ö
ç
ç
è
æ
=
´
=
Þ
2
2
1
1
1
2
2
2
2
2
1
1
2
1
u
M
u
M
u
u
)
u
M
(½
)
u
M
(½
t
t
(a) If
2
1
2
1
2
1
M
M
t
t
u
u =
Þ
=
(b) If
2
1
2
1
2
1
u
u
t
t
M
M =
Þ
=
(c) If M1u1 = M2u2 Þ t1 = t2
and
2
1 1 1 2 1 2
2
2 1 2 1
2 2
x (M u ) M x M
x M x M
(M u )
= ´ Þ =
(d) Consider two vehicles of masses M1 & M2
respectively.
Iftheyare moving with same velocities, then the ratio
of their stopping distances bythe application of same
retarding force is given by
2
1
2
1
M
M
x
x
= and let M2 > M1 then x1 < x2
Þ lighter mass will cover less distance then the
heavier mass
And the ratio of their retarding times are as follows :
2
1
2
1
M
M
t
t
= i.e
2
1
2
1
t
t
x
x
=
8. If kinetic energy of a body is doubled, then its momentum
becomes 2 times,
2
k k
p
E p E
2m
= Þ µ
9. If two bodies of masses m1 and m2 have equal kinetic
energies, then their velocities are inversely proportional
to the square root of the respective masses. i.e.
2 2
1 1 2 2
1 1
m v m v then
2 2
=
1
2
2
1
m
m
v
v
=
10. (a) The spring constant of a spring is inversely
proportional to the no. of turns i.e.
1
K
n
µ
or kn = const.
(b) Greater the no. of turns in a spring, greater will be the
work done i.e. W µ n
(c) The greater is the elasticity of the spring, the greater
is the spring constant.
11. Spring constant : The spring constant of a spring is
inversely proportional to length i.e.,
l
1
K µ or
Kl = constant.
(a) If a spring is divided into n equal parts, the spring
constant of each part = nK.
(b) If spring of spring constant K1, K2, K3 .......... are
connected in series, then effective force constant
......
K
1
K
1
K
1
K
1
3
2
1
eff
+
+
+
=
(c) If spring of spring constant K1, K2, K3........... are
connected in parallel, then effective spring constant
Keff = K1 + K2 + K3 +.............
POWER
Power of the body is defined as the time rate of doing work by
the body.
The average power Pav over the time interval Dt is defined by
av
W
P
t
D
=
D
...(i)
And the instantaneous power P is defined by
0
t
W dW
P Lim
t dt
d ®
D
= =
D
...(ii)
Power is a scalar quantity
The S.I. unit of power is joule per second
1 joule/sec = 1watt
The dimensions of power are [ML2T–3]
( . ) . .
dW d dS
P F S F F v
dt dt dt
= = = =
r
r
r r r r
(force is constant over a small time interval)
So instantaneous power (or instantaneous rate of working) of a
man depends not only on the force applied to body, but also on
the instantaneous velocity of the body.
Example1.
A metre stick of mass 600 mg, is pivoted at one end and
displaced through an angle of 60º. The increase in its
potential energy is (g = 10 ms–2)
(a) 1.5 J (b) 15 J
(c) 150 J (d) 0.15 J
Solution : (a)
G
M
O
h
60º
x
l/2
l
/
2
G´
A´
A
TheC.G. of stick rises from G toG´.
Increase in P.E. = mgh
= mg (l/2 –x) = (1 cos60º)
2
mg
-
l
J
5
.
1
2
1
1
2
1
x
10
x
6
.
0
=
ú
û
ù
ê
ë
é
-
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-146-320.jpg)
![150 PHYSICS
7. Amotor of 100 H.P. moves a load with a uniform speed of72
km/hr. The forward thrust applied by the engine on the car
is
(a) 1111N (b) 3550N
(c) 2222N (d) 3730N
8. Two bodies A and B having masses in the ratio of 3 : 1
possess the same kinetic energy. The ratio of linear
momentum of B toAis
(a) 1: 3 (b) 3: 1
(c) 1: 3 (d) 3 :1
9. When a U238 nucleus, originallyat rest, decays byemitting
an a-particle, saywith speed of vm/sec, the recoil speed of
the residual nucleus is (in m/sec.)
(a) – 4 v/234 (b) – 4 v/238
(c) 4v/238 (d) – v/4
10. Calculate the K.E and P.E. of the ball half way up, when a
ball of mass 0.1 kg is thrown vertically upwards with an
initial speed of 20 ms–1.
(a) 10 J, 20 J (b) 10 J, 10 J
(c) 15 J, 8 J (d) 8 J, 16 J
11. A spring of force constant 800 N/m has an extension of 5
cm. The work done in extending it from 5 cm to 15 cm is
(a) 16J (b) 8 J
(c) 32J (d) 24J.
12. Ifthe linear momentumisincreased by5%, the kineticenergy
will increase by
(a) 50% (b) 100%
(c) 125% (d) 10%
13. A cord is usedtolower verticallya block ofmass M, through
a distance d at a constant downward acceleration of g/8.
Then the work done by the cord on the block is
(a) Mg d/8 (b) 3 Mg d/8
(c) Mg d (d) – 7 mg d/8
14. Aforce F (5i 3j 2k)N
= + +
r r r
r
is appliedover a particlewhich
displaces it from its origin to the point r (2i j)m.
= -
r r
r
The
work done on the particle in joule is
(a) +10 (b) +7
(c) –7 (d) +13
15. A spring ofspring constant 5 × 103 N/m is stretched initially
by 5 cm from the unstretched position. Then the work
required to stretch it further by another 5 cm is
(a) 18.75J (b) 25.00J
(c) 6.25J (d) 12.50J
16. Two solid rubber balls A and B having masses 200 &
400 gm respectively are moving in opposite direction with
velocity of A equal to 0.3 m/sec. After collision the two
balls come to rest when the velocity of B is
(a) 0.15m/sec (b) 1.5m/sec
(c) –0.15m/sec (d) None of these
17. A bomb of mass 9 kg explodes into the pieces of masses
3 kg and 6 kg. The velocity of mass 3 kg is 16 m/s. The
kinetic energy of mass 6 kg in joule is
(a) 96 (b) 384
(c) 192 (d) 768
18. A long string is stretched by 2 cm and the potential energy
isV. If the spring is stretched by10 cm, its potential energy
will be
(a) V/25 (b) V/5
(c) 5V (d) 25V
19. When the kinetic energy of a body is increased to three
times, then the momentum increases
(a) 6 times (b) 1.732times
(c) 2 times (d) 2 times
20. Twobodies of masses 2 m and m have their KE in the ratio
8 : 1. What isthe ratio oftheir momenta ?
(a) 8: 1 (b) 4: 1
(c) 2: 1 (d) 1: 1
21. A body of mass 5 kg initially at rest explodes into 3
fragments with mass ratio3: 1 :1. Twoof fragments each of
mass ‘m’ are found tomove with a speed 60m/s in mutually
perpendicular directions. The velocityofthird fragment is
(a) 60 2 (b) 20 3
(c) 10 2 (d) 20 2
22. A machine, which is 75% efficient, uses 12 J of energy in
lifting up a 1 kg mass through a certain distance. The mass
is then allowed to fall through that distance. The velocity
at the end of its fall is (in m/s)
(a) 24 (b) 12
(c) 18 (d) 9
23. A body accelerates uniformly from rest to a velocity of 1
ms–1 in 15 seconds. The kinetic energyof the body will be
9
2
J when 't' is equal to [Take mass of body as 1 kg]
(a) 4s (b) 8s
(c) 10s (d) 12s
24. A machine gun fires a bullet of mass 40 g with a velocity
1200 ms–1. Theman holding it can exert a maximum forceof
144 N on the gun. Howmany bullets can he fire per second
at the most?
(a) 2 (b) 4
(c) 1 (d) 3
25. A crane is used to lift 1000 kg of coal from a mine 100 m
deep. The time taken bythe crane is 1 hour. The efficiency
of the crane is 80%. If g = 10 ms–2, then the power of the
crane is
(a) 104 W (b) 105 W
(c) W
8
36
104
´
(d) W
8
36
105
´
26. In figure, a carriagePis pulled up fromAto B. The relevant
coefficient of friction is 0.40. The work done will be
(a) 10kJ
(b) 23kJ
(c) 25kJ
P 50 kg
50
30 m
B
A
q
C
m
(d) 28kJ](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-154-320.jpg)
![151
Work, Energy and Power
27. A neutron with velocity v strikes a stationary deuterium
atom, its K.E. changes by a factor of
(a)
16
15
(b)
2
1
(c)
1
2
(d) None of these
28. A bodymoves a distance of10 m along a straight line under
the action of a force of 5 newtons. If the work done is 25
joules, the angle which the force makes with the direction
of motion of body is
(a) 0º (b) 30º
(c) 60º (d) 90º
29. A sphere of mass 8m collides elastically(in one dimension)
with a block ofmass 2m. If the initial energyof sphere is E.
What is the final energy of sphere?
(a) 0.8E (b) 0.36E
(c) 0.08E (d) 0.64E
30. Johnny and his sister Jane race up a hill. Johnny weighs
twice as much as jane and takes twice as long as jane to
reach the top . Compared to Jane
(a) Johnnydid more work and delivered more power.
(b) Johnnydid more work and delivered the same amount
of power.
(c) Johnny did more work and delivered less power
(d) Johnnydid less work and johnnydelivered less power.
31. A bodyof mass m moving with velocityv makes a head on
elastic collision with another body of mass 2m which in
initially at rest. The loss of kinetic energy of the colliding
body (mass m ) is
(a)
2
1
ofits initial kinetic energy
(b)
9
1 ofits initial kinetic energy
(c)
9
8
ofits initial kinetic energy
(d)
1
4
ofits initial kinetic energy
32. In the non-relativistic regime, ifthe momentum, is increase
by 100% , the percentage increase in kinetic energy is
(a) 100 (b) 200
(c) 300 (d) 400
33. A body is dropped from a height of 20m and rebounds to a
height 10m. The loss of energy is
(a) 10% (b) 45%
(c) 50% (d) 75%
34. A moving body with a mass m1 and velocity u strikes a
stationarybodyofmass m2. The masses m1 and m2 should
be in the ratio m1/m2 so as to decrease the velocity of the
first body to 2u/3 and giving a velocityofu to m2 assuming
a perfectly elastic impact. Then the ratio m1/m2 is
(a) 5 (b) 5
/
1
(c) 25
/
1 (d) 25.
35. A body of mass m is suspended from a massless spring of
natural length l. It stretches the spring through a vertical
distance y. The potential energy of the stretched spring is
(a) )
y
(
mg +
l (b) )
y
(
mg
2
1
+
l
(c) mgy
2
1
(d) mgy
36. Figure here shows the frictional force versus displacement
for a particle in motion. The loss of kinetic energy in
travelling over s = 0 to 20 m will be
15
10
5
0
0 5 10 20
x(m)
f(N)
(a) 250J (b) 200J
(c) 150J (d) 10J
37. Ten litreof water per second is lifted from a well through 10
m and delivered with a velocity of 10 ms–1. If
g = 10 ms–2 , then the power of the motor is
(a) 1 kW (b) 1.5kW
(c) 2 kW (d) 2.5kW
38. A nucleus ruptures into twonuclear parts which have their
velocity ratio equal to 1
:
2 . The ratio of their respective
nuclear sizes (nuclear radii )is
(a) 1: 2 (b) 2
:
1
(c) 1 : 21/3 (d) 1: 8
39. The rest energyofan electron is 0.511 MeV. The electron is
accelerated from rest to a velocity 0.5 c. The change in its
energywill be
(a) 0.026MeV (b) 0.051MeV
(c) 0.08MeV (d) 0.105MeV
40. A one-ton car moves with a constant velocity of
15 ms–1 on a rough horizontal road. The total resistance to
the motion of the car is 12% of the weight of the car. The
power required to keep the car moving with the same
constant velocity of 15ms–1 is
[Take g = 10 ms–2]
(a) 9 kW (b) 18kW
(c) 24kW (d) 36kW
41. Hail storms are observed to strike the surface of the frozen
lake at 300 with the vertical and rebound at 600 with the
vertical. Assume contact to be smooth, the coefficient of
restitution is
(a)
1
e
3
= (b)
1
e
3
=
(c) e 3
= (d) e = 3](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-155-320.jpg)
![152 PHYSICS
42. Abodystarts from rest and acquires a velocityV in time T.
Thework done on the bodyin time t will be proportional to
(a)
V
t
T
(b)
2
2
V
t
T
(c)
2
2
V
t
T
(d)
2
2
2
V
t
T
43. Aball isallowedtofall froma height of10 m. Ifthere is 40%
loss of energydue to impact, then after one impact ball will
go up to
(a) 10m (b) 8m
(c) 4m (d) 6m
44. The potential energy of a conservative system is given by
U = ay2 – by, where yrepresents the position of the particle
and a as well as bare constants. What is the force acting on
the system ?
(a) – ay (b) – by
(c) 2ay – b (d) b – 2ay
45. An automobile engine ofmass M accelerates and a constant
power p is applied bythe engine. The instantaneous speed
of the engine will be
(a) 1/ 2
[Pt / M] (b) 1/ 2
[2Pt / M]
(c) 1/ 2
[Pt / 2M] (d) 1/ 2
[Pt / 4M]
46. A bullet fired into a fixed target loses half of its velocity
after penetrating 3 cm. How much further it will penetrate
before coming to rest assuming that it faces constant
resistance to motion ?
(a) 2.0cm (b) 3.0cm
(c) 1.0cm (d) 1.5cm
47. A bomb of mass 16kg at rest explodes into two pieces of
masses 4 kg and 12 kg. The velolcity of the 12 kg mass is
4ms–1. The kinetic energy of the other mass is
(a) 144J (b) 288J
(c) 192J (d) 96J
48. Given that a force F̂ acts on a bodyfor time t, and displaces
the body by d̂ . In which of the following cases, the speed
of the body must not increase?
(a) F > d (b) F < d
(c) ˆ
F̂ d
= (d) ˆ
F̂ d
^
49. A body is attached to the lower end of a vertical helical
spring and it isgraduallyloweredto its equilibrium position.
This stretches the spring by a length x. If the same body
attached to the same spring is allowed to fall suddenly,
what would be the maximum stretching in this case?
(a) x (b) 2x
(c) 3x (d) x/2
50. A bag of mass M hangs by a long thread and a bullet (mass
m) comes horizontally with velocity V and gets caught in
the bag. Then for the combined (bag + bullet) system
(a)
mvM
momentum
M m
=
+
(b)
2
mV
kinetic energy
2
=
(c)
mV(M m)
momentum
M
+
=
(d)
2 2
m v
kinetic energy
2(M m)
=
+
51. Aparticle, initiallyat rest on a frictionless horizontal surface,
is acted upon by a horizontal force which is constant in
magnitude and direction. A graph is plotted of the work
done on the particle W, against the speed of the particle v.
Ifthere are no other horizontal forces acting on the particle,
the graph would look like
(a)
W
v
(b)
W
v
(c)
W
v
(d)
W
v
52. A particle of mass m moving eastward with a speed v
collides with another particle of the same mass moving
northwards with the same speed. If two particles coalesce
on collision, the new particle of mass 2 m will move in the
north-east direction with a velocity
(a) v / 2 (b) v 2
(c) v / 2 (d) None of these
53. A small block ofmass m is kept on a rough inclined surface
of inclination q fixed in an elevator. The elevator goes up
with a uniform velocity v and the block does not slide on
the wedge. The work done by the force of friction on the
block in time t as seen by the observer on the inclined p
lane will be
(a) zero (b) mgvt cos2q
(c) mgvt sin2 q (d) mgvt sin 2q
54. A nucleus moving with a velocity v
r
emits an a-particle.
Let the velocities ofthea-particle andtheremainingnucleus
be 1
v
r
and 2
v
r
and their masses be m1 and m2.
(a) v
r
, 1
v
r
and 2
v
r
must be parallel to each other
(b) None of the two of v
r
, 1
v
r
and 2
v
r
should be parallel to
each other
(c) 1 1 2 2
m v m v
+
r r
must be parallel to 1 2
(m m )v
+
r
.
(d) None of these
55. A shell is fired from a cannon with a velocityV at an angle
q with the horizontal direction. At the highest point in its
path, it explodes into two pieces of equal masses. One of
the pieces retraces its path to the cannon. The speed of the
other piece immediatelyafter the explosion is
(a) 3 V cos q (b) 2 V cos q
(c)
3
Vcos
2
q (d) V cos q](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-156-320.jpg)
![158 PHYSICS
17. Which ofthe diagrams in figure correctly shows the change
in kinetic energy of an iron sphere falling freely in a lake
having sufficient depth to impart it a terminal velocity?
(a)
KE
Depth
t
O
(b)
KE
Depth
t
O
(c)
KE
Depth
t
O
(d)
KE
Depth
t
O
18. A cricket ball of mass 150 g moving with a speed of 126 km/
h hits at the middle of the bat, held firmlyat its position by
the batsman. The ball moves straight back to the bowler
after hitting the bat. Assuming that collision between ball
and bat is completelyelastic and the two remain in contact
for 0.001s, the force that the batsman had to apply to hold
the bat firmly at its place would be
(a) 10.5N (b) 21N
(c) 1.05× 104 N (d) 2.1 × 104 N
NEET/AIPMT (2013-2017) Questions
19. A uniform force of ˆ ˆ
(3 )
i j
+ newton acts on a particle of
mass 2 kg. The particle is displaced from position $
(2 )
i k
$ +
meter to position $ $
(4 3 )
i j k
$ + - meter. The work done by
the force on the particle is [2013]
(a) 6 J (b) 13J
(c) 15J (d) 9 J
20. An explosion breaks a rock into three parts in a horizontal
plane. Two of them go off at right angles to each other. The
first part of mass 1 kg moves with a speed of 12 ms–1 and
the second part of mass 2 kg moves with speed 8 ms–1. If
the third part flies off with speed 4 ms–1 then its mass is
(a) 5 kg (b) 7 kg [2013]
(c) 17kg (d) 3 kg
21. A person holding a rifle (mass of person and rifle together
is 100 kg) stands on a smooth surface and fires 10 shots
horizontally, in 5 s. Each bullet has a mass of 10 g with a
muzzle velocityof 800 ms–1. The final velocityacquired by
the person and the average force exerted on the person are
[NEET Kar. 2013]
(a) –1.6 ms–1; 8 N (b) –0.08 ms–1;16N
(c) – 0.8 ms–1; 8 N (d) –1.6 ms–1; 16 N
22. Aparticlewith total energyE ismoving in apotentialenergy
region U(x). Motion ofthe particle is restricted tothe region
when [NEET Kar. 2013]
(a) U(x)> E (b) U(x)< E
(c) U(x)= O (d) U(x) £E
23. One coolie takes 1 minute to raise a suitcase through a
height of 2 m but the second coolie takes 30 s to raise the
same suitcase tothe same height. The powers of twocoolies
are in the ratio of [NEET Kar. 2013]
(a) 1: 2 (b) 1: 3
(c) 2: 1 (d) 3: 1
24. A bodyof mass (4m) is lying in x-yplane at rest. It suddenly
explodes into three pieces. Two pieces, each of mass (m)
move perpendicular to each other with equal speeds (v).
The total kinetic energy generated due to explosion is :
(a) mv2 (b)
2
3
mv
2
[2014]
(c) 2 mv2 (d) 4 mv2
25. A particle of mass m is driven bya machine that delivers a
constant power of k watts. If theparticle starts from rest the
force on the particle at time t is [2015]
(a) –1/2
mk t (b) –1/2
2mk t
(c)
–1/2
1
mk t
2
(d)
–1/2
mk
t
2
26. Two similar springs Pand Q have spring constants KP and
KQ, such that KP > KQ. They are stretched, first by the
same amount (case a,) then by the same force (case b). The
work done by the springs WP and WQ are related as, in
case (a) and case (b), respectively [2015]
(a) WP = WQ ; WP = WQ
(b) WP > WQ ; WQ > WP
(c) WP < WQ ; WQ < WP
(d) WP = WQ ; WP > WQ
27. A blockof mass 10 kg, moving in x direction with a constant
speed of 10ms–1, is subject toa retarding forceF =0.1 × J/m
during its travel from x = 20 m to 30 m. Its final KE will be:
(a) 450J (b) 275J [2015]
(c) 250J (d) 475J
28. The heart ofman pumps 5 litresofbloodthrough the arteries
per minute at a pressure of150 mm ofmercury. If the density
of mercury be 13.6 ×103 kg/m3 and g = 10m/s2 then the
power of heart in watt is : [2015 RS]
(a) 2.35 (b) 3.0
(c) 1.50 (d) 1.70](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-162-320.jpg)
![159
Work, Energy and Power
29. A ball isthrown verticallydownwards from a height of 20 m
with an initial velocity v0. It collides with the ground loses
50 percent of its energy in collision and rebounds to the
same height. The initial velocityv0 is : [2015 RS]
(Take g = 10 ms–2)
(a) 20 ms–1 (b) 28 ms–1
(c) 10 ms–1 (d) 14 ms–1
30. On a frictionless surfacea block of mass M moving at speed
v collides elastically with another block of same mass M
which is initiallyat rest.After collision thefirst block moves
at an angle q to its initial direction and has a speed
v
3
. The
second block's speed after the collision is : [2015 RS]
(a)
3
v
4
(b)
3
v
2
(c)
3
v
2
(d)
2 2
v
3
31. Two particles A and B, move with constant velocities 1
v
r
and 2
v
r
. At the initial moment their position vectors are 1
r
r
and 2
r
r
respectively. The condition for particlesAand Bfor
their collision is: [2015 RS]
(a) 1 1 2 2
r .v r .v
=
r r
r r
(b) 1 1 2 2
r v r v
´ = ´
r r
r r
(c) 1 2 1 2
r r v v
- = -
r r r r
(d) 1 2 2 1
1 2 2 1
r r v v
| r r | | v v |
- -
=
- -
r r r r
r r r r
32. A body of mass 1 kg begins to move under the action of a
time dependent force 2
ˆ ˆ
F (2ti 3t j)
= +
r
N, where î and ĵ are
unit vectors alogn x and y axis. What power will be
developed by the force at the time t? [2016]
(a) (2t2 + 3t3)W (b) (2t2 + 4t4)W
(c) (2t3 + 3t4) W (d) (2t3 + 3t5)W
33. Aparticle of mass10 gmovesalong acircleofradius 6.4cm
with a constant tangential acceleration. What is the
magnitude of this acceleration if the kinetic energy of the
particle becomes equal to 8 × 10–4 J by the end of the
second revolution after thebeginning ofthe motion ? [2016]
(a) 0.1m/s2 (b) 0.15 m/s2
(c) 0.18 m/s2 (d) 0.2m/s2
34. Consider a drop ofrain water having mass 1 g falling from a
height of 1 km. It hits the ground with a speed of 50 m/s.
Take 'g' constant with a value 10 m/s2.
The work done by the (i) gravitational force and the (ii)
resistive force of air is [2017]
(a) (i)1.25J (ii) –8.25J
(b) (i)100J (ii) 8.75J
(c) (i) 10J (ii) – 8.75 J
(d) (i) – 10 J (ii) –8.25J](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-163-320.jpg)
![162 PHYSICS
24. (d) Let n be the number of bullets that the man can fire in
one second.
change in momentum per second mv
n ´
= = F
[ m= mass of bullet, v = velocity] (Q F is the force)
1200
40
1000
144
mv
F
n
´
´
=
=
= 3
25. (d) Power supplied =
t
mgh
Power used by crane =
80
100
t
mgh
´
W
8
36
10
80
100
3600
100
10
1000 5
´
=
´
´
´
=
26. (b) Work done against gravity
Wg = 50 × 10 × 30 = 15 kJ
Work done against friction
Wf mgcos s
= m q´ =
4
0.4 50 10 50 8 kJ
5
´ ´ ´ ´ =
Total work done kJ
23
kJ
8
kJ
15
W
W f
g =
+
=
+
=
27. (d) Let mass of neutron = m
then mass of deuterium = 2m
[Q it has double nuclides thus has neutron].
Let initial velocityofneutron = v and final velocities
of neutron and deuterium are v1and v2 respectively.
m 2m
v=0
m 2m
v
1
v
® 2
v
®
neutron deuterium
Applying conservation of momentum
2
1 mv
2
mv
)
0
(
m
2
mv +
=
+
)
i
.(
..........
v
2
v
v 2
1 +
=
Þ
applying conservation of energy
2 2 2
1 2
1 1 1
mv mv 2m.v
2 2 2
= +
2 2 2
1 2
v v 2v ......(ii)
Þ = +
from (i)and (ii), 2
2
2
2
2
v
2
)
v
2
v
(
v +
-
=
2
2
2
2
2
2
2
v
2
v
v
4
v
4
v
v +
-
+
=
Þ
0
v
v
4
v
6 2
2
2 =
-
3
v
v
&
3
v
2
v 1
2 -
=
=
Þ
now fractional change in kinetic energy
i f
i
K K
K
-
=
2
2
1
2
mv
2
1
mv
2
1
mv
2
1
-
=
2
2
2
v
v
8
9
9
v
-
= =
28. (c) W = F s cos q, ,
2
1
10
5
25
s
F
W
cos =
´
=
=
q q = 60º.
29. (b) For elastic collision in one dimension
1
v =
2
1
2
2
m
m
u
m
2
+
+
)
m
m
(
u
)
m
m
(
2
1
1
2
1
+
-
As mass 2m, is at rest, So 2
u = 0
Þ 1
v =
m
2
m
8
u
)
m
2
m
8
(
+
-
= u
5
3
Final energy of sphere = f
.)
E
.
K
(
=
2
5
u
3
)
m
8
(
2
1
÷
ø
ö
ç
è
æ
=
2
2
5
3
u
)
m
8
(
2
1
÷
ø
ö
ç
è
æ
´
= E
25
9
=0.36E
30. (b) The work is done against gravity so it is equal to the
change in potential energy. W= Ep = mgh
For a fixed height, workisproportional toweight lifted.
Since Johnnyweighs twice as much as Jane he works
twice as hard to get up the hill.
Power is work done per unit time. For Johnny this is
W/Dt. Jane did half the work in half the time, (1/2 W)/
(1/2 Dt) = W/Dt which is the same power delivered by
Johnny.
31. (c) Fraction of energytransferred =
9
8
)
2
1
(
2
4
2
=
+
´
32. (c)
m
2
p
E
2
= Þ
2
2
2
1
2
1
p
p
E
E
= Þ 4
E
E 1
2 ´
=
1
1
2 E
3
E
E =
-
33. (c) Since the new height gained is half, therefore there is
50% loss of energy.
34. (a) u
m
3
u
2
m
u
m 2
1
1 +
= (By condition of linear
momentum)
Þ 1 2
1
m u m v
3
= ......(i)
Also
|
u
u
|
|
v
v
|
e
1
2
2
1
-
-
=
Þ
2u
v u
3
- = Þ
5
v u
3
= ......(ii)
From (i) and (ii), u
m
3
5
u
m
3
1
2
1 = Þ 5
m
m
2
1
=
35. (c) At Equilibrium , ky= mg Þ
y
mg
k =
mgy
2
1
y
y
mg
2
1
U 2
=
÷
÷
ø
ö
ç
ç
è
æ
=
36. (a) Loss in K.E =Area under the curve](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-166-320.jpg)
![163
Work, Energy and Power
37. (b) In this case, P =
t
mv
2
1
mgh 2
+
Þ
ú
ú
û
ù
ê
ê
ë
é
+
=
2
v
gh
t
m
P
2
Þ W
2
10
10
10
10
1
10
P ú
û
ù
ê
ë
é ´
+
´
= =1500W
38. (c) Byconservation of linear momentum,
1 2
2 1
m v
m v
=
or
2
1
r
3
4
r
3
4
3
2
x
3
x 1
=
p
r
p
r
Þ 3
/
1
2
1
2
1
r
r
=
39. (c)
2 2 2
0 0
2 2
2 2
m m
mc c c
v (0.5c)
1 1
c c
= =
- -
2
0
2
0
2
0
c
m
15
.
1
c
75
.
0
m
c
25
.
0
1
m
=
=
-
=
Change in energy = 1.15 m0c2 – m0c2
= 0.15 × 9.1 × 10–31 × (3× 108)2
= 12.285 × 10–15 J
MeV
10
6
.
1
10
285
.
12
13
15
-
-
´
´
=
=0.07678MeV
40. (b)
12
F 1000 10 N 1200 N
100
= ´ ´ =
P= Fv= 1200 N × 15 ms–1 = 18 kW.
41. (b) Components of velocity before and after collision
parallel to the plane are equal, So
vsin 60° = u sin 30°.......(1)
Componentsofvelocitynormal tothe plane are related
to each other
v cos 60° = e u (cos 30°) ........(2)
Þ cot 60° = e cot 30° Þ
cos60
e
cot30
°
=
°
Þ
3
3
1
e = Þ .
3
1
e =
42. (d) Work done on the body is gain in the kinetic energy.
Acceleration of the body is a = V/T.
Velocityacquired in time t is v = at
V
t
T
=
K.E. acquired µ v2. That is work done
2 2
2
V t
T
µ
43. (d) Kinetic energyof ball when reaching the ground
= mgh = mg× 10
Kinetic energyafter the impact
60
mg 10 6mg
100
= ´ ´ =
If the ball rises to a height h, then mgh = 6 mg.
Hence, h = 6 m.
44. (d)
dU
F b 2ay
dy
= - = -
45. (b) Fv = P . or
dv
M v P
dt
=
That is
P
vdv dt
M
=
ò ò
Hence 1/ 2
v [2Pt / M]
=
46. (c)
cm
3
u
2
u 0
v =
x
Case I :
3
.
a
.
2
u
2
u 2
2
=
-
÷
ø
ö
ç
è
æ
or –
4
u
3 2
= 2. a. 3 Þ a = –
8
u2
Case II :
2
2
u
0 ÷
ø
ö
ç
è
æ
- = 2. a. x or –
4
u2
= 2.
÷
÷
ø
ö
ç
ç
è
æ
-
8
u2
× x
Þ x=1cm
Alternative method : Let K be the initial energy and F
be the resistive force. Then according towork-energy
theorem,
W = DK
i.e.,
2
2
1 1 v
3F mv m
2 2 2
æ ö
= - ç ÷
è ø
2
1 1
3F mv 1
2 4
æ ö
= -
ç ÷
è ø
2
3 1
3F mv
4 2
æ ö
= ç ÷
è ø
...(1)
and
2
2
1 v 1
Fx m m(0)
2 2 2
æ ö
= -
ç ÷
è ø
i.e.,
2
1 1
mv Fx
4 2
æ ö
=
ç ÷
è ø
...(2)
Comparing eqns. (1) and (2)
F=Fx
or x=1cm](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-167-320.jpg)
![164 PHYSICS
47. (b) Let the velocity and mass of 4 kg piece be v1 and m1
and that of 12 kg piece be v2 and m2.
Applying conservation of linear momentum
m2v2 = m1v1
1
1 ms
12
4
4
12
v -
=
´
=
Þ
J
288
144
4
2
1
v
m
2
1
.
E
.
K 2
1
1
1 =
´
´
=
=
48. (d) Velocitywill increase when forceis along the direction
ofdisplacement i.e. d̂
F̂ = .
49. (b) When body is lowered gradually, its weight acts at
C.G. of the spring. When same bodyis allowed to fall
freely, the same weight acts at lower end of the spring.
In the latter case, original length (L) ofspring is double.
As DL µ L, therefore, DL becomes twice in second
casei.e. 2x.
50. (d) If V is velocity of combination (bag + bullet), then
from principle ofconservation of linear momentum
m v
(m M)V m v or V
(m M)
+ = =
+
2 2
2
1 m v
K.E. (m M)V
2 2(m M)
= + =
+
51. (c) Workdone W = [ML2 T–2]
It shows that 1 2 2
W (LT ) i.e. W v .
-
µ µ
graph between W & v is a parabola.
Alternatively:
According to work energy theorem
2
2
2
mv
2
1
mu
2
1
mv
2
1
W =
-
= (Q u = 0)
Þ W µ v2
52. (c) Applyingprinciple ofconservation of linear momentum
2 2
(2m)V (mv) (mv) mv 2
= + =
v 2 v
V
2 2
= =
53. (a) Since block does not slide on wedge so displacement
is zero & hence work done by force is zero.
f
m
gsinq
q
u
mg
54. (c) If mass of nucleus is m, mass of a particle is m1 &
mass of remaining nucleus is m2, then from the lawof
conservation of momentum.
nucleus
remaining
of
momentum
2
2
particle
of
momentum
1
1
nucleus
of
momentum
initial
v
m
v
m
v
m
r
r
r
+
=
a
55. (a) Let M be the mass of shell. Applying law of
conservation of linear momentum
MV cosq =
MV M
cos v
2 2
æ ö
- q +
ç ÷
è ø
ie, MV cos q +
M
2
Vcosq =
M
2
v
or v = 3 Vcosq .
56. (d) As bob B is of same material and same mass as the
bob A, therefore, on elastic collision, their velocities
are exchanged. BobAcomes to rest and B moves with
the velocity of A.
57. (c) As velocities are exchanged on perfectly elastic
collision, therefore masses of two objects must be
equal.
.
m
m
or
1
m
m
b
a
b
a
=
=
58. (c) a kx
= -
dv
kx
dt
Þ = -
Also
dv
dx
dt
or
v
dt
dx
=
=
v x
2
v 0
1
vdv
kx v dv kxdx
dx
= - Þ = -
ò ò
( )
2
2 2
2 1
kx
v v
2
- = - ( )
2
2 2
2 1
1 1 kx
m v v m
2 2 2
æ ö
-
Þ - = ç ÷
è ø
2
K x
D a
59. (c)
2
i 1 1
1
K m u ,
2
=
1
2
1
2
1
1
2
1
1
f u
m
m
m
m
v
,
v
m
2
1
K
+
-
=
=
Fractional loss
2
1
1
2
1
1
2
1
1
i
f
i
u
m
2
1
v
m
2
1
u
m
2
1
K
K
K
-
=
-
( )
( )2
2
1
2
2
1
2
1
2
1
m
m
m
m
1
u
v
1
+
-
-
=
-
=
( )2
2
1
2
1
m
m
m
m
4
+
=
m
m2 = ; nm
m1 =
( )2
i
f
n
1
n
4
K
K
1
+
=
-
Energytransfer is maximum when 0
Kf =
( )2
4n
1
1 n
=
+
2
4n 1 n 2n
Þ = + + Þ 0
n
2
1
n2
=
-
+
( ) 0
1
n 2
=
- 1
n = ie. m
m2 = , m
m1 =
Transfer will be maximumwhen both masses are equal
and one is at rest.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-168-320.jpg)
![165
Work, Energy and Power
60. (c) s
/
m
10
4
10
0
4
14
10
Vc =
+
´
+
´
= ; since spring force is
internal force.
61. (a) For inelastic collision, linear momentum is conserved
Þ 1
mv = 2
mv
2 Þ
2
v
v 1
2 =
Lossin K.E. = Gain in P.E.
= mgh
2
v
)
m
2
(
2
1
mv
2
1 2
2
2
1 =
-
Þ 4 mgh =
2
1
mv –
2
mv2
1 =
2
mv2
1
=
2
mv2
Þ h =
g
8
v2
62. (d) P F v
= ´ Þ P = F a t P t
µ
63. (c) g
h
2
tAB =
1
BC CB
2
2h
t t 2
g
2e h 2h
2 2e
g g
+ =
= =
h
h
A
B
h
1
2
D
C
g
h
2
e
2
t
t 2
DB
BD =
+
Total time taken by the body in coming to rest
.........
g
h
2
e
2
g
h
2
e
2
g
h
2 2
+
+
+
=
2
2h 2h
2e [1 e e .........]
g g
= + + + +
=
e
1
1
g
h
2
e
2
g
h
2
-
´
+
2h 1 e 1 e
t
g 1 e 1 e
+ +
é ù æ ö
= = ç ÷
ê ú
- -
ë û è ø
64. (d) Initially, mg
k
3 = or
3
mg
k =
New force constant of longer part
2
mg
3
mg
2
3
k
2
3
'
k =
´
=
=
Finally, mg
y
'
k =
cm
2
2
mg
mg
'
k
mg
y =
´
=
=
65. (d) kg
8
kg
8
.
0
10
m =
´
=
height ofiron chain = 5m
mgh 8 10 5
P W 40W
t 10
´ ´
= = =
66. (c) Bylaw of conservation ofmomentum,
Mu = MV + mv ....(i)
Also e = MV
Mv
Mu
|
u
u
|
|
v
v
|
2
1
2
1
-
=
Þ
-
-
....(ii)
From (i) and (ii), 2Mu = (M + m)v
Þ
2uM 2u
v v
m
M m
1
M
= Þ =
+
+
67. (d) Centripetal force =
R
mv2
R
aS
2
R
K
2
R
2
mv
2
1 2
=
=
÷
ø
ö
ç
è
æ
=
68. (a) As the block is at rest at P.
q
m
=
q cos
mg
sin
mg
dx
dy
slope
tan =
=
q
=
m
L
x
L
2
x
dx
d 2
=
÷
÷
ø
ö
ç
ç
è
æ
=
mg
mg cosq
mg sinq
q
N P
x
y
8
.
0
L
x
=
Þ x = 0.8× 10 = 8m
69. (d) Work done by centripetal force is zero.
70. (a)
dU
F
dx
= -
71. (c) Thechangein momentum in the ball after the collision
with surface is m(0–v) = –mv
Since n balls impinge elastically each second on the
surface, then rate ofchange of momentum of ball per
second is
mvn (consider magnitudeonly)
Now According to Newton’s second law
rate of change ofmomentum per second ofball = force
experienced by surface.
72. (d)
h=1.5m
The mass of water is m = 1 × 103 kg
The increase in potential energy of water is
= mgh =(1 × 103) (10)(1.5) = 15 kJ](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-169-320.jpg)
![167
Work, Energy and Power
85. (b) From law of conservation of momentum, before
collision and after collision linear momentum (p) will
be same.
or initial momentum = final momentum.
2
p
E
2m
=
According to question,
2
1 1 2
2
2 1 2
E p 2m
E 2m p
= ´
1
2
2
1
m
m
E
E
=
Þ [p1 = p2]
86. (b) 2
2
kx
2
1
mv
2
1
= Þ mv2 = kx2 or 0.5 × (1.5)2 = 50×x2
x=0.15m
87. (a) Clearly v1 = 2 ms–1, v2 = 0
m1 = m (say), m2 =2m
v1' = ?, v'2 = ?
e =
1 2
2 1
v ' v '
v v
-
- ....(i)
Byconservation of momentum,
2m= mv1' + 2mv2' ...(ii)
From (i), 0.5 =
2 1
v ' v '
2
-
v2' = 1 + v1'
From (ii), 2 = v1'+ 2 + 2 v1'
Þ v1 = 0 and v2 = 1 ms–1
88. (b) Power exerted by a force is given by
P= F.v
When the body is just above the earth’s surface, its
velocityis greatest.At this instant, gravitational force
is also maximum. Hence, the power exerted by the
gravitational force is greatest at the instant just before
the body hits the earth.
89. (a) As the two masses stick together after collision, hence
it is inelastic collision. Therefore, only momentum is
conserved.
2v
3m
v x
m
ˆ ˆ
mvi 3m(2v)j (4m)v
r
+ =
v 6
ˆ ˆ
v i vj
4 4
= +
r
=
v 3
ˆ ˆ
i vj
4 2
+
90. (b) For equilibrium
0
dU
dr
= Þ 3 2
2
0
A B
r r
-
+ =
r =
2A
B
for stable equilibrium
2
2
d U
dr
should be positive for the value of r.
here
2
2 4 3
6 2
d U A B
dr r r
= - is +ve value for
2A
r
B
= So
91. (d) Given : F
r
= 3i j
+
$ $
1
r
u
r
= $
( )
2i k
+
$ , 2
r
uu
r
= ( )
4i 3j k
+ -
r
$ $
r
r
= 2 1
r r
-
uu
r u
r
= ( )
4i 3j k
+ -
r
$ $ – $
( )
2i k
+
$
or r
r
= 2i 3j
+
$ $ – $
2k
So work done by the given force w = f.r
r r
= ( ) $
( )
3i j . 2i 3j 2k
+ + -
$ $ $ $ = 6 + 3 = 9J
92. (a)
4m
/sec
m
3
2 kg m2
Presultant
1 kg
x
y
8 m/sec
12 m/sec
m1
Presultant = 2 2
12 16
+
= 144 256
+ = 20
m3v3 = 20 (momentum ofthird part)
or, m3 =
20
4
= 5 kg
93. (c) p 2mK
=
p' 2m[K 3K]
= + = 2p
p 100 2p p
100 100%
p p
D ´ -
= ´ =
94. (c) Let m and v be the mass and the velocity of the body.
Then initial K.E. ,
Ki =
2
1
2
mv
Now, mass =
2
m
velocity = 2v](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-171-320.jpg)
![169
Work, Energy and Power
10. (b) As we know that,
W.D.
2 2
1 1
0
x x
x x
F dx ma dx
= × = ×
ò ò
uu
r uu
r
r r
As given that,
m = 0.5 kg, a = 5 m–1/2 s–1,
work done (W) = ?
3/2
v ax
=
We also know that,
Acceleration,
3/ 2 3/2
0 ( )
dv dv d
a v ax ax
dt dx dx
= = × =
3/ 2 1/2 2 2
3 3
2 2
ax a x a x
= ´ ´ ´ =
Now, Force =
2 2
0
3
2
ma m a x
=
From(i),
Work done
2
0
x
x
Fdx
=
=
= ò
2 2 2
0
3
2
ma x dx
é ù
= ê ú
ë û
ò
2
3
2
0
3
2 3
x
ma
æ ö
= ´ç ÷
ç ÷
è ø
2
1
8
2
ma
= ´
1
(0.5) (25) 8 50 J
2
= ´ ´ ´ =
11. (b) As given that power = constant
As we know that power (P)
dW F dx F dx
P
dt dt dt
×
= = =
uu
r
r
As the body is moving unidirectionally.
Hence, cos 0
F dx Fdx Fdx
× = ° =
constant
Fdx
P
dt
= = ( P =
Q constant by question)
Now, by dimensional formula
0
F v
× =
[F] [v] = constant
[MLT–2] [LT–1] = constant
[ML2T–3] = constant
3
2 T
L
M
=
(As mass of body constant)
2 3 3/ 2
L T L T
µ Þ µ Þ Displacement 3/2
( )
d t
µ
Verifies the graph (b).
12. (d) The speed of earth around the sun can never be zero
or negative, so the kinetic energy of earth cannot be
zero and negative.
So, option (b) and (c) represents wrongly the variation
in kinetic energy of earth.
A B
SUN
When the earth is closest to the sun, speed of the
earth is maximum, hence, KE is maximum. When the
earth isfarthest from the sun speed isminimum hence,
KEisminimum.
So, K.E. of earth increases (B to A) and then decrease,
variation is correctly represented by option (d).
13. (c) When a pendulum oscillates in air,due toair resistance
the force of friction acts between bob of pendulum
and air, so it will lose energy continuously in
overcoming. Therefore, total mechanical energy(KE
+ PE) of the pendulum decreases continuously with
time and finallybecomes zero.
Sum of KE and PE can never be negative. So, option
(a) and (d) are incorrect. Hence option (c) is verifies.
14. (a) As given that, mass (m) = 5 kg,
n = 300 revolution
Radius (R) =1 m
t = 60 sec
2
(300 2 )rad / 60s
n
t
p
æ ö
w = = ´ ´p
ç ÷
è ø
600
rad/s = 10 rad/s
60
´p
= p
linear speed (v) = wR = (10p× 1)
v = 10pm/s
KE
2
1
2
mv
=
2
1
5 (10 )
2
= ´ ´ p
2 1
100 5
2
= p ´ ´
2
250 J
= p
So, verifies the option (a).
15. (b) P.E. is maximum when drop start falling at
t = 0 as it fall is P.E. decrease gradually to zero. So, it
rejects the graph (a), (c) and (d).
K.E. at t = 0 is zero as drop falls with zero velocity, its
velocity increases (gradually), hence, first KE also
increases. After sometime speed (velocity) is constant
this is called terminal velocity, so, KE also become
constant. It happens when it falls
3
4
æ ö
ç ÷
è ø
height or
remains at
4
4
æ ö
ç ÷
è ø
from ground, then PE decreases
continuously as the drop is falling continuously.
Thevariation in PE and KE is best represented by(b).](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-173-320.jpg)
![179
System of Particles andRotationalMotion
(5)Annular disc (a) Perpendicular to plane
passing through centre
M
2
[ 2 2
1 2
R R
+ ]
2 2
1 2
R R
2
+
of mass
(b) Diameter in the plane
2 2
1 2
M[R R ]
4
+ 2 2
1 2
R R
4
+
(6) Hollowcylinder (a)Geometrical axis
2 2
1 2
R R
M
2
é ù
+
ê ú
ê ú
ë û
2 2
1 2
R R
2
+
(b) Perpendicular to length
passing through centre
2 2
2
1 2
(R R )
L
M
12 4
é ù
+
+
ê ú
ê ú
ë û
2 2
2
1 2
R R
L
12 4
+
+
of mass
(7) Solid sphere (a)Along the diameter
2
5
MR2
2
5
R
(b) Along the tangent
7
5
MR2
7
5
R
Ic
R1
R2 C
M
Ic
R1
R2
C
M
I
M
cm
L
R1
R2
M
cm
L
R1
R2
I
Ic
M
cm
R
Ic
M
cm
R
Id](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-183-320.jpg)
![182 PHYSICS
TORQUE, ANGULAR MOMENTUM AND ANGULAR
IMPULSE
Torque :
The moment of force is called torque. It is defined as the product
of force and the perpendicular distance of the force from the
axis of rotation.
O X
Y
q
q
r
F
i.e., r F
t = ´
r
r r
or, sin
rF
t = q
where q is the angle between r
r
and F
r
.
Its S.I. unit is (N-m). The dimensions of torque [ML2T–2] are
the same as that of energy but it is not energy.
Note : If the line of action of a force passes through axis of
rotation then no torque will be formed.
Angular Momentum:
The angular momentum ofa particle about an arbitrrypoint 'O' is
themoment of linear momentum taken about that point.
r
p
O X
Y
q
q
It is given as L r p
= ´
r r r
or, sin
L rp
= q
where q is the angle between r
r
and F
r
.
Angular Impulse :
2
1
2 1
t
t
J dt L L
= t = -
ò (= Change in angular momentum)
CONSERVATION OF ANGULAR MOMENTUM
From equation
dt
L
d
ext =
t
r
.
If constant
L
0
ext =
Þ
=
t
r
...(i)
This is called law of conservation of angular momentum.
According to this “if resultant external torque ext
t
r
acting on
the system is zero then total angular momentum of the system is
constant.
The magnitude of angular momentum for a system is given by
i
i
i
n
1
i
v
)
r
m
(
|
L
|
=
å
= w
å
=
=
)
r
m
( 2
i
i
n
1
i
[Q v = rw]
Þ w
=
w
´
å
=
=
I
)
I
(
|
L
| i
n
1
i
...(ii)
Where Ii is the moment ofinertia of the ith particle ofthat system
and I is total moment of inertia of the system
n
3
2
1
i
n
1
i
I
..........
..........
I
I
I
I
I +
+
+
+
=
å
=
=
or 2
n
n
2
2
2
2
1
1 r
m
..........
..........
r
m
r
m
I +
+
+
= ...(iii)
So if a system undergoes a redistribution of its mass, then its
moment of inertia changes but since noexternal torque is applied
on the system so total angular momentum is constant before and
after the distribution of mass, even if moment of inertia of the
system is changed.
Initial Final
L L
=
uu
r uu
r
...(iv)
or, I1w1 = I2w2 ...(v)
where L initial denote the state previous to the redistribution of
mass and final denote the state after the redistribution of mass in
that system.
Acomparisonof useful relationsin rotational andtranslational
or linear motion :
Rotational motion about Linear motion
a fixed axis
Angular velocity
d
dt
q
w = Linear velocity
dx
v
dt
=
Angular acceleration
d
dt
w
a = Linear acceleration
dv
a
dt
=
Resultant torque t = Ia Resultant force F = ma
Equations of rotational Equations of linear motion
motion
0
2
0 0
2 2
0 0
t
constant then t ½ t
2 ( )
w =w + a
ì
ï
ï
a = q -q = w + a
í
ï
w =w + a q -q
ï
î
2
2 2
v u at
1
a constant then s ut at
2
v u 2as
ì = +
ï
ï
= = +
í
ï
ï - =
î
where (s = x – x0)
Work
o
W d
q
q
= t q
ò
r
r
work
o
x
x
W Fdx
= ò
r r
Kinetic energy Ek = ½ Iw2 Kinetic energy Ek=½ mv2
Power P = tw Power P = Fv
rr
Angular momentum L = Iw
rr
Linear momentum p= mv
Torque
dL
dt
t = Force
dp
F
dt
=
Work EnergyTheoreminRotational Motion:
According to this theorem “the work done by external forces in
rotating a rigid body about a fixed axis is equal to the change
in rotational kinetic energy of the body.”](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-186-320.jpg)
![193
System of Particles andRotationalMotion
32. A composite disc is to be made using equal masses of
aluminiumandiron sothat it has ashigh a moment ofinertia
as possible. This is possible when
(a) thesurfaces ofthediscare madeofiron with aluminium
inside
(b) the whole ofaluminium is kept in the core and the iron
at the outer rim of the disc
(c) the whole of the iron is kept in the core and the
aluminium at the outer rim of the disc
(d) the whole disc is made with thin alternate sheets of
iron and aluminium
33. A ball rolls without slipping. The radius of gyration of the
ball about an axis passing through its centre of mass is K. If
radius of the ball be R, then the fraction of total energy
associated with its rotational energywill be
(a) 2
2
2
R
K
R
+
(b) 2
2
2
R
R
K +
(c) 2
2
R
K
(d) 2
2
2
R
K
K
+
34. When a bucket containing water is rotated fast in a vertical
circle of radius R, the water in the bucket doesn't spill
provided
(a) thebucket is whirled with a maximum speedof 2gR
(b) the bucket is whirled around with a minimum speed of
]
gR
)
2
/
1
[(
(c) the bucket is having a rpm of )
R
/(
g
900 2
p
(d) the bucket is having a rpm of )
R
/(
g
3600 2
p
35. A coin placed on a gramophone record rotating at
33 rpm flies off the record, if it is placed at a distance of
more than 16 cm from the axis of rotation. If the record is
revolving at 66 rpm, the coin will fly off if it is placed at a
distance not less than
(a) 1cm (b) 2cm
(c) 3cm (d) 4cm
36. Two fly wheels A and B are mounted side by side with
frictionless bearings on a common shaft. Their moments of
inertia about the shaft are 5.0 kg m2 and 20.0 kg m2
respectively. Wheel Ais madeto rotate at 10 revolution per
second. Wheel B, initially stationary, is now coupled to A
with the help of a clutch. The rotation speed of the wheels
will become
(a) 2 5 rps (b) 0.5rps
(c) 2 rps (d) None of these
37. A wheel is rolling straight on ground without slipping. If
the axis of the wheel has speed v, the instantenous velocity
of a point P on the rim, defined by angle q, relative to the
ground will be
(a) ÷
ø
ö
ç
è
æ
q
2
1
cos
v
(b) ÷
ø
ö
ç
è
æ
q
2
1
cos
v
2
q
P
(c) )
sin
1
(
v q
+
(d) )
cos
1
(
v q
+
38. Acertain bicycle can go up a gentle incline with constant
speed when the frictional force of ground pushing the rear
wheel is F2 = 4 N. With what forceF1 must the chain pull on
the sprocket wheel if R1=5 cm and R2 = 30 cm?
Road
Chain F1
R2
R1
F = 4N
2
Horizontal
(a) 4N (b) 24N
(c) 140N (d)
4
35
N
39. Auniform rod of length l is free torotate in a vertical plane
about a fixed horizontal axis through O. The rod begins
rotating from rest from its unstable equilibrium position.
When it has turned through an angle q, its angular velocity
w is given as
q
O
(a) q
sin
g
6
l
(b)
2
sin
g
6 q
l
(c)
2
cos
g
6 q
l
(d) q
cos
g
6
l
40. A sphere of mass 2000 g and radius 5 cm is rotating at the
rate of 300 rpm .Then the torque required to stop it in 2p
revolutions, is
(a) 1.6 × 102 dyne cm (b) 1.6 × 103 dyne cm
(c) 2.5 × 104 dyne cm (d) 2.5 × 105 dyne cm
41. Five masses are placed in a plane as shown in figure. The
coordinates of the centre of mass are nearest to
y
2 3 kg
5 kg
2 kg
1 kg
4 kg
1
0
0 2
x
1
(a) 1.2,1.4 (b) 1.3,1.1
(c) 1.1,1.3 (d) 1.0,1.0](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-197-320.jpg)
![197
System of Particles andRotationalMotion
Exemplar Questions
1. For which ofthe following does the centre ofmass lie outside
the body?
(a) A pencil (b) A shotput
(c) A dice (d) A bangle
2. Which of the following points is the likely position of the
centre of mass of the system shown in figure?
A
B
C
D
Hollow sphere
Air
R/2
R/2
Sand
(a) A (b) B
(c) C (d) D
3. A particle of mass m is moving in yz-plane with a uniform
velocity v with its trajectory running parallel to +ve y-axis
and intersecting z-axis at z = a in figure. The change in its
angular momentum about the origin as it bounces elastically
from a wall at y = constant is
(a) ˆ
mva x
e
v
z
a
y
(b) ˆ
2mva x
e
(c) ˆ
ymv x
e
(d) ˆ
2ymv x
e
4. When a disc rotates with uniform angular velocity, which of
the following is not true?
(a) The sense of rotation remains same
(b) The orientation of the axis of rotation remains same
(c) The speed of rotation is non-zero and remains same
(d) The angular acceleration is non-zero and remains same
5. A uniform square plate has a small piece Q of an irregular
shape removed and glued to the centre of the plate leaving
a hole behind in figure. The moment of inertia about the z-
axis is then,
Q
P
y
x
Q
P
y
x
hole
(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner
6. In problem-5, the CM of the plate is now in the following
quadrant of x-y plane.
(a) I (b) II
(c) III (d) IV
7. The density of a non-uniform rod of length 1 m is given by
r(x) = a (1 + bx2) where, a and b are constants and
0 1.
x
£ £ The centre of mass of the rod will be at
(a)
3(2 )
4(3 )
b
b
+
+
(b)
4(2 )
3(3 )
b
b
+
+
(c)
3(3 )
4(2 )
b
b
+
+
(d)
4(3 )
3(2 )
b
b
+
+
8. A merry-go-round, made of a ring-like platform ofradius R
and mass M, is revolving with angular speed w.Aperson of
mass M is standing on it. At one instant, the person jumps
off the round, radially awayfrom the centre ofthe round (as
seen from the round). The speed ofthe round of afterwards
is
(a) 2w (b) w
(c)
2
w
(d) 0
NEET/AIPMT (2013-2017) Questions
9. A small object of uniform densityrolls up a curved surface
with an initial velocity ‘n’.Itreachesuptoamaximum height
of
2
3
4g
n
with respect to the initial position. The object is a
(a) solid sphere (b) hollow sphere [2013]
(c) disc (d) ring
10. A rod PQ of mass M and length L is hinged at end P. The
rod is kept horizontal bya massless string tied topoint Q as
shown in figure. When string is cut, the initial angular
acceleration of the rod is [2013]
(a) g /L (b) 2g/L
(c)
2
3
g
L
(d)
3
2
g
L](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-201-320.jpg)
![198 PHYSICS
11. Twodiscs are rotating about their axes, normal to the discs
and passing through the centres of the discs. Disc D1 has
2 kg mass and 0.2 m radius and initial angular velocityof 50
rads–1.DiscD2 has4kgmass,0.1mradiusandinitialangular
velocityof 200 rad s–1. The twodiscs are brought in contact
faceto face, with their axes of rotation coincident. The final
angular velocity (in rad s–1) of the system is
(a) 40 (b) 60 [NEET Kar. 2013]
(c) 100 (d) 120
12. The ratioof radii ofgyration of a circular ring and a circular
disc, of the same mass and radius, about an axis passing
through their centres and perpendicular to their planes are
(a) 2 :1 (b) 1: 2 [NEET Kar. 2013]
(c) 3: 2 (d) 2: 1
13. The ratio of the accelerations for a solid sphere (mass ‘m’
and radius ‘R’) rolling down an incline ofangle‘q’ without
slipping and slipping down the incline without rolling is :
(a) 5: 7 (b) 2: 3 [2014]
(c) 2: 5 (d) 7: 5
14. A solid cylinder of mass 50 kg and radius 0.5 m
is free to rotate about the horizontal axis.A massless string
is wound round the cylinder with one end attached to it
and other hanging freely. Tension in the string required to
produce an angular acceleration of 2 revolutions s– 2 is :
(a) 25N (b) 50N [2014]
(c) 78.5N (d) 157N
15. Threeidentical spherical shells, each ofmass m and radius
r are placed as shown in figure. Consider an axis XX' which
is touching to two shells and passing through diameter of
third shell. Moment of inertia of the system consisting of
these three spherical shells about XX' axis is [2015]
(a) 3mr2
(b) 2
16
mr
5
X
X¢
(c) 4mr2
(d)
2
11
mr
5
16. A rod of weight W is supported bytwo parallel knife edges
A and B and is in equilibrium in a horizontal position. The
knives are at a distance d from each other. The centre of
mass ofthe rod is at distance x fromA. The normal reaction
on A is [2015]
(a)
Wd
x
(b)
W(d – x)
x
(c)
W(d – x)
d
(d)
Wx
d
17. A mass m moves in a circle on a smooth horizontal plane
with velocity v0 at a radius R0. The mass is attached to
string which passes through a smooth hole in the plane as
shown.
v0
m
The tension in the string is increased graduallyand finally
m moves in a circle of radius 0
R
2
. The final value of the
kineticenergy is [2015]
(a)
2
0
1
mv
4
(b) 2
0
2mv
(c)
2
0
1
mv
2
(d) 2
0
mv
18. A force ˆ ˆ ˆ
F i 3j 6k
= a + +
r
is acting at a point
ˆ ˆ ˆ
r 2i 6j 12k
= - -
r
. The value of a for which angular
momentum about origin is conserved is : [2015 RS]
(a) 2 (b) zero
(c) 1 (d) –1
19. Point masses m1 and m2 are placed at the opposite ends of
a rigid rod oflength L, and negligible mass. The rod is to be
set rotating about an axis perpendicular to it. The position
of point P on this rod through which the axis should pass
sothat the work required to set the rod rotating with angular
velocity w0 is minimum, is given by: [2015 RS]
m1 m2
w0
P
x (L–x)
(a)
1
2
m
x L
m
= (b)
2
1
m
x L
m
=
(c)
2
1 2
m L
x
m m
=
+ (d)
1
1 2
m L
x
m m
=
+
20. An automobile moves on a road with a speed of 54 km h-1.
The radius of its wheels is 0.45 m and themoment ofinertia
of the wheel about its axis of rotation is 3 kg m2. If the
vehicle is brought to rest in 15s, the magnitude of average
torque transmitted byits brakes to the wheel is : [2015 RS]
(a) 8.58 kg m2 s-2 (b) 10.86 kg m2 s-2
(c) 2.86 kg m2 s-2 (d) 6.66 kg m2 s-2](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-202-320.jpg)
![199
System of Particles andRotationalMotion
21. A disk and a sphere of same radius but different massesroll
off on two inclined planes of the same altitude and length.
Which one of the two objects gets to the bottom of the
plane first ?
(a) Disk [2016]
(b) Sphere
(c) Both reach at the same time
(d) Depends on their masses
22. A uniform circular disc of radius 50 cm at rest is free toturn
about an axis which is perpendicular to its plane and passes
through its centre. It is subjected toa torque which produces
a constant angular acceleration of 2.0 rad s–2. Its net
acceleration in ms–2 at the end of2.0s is approximately:
(a) 8.0 (b) 7.0 [2016]
(c) 6.0 (d) 3.0
23. FromadiscofradiusRandmassM,acircular holeofdiameter
R, whose rim passes through the centre is cut. What is the
moment of inertia of the remaining part of the disc about a
perpendicular axis, passing through the centre ? [2016]
(a) 15 MR2/32 (b) 13 MR2/32
(c) 11 MR2/32 (d) 9 MR2/32
24. Which of the following statements are correct ? [2017]
(A) Centre of mass of a body always coincides with the
centre of gravity of the body
(B) Centre ofmass ofa bodyis the point at which the total
gravitational torque on the body is zero
(C) A couple on a body produce both translational and
rotation motion in a body
(D) Mechanical advantage greater than one means that
small effort can be used to lift a large load
(a) (A) and (B) (b) (B) and(C)
(c) (C) and(D) (d) (B) and (D)
25. Two discs of same moment of inertia rotating about their
regular axis passing through centre and perpendicular to
the plane of disc with angular velocities w1 and w2 . They
are brought into contact face to face coinciding the axis of
rotation. The expression for loss of energy during this
process is:- [2017]
(a)
2
1 2
1
I( )
4
w - w (b) 2
1 2
I( )
w - w
(c)
2
1 2
1
( )
8
w - w (d) 2
1 2
1
I( )
2
w + w
26. A rope is wound around a hollow cylinder ofmass 3 kg and
radius 40 cm. What is the angular acceleration of the
cylinder if the rope is pulled with a force of 30 N ? [2017]
(a) 0.25 rad/s2 (b) 25 rad/s2
(c) 5 m/s2 (d) 25 m/s2](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-203-320.jpg)
![200 PHYSICS
EXERCISE - 1
1. (a) 2. (b) 3. (c) 4. (c)
5. (b) 6. (d) 7. (b)
8. (d) The pull of earth changes onlywhen the body moves
sothat g changes. It is an exceptional case and should
not be considered unless otherwise mentioned.
9. (b) Ifwe applya torqueon a body, then angular momentum
of the body changes according to the relation
dL
τ
dt
=
r
r
Þ if t
r
= 0 then, L = constant
10. (b) Angular momentum L
r
is defined as L r m(v)
= ´
r r r
so L
r
is, an axial vector.
.
11. (c) The boy does not exert a torque to rotating table by
jumping, so angular momentum is conserved i.e.,
dL
0 L
dt
= Þ
r
r
= constant
12. (b) Since no external torque act on gymnast, so angular
momentum (L=Iω)isconserved.After pullingher arms
& legs, the angular velocity increases but moment of
inertia of gymnast, decreases in, such a way that
angular momentum remains constant.
13. (a)
14. (b) Angularmomentum willremainthesamesinceexternal
torque is zero.
15. (a)
2 2
B A
1
I MR I MR
2
= =
A B
I I
<
16. (a) Angular momentum ( )
r linear momentum
= ´
r
17. (b) We know that ext
dL
τ
dt
=
ifangular momentum is conserved, it means change in
angular momentum = 0
0
dL
,
or = , ext
dL
0 τ 0
dt
= Þ =
Thus total external torque = 0.
18. (a) Analogue of mass in rotational motion is moment of
inertia. It plays the same role as mass plays in
translational motion.
19. (a) Basic equation of moment of inertia is given
by 2
i
i
n
1
i
r
m
I
=
S
=
i
m
i
r
where i
m is the mass of th
i particle at a distance of
i
r from axis ofrotation.
Thus it does not depend on angular velocity.
20. (d) Initially centre of gravity is at the centre. When sand
is poured it will fall and again after a limit, centre of
gravitywill rise.
21. (d) Here, the centre of mass of the system remains
unchanged. in the horizontal direction. When the mass
m moved forward by a distance L cos q, let the mass
(m + M) moves by a distance x in the backward
direction. hence
(M + m) x – m L cos q = 0
x = (m L cos q)/(m + M)
22. (b) Acceleration of solid sphere is more than that of
hollow sphere, it rolls faster, and reaches the bottom
oftheinclined planeearlier.
Hence, solid sphere and hollow sphere can be
distinguished by rolling them simultaneously on an
inclined plane.
23. (d) Applying the law of conservation of momentum
m v= M V ...(1)
Byconservation of angular momentum
2
M L
m v(L / 2) ω
12
æ ö
= ç ÷
ç ÷
è ø
...(2)
As the collision is elastic, we have
2 2 2
1 1 1
m v M V Ι ω
2 2 2
= + ...(3)
Substituting the values, we get m = M/4
24. (b) a is propotional to w
Let a= kw (Qk is a constant)
dω
kω
dt
= [also
dθ dθ
ω dt
dt ω
= Þ = ]
ωdω
kω
dθ
= Þ dw= kdq
Now
ω/ 2
ω
dω k dθ
=
ò ò
0 θ
1
ω/2 0
ω ω
dω k dθ kθ kθ
2 2
= Þ - = Þ - =
ò ò
(Qq1 = 2pn)
θ = 1
θ or 2pn1 = 2pn
n1 = n
25. (b) Sorawegg islikea spherical shell &hard bioled egg is
like solid sphere. Let 1
I , 2
I be M. I. of raw egg and
boiled egg respectively.
Given that angular momentum L, is same
I1w1= I2w2 Þ w2>w1 QI1>I2
Nowfrom first equation ofangular motion (wf=wi+at)
here a is retarding decceleration for both cases &
wf = 0 for both case.
So 1 1 1
2 2 2
t (raw egg) ω / α t
1
t (hard egg) ω / α t
= Þ <
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-204-320.jpg)
![202 PHYSICS
11. (b) Moment of inertia (M.I) ofcircular ring I1 = mr2
moment of inertia (M.I) of circular disc I2 = ½mr2
1
2
I
I
2
1
=
Þ
12. (a) Tension provides the necessary centripetal force for
the rest of rod.
13. (d) Centrifugal force is observed when the observer is in
a frameundergoing circular motion.
14. (d)
2
1
mR
2
I =
2
M t R
µ µ
For disc X,
2
X
1
Ι (m)(R)
2
= =
2 2
1
(πr t).(R)
2
for discY, 2 2
Y
1
Ι [π(4R) .t / 4][4R]
2
=
Þ X
3
Y
Ι 1
Ι (4)
= Þ Y X
I 64 I
=
15. (a)
1
Angular momentum
Angular frequency
µ
Kinetic energy
µ
Þ
K.E.
L
ω
=
r
1 1 2
2 1 2
L K.E ω
4
L ω K.E
æ ö
= ´ =
ç ÷
è ø Þ 2
L
L
4
=
16. (a) When q = 0°, the net force is directed vertically
upwards.
17. (a) Let velocity at A
v
A = and velocity at B
v
B =
A
T
r
B
B
v
A
v
Applying conservation of energy at A & B
2
B
2
A mv
2
1
gmr
2
mv
2
1
=
+ [Q (P.E)A = mg (2r)]
)
i
(
..........
gr
4
v
v 2
A
2
B +
=
Nowas it is moving in circular path it has centripetal
force.
At point A
r
mv
mg
T
2
A
=
+
Þ
0
T
velocity
minimum
for ³
2
2
A
A
mv
or mg v gr
r
³ Þ ³ gr
vA ³
Þ
18. (a) A couple consists of two equal and opposite forces
whose lines of action are parallel and laterally
separated by same distance. Therefore, net force (or
resultant) of a couple is null vector, hence no
translatory motion will be produced and only
rotational motion will be produced.
19. (b) Angular momentum of mass mmoving with a constant
velocity about origin is
line parallel
to x-axis
v
m
O
y
x
L = momentum × perpendicular distance of line of
action ofmomentum from origin
L= mv × y
In the given condition mvy is a constant. Therefore
angular momentum is constant.
20. (c) Since the spheres are smooth, there will benotransfer
of angular momentum from the sphereA tosphere B.
The sphereAonlytransfers its linear velocityv to the
sphere B and will continue to rotate with the same
angular speed w.
21. (c)
2
)
ŷ
x̂
(
n̂
+
-
= –a cos45 x
o
–a sin45 y
o
45o
a
2
ˆ ˆ
v 4 (x y)
ˆ
a n
r 4 2
- +
= = ´
u
r
22. (b) Since v is changing (decreasing), L is not conserved
in magnitude. Sinceit isgiven that a particle isconfined
to rotate in a circular path, it can not have spiral path.
Since the particle has two accelerations ac and at
thereforethenet acceleration is not towards thecentre.
L ac
at
v
The direction of L remains same even when the
speed decreases.
23. (a) Themoment ofinertia (I) ofcircular ring whose axis of
rotation is passing through its center, 2
1
1 R
m
I =
Also, 2
2
2 )
nR
(
m
I =
Since both rings have same density,
1
1
2
2
A
R
2
m
A
)
nR
(
2
m
´
p
=
´
p
Þ](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-206-320.jpg)
![205
System of Particles andRotationalMotion
43. (a) So couple about the C.M must be zero for rotational
equilibrium, It means that
6
L
x
x
4
W
3
2
L
4
W
=
Þ
´
=
´
C.M
x
L/2
L/2
W
4
3W
4
boy man
and the distance of man from other end is
2
L
– x
3
L
6
L
2
L
=
-
=
44. (b,c)
45. (b) Position of C.M w.r. toA
A B C
1 0 1 AB 1 AC AB AC
1 1 1 3
´ + ´ + ´ +
= =
+ +
46. (b) Use theorem of parallel axes.
47. (a) Angular momentum
2
2 2
I MR .
5 T
p
= w =
2
4 MR
5T
p
=
48. (d)
2 1
2 1
t t
w -w
a =
-
1
2 1200
40
60
p´
w = = p ; 2
2 4500
150
60
p´
w = = p
2
110
rad /sec
10
p
a =
Now, radian 180
p = °
180
1 rad degree
=
p
2
11 180
degree/sec
p´
a =
p
= 1980 degree/sec2
49. (b) 2 2 2
I 2 5 (0.2) 2 2 (0.4) 1kg m
= ´ ´ + ´ ´ = ´
50. (b) E = 1500= 1/2× 1.2w2
2 3000
2500
1.2
w = =
50rad / sec
w =
50
t 2sec
25
w
= = =
a
51. (c) Velocity of P (NP) (NM MP)
= w = + w
r(r sin ) v(1 sin )
= + q w = + q
52. (d) Velocity at the bottom and top of the circle is 5gr
and gr . Therefore (1/2)M(5gr) = MgH and
(1/2) M (gr) = Mgh.
53. (a) Acceleration of a bodyrolling down an inclined plane
is given by , 2
2
gsin
a
K
1
R
q
=
+
In case of a solid sphere , we have
2 2
2 2 2 2
K [(I / M)] I (2 /5)MR 2
5
R R MR MR
= = = =
Substituting
2
2
K 2
5
R
= in equation (1) we get
5
a gsin
7
= q
54. (c) The velocity of the top point of the wheel is twice that
of centre of mass. And the speed of centre of mass is
same for both the wheels.
1
a
2
a
1
1 R
v w
=
1
R 2
2 R
v w
=
2
R
55. (c) 1
2
1
2
1
2
1
2
1
1 R
R
R
R
v
a w
=
w
=
=
2
2
2
2
2
2 R
R
v
a w
=
=
Taking particle masses equal
2
1
2
1
2
1
2
1
R
R
a
a
ma
ma
F
F
=
=
=
Alternative method : The force experienced by any
particle is onlyalong radial direction or the centripetal
force.
Force experienced bythe particle, F = mw2R
1 1
2 2
F R
F R
=
56. (d) t × Dt = L0 {Q since Lf = 0}
Þ t × Dt = Iw
or t× 60 = 2 × 2 × 60p/60
( f 60rpm
=
Q
60
2 f 2
60
ö
w = p = p´ ÷
ø
m
N
15
-
p
=
t
57. (c) m1d = m2 d2
2
1
2
m
d
m
d =
Þ
58. (d)
y1 y¢
1
Circular disc (1)
4
MR
I
2
1
y =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-209-320.jpg)
![NEWTON'S UNIVERSAL LAW OF GRAVITATION
Gravitational force is an attractive force between any two point
masses M1 and M2 separated by any distance r.
It is given by F = G 1 2
2
æ ö
ç ÷
è ø
M M
r
where G is the universal gravitational constant.
Its value, G = 6.67 × 10–11 Nm2 kg–2
Dimensions of G are [M–1 L3 T–2]
(a) The gravitational force is the central force and follows
inversesquare law. It acts along the linejoining theparticles.
(b) Since the work done by the gravitational force is
independent of the path followed and hence it is a
conservative force.
(c) It is the weakest force in nature. It is 1038 times smaller
than nuclear force and 1036 timessmaller than electricforce.
Strongest force being nuclear force (for small range)
followed by electric force.
(d) Gravitation is independent of the presence of other bodies
around it.
ACCELERATION DUE TO GRAVITY (g)
The force of attraction exerted by earth on a body of mass m is
the force of gravity.
So the force of gravityfrom Newton’s gravitational law is
2
= e
GM m
F
r
……(i)
where Me =mass of earth
r=distance of the body from the centre of earth.
The force of gravity can be written as F = mg ……(ii)
where g is called the acceleration due togravity.
From the expression (i) and (ii), we get
2
= e
GM
g
r
……(iii)
If body is located at the surface of earth i.e., r = Re (radius of
earth) then
3
2 2
(4 / 3 ) 4
3
p r
= = = pr
e e
e
e e
GM G R
g GR
R R
……(iv)
where r is the density of earth.
This is the relation between universal gravitational constant
'G' and acceleration due to gravity 'g'.
Change in the Value of Acceleration due to Gravity (g)
(i) Due to rotation or latitude of earth : Let us consider a
particle P at rest on the surface of earth at a latitude f.
( ) 2 2
cos
f = - w f
e
g g R
f
r
R g¢
w
At poles f= 90º, so g(f) = g ...(v)
At equator f = 0º so g(f) = g –w2Re ...(vi)
(ii) Due to shape of earth : The shape of the earth is not
perfectlyspherical. It is flattened at poles. The polar radius
is 21 km less than the equatorial radius. Hence acceleration
due to gravity at poles is greater than at the equator.
gp = 9.83 ms–2 = value of g at poles
ge = 9.78 ms–2 = values of g at equator
i.e. gp > ge
(iii) At a depth 'd' below the earth surface : The acceleration
due to gravity at the surface of earth.
G
R
3
4
R
GM
g e
2
e
e
ú
û
ù
ê
ë
é
r
p
÷
ø
ö
ç
è
æ
=
= (from eqn. iv) ...(vii)
d
R-d
O R
P
Earth
Surface of earth
8 Gravitation](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-215-320.jpg)
![212 PHYSICS
If a body is taken at a depth d belowthe earth surface, then
the body is attracted by inner core of mass M' of earth
radius (Re– d). Then acceleration due to gravityat point P
is
3
e
2 2
e e
G (4 /3) (R d)
GM
g
(R d) (R d)
é ù
p - r
¢ ë û
¢ = =
- -
( )
( )
r
-
p
= )
d
R
(
3
/
4
G e ...(viii)
From eqns. (vii) and (viii) we have
1
æ ö æ ö
-
¢ = = -
ç ÷ ç ÷
è ø è ø
e
e e
R d d
g g g
R R
...(ix)
The value of acceleration due to gravity at the centre of the
earth is zero.
(iv) At a height 'h' above the earth surface : The acceleration
due to gravity at the surface of earth from expression (iv)
is defined as
2
e
e
R
GM
g =
At a height h above the earth surface, the acceleration due
to gravity is
2
e
2
e
e
r
GM
)
h
R
(
GM
g =
+
=
¢ ...(x)
Þ
2
e
e
R
h
R
g
g
-
÷
÷
ø
ö
ç
ç
è
æ +
=
¢
Þ [from eqs.(ix) &(x)]
If h << Re
e
2
1
R
æ ö
» -
¢ ç ÷
è ø
h
g g ...(xi)
Keep in Memory
1. The value of the acceleration due to gravity on the moon
is about one sixth of that on the earth and on the sun is
about 27 times that on the earth.
E
s E M
g
g 27g and g
6
æ ö
= =
ç ÷
è ø
2. The value of g is minimum on the mercury, among all
planets.
3. For h<<R, the rate of decrease of the acceleration due to
gravitywith height istwice as compared tothat with depth.
4. The value of g increases with the increase in latitude. Its
value at latitude q is given by : gq= g – Rw2 cos2q.
5. Rotation of the earth about its own axis is responsible for
decrease in the value of g with latitude.
6. Theweight ofthebodyvaries along with the valueof g (i.e.
W = mg)
7. Inertial mass and gravitational mass-
(a) Inertial mass (Mi) defined byNewton's law of motion
Mi = F/a =
External force applied on the body
Acceleration produced in it
i 1 2
i 2 1
(M ) a
(M ) a
=
(ff applied force on (Mi)1 and (Mi)2 is same.)
(b) Gravitational mass Mg defined by Newton's law of
gravitation
Mg = gravity
to
due
on
Accelerati
body
of
Weight
g
W
g
Fg
=
=
8. If two spheres ofsame material, mass and radiusare put in
contact, the gravitational attraction between them isdirectly
proportional to the fourth power of the radius.
4
2
R
3
4
G
4
1
F ÷
ø
ö
ç
è
æ
ps
=
A
R
R
B
m 2R
¬ ® m
This system maybe considered to be made up of two mass
each separated by 2R distance.
9. If the earth stops rotating about its axis, the value of g at
the equator will increase by about 0.35%, but that at the
poles will remain unchanged.
If the earth starts rotating at the angular speed of about 17
times the present value, there will be weightlessness on
equator, but g at the poles will remain unchanged. In such a
case, the duration of the daywill be about 84 min.
10. If the radius of planet decreases by n%, keeping the mass
unchanged, the acceleration due to gravity on its surface
increases by 2n%. i.e.,
R
R
2
g
g D
-
=
D
11. If the mass of the planet increases by m% keeping the
radius constant, the acceleration due to gravity on its
surfaceincreases bym% i.e.,
g M
g M
D D
= where R= constant.
12. Ifthe densityof planet decreases by r % keeping the radius
constant, the acceleration due to gravitydecreases by r%.
13. If the radius of the planet decreases by r% keeping the
densityconstant, the acceleration due to gravitydecreases
by r%.
Example1.
What will be the acceleration due to gravity on the surface
of the moon with its radius as 1/4th the radius of the earth
and its mass as 1/80th the mass of earth?
Solution :
2
e
e
R
GM
g = and 2
m
m
R
GM
'
g = ;
5
1
1
4
80
1
R
R
M
M
g
'
g
2
2
m
e
e
m
=
÷
ø
ö
ç
è
æ
´
=
÷
÷
ø
ö
ç
ç
è
æ
= ; 5
/
g
'
g =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-216-320.jpg)
![213
Gravitation
Example2.
A mass M splits into two parts m and (M–m), which are
then separated by a certain distance. What will be the
ratio m/M which maximizes the gravitational force
between the two parts?
Solution :
2
x
)
m
M
(
Gm
F
-
= ;For maxima,
2
1
M
m
or
0
)
m
2
M
(
x
G
dm
dF
2
=
=
-
=
Example3.
Identical point masses each equal to m are placed at
x = 0, x = 1, x = 2, x = 4, ... The total gravitational force
on mass m at x = 0 due to all other masses is
(a) infinite (b) (4/3)Gm2
(c) (4/3GM2) (d) zero
Solution : (c)
o m m m
x=4
x=2
x=1
x=0
....
4
Gm
2
Gm
1
Gm
F
2
2
2
2
2
2
0 +
+
+
=
ú
û
ù
ê
ë
é
+
+
+
= ...
4
1
2
1
1
1
Gm
F
2
2
2
2
0
It is a G.P. with common ratio1/4.
ú
û
ù
ê
ë
é
-
=
4
/
1
1
1
Gm
F 2
0
3
Gm
4
F
2
0 =
Example4.
Four identical point masses each equal to m are placed
at the corners of a square of side a. The force on a point
mass m’ placed at the point of intersection of the two
diagonals is
(a) (4Gmm¢)/a2 (b) (2Gmm¢)/a2
(c) (Gmm¢)/a2 (d) zero
Solution : (d)
The forces acting on the mass m¢ placed at the point of
intersection of diagonals are balanced by each other as
shown in figure. Therefore net force is zero.
m m
m
m
m¢
F
F
F
F
Example5.
A tunnel is dug along the diameter of the earth. Determine
the force on a particle of mass m placed in the tunnel at a
distance x from the centre.
Solution :
Force on a mass m placed at a distance x from the centre is
equal to the force of gravity due to a mass of spherical
volume of earth of radius x.
m
R
x
Center of earth
2
3
x
m
d
3
x
4
G
F
÷
÷
ø
ö
ç
ç
è
æ p
=
where d = density of earth = 3
R
4
M
3
p
;
x
.
R
GMm
F
3
= [M is total mass of earth]
Example6.
If the radius of the earth were to shrink by two per cent,
its mass remaining the same, the acceleration due to
gravity on the earth’s surface would
(a) decrease by 2% (b) increase by 2%
(c) increase by 4% (d) decrease by 4%
Solution : (c)
As 2
R
GM
g = ; So, if R decreases then g increases.
Taking logarithm of both the sides;
log g = log G + log M – 2 log R
Differentiating it, we get
g
dg
= 0 + 0 –
R
dR
2
;
g
dg
100
4
100
2
2 =
÷
ø
ö
ç
è
æ -
-
=
% increase in 100
g
dg
g ´
= = %
4
100
100
4
=
´
Example7.
An apple of mass 0.25 kg falls from a tree. What is the
acceleration of the apple towards the earth? Also calculate
the acceleration of the earth towards the apple.
Given : Mass of earth = 5.983 × 1024 kg,
Radius of earth = 6.378 × 106 metre,
G = 6.67 × 10–11 Nm2kg–2.
Solution :
Mass of apple, m = 0.25 kg,
Mass of earth, M = 5.983 × 1024 kg,
Radius of earth, R= 6.378 × 106 metre
Gravitational constant, G = 6.67 × 10–11 Nm2kg–2
Let F be the gravitational force of attraction between the
apple and earth.
Then, 2
GmM
F
R
=
Let a be the acceleration of apple towards the earth.
2 2
F GmM GM
a
m mR R
= = =
Þ
11 24
2 2
6 2
6.67 10 5.983 10
a ms 9.810ms
(6.378 10 )
-
- -
´ ´ ´
= =
´](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-217-320.jpg)
![214 PHYSICS
Let a¢ be the acceleration of the earth towards the apple.
2
11
2
6 2
25 2
F Gm
a
M R
6.67 10 0.25
ms
(6.378 10 )
4.099 10 ms .
-
-
- -
= =
¢
´ ´
=
´
= ´
Example8.
The acceleration due to gravity at the moon’s surface is
1.67 ms–2. Iftheradius ofthemoon is 1.74× 106 m, calculate
the mass of the moon. Use the known value of G.
Solution :
2
GM
g
R
= or
2
gR
M
G
=
This relation is true not only for earth but for any heavenly
body which is assumed to be spherical.
Here,
2 6
g 1.67ms , R 1.74 10 m
-
= = ´ and
11 2 2
G 6.67 10 Nm kg
- -
= ´
Mass of moon,
6 2
22
11
1.67 (1.74 10 )
M kg 7.58 10 kg
6.67 10-
´ ´
= = ´
´
GRAVITATIONAL FIELD
(or Gravitational Field Intensity)
The region or space around a body in which its gravitational
influence is experienced by other bodies is the gravitational
field of that body.
The gravitational field strength Eg, produced by a mass M at
any point P is defined as the force exerted on the unit mass
placed at that point P. Then
2 2
æ ö
= = =
ç ÷
è ø
g
F G Mm GM
E m
m r r
...(i)
where m= test mass
r=distance between M and m
(i) The direction of Eg always points towards the mass
producing it.
(ii) The gravitational field can be represented bygravitational
lines of force.
(iii) The S.I unit of Eg is newton/kg. Its dimensions are
[M0LT–2].
Gravitational intensity (Eg) for spherical shell
r
|E |=0
g
Radius of
spherical shell
|E |
g
O
R
2
r
GM
|
Eg
| =
(i) |Eg|= 0 at points inside the spherical shell (i.e. r < R)
(ii) 2
| |
-
=
g
GM
E
R
at the surface of shell (i.e. r = R)
(iii) 2
| |
-
=
g
GM
E
r
for outside the spherical shell (r >R).
It is clear that as r increases, Eg decreases.
Gravitational intensity for solid sphere :
R
r
|E |
g
Solid
sphere
Eg
(i) 3
.
| |= -
g
GM r
E
R
for points inside the solid sphere (r<R)
(ii) At the surface of solid sphere (r = R) 2
| |= -
g
GM
E
R
(iii) Outside the solid sphere
2
| |= -
g
GM
E
r
(r > R)
GRAVITATIONAL POTENTIAL
Gravitational potential at any point is the work done by
gravitational force in carrying a body of unit mass from infinity
to that point in gravitational influence of source.
i.e., A
o
W GM
V
m r
-
= =
where r is the distance of point from source mass M.
The S.I. unit of V is joule/kg. Its dimensions are [M0L2T–2]. It is
a scalar quantity.
Gravitational potential at the earth surface is
-
= = -
e
A e
e
GM
V gR
R
Gravitational potential due to a uniform ring at a point on its
axis
2 2
= -
+
g
GM
V
a r
a
O r P
q
Gravitational potential due to a uniform thin spherical shell
O
Vg
r
R
GM
-](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-218-320.jpg)
![216 PHYSICS
1 Kg 1 Kg
1 Kg
1 Kg
1m
1m
1m 1m
2
m
2
m
When one mass is removed, then
P.E. = U2 = ú
û
ù
ê
ë
é ´
´
+
´
´
-
2
1
1
G
1
1
1
G
.
2
G
2
1
2
2
U2
ú
ú
û
ù
ê
ê
ë
é +
-
= ;
Work done =
ú
ú
û
ù
ê
ê
ë
é -
-
+
2
1
2
2
2
2
4
G =
ú
ú
û
ù
ê
ê
ë
é +
2
1
2
2
G
[Q Work done = change in P.E. = –U2–[–U1] = U1–U2]
Example10.
A solid sphere of uniform density and mass M has a radius
of 4m. Its centre is at the origin of coordinate system. Two
spheres of radii 1m are taken out so that their centres are
at P(0, –2, 0) and Q(0, 2, 0) respectively. This leaves two
spherical cavities. What is the gravitational field at the
centre of each cavity?
O
P Q
R
z
y
Solution :
If the cavities are not made, then the intensityat the point
P (or Q), due to whole sphere
32
GM
2
64
GM
x
.
R
GM
I 3
=
´
=
= where P Q R
I I I I
= + +
r r r r
Q
P
R I
I
I
I
r
r
r
r
-
-
= ……(i)
Where IR is the intensity of gravitational field at a point,
distance x from O due toremainder sphere (exclude cavity)
Mass of big sphere
3
d
)
4
(
4
M
3
´
p
=
Mass of small sphere P or Q,
3
4π (1)
3
´
=
d
m
M
m
64
=
At P, Ip = 0, ÷
ø
ö
ç
è
æ
=
=
64
M
.
4
G
r
Gm
I 2
2
Q
IR =
GM GM
32 1024
- [From eq (i)]
R
31GM
I
1024
=
Example11.
Find the gravitational potential energy of a system of four
particles, each having mass m, placed at the vertices of a
square of side l. Also obtain the gravitational potential at
the centre of the square.
Solution :
The system has four pairs with distance l and two diagonal
pairs with distance 2 l .
m
l
2l
m
m m
2 2 2 2
Gm Gm 2Gm 1 Gm
U 4 2 2 5.41
2 2
æ ö
= - - = - + = -
ç ÷
è ø
l l l
l
The gravitational potential at the centre of the square is
( )
r 2 / 2
= l
V = Algebraic sum of potential due to each particle
Þ
4 2Gm
V = -
l
Example12.
The radius of earth is R. Find the work done in raising a
body of mass m from the surface of earth to a height R/2.
Solution :
Ifa bodyofmass m is placed at a distance x from the centre
of earth, the gravitational force of attraction F between the
body and the earth is
2
x
m
GM
F =
Small amount of work done in raising a body through a
small distance dx is given by
dx
x
m
GM
dx
F
dW 2
=
=
Total work done in raising the body from the surface of
earth to a height R/2 is given by
W =
R R/2
2
R
GM m
dx
x
+
ò
=
3 R/2 3 R/2
2
R
R
1
GMm x dx GMm
x
- æ ö
= - ç ÷
è ø
ò
=
2 1
GMm
3R R
é ù
- -
ê ú
ë û
=
2
M m gR m
3R 3R
G
=
=
1
mg R
3
[ GM = gR2]](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-220-320.jpg)
![217
Gravitation
SATELLITES
A satellite is any body revolving around a large body under
the gravitational influence of the latter.
The period of motion T of an artificial satellite of earth at a
distance h above the surface of the earth is given by,
p
3
2
( + )
= 2 e
e
R h
T
gR
where, Re = radius of the earth
where, g = acceleration dueto gravityon thesurface ofthe earth.
If Re > > h, then e
R
T 2
g
= p ; Re = 6.4 × 106 m ;
g = 9.8 ms–2
i.e. T 84.58 minutes 5075sec
»
;
Assuming Re + h = r, the distance of the satellite from the centre
of the earth, 3 / 2
T (r)
µ
Orbital velocity (v0 )
Let a satellite ofmass m revolve around the earth in circular orbit
of radius r with speed v0. The gravitational pull between satellite
and earth provides the necessary centripetal force.
Centripetal force required for the motion =
2
0
mv
r
Gravitational force =
GMm
r²
2
0
2
mv GMm
r r
= or
2
0
GM
v
r
=
or 0 =
GM
v
r
..........(1)
or 0 =
+
g
v R
R h
[Q 2
GM
g
R
= and r = (R + h)]
Orbital velocity,
e
R
= =
e
e
GM
Vo gR
(i) Valueof orbital velocitydoes not depend on the mass of the
satellite.
(ii) Around the earth the value of orbital velocity is
7.92 km/sec.
(iii) Greater is the height of the satellite, smaller is the orbital
velocity.
(iv) The direction of orbital velocity is along the tangent to the
path.
(v) The work done by the satellite in a complete orbit is zero.
Angularmomentum (L):For satellite motion, angular momentum
will be given by
GM
L mvr mr
r
= = i.e., L = [m2GMr]1/2
Angular momentum of a satellite depends on both, the mass of
orbiting and central body. It also depends on the radius of the
orbit.
Energy of a satellite
Total energy of satellite revolving in an orbit of radius r around
the earth can be calculated as follows :
(i) The gravitational potential energy of a satellite of mass m
is –
=
g
GMm
U
r
, where r is the radius of the orbit.
(ii) Kinetic energy of the satellite is
2
0
1
2 2
= =
k
GMm
E mv
r
(iii) So, total energy of the satellite
2
= + = -
g k
GMm
E U E
r
The negative sign shows that satellite cannot leave the
orbit itself. It requires an energyequal to
2
GMm
R
, which is
called BindingEnergy (B.E.) ofsatellite.
(iv) Total energy of a satellite at a height equal to the radius of
the earth
1
– –
2( ) 4 4
= = =
+
GMm GMm
mgR
R R R
GEO-STATIONARY SATELLITE
A satelite which appears to be stationary for a person on the
surface of the earth is called geostationary satellite.
It is also known as parking satellite or synchronous satellite.
(i) The orbit ofthesatellitemust becircular and in theequatorial
plane of the earth.
(ii) The angular velocity of the satellite must be in the same
direction as the angular velocityof rotation ofthe earth i.e.,
from west to east.
(iii) The period ofrevolution ofthe satellite must be equal tothe
period of rotation of earth about its axis.
i.e. 24 hours = 24 × 60 × 60 = 86400 sec.
2 4 ²
T r³
GM
p
= or
1 1
2 2 2
3 3
2 2 2
GMT GM R T
r
4 R 4
æ ö æ ö
= = ´
ç ÷ ç ÷
p p
è ø è ø
=
1
3
6 2 (86400)²
9.8 (6.38 10 )
4 ²
é ù
´ ´ ´
ê ú
p
ë û
=42237 ×103 m = 42,237 km.»42000km.
h= r –R= 42000–6400= 35600km.
(a) Height of geostationary satellite from the surface of
the earth is nearly35600 km.
(b) The orbital velocity of this satellite is nearly
3.08km/sec.
(c) The relative velocity of geostationary satellite with
respect to earth is zero.
This type of satellite is used for communication
purposes. The orbit of a geostationary satellite is called
‘ParkingOrbit’.
Polar Satellite :
Polar satellites travel around the earth in an orbit that travels
around the earth over the poles. The earth rotates on its axis as
the satellite goes around the earth. Thus over a period of many
orbits it looks down on every part of the earth.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-221-320.jpg)
![218 PHYSICS
Different orbital shapes correspondingtodifferent velocitiesof
a satellite :
[1] When v< v0
(i) Thepath is spiral. The satellite finallyfalls on the earth
(ii) Kinetic energy is less then potential energy
(iii) Total energy is negative
[2] When v= v0
(i) The path is circular
(ii) Eccentricityis zero
(iii) Kinetic energy is less than potential energy
(iv) Total energy is negative
[3] When v0 < v< ve
(i) The path is elliptical
(ii) Eccentricity< 1
(iii) Kinetic energy is less than potential energy
(iv) Total energy is negative
[4] When v = ve
(i) The path is a parabola
(ii) Eccentricity= 1
(iii) Kinetic energy is equal to potential energy
(iv) Total energy is zero
[5] When v > ve
(i) The path is a hyperbola
(ii) Eccentricity> 1
(iii) Kinetic energy is greater than potential energy
(iv) Total energy is positive
If the orbit of a satellite is elliptical
(1) The energy
GMm
E
2a
= - = const. with ‘a’as semi-major axis;
(2) KE will bemaximum when the satelliteisclosestto thecentral
body(at perigee) and minimum when it is farthest from the
central body (at apogee) [as for a given orbit L = const.,
i.e., mvr = const., i.e., vµ 1/r]
(3) PE = (E – KE) will be minimum when KE = max, i.e., the
satellite is closest to the central body (at perigee) and
maximum when KE = min, i.e., the satellite is farthest from
the central body (at apogee).
ESCAPE SPEED (Ve)
It is the minimum speed with which a body should be projected
from the surface of a planet so as to reach at infinity i.e., beyond
the gravitational field of the planet.
If a bodyof mass m is projected with speed v from the surface of
a planet of mass M and radius R, then
as 2
1
2
=
K mv and
GMm
U
R
= - , ES = 2
1
2
mv –
GMm
R
Now if v' is the speed of body at ¥, then
2 2
1 1
( ) 0 ( )
2 2
¥ = + =
¢ ¢
E m v m v [as U 0
¥ = ]
So by conservation of energy
2 2
1 1
( )
2 2
= - = ¢
GMm
mv m v
R
i.e.
2 2
1
( )
2 2
1
= + ¢
GMm
mv m v
R
sov will be minimum when v¢ ® 0,
i.e. min
2
2
= = =
e
GM
v v gR
R 2
GM
as g
R
é ù
=
ê ú
ë û
• The value of escape velocitydoes not depend upon the mass
of the projected body, instead it depends on the mass and
radius of the planet from which it is being projected.
• The value of escape velocity does not depend on the angle
and direction of projection.
• The value of escape velocityfrom the surface of the earth is
11.2 km/sec.
• The minimum energyneeded for escape is = GMm/R.
• If the velocity of a satellite orbiting near the surface of the
earth is increased by 41.4%, then it will escape away from
the gravitational field of the earth.
• Ifa bodyfalls freelyfrom infinite distance, then it will reach
the surface ofearth with a velocityof 11.2 km/sec.
Relationbetweenorbital velocity(V0) andescape speed(Ve)
0
2 2
= =
e
V gR V
Keep in Memory
1. The escape velocity on moon is low
6
æ ö
=
ç ÷
è ø
E
m
g
as g hence
there is no atmosphere on moon.
2. If the orbital radius of the earth around the sun be one
fourth of the present value, then the duration of the year
will be one eighth of the present value.
3. The satellites revolve around the earth in a plane that
coincides with the great circle around the earth.
KEPLER'S LAWS OF PLANETARY MOTION
1. The law of orbits : Each planet revolves about the sun in
an elliptical orbit with the sun at one of the foci of the
ellipse.
ea
ea
P
x
Perihelion Aphelion
Sun
S¢
y
a a
S
a = Semi-major axis
A planet ofmass m moving in an elliptical orbit around the
sun. The sun of mass M, is at one focus S¢ of the ellipse.
The other focus is S, which is located in empty space. Each
focus is at distance ‘ea’ from the ellipse’s centre, with ‘e’
being the eccentricity of the ellipse and ‘a’ semimajor axis
of the ellipse, the perihelion (nearest to the sun) distance
rmin., and the aphelion (farthest from the sun) distance rmax.
are also shown.
rmax = a + e a = (1 + e) a
rmin = a – e a = (1 – e) a](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-222-320.jpg)
![227
Gravitation
49. A diametrical tunnel is dug across the Earth. A ball is
dropped into the tunnel from one side. The velocity of the
ball when it reaches the centre of the Earth is .... (Given :
gravitational potential at the centre of Earth =
R
GM
2
3
– )
(a) R (b) gR
(c) gR
5
.
2 (d) gR
1
.
7
50. A man of mass m starts falling towards a planet of mass M
and radius R.As he reaches near to the surface, he realizes
that he will pass through a small hole in the planet. As he
enters the hole, he seesthat the planet is reallymade of two
pieces a spherical shell ofnegligible thickness ofmass 2M/
3 and a point mass M/3 at the centre. Change in the force of
gravity experienced by the man is
(a) 2
2 GMm
3 R
(b) 0
(c) 2
1 GMm
3 R
(d) 2
4 GMm
3 R
51. In a region ofonlygravitational field ofmass 'M' a particle
is shifted from AtoB via three different paths in the figure.
The work done in different paths are W1, W2, W3
respectively then
M
B
A
3
2
1
C
(a) W1 = W2 = W3 (b) W1 > W2 > W3
(c) W1 = W2 > W3 (d) W1 < W2 < W3
52. Four similar particles of mass m are orbiting in a circle of
radiusr in thesameangular direction becauseoftheir mutual
gravitational attractiveforce. Velocityofa particle isgiven by
m
m
m
m r
(a)
1/ 2
GM 1 2 2
r 4
é ù
æ ö
+
ê ú
ç ÷
è ø
ê ú
ë û
(b) 3
GM
r
(c) ( )
GM
1 2 2
r
+ (d)
1/ 2
1 GM 1 2
2 r 2
é ù
æ ö
+
ê ú
ç ÷
è ø
ê ú
ë û
53. The percentage change in the acceleration of the earth
towards the sun from a total eclipse of the sun to the point
where the moon is on a side of earth directly opposite to
the sun is
(a)
s 2
m 1
M r
100
M r
´ (b)
2
s 2
m 1
M r
100
M r
æ ö
´
ç ÷
è ø
(c)
2
1 m
2 s
r M
2 100
r M
æ ö
´
ç ÷
è ø
(d)
2
s
1
2 m
M
r
100
r M
æ ö
´
ç ÷
è ø
54. A satellite is revolving round the earth in an orbit of radius
r with time period T. If the satellite is revolving round the
earth in an orbit ofradius r + Dr (Dr << r) with time periodT
+ D T then,
(a)
T 3 r
T 2 r
D D
= (b)
T 2 r
T 3 r
D D
=
(c)
T r
T r
D D
= (d)
T r
T r
D D
= -
55. A cavity of radius R/2 is made inside a solid sphere of
radius R. The centre ofthe cavity is located at a distance R/
2 from the centre of the sphere. The gravitational force on a
particle of mass ‘m’ at a distance R/2 from the centre of the
sphere on the line joining both the centres of sphere and
cavity is – (opposite to the centre of gravity)
[Here g = GM/R², where M is the mass of the sphere]
(a)
mg
2
(b)
3mg
8
(c)
mg
16
(d) None of these
56. A spherical uniform planet is rotating about its axis. The
velocityof a point on its equator is V. Due to the rotation of
planet about its axis the acceleration due to gravity g at
equator is 1/2 ofg at poles. The escape velocityof a particle
on the pole of planet in terms of V is
(a) Ve=2V (b) Ve=V
(c) Ve=V/2 (d) e
V 3V
=
57. The escape velocity from a planet is ve. A tunnel is dug
along a diameter of the planet and a small bodyis dropped
into it at the surface. When the body reaches the centre of
the planet, its speed will be
(a) ve (b) e
v / 2
(c) ve/2 (d) zero](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-231-320.jpg)
![230 PHYSICS
4. Satellites orbitting the earth have finite life and sometimes
debris of satellites fall to the earth. This is because
(a) the solar cells and batteries in satellites run out
(b) the laws of gravitation predict a trajectory spiralling
inwards
(c) of viscous forces causing the speed of satellite and
hence height to gradually decrease
(d) of collisions with other satellites
5. Both the earth and the moon are subject to the gravitational
force of the sun. As observed from the sun, the orbit of the
moon
(a) will be elliptical
(b) will not be strictly elliptical because the total
gravitational force on it is not central
(c) is not elliptical but will necessarily be a closed curve
(d) deviates considerably from being elliptical due to
influence of planets other than the earth
6. In our solar system, the inter-planetary region has chunks
ofmatter (much smaller in size compared toplanets) called
asteroids. They
(a) will not move around the sun, since they have very
small masses compared to the sun
(b) will move in an irregular way because of their small
masses and will drift away into outer space
(c) will move around the sun in closed orbits but not obey
Kepler's laws
(d) will move in orbits likeplanets and obeyKepler's laws
7. Choose the wrong option.
(a) Inertial mass is a measure of difficulty of accelerating
a bodyby an external force whereas the gravitational
mass is relevant in determining the gravitational force
on it byan external mass
(b) That thegravitational mass andinertial mass areequal
is an experimental result
(c) That the acceleration due togravityon the earth is the
samefor all bodies isduetotheequalityofgravitational
mass and inertial mass
(d) Gravitational mass of a particlelikeproton can depend
on the presence of neighbouring heavy objects but
the inertial mass cannot
8. Particles ofmasses 2M, mandM arerespectivelyat pointsA,
B and C with
1
( )
2
AB BC m
= × is much-much smaller than
M and at time t = 0, theyare all at rest as given in figure.
At subsequent times before any collision takes place.
r
A B C
M
m
2M
(a) m will remain at rest
(b) m will move towards M
(c) m will move towards2M
(d) m will have oscillatorymotion
NEET/AIPMT (2013-2017) Questions
9. Abody of mass ‘m’is taken from the earth’s surface to the
height equal to twice the radius (R) ofthe earth. The change
in potential energy of body will be [2013]
(a)
2
3
mgR (b) 3mgR
(c)
1
3
mgR (d) mg2R
10. Infinite number ofbodies, each of mass 2 kg are situated on
x-axisat distances1m,2m, 4m, 8m, .....respectively, from the
origin. The resulting gravitational potential due to this
system at the origin will be [2013]
(a)
8
3
- G (b)
4
3
- G
(c) – 4 G (d) –G
11. The radius of a planet is twice the radius of earth. Both
have almost equal average mass-densities. If VP and VE are
escape velocities of the planet and the earth, respectively,
then [NEET Kar. 2013]
(a) VE = 1.5VP (b) VP = 1.5VE
(c) VP = 2VE (d) VE = 3VP
12. A particle of mass ‘m’ is kept at rest at a height 3R from the
surface ofearth, where ‘R’ is radius of earth and ‘M’ is mass
of earth. The minimum speed with which it should be
projected, sothat it does not return back, is (g is acceleration
due to gravity on the surface of earth) [NEET Kar. 2013]
(a)
1
2
GM
R
æ ö
ç ÷
è ø
(b)
1
2
2
GM
R
æ ö
ç ÷
è ø
(c)
1
2
4
gR
æ ö
ç ÷
è ø
(d)
1
2
2
4
g
æ ö
ç ÷
è ø
13. Dependence of intensity of gravitational field (E) of earth
with distance(r) from centreofearth is correctlyrepresented
by: [2014]
(a)
E
O R
r
(b)
E
O
R r
(c)
E
O
R
r (d)
E
O
R r
14. A black hole is an object whose gravitational field is so
strong that even light cannot escape from it. To what
approximate radius would earth (mass = 5.98 × 1024 kg)
have to be compressed to be a black hole? [2014]
(a) 10– 9 m (b) 10– 6 m
(c) 10– 2 m (d) 100m](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-234-320.jpg)
![231
Gravitation
15. Twospherical bodies of mass M and 5 M and radii R and 2
Rreleased in free spacewith initial separation between their
centres equal to 12 R. If they attract each other due to
gravitational force only, then the distance covered by the
smaller bodybefore collision is [2015]
(a) 4.5R (b) 7.5R
(c) 1.5R (d) 2.5R
16. Kepler's third law states that square of period of revolution
(T)of a planet aroundthe sun, is proportional tothird power
of average distance r between sun and planet i.e. T2 = Kr3
here K is constant. If the masses of sun and planet are M
and m respectively then as per Newton's law ofgravitation
force of attraction between them is F = 2
GMm ,
r
here G is
gravitational constant. The relation between G and K is
described as [2015]
(a) GMK= 4p2 (b) K=G
(c) K =
1
G
(d) GK=4p2
17. A remote - sensing satellite of earth revolves in a circular
orbit at a height of0.25 × 106 m above the surface of earth.
If earth's radius is 6.38 × 106 m and g = 9.8 ms–2, then the
orbital speed of the satellite is: [2015 RS]
(a) 8.56 km s–1 (b) 9.13 km s–1
(c) 6.67 km s–1 (d) 7.76 km s–1
18. A satellite S ismovingin an elliptical orbit around theearth.
Themass ofthe satellite is verysmall compared to themass
of the earth. Then, [2015 RS]
(a) the total mechanical energy of S varies periodically
withtime.
(b) the linear momentum of S remains constant in
magnitude.
(c) the acceleration of S is always directed towards the
centre of the earth.
(d) the angular momentum of S about the centre of the
earth changes in direction, but its magnitude remains
constant.
19. The ratio of escape velocity at earth (ve) to the escape
velocityat a planet (vp) whoseradius and mean densityare
twice as that of earth is :
(a) 1: 2 (b) 1 : 2 2 [2016]
(c) 1: 4 (d) 1: 2
20. At what height from the surface of earth the gravitational
potential andthe value ofg are –5.4× 107 J kg–1 and 6.0 ms–2
respectively ?
Take the radius of earth as 6400 km : [2016]
(a) 2600km (b) 1600km
(c) 1400km (d) 2000km
21. Two astronauts are floating in gravitation free space after
having lost contact with their spaceship. The two will
(a) move towards each other. [2017]
(b) move away from each other.
(c) become stationary
(d) keep floating at the same distance between them.
22. The acceleration due to gravity at a height 1 km above the
earth is the same as at a depth d below the surface of earth.
Then [2017]
(a) d = 1 km (b) d =
3
km
2
(c) d = 2 km (d) d =
1
km
2](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-235-320.jpg)
![234 PHYSICS
9. (c) 2 3
T R
µ
Þ
3
2
1
2
2
1
R
R
T
T
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
Þ
2
/
3
2
/
3
2
1
2
1
4
1
R
R
T
T
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
Þ 8
)
4
(
T
T 2
/
3
1
2
=
=
Þ 2 1
T 8 T 8 5 40
= ´ = ´ = hours
10. (b) At poles, the effect of rotation is zero and also the
distance from the centre of earth is least.
11. (c) 2 3
T R
µ (According to Kepler’s law)
2 13 3 2 12 3
1 2
T (10 ) and T (10 )
µ µ
2
3
1 1
2
2
2
T T
(10) or 10 10
T
T
= = .
12. (c) According to Kepler’s law of period T2 µ R3
2 3 3
1 1
2 3 3
2 2
(6 )
8
(3 )
T R R
T R R
= = =
2
2
24 24
8
T
´
=
2
2
24 24
8
T
´
= = 72=36 ×2
2 6 2
T =
13. (b) Since escape velocity( )
e e
v 2gR
= is independent
of angle of projection, so it will not change.
14. (c) Given that l2 = 1.02 l1
We know that T 2π or T
g
æ ö
= µ
ç ÷
è ø
l
l
2 2 1
1 1 1
T (1.02 )
101
T
æ ö
= = =
ç ÷
è ø
l l
l l
Thus time period increases by 1%.
15. (b) We know that e 0
v 2 v
= , where v0 is orbital
velocity.
K.E. in the orbit,
2
0
1
E M v
2
=
K.E. to escape
2 2
e 0
1 1
E M v M (2 v )
2 2
= =
2
0
1
M v 2 2E
2
= ´ =
16. (c)
1/ 2
3
2π (R h)
T
R g
é ù
+
= ê ú
ê ú
ë û
Here (R + h) changes from R to 4 R. Hence period of
revolution changes from T to (43)1/2 T = 8 T.
17. (b) h
2h
g g 1
R
æ ö
= -
ç ÷
è ø
x
x
g g 1
R
æ ö
= -
ç ÷
è ø
Given that, h x
g g
=
x= 2h
18. (d) 2
2
)
x
R
(
GmM
)
x
R
(
mv
+
=
+ also 2
R
GM
g =
2 2
2 2
mv GM R
m
(R x) R (R x)
æ ö
= ç ÷
è ø
+ +
2
2
2
)
x
R
(
R
mg
)
x
R
(
mv
+
=
+
x
R
gR
v
2
2
+
=
Þ
2
/
1
2
x
R
gR
v
÷
÷
ø
ö
ç
ç
è
æ
+
=
19. (c) 2
2
)
x
R
(
GmM
x
R
mv
+
=
+
x = height ofsatellite from earth surface
m = mass of satellite
2 GM GM
v or v
(R x) R x
Þ = =
+ +
2 (R x) 2 (R x)
T
v GM
R x
p + p +
= =
+
which is independent of mass of satellite
20. (c) 2
n
MR
KR
F w
=
= - )
1
n
(
2
KR +
-
=
w
Þ
or 2
)
1
n
(
R
'
K
+
-
=
w [where K' = K1/2, a constant]
2
)
1
n
(
R
T
2
+
-
a
p
(n 1)
2
T R
+
a
21. (d) According to Newton gravitational force
mg
R
m
GM
2
e
e
= & g
m
)
h
R
(
m
GM
2
e
e ¢
=
+
Þ g'
2
e
2
e
2
e )
h
R
(
gR
)
R
/
h
1
(
1
g
+
=
+
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-238-320.jpg)
![235
Gravitation
where Me is mass ofearth, G is gravitational constant,
Re is radius & earth, h is height of satellite above the
surfaceof earth, g is value at the surface of earth, g' is
value at height h above the surface of earth.
so g
m
)
h
R
(
mgR
)
h
R
(
mv
2
e
2
e
e
2
¢
=
+
=
+ h
R
gR
v
e
2
e
+
=
Þ
h
Re
+Re
O
Earth
earth
surface
satellite
orbit
v
22. (c)
23. (d) Both decreases but variation are different.
24. (b) There is no gravitational field in the shell.
25. (c) Thegravitational field duetothe ring at a distance r
3
is given by
2
2
/
3
2
2
r
8
Gm
3
E
]
)
r
3
(
r
[
)
r
3
(
Gm
E =
Þ
+
=
Attractive force = 2
r
8
GmM
3
26. (a) For uniform gravitational field
Eg =
r
V
-
2
ms
10
1
20
2 -
=
-
-
=
Now, W = mgh =
1
5 4 2J
10
´ ´ =
27. (b)
2
0
2
2
e
1 mv
2
1
E
,
mv
2
1
E =
=
0
e
2
0
e
2
1
v
2
v
2
v
v
E
E
=
=
÷
÷
ø
ö
ç
ç
è
æ
=
Q
28. (d) g R
µ r
29. (c)
30. (b) As buoyant force involves ‘g’ in it.
31. (d) Work done = Uf – Ui
mgR
3
2
R
GmM
3
2
R
Gmm
R
R
2
GMm
=
=
÷
ø
ö
ç
è
æ
-
-
+
-
=
32. (b) 33. (a)
34. (d) The gravitational force due to the whole sphere at A
point is
e o
1 2
GM m
F
(2R)
= , where m0 is the assumed rest mass at
point A.
In the second case, when we made a cavity of radius
(R/2), then gravitational force at point Ais
2
o
e
2
)
2
/
R
R
(
m
GM
F
+
= F2/F1= 1/9
35. (b)
R
2
R0 0
GMm 1 1
P.E. dr GMm
R R
r
é ù
= = - -
ê ú
ë û
ò
TheK.E. acquired bythe bodyat the surface
2
1
m v
2
=
2
0
1 1 1
mv GMm
2 R R
é ù
= - -
ê ú
ë û
0
1 1
v 2G M
R R
æ ö
= -
ç ÷
è ø
36. (c) Applying the properties of ellipse, we have
1 2
1 2 1 2
r r
2 1 1
R r r r r
+
= + =
R
Instant position
of satellite
major axis
r2
r1
Sun
1 2
1 2
2r r
R
r r
=
+
37. (a)
GM
v
r
æ ö
= ç ÷
è ø
where r is radius of the orbit of the
satellite.
herer =Re + h = e
e e
R 3
R R
2 2
+ =
So, 0
e
2GM 2
v v
3R 3
= = ,
where v0 is the orbital velocity of the satellite, which
is moving in circular orbit of radius, r = Re
38. (d) 2 2
1 1 2 2 1 1 2 2
v r v r or r ω r ω
= =
(QL=mrv= constant)
or 2 2
min max
r ω r ω¢
=
2 2
min max
ω (r / r )ω
¢
=
39. (c) Distance between the surface of the spherical bodies
= 12R – R– 2R = 9R
Force Mass
µ
Acceleration Mass
µ
Distance Acceleration
µ
Þ 1
2
a M 1
a 5M 5
= = Þ 1
2
2
1
S
5
S
5
1
S
S
=
Þ
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-239-320.jpg)
![236 PHYSICS
S1 + S2 = 9
Þ 6S1 = 9Þ 5
.
1
6
9
S1 =
=
S2 = 1.5×5 = 7.5
Note: Maximum distance will be travelled bysmaller
bodies due to the greater acceleration caused by the
same gravitational force
40. (c) 2
2
)
R
99
.
0
(
GM
'
g
;
R
GM
g =
=
2
2
g' R
g' g
g 0.99R
æ ö
= Þ >
ç ÷
è ø
41. (b) Acceleration due to gravity at lattitude’ l ’ is given by
l
w
-
=
l
2
2
e
e cos
R
g
g
At equator, l = 90° Þ l
cos = cos 90° = 0
or l
g = e
g = g (as given in question)
At 30°,
2
2
2
30 R
4
3
g
30
cos
R
g
g w
-
=
w
-
=
or, 2
30 R
4
3
g
g w
=
-
42. (b) The orbital speed of satellite is independent of mass
of satellite, so ball will behave as a satellite & will
continue to move with same speed in original orbit.
43. (a) The total momentum will be zero and hence velocity
will be zero just after collisiion. The pull of earth will
makeit fall down.
44. (d) Weight of liquid displaced = mg.
45. (a) The force of attraction between sphere and shaded
position
2
x
l
÷
ø
ö
ç
è
æ
=
dx
m
GM
dF
x
r l
m
M
r + l
2 2
2 1
2
1
1
2 1
1
( )
+ +
+
+ - +
-
+
+
-
= =
é ù
= = ê ú
- +
ë û
é ù
é ù
= - = - =
ë û ê ú +
ë û
ò ò
ò
r l r l
r r
r l
r l
r r
r l
r l
r
r
GMm GMm
F dx dx
lx l x
GMm GMm x
x dx
l l
GMm GMm GMm
x
l l x r r l
46. (a)
2 4 3
1 since
2
æ ö æ ö
= - = = +
ç ÷ ç ÷
è ø è ø
h g R
g g h R
h R h
Force on the satellite = mg
9
4
mgh =
N
889
10
200
9
4
»
´
´
=
47. (c)
2
R
1
g µ so we will not get a straight line.
AlsoF = 0 at a point where Force due to Earth = Force
due to mars
48. (c) Applying conservation of energy principle, we get
r
GMm
R
GMm
v
mk
2
1 2
e
2
-
=
-
r
GMm
R
GMm
R
GM
2
mk
2
1 2
-
=
-
Þ
R
k
R
1
r
1
r
1
R
1
R
k 2
2
-
=
Þ
-
=
-
Þ
2
2
k
1
R
r
)
k
1
(
R
1
r
1
-
=
Þ
-
=
Þ
49. (b) Loss in potential energy= Gain in kineticenergy
2
mv
2
1
R
GMm
2
3
R
GMm
=
÷
ø
ö
ç
è
æ
-
-
-
gR
R
GM
v
mv
2
1
R
2
GMm 2
=
=
Þ
=
Þ
50. (a) Change in force of gravity
=
2 2
M
G m
GMm 3
R R
- (onlydue tomass M/3 due toshell
gravitational field is zero(inside the shell))
= 2
2GMm
3R
51. (a)
52. (a) Centripetal force = net gravitational force
2
0
1
mv
2Fcos45 F
r
= ° + =
2 2
2 2
2GM 1 Gm
2
( 2r) 4r
+
r
V0
F F
F1
2r
45°
2 2
0
2
mv Gm
[2 2 1]
r 4r
= +
1/2
GM(2 2 1
4r
æ ö
+
Þ ç ÷
è ø](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-240-320.jpg)
![237
Gravitation
53. (c) During total eclipse :
Total attraction due to sun and moon,
s e m e
1 2 2
1 2
GM M GM M
F
r r
= +
When moon goes on the opposite side of earth.
Effective force of attraction,
s e m e
2 2 2
1 2
GM M GM M
F
r r
= -
Change in force,
m e
1 2 2
2
2GM M
F F F
r
D = - =
Change in acceleration ofearth
Da =
m
2
e 2
2GM
F
M r
D
=
Average force on earth, Fav =
av s
2
e 1
F GM
M r
=
%age change in acceleration
=
2
m 1
2
av s
2
2GM r
a
100 100
a GM
r
D
´ = ´ ´
2
1 m
2 s
r M
2 100
r M
æ ö
= ´
ç ÷
è ø
54. (a) Since, 3
2
kr
T =
Differentiating the above equation
Þ
T r
2 3
T r
D D
= Þ
T 3 r
T 2 r
D D
=
55. (b) Gravitational field at mass m due to full solid sphere
1 0
0 0
r R 1
E ........
3 6 4 G
r r é ù
= = e =
ê ú
e e p
ë û
r
r
Gravitational field at mass m due to cavity(–r)
3 3
2 2 2
0 0
( ) (R / 2) r a
E ........ using E
3 R 3 r
é ù
-r r
= =
ê ú
e e
ê ú
ë û
r
r
=
3
2
0
0
( )R R
24
24 R
-r -r
- = -
e
e
R
m
R/2
CM
Net gravitational field
1 2
0 0
R R
E E E
6 24
r r
= + = -
e e
r r r
0
R
8
r
=
e
Net force on m
0
m R
F mE
8
r
® = =
e
r
Here, 0
3
M 1
&
4 G
(4 /3) R
r = e =
p
p
then
3mg
F
8
=
56. (a) V= wR
2
0
g g R
= - w [g = at equator, g0 = at poles]
2
0
0
g
g R
2
= - w ;
2 0
g
R
2
w = ;
2 0
g R
V
2
=
2
e 0
V 2g R 4V 2V
= = =
57. (b)
2
e
1 GMm
mv
2 R
= ; e
GMm
v 2gR
R
= =
In tunnel bodywill perform SHM at centre
Vmax = Aw (see chapter on SHM)
=
e
v
R2
gR
2 R / g 2
p
= =
p
58. (b) 2
1
g
R
µ
R decreasing g increase hence, curve b represents
correct variation.
59. (a) Here,
2GM
v
R
= and
2GM
kv
R R
=
+
.
Solving
1
k
2
=
60. (b) Potential energy of particle at the centre of square
GMm
4
a
2
æ ö
ç ÷
= - ç ÷
ç ÷
è ø
2 2
GMm 1 8 2 GM
4 mv 0 v
a 2 a
2
æ ö
ç ÷
- + = Þ =
ç ÷
è ø
61. (b) T2 a r3, where r = mean radius =
2
r
r 2
1 +
62. (d) ÷
ø
ö
ç
è
æ
-
-
+
-
=
-
=
D
R
GMm
h
nR
GMm
U
U
U i
f
mgR
1
n
n
R
GMm
.
1
n
n
+
=
+
=
63. (c)
GM
)
h
R
(
2
T
3
+
p
=
GM
)
R
01
.
1
(
2
T
,
GM
R
2
T
3
2
3
1 p
=
p
=
%
5
.
1
100
T
T
T
1
1
2
=
´
-
64. (b) Electronic charge does not depend on acceleration
due to gravity as it is a universal constant.
So, electronic charge on earth
= electronic charge on moon
Required ratio = 1.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-241-320.jpg)
![239
Gravitation
6. (d) Asteroids are also being acted upon by central
gravitational forces, hence Asteroid will move in
circular orbits like planets and obey Kepler's laws.
7. (d) Gravitational mass of proton is equivalent to its inertial
mass and is independent of presence of neighbouring
heavy objects so verifies the option (d).
8. (c) Force of Gravitation,
2
g
GMm
F
r
=
Let AB = r
So, force on B due to A
2
(2 )
( )
BA
G Mm
F
AB
= = towards BA.
2
(2 )
2 g
G Mm
F
r
= =
and force on B due to C
2
( )
BC
GMm
F
BC
= = towards BC
As, (BC) = 2AB
2 2 2
4
(2 ) 4( ) 4
g
BC
F
GMm GMm GMm
F
AB AB r
Þ = = = =
As FBA > FBC,
hence, m will move towards (BA) i.e., (2M).
NEET/AIPMT (2013-2017) Questions
9. (a) Initial P
.E.,Ui =
GMm
R
-
,
Final P.E., Uf =
GMm
3R
-
[QR' = R+ 2R= 3R]
Change in potential energy,
DU=
GMm
3R
-
+
GMm
R
=
GMm
R
1
1
3
æ ö
-
ç ÷
è ø
=
2
3
GMm
R
=
2
3
mgR
GMm
mgR
R
æ ö
=
ç ÷
è ø
Q
ALTERNATE:DU=
mgh
h
1
R
+
By placing the value of h = 2R we get
DU=
2
3
mgR.
10. (c)
m
Gravitational potential V=
Gm
r
-
V0 = –
G 2
1
´
–
G 2
2
´
–
G 2
4
´
–
G 2
8
´
– 2G
1 1 1
1 ....
2 4 8
é ù
+ + + + ¥
ê ú
ë û
= – 2G ×
1
1
1
2
-
= – 2G ×
1
1
2
= – 4 G.
.
11. (c) Escape velocity,
8
3
e
V R GP
= p
Þ Ve µ R Þ 2
P P
E E
V R
V R
= =
Þ VP = 2VE.
12. (b) As we know, the minimum speed with which a bodyis
projected so that it does not return back is called
escape speed.
2 2 2
4
= = =
+
e
GM GM GM
V
r R h R
1
2
( 3 )
2
æ ö
= =
ç ÷
è ø
Q
GM
h R
R
13. (b) First when (r < R) E µ r and then when r > R
2
1
E
r
µ .
Hence graph (b) correctly dipicts.
14. (c) From question,
Escape velocity
=
2GM
R
= c = speed of light
Þ R=
2
2GM
c
=
11 24
8 2
2 6.6 10 5.98 10
m
(3 10 )
-
´ ´ ´ ´
´
= 10– 2 m
15. (b)
12R–3R=9R
12R
M
R 2R
5M
Before collision
M
5M
R 2R
At the time of collision
Let the distance moved by spherical body of mass M
is x1 and by spherical body of mass 5m is x2
As their C.M. will remain stationary
So, (M) (x1) = (5M)(x2)or, x1 =5x2
and for tauching x1 + x2 = 9R
So, x1 =7.5R
16. (a) As we know, orbital speed, orb
GM
V
r
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-243-320.jpg)
![242 PHYSICS
Its S.I. Unit is Nm-2
Stress is a tensor as its value changes when direction changes.
Types of Stress :
Longitudinal stress or tensile stress, volumetric stress and
tangential stress are the types of stress.
Strain :
It is defined as the ratio of the change in shape or size to the
original shape or size of the body.
=
Change in dimension
Strain
Original dimension
Strain has no units or dimensions.
Types of Strain :
Longitudinal strain : It is defined as the ratio of the change in
length to the original length.
Longitudinal strain =
Dl
l
Volume strain : It is the ratio of the change in volume to the
original volume.
Volume strain =
DV
V
Shearing strain : It is the angular deformation produced in a
body.
Shearing strain (q) =
Dx
h
HOOKE’S LAW
It is the fundamental law of elasticity given byRobert Hooke in
1679. It states that “the stress is directly proportional to strain
provided the strain is small “.
i.e. Stress µ strain constant
Þ = =
stress
E
strain
This proportionality constant is called modulus of elasticity
(name given byThomasYoung) or coefficient of elasticity (E).
Since stress has same dimensions as that of pressure and strain
is dimensionless. So the dimensions of E is same as, that of
stress or pressure i.e. [ML–1T–2]
The modulus of elasticity depends on the material and on the
nature of deformation. There are three type of deformations and
therefore three types of modulus of elasticity.
(i) Young’s modulus (Y) : It measuresthe resistance ofa solid
to elongation.
(ii) Shear modulus )
(h or modulus of rigidity : It measures
the resistance to motion of the plane of a solid sliding part
on each other.
(iii) Bulk modulus (B) : It measure the resistance that solid or
liquid offer to their volume change.
The stress under which the system breaks is called breaking
stress.
(i) Young's modulus (Y) : Let us consider a long bar (shown
in fig.) of cross-sectional area A and length lo, which is
clamped at one end. When we apply external force Fl
longitudinally along the bar, internal forces in bar resist
distortion, but bar attains equilibrium in which its length is
greater and in which external force is exactlybalanced by
internal forces. The bar is said to be stressed in this
condition.
l0
Fl
Fixed
Dl
Now if the solid bar obey the Hooke’ law, the Young's
modulus, Y is defined as
Tensile stress
= =
D
l
l l
/
/
F A
Y
Tensile strain
Where Dl is change in the length of bar, when we
apply Fl.
(ii) Shear modulus (h) : Shear modulus or modulus of
rigidityh is
h = =
D
/
/
t
F A
Shearing stress
Shearing strain x h
Ft
h
Dx
q
Fixed face
(a) As we see, there is no change in volume under
this deformation, but shape changes.
(b ) q
»
q
=
D
tan
h
x
(see the figure), where q is shear
angle.
(iii) Bulk modulus (B) : The Bulk modulus B is defined
as
-D
= = =
D D
/
/ /
n
F A
Volume stress P
B
Volume stain V V V V
Fn
V
V– V
D
Negative sign comes to make B positive, because with the
increase of pressure, the volume of body decreases or vice
versa.
The reciprocal of the Bulk modulus is called
compressibility of material
i.e., Compressibility = 1/B.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-246-320.jpg)
![248 PHYSICS
15. A and B are two wires. The radius of A is twice that of B.
They are stretched by the same load. Then the stress on B
is
(a) equal to that on A (b) four times that on A
(c) two times that on A (d) half that on A
16. Hooke's law defines
(a) stress
(b) strain
(c) modulus of elasticity
(d) elasticlimit
17. In case of steel wire (or a metal wire), the limit is reached
when
(a) the wire just break
(b) the load is more than the weight of wire
(c) elongation is inversely proportional to the tension
(d) None of these
18. A steel ring of radius r and cross sectional area A is fitted
onto a wooden disc of radius R (R > r). If the Young’s
modulus of steel is Y, then the force with which the steel
ring is expanded is
(a) AY(R/r) (b) AY(R – r)/r
(c) (Y/A)[(R– r)/r] (d) Yr/A R
19. Which of the following relation is true ?
(a) 3 (1 )
Y K
= - s (b)
9 Y
K
Y
h
=
+ h
(c) (6 )
K Y
s = + h (d)
05.Y - h
s =
h
20. For a constant hydraulic stress on an object, the fractional
change in the object volume
V
V
D
æ ö
ç ÷
è ø
and its bulk modulus
(B) are related as
(a)
V
B
V
D
µ (b)
1
V
V B
D
µ
(c)
2
V
B
V
D
µ (d)
2
V
B
V
-
D
µ
21. The diagram shown below represents the applied forces
per unit area with the corresponding change X (per unit
length) produced in a thin wire of uniform cross section in
the curve shown. Theregion inwhich the wirebehaveslikea
liquidis
(a) ab
(b) bc
(c) cd
F
O X
a
b
c
d
(d) Oa
22. A steel wire is suspended vertically from a rigid support.
When loaded with a weight in air, it extends byla and when
theweight is immersed completelyin water, the extension is
reduced to lw. Then the relative density of material of the
weight is
(a) w
a / l
l (b)
w
a
a
l
l
l
-
(c) )
/( w
a
w l
l
l - (d) a
w / l
l
23. When an elastic material with Young’s modulus Y is
subjected to stretching stress S, elastic energy stored per
unit volumeof the material is
(a) YS/ 2 (b) S2Y/ 2
(c) S2 /2Y (d) S/ 2Y
24. The ratio of shearing stress to the corresponding
shearing strain is called
(a) bulk modulus (b) Young's modulus
(c) modulus of rigidity (d) None of these
25. The Young’s modulus of a perfectly rigid body is
(a) unity
(b) zero
(c) infinity
(d) some finite non-zero constant
1. Two wires of same material and length but cross-sections
in the ratio 1 : 2 are used to suspend the same loads. The
extensions in them will be in the ratio
(a) 1: 2 (b) 2: 1
(c) 4: 1 (d) 1: 4
2. Abodyof mass 10 kg is attached to a wire ofradius 3cm. It’s
breakingstressis4.8× 107 Nm–2, thearea ofcross-section of
thewireis10–6 m2.Whatisthemaximumangularvelocitywith
which it can be rotated in the horizontal circle?
(a) 1 rad sec–1 (b) 2 rad sec–1
(c) 4 rad sec–1 (d) 8 rad sec–1
3. The Young’s modulus of brass and steel are respectively
1010 N/m2. and2 × 1010 N/m2.Abrass wire and a steel wire
of the same length are extended by 1 mm under the same
force, the radii of brass and steel wires are RB and RS
respectively. Then
(a) B
S R
2
R = (b) 2
/
R
R B
S =
(c) B
S R
4
R = (d) 4
/
R
R B
S =
4. A steel wire of length 20 cm and uniform cross-section
1 mm2 is tied rigidly at both the ends. The temperature of
the wire is altered from 40ºC to 20ºC. Coefficient of linear
expansion for steel a = 1.1 × 10–5/ºC and Y for steel is
2.0 × 1011 N/m2. The change in tension of the wire is
(a) 2.2 × 106 newton (b) 16 newton
(c) 8 newton (d) 44 newton](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-252-320.jpg)
![252 PHYSICS
Exemplar Questions
1. Modulus of rigidityof ideal liquids is
(a) infinity
(b) zero
(c) unity
(d) some finite small non-zero constant value
2. The maximum load a wire can withstand without breaking,
when its length is reduced to half of its original length, will
(a) be double (b) be half
(c) be four times (d) remain same
3. The temperature of a wire is doubled. TheYoung's modulus
of elasticity
(a) will also double (b) will become four times
(c) will remain same (d) will decrease
4. A spring is stretched byapplying a load to its free end. The
strain produced in the spring is
(a) volumetric
(b) shear
(c) longitudinal and shear
(d) longitudinal
5. A rigid bar of mass M is supported symmetrically bythree
wires each of length l. Those at each end are of copper and
themiddle one is ofiron. The ratiooftheir diameters, ifeach
is to have the same tension, is equal to
(a) Ycopper / Yiron (b)
iron
copper
Y
Y
(c)
2
iron
2
copper
Y
Y
(d)
iron
copper
Y
Y
6. A mild steel wire of length 2L and cross-sectional area A is
stretched, well within elastic limit, horizontallybetween two
pillars (figure ). Amass m is suspended from the mid-point
ofthe wire. Strain in the wire is
m
x
2L
(a)
2
2
2
x
L
(b)
x
L
(c)
2
x
L
(d)
2
2
x
L
7. A rectangular frame is to be suspended symmetrically by
two strings ofequal length on two supports (figure). It can
be done in one of the following three ways;
(a) (b) (c)
The tension in the strings will be
(a) the same in all cases (b) least in (a)
(c) least in (b) (d) least in (c)
8. Consider two cylindrical rods of identical dimensions, one
of rubber and the other of steel. Both the rods are fixed
rigidlyat one end to the roof. A mass M is attached to each
of the free ends at the centre of the rods.
(a) Both the rods will elongate but there shall be no
perceptible change in shape
(b) The steel rod will elongate and change shape but the
rubber rod will onlyelongate
(c) The steel rod will elongate without any perceptible
change in shape, but the rubber rod will elongate and
the shape of the bottom edge will change to an ellipse
(d) The steel rod will elongate, without any perceptible
changein shape, but the rubber rod will elongate with
the shape of the bottom edge tapered to a tip at the
centre
NEET/AIPMT (2013-2017) Questions
9. The following four wires are made of the same material.
Which of these will have the largest extension when the
same tension is applied ? [2013]
(a) Length = 100 cm, diameter = 1mm
(b) Length = 200 cm, diameter = 2 mm
(c) Length = 300 cm, diameter = 3 mm
(d) Length = 50cm, diameter = 0.5mm](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-256-320.jpg)
![253
Mechanical Properties of Solids
10. If the ratio of diameters, lengths and Young’s modulus of
steel and copper wires shown in the figure are p, q and s
respectively, then the corresponding ratio of increase in
their lengths would be [NEET Kar. 2013]
(a)
7
(5 )
q
sp
(b) 2
5
(7 )
q
sp
Steel
2m
Copper
5m
(c) 2
7
(5 )
q
sp
(d)
2
(5 )
q
sp
11. Copper of fixed volume ‘V; is drawn into wire of length ‘l’.
When this wire is subjected to a constant force ‘F’, the
extension produced in the wire is ‘Dl’. Which of the
following graphs is a straight line? [2014]
(a) Dl versus
1
l
(b) Dl versus l2
(c) Dl versus 2
1
l
(d) Dl versus l
12. The approximate depth of an ocean is 2700 m. The
compressibilityof water is 45.4 × 10–11 Pa–1 and densityof
water is 103 kg/m3.What fractional compression of water
will be obtained at the bottom of the ocean ? [2015]
(a) 1.0 × 10–2 (b) 1.2 × 10–2
(c) 1.4 × 10–2 (d) 0.8 × 10–2
13. The Young's modulus of steel is twice that of brass. Two
wires of same length and of same area of cross section, one
of steel and another of brass are suspended from the same
roof. If we want the lower ends of the wires to be at the
same level, then the weights added to the steel and brass
wires must be in the ratio of : [2015 RS]
(a) 2 :1 (b) 4 : 1
(c) 1 :1 (d) 1 : 2
14. Thebulkmodulus ofa sphericalobject is'B'. Ifit is subjected
to uniform pressure 'p', the fractional decrease in radius is
(a)
B
3p
(b)
3p
B
[2017]
(c)
p
3B
(d)
p
B](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-257-320.jpg)
![254 PHYSICS
EXERCISE - 1
1. (c) 2. (b) 3. (d) 4. (a) 5. (d)
6. (c) 7. (a)
8. (c) Couple per unit angle of twist,
l
2
r
C
4
h
p
=
Couple t =
l
2
r
C
4
q
h
p
=
q
Here h, l, C & t are same. So, r4q = constant
÷
÷
ø
ö
ç
ç
è
æ
=
q
q
4
1
4
2
2
1
r
r
9. (a)
strain
stress
)
90
tan( =
q
-
10. (c) If l isthe original length ofwire, then changein length
offirst wire, )
( 1
1 l
l
l -
=
D
change in length of second wire, )
( 2
2 l
l
l -
=
D
Now,
2
2
1
1
A
T
A
T
Y
l
l
l
l
D
´
=
D
´
=
or
2
2
1
1 T
T
l
l D
=
D
or
l
l
l
l -
=
- 2
2
1
1 T
T
or T1 l 2 – T1 l = T2 l 1 – l T2 or
1
2
2
1
1
2
T
T
T
T
-
-
=
l
l
l
11. (c) F F mg
Y
A YA YA
= Þ D = =
D
l l l
l
l
12. (d) The elasticity of a material depends upon the
temperature of the material. Hammering & annealing
reduces elastic property of a substance.
13. (b) The ratio of stress to strain is always constant. If
stress is increased, strain will also increase so that
their ratio remains constant.
14. (c)
1
FL
l l
YA A
= Þ µ
15. (b) Stress = 2
Force 1
Stress
Area πr
µ
2
2
(2) 4
B A
B A
A B
S r
S S
S r
æ ö
= = Þ =
ç ÷
è ø
16. (c)
17. (d) According to Hooke's Law, within the elastic limits
stress is directly proportional to strain.
18. (b) Let T be the tension in the ring, then
)
r
R
(
A
r
T
)
r
R
(
2
.
A
r
2
.
T
Y
-
=
-
p
p
=
r
)
r
R
(
A
Y
T
-
=
19. (d)
0.5
2 (1 )
Y
Y
- h
= h + s Þ s =
h
20. (b)
1
/
p V
B
V V B V
D D
= Þ µ
D
[Dp = constant]
21. (b) The wire starts behaving like a liquid at point b. It
behaves like a viscous liquid in the region bc of the
graph.
22. (b) Let V be the volume of the load and r its relative
density
So,
a
a A
L
g
V
A
L
F
Y
l
l
r
=
= .....(1)
When the load is immersed in the liquid, then
w
w A
L
)
g
1
V
g
V
(
A
L
F
Y
l
l
´
´
-
r
=
¢
= ....(2)
(Q Now net weight = weight – upthrust)
From eqs. (1) and (2), we get
w
a
)
1
(
l
l
-
r
=
r
or
)
( w
a
a
l
l
l
-
=
r
23. (c) Energy stored per unit volume
= strain
stress
2
1
´
´
= )
modulus
s
'
Young
/
stress
(
stress
2
1
´
´
= )
modulus
s
'
Young
/(
)
stress
(
2
1 2
´ =
Y
2
S2
24. (c)
25. (c) For a perfectly rigid body strain produced is zero for
the given force applied, so
Y = stress/strain = ¥
EXERCISE - 2
1. (b) Let W newton be the load suspended. Then
1
1
1
1
A
L
W
)
L
/
(
)
A
/
W
(
Y
l
l
=
= ...(1)
and
2
2
2
2
A
L
W
)
L
/
(
)
A
/
W
(
Y
l
l
=
= ....(2)
Dividing equation (1) byequation (2), we get
÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
=
1
2
A
A
1
1
2
1
2
1
2
l
l
l
l
1
2
2
1 =
l
l
or l1 : l2 = 2 : 1
2. (c) Given that F/A = 4.8 × 107 Nm–2
F = 4.8 ×107 ×Aor
2
7 6
mv
4.8 10 10 48
r
-
= ´ ´ =
or 48
r
r
m 2
2
=
w
or
r
m
48
2
=
w
sec
/
rad
4
16
3
.
0
10
48
=
=
÷
ø
ö
ç
è
æ
´
=
w
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-258-320.jpg)
![255
Mechanical Properties of Solids
3. (b) We know that Y= F L/pr2 l or r2 = F L/(Yp l)
)
Y
/(
L
F
R B
2
B l
p
= and )
Y
/(
L
F
R S
2
S l
p
=
or 2
10
10
2
Y
Y
R
R
10
10
B
S
2
S
2
B
=
´
=
=
or 2
S
2
B R
2
R = or S
B R
2
R =
2
/
R
R B
S =
4. (d) F =YAat = (2.0 × 1011) (10–6) (1.1 × 10–5) (20)
= 44 newton
5. (d)
6. (d) Elastic energy = x
F
2
1
´
´
F=200 N, x=1 mm=10–3 m
J
1
.
0
10
1
200
2
1
E 3
=
´
´
´
= -
7. (a) U/ volume =
2
strain
Y
2
1
´ = 3600 J m–3
[Strain = 0.06 × 10–2]
8. (b)
2
l
F Y A F r
L
= ´ ´ Þ µ (Y, l and and Lare constant)
Ifdiameter is madefour times then force required will
be 16 times, i.e., 16 × 103 N
9. (b)
2
8
.
0
100
1
r
´
=
q
=
f
l
= 0.004 radian
10. (d) Q Both wires are same materials so both will have
sameYoung’s modulus, and let it beY.
( )
ΔL/L
A.
F
strain
stress
Y =
= , force
applied
F =
A area of cross-section of wire
=
Now, 2
1 Y
Y =
( )( ) ( )( )
2
2
1
1 L
A
FL
L
A
FL
D
=
D
Þ
Since load and length are same for both
2
2
2
1
2
1 L
r
L
r D
=
D
Þ , 4
r
r
L
L
2
1
2
2
1
=
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
D
D
1 2
L : L 4:1
D D =
11. (d) 2 (1 )
Y = h + s
2.4 2 (1 ) 1.2 1 0.2
h = h + s Þ = + s Þ s =
12. (b) Bulk modulus
V
PV
)
V
/
V
(
P
B
D
-
=
D
-
= ....(1)
and T
.
V
.
3
T
V
V a
=
D
g
=
D or
.
T
.
3
1
V
V
a
=
D
-
...(2)
From eqs. (1) and (2), )
T
.
3
/(
P
B a
= or
B
3
P
T
a
=
13. (a)
P
V
/
V
B
1
K
D
=
= . Here, P = 100 atm,
K = 4 ×10–5 andV =100cm3.
Hence, DV=0.4cm3
14. (a)
I
=
d
Y
3
W 3
l
, where W= load, l= length of beam and I is
geometrical moment ofinertia for rectangular beam,
3
bd
Ι
12
= where b = breadth and d = depth
For square beam b = d
4
1
b
Ι
12
=
For a beam of circular cross-section,
4
2
πr
Ι
4
æ ö
=ç ÷
è ø
4
3
4
3
1
b
Y
W
4
b
Y
3
12
W l
l
=
´
=
d (for sq. cross section)
and
)
r
(
Y
3
W
4
)
4
/
r
(
Y
3
W
4
3
4
3
2
p
=
p
=
d
l
l
(forcircular cross-section)
Now
p
=
p
p
=
p
=
d
d 3
)
r
(
r
3
b
r
3
2
2
4
4
4
2
1
( 2 2
b π r
=
Q i.e., theyhave same cross-sectional area)
15. (d)
Weight of load Weight of puller
Compressive stress
area
+
=
5
6
20A d 10 5 10
1.6 10
area
´ ´ + ´
Þ = ´
Where d is the density.
3 5
6
20A 2.5 10 10 5 10
1.6 10
A
´ ´ ´ + ´
= ´
ie, 5× 105 = 1.1× 106A or A= 0.45m2
16. (a)
l
l 2
r
C
,
2
)
r
r
(
C
4
2
4
1
4
2
1
ph
=
-
ph
=
Initial volume= Final volume
r
p
=
r
-
p l
l 2
2
1
2
2 r
]
r
r
[
)
r
r
)(
r
r
(
r
r
r
r 1
2
1
2
2
2
1
2
2
2
-
+
=
Þ
-
=
Þ
)
98
.
7
02
.
8
)(
98
.
7
02
.
8
(
r2
-
+
=
Þ
cm
8
.
0
r
cm
64
.
0
04
.
0
16
r2
=
Þ
=
´
=
Þ
4
4
4
4
4
1
4
2
2
1
]
8
.
0
[
]
98
.
7
[
]
02
.
8
[
r
r
r
C
C -
=
-
=
17. (a) Weights without and with liquid proportional to rb
and rb – rl.
18. (b)
19. (c) d
For a beam, the depression at the centre is given by,
3
f L
4Ybd
æ ö
d = ç ÷
è ø
[f, L, b, d are constants for a particular beam]
i.e.
1
Y
d µ](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-259-320.jpg)
![257
Mechanical Properties of Solids
34. (b) The atoms when brought from infinity are attracted
due to interatomic electrostatic force of attraction. At
point B, the potential energyis minimum and force of
attraction is maximum. But if we bring atoms closer
than x = B, forceof repulsion between two nuclei starts
and P.E. increases.
x
B
C
P.E.
A
35. (c)
F
x
Fcom Fext
From the figure, it is clear that
Fcom < Fext.
36. (d)
4
. Constant
2
r
r C
L
ph q
= q = =
4 4
0 0
( ) ( / 2) ( ')
2 2( / 2)
r r
l l
ph q -q ph q -q
Þ =
l/2
l
q0
q
q¢ = 0
0 0
0
( ) 8
2 16 9
q- q q
Þ = Þ q = q
37. (a)
6
3
10 100
34
3
3 10 10
S
L m
dg
= = = =
´ ´
38. (b)
r 6mm 30
r 0.18
1m
q ´ °
q = f Þ f = = = °
l
l
39. (a) F YA YA
D
= = aDq
l
l
= 2 × 1011 × (1 × 10–6) × 1.1 × 10–5 × (10) = 22 N
40. (a) 2
mg
Y
( / 4) d x
=
p
l
.........(1)
max
dY m d x
2
Y m d x
D D D D
æ ö
= + + +
ç ÷
è ø
l
l
m = 20.0 kg Þ Dm= 0.1 kg
l = 125cm Þ Dl = 1 cm.
d =0.050 cm. Þ Dd = 0.001 cm
x= 0.100 cm. Þ Dx=0.001cm.
max
0.1kg 1cm
dY 20.0kg 125cm 100% 6.3%
0.001cm 0.001cm
Y 2
0.05cm 0.100cm
æ ö
+
ç ÷
æ ö
= ´ =
ç ÷ ç ÷
è ø
+ ´ +
ç ÷
è ø
41. (d) Given, 1
2
,
l
a
l
= 1
2
r
b
r
= , 1
2
Y
c
Y
=
Brass
T
T
2g
Steel
T
2g
Let Young’s modulus of steel beY1, and that of brass
be Y2
1 1
1
1 1
F l
Y
A l
=
D …(i)
and
2 2
2
2 2
F l
Y
A l
=
D …(ii)
Dividing equation (i) byequation (ii), we get
1 1 2 1 2
2 2 1 2 1
Y F A l l
Y F A l l
D
=
D …(iii)
Force on steel wire from free bodydiagram
1 (2 ) Newton
T F g
= =
Force on brass wire from free bodydiagram
2 1 2 4 Newton
F T T g g
= = + =
Now putting the value of F1, F2 in equation (iii), we
get
2
1 2 1 2 2
2 2
2 2 1 1
1
2 1 1
. . .
4 2
Y r l l l
g
a
Y g l l l
r b
æ ö é ù æ ö æ ö
æ ö p D D
æ ö
= =
ç ÷ ê ú ç ÷
ç ÷ ç ÷ ç ÷
è ø
è ø D D
è ø è ø
p
è ø ë û
42. (d) If side of the cube is L then
3
3
dV dL
V L
V L
= Þ =
% change in volume = 3 × (% change in length)
= 3 × 1% = 3%
Bulk strain 0.03
V
V
D
=
43. (b)
2
2
4 9.8
19.6 10
2 10
F
K
x -
´
= = = ´
´
Work done =
2 2
1
19.6 10 (0.05) 2.45
2
J
´ ´ =
44. (d) F =
YA
L
× l
So, extension, lµ
L
A
µ 2
L
D
[Q F and Y are
constant]
l1 µ 2
100
1
µ 100 and l2 µ 2
200
2
µ 50
l3µ 2
300
3
µ
100
3
and l4 µ
50
1
4
µ 200
The ratio of
2
L
D
is maximum for case (d).
Hence, option (d) is correct.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-261-320.jpg)
![258 PHYSICS
45. (a) Percentage elongation in the wire
F
AY
=
l
4
2 11
6.28 10 (1) 1
1000
(0.01) 2 10
´ ´
= =
p ´ ´
46. (a) Work done 2
1 1
Stress Strain (Strain)
2 2
Y
= ´ ´ = ´ ´
Since, elasticity of steel is more than copper, hence
more work has to be done in order to stretch the steel.
47. (b) Stress is defined as internal force (restoring force) per
unit area of a body. Also, rubber is less elastic than
steel, because restoring force is less for rubber than
steel.
48. (d) 49. (d) 50. (a)
EXERCISE - 3
Exemplar Questions
1. (b) As liquid is ideal so no. frictional force exists hence,
tangential forces are zero so there is no stress
developed.
2. (d) As we know that,
Breaking stress =
Breaking force
Area of cross-section
..(i)
When length of the wire changes (or byreducing half)
area of cross-section remains same.
Hence, breaking force will be same because breaking
stress does not depend on length.
3. (d) Aswe knowthat, length of awirewhen thetemperature
increased
0 (1 )
t
L L T
= + aD
where DT is change in the temperature.
L0 is original length,
a is coefficient oflinear expansion and
Lt is length at temperature T.
Now, 0 0
t
L L L L T
D = - = aD
Now, Young's modulus (Y)
Stress
Strain
=
0 0
0
1
FL FL
A L AL T T
= = µ
´D aD D
As,
1
Y
T
µ
D
So, if temperature increass DT increases, hence
Young's modulus of elasticity (Y) decreases.
4. (c) When a spring is stretched by applying a load to its
free end. Clearly the length and shape of the spring
changes. So strain produced when change in length
corresponds to longitudinal strain andchange in shape
corresponds to shearing strain.
5. (b) As we know that,
The Young's modulus
Stress /
Strain /
F A F L
Y
L L A L
= = = ´
D D
2 2
4
( / 2)
F L FL
L
D D L
= ´ =
D
p p D
2 4 4
FL FL
D D
LY LY
= Þ =
pD pD
If F and
L
L
D
are constants.
So,
1
D
Y
µ
Hence, we can find ratio as
copper iron
iron copper
D Y
D Y
=
6. (a) Consider the given diagram
m
90°–q
90°–q
q q
x
O
B
A C
L L
So, change in length
0 0
( ) ( )
L A B AC CB
D = + - +
= 2BO – 2AC
= 2 [BO – AC] ( AO = BO, AC = CB)
2 2 1/ 2
2[( ) ]
x L L
= + -
1/ 2
2
2
2 1 1
x
L
L
é ù
æ ö
ê ú
= + -
ç ÷
ç ÷
ê ú
è ø
ë û
2 2
2
1
2 1 1
2
x x
L L
L
L
é ù
D » + - =
ê ú
ë û
[ ]
x L
<<
Q
2 2
2
/
Strain =
2 2 2
L x L x
L L L
D
= =
7. (c) Let us consider the free body diagram of the
rectangular frame
Tsinq Tsinq
T
Tcosq
Tcosq
q
q
m
mg
Net forces acting on frame will be zero.
So, Balancing vertical forces
2T sinq – mg = 0 [T is tension in the string]
2T sinq = mg ...(i)
Total horizontal force
cos cos 0
T T
= q - q =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-262-320.jpg)
![259
Mechanical Properties of Solids
Nowfrom Eq. (i),
2sin
mg
T =
q
max
min
2sin
mg
T =
q
As mg is constant then
1
sin
T µ
q
min min
sin 0 0
q = Þ q =
No option matches with q = 0°
min
max
2sin
mg
T =
q
(since, sinqmax = 1)
max
sin 1 90
q = Þ q = °
So tension is all three cases are different rejects option
(a)
For minimum tension q must be 90° i.e. sin q = 1
Hence, tension is the least for the case (b).
8. (d) A mass M is attached at the centre or midpoint of rod
of rubber and steel. As the mass is attached to both
the rods, both rod will be elongated as shown in
figuresbut due todifferent elasticproperties ofmaterial
rubber changes shape also.
M M
(Steel)
(Rubber)
As the Young's modulus of rigidity for steel, is larger
than rubber, so strain
L
L
D
for rubber is larger than
steel for same stress.
NEET/AIPMT (2013-2017) Questions
9. (d) F =
YA
L
× l! So, extension, lµ
L
A
µ 2
L
D
[Q F and Y are constant]
l1 µ 2
100
1
µ 100 and l2 µ 2
200
2
µ 50
l3µ 2
300
3
µ
100
3
and l4 µ
50
1
4
µ 200
The ratio of
2
L
D
is maximum for case (d).
Hence, option (d) is correct.
10. (c) From formula,
Increase in length 2
4
FL FL
L
AY D Y
D = =
p
2
S S C C S
C C S S C
L F D Y L
L F D Y L
æ ö
D
= ç ÷
D è ø
=
2
7 1 1
5
q
p s
æ ö æ ö
´ç ÷ ç ÷
è ø
è ø
= 2
7
(5 )
q
sp
11. (b) As Y=
F
F
A
AY
Þ D =
D
l
l
l
l
But V = Al soA=
V
l
Therefore Dl =
2
2
F
VY
µ
l
l
Hence graph of Dl versus l2 will give a straight line.
12. (b) Compressibilityof water,
K = 45.4 × 10–11 Pa–1
densityof water P = 103 kg/m3
depth of ocean, h = 2700 m
We have to find
V
?
V
D
=
As we know, compressibility,
K=
1 ( V / V)
(P gh)
B P
D
= = r
So, (DV/V)= Krgh
= 45.4 × 10–11 × 103 ×10 ×2700 = 1.2258× 10–2
13. (a) Young's modulusY =
W
.
A
l
l
D
1 2
1 2
W W
Y Y
= [Q A, l, Dl same for both brass and steel]
Brass
A, ,
l Dl
Steel
l, A, Dl
Y2 Y1
1 1
2 2
W Y
2
W Y
= = [Ysteel/Ybrass = 2 given]
14. (c) Bulk modulus is given by
B =
P
V
V
D
æ ö
ç ÷
è ø
or
V P
V B
D
=
R P
3
R B
D
= (here,
R
R
D
= fractional decreases in radius)
R P
R 3B
D
Þ =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-263-320.jpg)
![260 PHYSICS
FLUIDS
Fluid is something that can flow. All liquids and gases are fluids.
The force exerted normally at a unit area of the surface of a fluid
is called fluid pressure.
i.e., P =
F
a
Its S.I. unit is Nm–2 or Pascal. Its dimensions are [ML–1T–2].
PASCAL'S LAW AND ITS APPLICATIONS
Pascal's law : Pressure in a fluid in equilibrium is the same
everywhere, if the effect of gravity is neglected.
Another form of Pascal's law : The excess pressure, applied
anywhere in a mass of confined incompressible fluid is
transmitted by the fluid in all directions without being
diminished in magnitude.
Applications of Pascal's Law
Hydraulic lift : Its working is based on Pascal's law. Apiston of
small cross-sectional area (a) exerts a force (f) on the liquid.
Applied force = f
F = pA
f = pa
Area = a
Area = A
The pressure is transmitted undiminished to the larger cylinder
of cross-sectional area A.
= =
f F
P
a A
Þ =
A
F f
a
Hydraulic brakes also work on pascal's law.
ATMOSPHERIC, HYDROSTATIC AND GUAGE PRESSURE
Atmospheric pressure : The atmosphere exerts pressure on the
earth's surface. The atmospheric pressure at sea level is given by
P0 = 1.01 × 105 Pa
Hydrostatic pressure : The hydrostatic pressure at a depth h
below the surface of a fluid is given by
P = hrg
where r is the density of the fluid, g acceleration due to gravity
and h is the depth of the liquid column.
Gauge pressure : The pressure at anypoint in a fluid is equal to
the sum of the atmospheric pressure acting on its surface and the
hydrostatic pressure due to the weight of the fluid above that
point which is at a depth h below the surface of the fluid.
The gauge pressure is given by
P = P0 + hrg
or, P – P0 = hrg
BUOYANCY AND ARCHIMEDES' PRINCIPLE
Buoyancy
If a body is partiallyor wholly immersed in a fluid, it experiences
an upward force due tothe fluid surrounding it. The phenomenon
of force exerted by fluid on the body called buoyancy and the
force is called buoyant force or upthrust.
A bodyexperiences buoyant force whether it floats or sinks, under
its own weight or due to other forces applied on it.
Archimedes' Principle
When any body is immersed (totally or partially) in a liquid it
appears to lose part of its weight and the apparent loss of weight
is equal to the weight of liquid displaced.
Let a bodyof weight W is immersed in a fluid and W' is upthrust
on it then
(i) if W > W', then body will sink.
(ii) if W = W', then the bodyfloats with whole or some part of
its volume inside the fluid.
Let V be the volume of a body of density d and V' be the
volume of liquid of density r displaced. If the body floats
then Vd = V¢r
V
'
V
d
=
r
gives the fraction of the volume inside the liquid
in which the body floats.
Also, body immersed in a fluid experiences an upward buoyant
force equivalent to the weight of the fluid displaced by it.
The proof of this principle is very simple. Imagine a body of
arbitrary shape completely immersed in a liquid of density r as
10
Mechanical
Properties of Fluids](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-264-320.jpg)
![261
Mechanical Properties of Fluids
shown in the figure (a). Abody is being acted upon by the forces
from all directions. Let us consider a vertical element of height h
and cross-sectional area dA (as shown in the figure (b)).
h2
dA
h
h1 F1
F2
(a) (b)
The force acting on the upper surface of the element is F1
(downward) and that on the lower surface is F2 (upward). Since
F2 > F1, therefore, the net upward force acting on the element is
dF = F2 – F1
It can be easily seen from the figure (b), that
1 1
F ( gh )dA
= r and 2 2
F ( gh )dA
= r so dF g(h)dA
= r
Also, h2 – h1 = h and d (dA) = dV
The net upward force is F gdV Vg
= r = r
ò
Hence, for the entire body, the buoyant force is the weight of the
volume of the fluid displaced.
The buoyant force acts through the centre of gravity of the
displaced fluid.
Keep in Memory
1. The pressure is perpendicular to the surface of the fluid.
2. The upthrust on a body immersed in a liquid does not
depend on the mass, density or shape of the body. It only
depends on the volume of the body.
3. The weight of the plastic bag full of air is same as that of
the empty bag because the upthrust is equal to the weight
of the air enclosed.
4. The cross-section of the water stream from a top decreases
as it goes down in accordance with the equation of
continuity.
5. We cannot sip a drink with a straw on the moon, because
there is no atmosphere on the moon.
6. The line joining the centre of gravity and centre of
buoyancy is called central line.
7. Metacenter - is a point where the vertical line passing
through the centre of buoyancy intersects the central line.
8. The floating body is in stable equilibrium when the
metacenter is above the centre of gravity(centre of gravity
is below the centre of buoyancy).
9. The floating bodyis in the unstable equilibrium when the
metacenter lies below thecentre ofgravity(centre of gravity
is above the centre of buoyancy).
10. The floating bodyis in the neutral equilibriumwhen centre
of gravity coincides with the metacenter (centre of gravity
coincides with the centre of buoyancy).
11. The wooden rod cannot float vertically in a pond of water
because centre of gravity lies above the metacenter.
12. (i) If a body just floats in a liquid (density of the body is
equal to the densityof liquid) then the bodysinks if it
is pushed downwards.
(ii) If two bodies have equal upthrust when just immersed
in a liquid, both will have the same volume.
(iii) If a person floats on his back on the surface of water,
the apparent weight of person is zero.
13. The hydrometer can be used to measure density of the
liquid or fluid.
Relative Density (or Specific Gravity)
Liquids may be treated as incompressible. Hence their density
may be assumed to be constant throughout.
Relative density =
Weight of substance in air
Weight of equal volume of water
=
Weight of substance in air
Loss of weight in water
4º
=
Density of substance
Density of water at C
Density in SI system = 1000 × density in the cgs system.
(i) The density of liquid of bulk modulus B at a depth h is
given by 0 1
h
gh
B
r
æ ö
r = r +
ç ÷
è ø
where 0
r is the density of
liquid on its surface and r is the average densityof liquid.
(ii) The density of liquid changes with pressure as
0 1
h
P
P P
B
D
æ ö
= +
ç ÷
è ø
where =
DP change in pressure and B = bulk modulus of
liquid.
(iii) If two liquids of masses m1, m2 and densities r1, r2 are
mixed together, then the density of the mixture is given by
1 2
1 2
1 2
m m
m m
+
r =
+
r r
And if m1 = m2 but different densities are mixed together,
then the density of the mixture is harmonic mean of the
densities.
i.e.,
2
1
2
1
2
r
+
r
r
r
=
r or ú
û
ù
ê
ë
é
r
+
r
=
r
1
2
1
1
1
2
1
(iv) If two drops of same volume but different densities are
mixed together, then the density of the mixture is the
arithmetic mean of the densities.
i.e.,
2
2
1 r
+
r
=
r (as
1 0 2 0
0 0
V V
V V
r + r
r =
+
)
SURFACE TENSION
It is defined as the force per unit length acting at right angles
on either side of an imaginary line drawn on the free surface of
the liquid.
i.e., =
l
F
S
The surface tension is also defined as the work required to
increase unit area of that liquid film.
Its SI unit is N/m or J/m2 and dimensions are [ML0T–2].](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-265-320.jpg)
![264 PHYSICS
From eqns. (1) and (2), V = V1
Hence, no change in the level of water.
Example3.
Two substances of densities r1 and r2 are mixed in equal
volume and the relative density of mixture is 4. When they
are mixed in equal masses, the relative density of the
mixture is 3. Determine the values of r1 and r2 .
Solution :
When the substances are mixed in equal volumes, then
4
V
2
V
V 2
1 ´
=
r
+
r ... (1)
When the two substances are mixed in equal masses, then
3
m
2
m
m
2
1
=
r
+
r
...(2)
From eq. (1), 8
2
1 =
r
+
r ... (3)
From eqn. (2)
3
2
or
3
2
1
1
2
1
2
1
2
1
=
r
r
r
+
r
=
r
+
r
or 12
or
3
2
8
2
1
2
1
=
r
r
=
r
r
. ...(4)
Now
2
/
1
2
1
2
2
1
2
1 ]
4
)
[( r
r
-
r
+
r
=
r
-
r
4
]
48
64
[ 2
/
1
=
-
= ...(5)
Solving eqns. (3) and (5), we get 6
1 =
r and 2
ρ 2
= ]
Example4.
A sealed tank containing a liquid of density r moves with
a horizontal acceleration a, as shown in fig. Find the
difference in pressure between the points A and B.
C A
B
h
a
l
Solution :
Since points A and C are in the same horizontal line but
separated by distance l and liquid tank is moving
horizontallywith acceleration a, hence
PC – PA = lra or PC = PA + lra
Points B and C are vertically separated by h
PB – PC = h r g
or PB – (PA + l r a) = h r g
or PB – PA = h r g + l ra
Example5.
Calculate the excess pressure within a bubble of air of
radius 0.1 mm in water. If the bubble had been formed 10 cm
below the water surface when the atmospheric pressure
was 1.013 × 105 Pa, then what would have been the total
pressure inside the bubble?
Solution :
Excess pressure within air bubble
=
2T
r
=
3
3
2 73 10
0.1 10
-
-
´ ´
´
= 1460 Pa
The pressure at a depth d, in liquid P = hdg. Therefore, the
total pressure inside the air bubble is
Pin = Patm + hdg +
2T
r
or Pin = 1.013× 105 + 10 ×10–2 × 103 × 9.8 + 1460
= 101300 + 980 + 1460
=103740 =1.037 ×105 Pa.
Example6.
A capillary of the shape as shown is dipped in a liquid.
Contact angle between the liquid and the capillary is 0°
and effect of liquid inside the meniscus is to be neglected. T
is the surface tension of the liquid, r isradius of the meniscus,
g is acceleration due to gravity and r is density of the
liquid then determine the height h in equilibrium.
h
Solution :
As weight ofliquidin capillaryisbalancedbysurfacetension,
then 2
1
T 2 r r h g
´ p = p r (for a tube ofuniform radius r)
1
2T
h
r g
=
r
r
h1
But weight ofliquid in tapered tube ismore than uniform tube
of radius r, then in order to balance h < h1.
2T
h
r g
<
r
r
Example7.
A hydraulic automobile lift is designed to lift car with a
maximum mass of 3000 kg. The area of cross-section of the
piston carrying the load is 425 cm2. What maximumpressure
would the smaller piston have to bear ?
Solution :
Here mass ofcar = 3000 kg.
Area of cross section of larger piston
= 425cm2 = 425 ×10–4 m2.
The maximum pressure that the smaller piston would
have to bear
= 4
Weight of car 3000 9.8
Area of cross-section 425 10-
´
=
´
5 2
6.92 10 Nm-
= ´](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-268-320.jpg)
![267
Mechanical Properties of Fluids
VELOCITY OF EFFLUX AND TORRICELLI'S THEOREM
(i) Torricelli's theorem : For liquid filled in a tank upto a
height H having a hole O at a depth h from free level of
liquid through which the liquid is coming out, velocity of
efflux of liquid v = 2gh .
(ii) Time taken by the liquid in falling from hole to ground
level
2( )
-
=
H h
t
g
H
h
Free
surface
(iii) The horizontal range covered by the liquid
x = horizontal velocity (v) ×time (t)
2( )
2
-
= ´
H h
gh
g
2 ( )
= -
h H h .
For maximum horizontalrange, differentiating both sidew.r.t.
h and we get h = H/2, the range is maximum.
Keep in Memory
1. The cross-section of the water stream from a top decreases
as it goes down in accordance with the equation of
continuity.
2. When a hale blows over a roof, the force on the roof is
upwards.
3. Sudden fall in atmospheric pressure predicts possibility of
a storm.
4. Venturimeter is a device used for the measurement of the
rate of flow of incompressible fluid through a tube. Its
working is based on Bernoulli’s principle.
Example8.
The vessel of area of cross-section A has liquid to a height
H. There is a hole at the bottom of vessel having area of
cross-section a. Find the time taken to decrease the level
from H1 to H2.
Solution :
The average velocity of efflux, 1 2
2g H 2g H
v
2
+
=
Let t be the time taken to empty the tank from level H1 to
H2.
Then, ]
H
H
[
A
t
a
2
H
g
2
H
g
2
2
1
2
1
-
=
´
´
+
or t
( ) ( )
( )( )ú
ú
û
ù
ê
ê
ë
é
-
+
-
´
-
´
÷
÷
ø
ö
ç
ç
è
æ
´
=
2
1
2
1
2
1
2
1
H
H
H
H
H
H
H
H
g
2
a
A
( )
2
1 H
H
g
2
a
A
-
´
÷
÷
ø
ö
ç
ç
è
æ
´
=
( )
1 2
A 2
H H
a g
æ ö
= ´ ´ -
ç ÷
è ø
Example9.
In a test experiment on a model aeroplane in a wind tunnel,
the flow speeds on the upper and lower surfaces of the
wing are 70 m/s and 63 m/s respectively. What is the lift on
the wing if its area is 2.5 m2 ? Take the density of air to be
1.3 kg m–3.
Solution :
Let v1 be the speed and P1 be the pressure on the upper
surface of the wing, and corresponding values on the lower
surface be v2 and P2 respectively.
v1 = 70 m/s, v2 = 63 m/s, A= 2.5 m2,
P =1.3 kg m–3.
According to Bernoulli’s theorem
2 2
1 1 2 2
1 1
P v P v
2 2
+ r = + r
2 2
2 1 1 2
1
P P (v v )
2
- = r - =
2 2
2 1
1
P P 3 (70 63 )
2
- = ´ ´ -
Force (lift) on the wing = 2 1
A (P P )
-
=
2 2
1
2.5 1.3 (70 63 )
2
´ ´ ´ -
=
1
2.5 1.3 133 7
2
´ ´ ´ ´
3
1.5 10 N
= ´
Example10.
In Millikan’s oil drop experiment, what is the terminal
speed of an uncharged drop of radius 2.0 × 10–5 m and
density 1.2 × 103 kg m–3. Take the viscosity of air at the
temperature of the experiment to be 1.8 × 10–5 Pa s. How
much is the viscous force on the drop at that speed ? Neglect
buoyancy of the drop due to air.
Solution :
Here, r = 2.0 × 10–5 m, r = 1.2 × 103 kg m–3,
h = 1.8 × 10–5 Pa s.
From formula, terminal velocity
2
2 r ( ) g
V
9
r - s
=
h
Þ
5 2 3
5
2 (2 10 ) (1.2 10 0) 9.8
V
9 1.8 10
-
-
´ ´ ´ - ´
=
´ ´
2 1
5.8 10 ms
- -
= ´
Now viscous force on the drop
F= 6phrv
Þ 5 5 2
22
F 6 (1.8 10 ) (2 10 ) 5.8 10
7
- - -
= ´ ´ ´ ´ ´ ´ ´
= 3.93 × 10–19 N](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-271-320.jpg)
![274 PHYSICS
Exemplar Questions
1. A tall cylinder is filled with viscous oil. A round pebble is
dropped from the top with zero initial velocity. From the plot
shown in figure, indicatethe one that represents the velocity
(v) of the pebble as a function of time (t) .
v
t
(a)
v
t
(b)
v
t
(c)
v
t
(d)
2. Which of the following diagrams does not represent a
streamline flow?
(b)
(a)
(d)
(c)
3. Along a streamline,
(a) the velocityof a fluid particle remains constant
(b) the velocity of all fluid particles crossing a given
position is constant
(c) the velocity of all fluid particles at a given instant is
constant
(d) the speed of a fluid particle remains constant
4. An ideal fluid flows through a pipe of circular cross-section
made of two sections with diameters 2.5 cm and 3.75 cm.
The ratio of the velocities in the two pipes is
(a) 9: 4 (b) 3: 2 (c) 3 : 2 (d) 2 : 3
5. The angle of contact at the interface of water-glass is 0°,
ethyl alcohol-glass is 0°, mercury-glass is 140° and
methyliodide-glass is 30°.Aglass capillaryis put in a trough
containing one of these four liquids. It is observed that the
meniscus is convex. The liquid in the trough is
(a) water (b) ethylalcohol
(c) mercury (d) methyliodide
NEET/AIPMT (2013-2017) Questions
6. The wetability ofa surface by a liquid depends primarilyon
(a) surface tension [2013]
(b) density
(c) angle of contact between the surface and the liquid
(d) viscosity
7. A fluid is in streamline flow across a horizontal pipe of
variable area ofcross section. For this which of the following
statements is correct? [NEET Kar. 2013]
(a) The velocity is minimum at the narrowest part of the
pipe and the pressure is minimum at the widest part of
the pipe
(b) The velocity is maximum at the narrowest part of the
pipe and pressure is maximum at the widest part of the
pipe
(c) Velocity and pressure both are maximum at the
narrowest part of the pipe
(d) Velocityand pressure both are maximum at the widest
part of the pipe
49. Statement 1 : Falling raindrops acquire a terminal velocity.
Statement 2 : A constant force in the direction of motion
and a velocity dependent force opposite to the direction of
motion, always result in the acquisition of terminal velocity.
50. Statement 1 : Thebuoyant force on a submerged rigid object
can be considered to be acting at the centre of mass of the
object.
Statement 2 : For a rigid body a force field distributed
uniformlythrough its volume can be considered tobe acting
at the centre of mass of the body.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-278-320.jpg)
![275
Mechanical Properties of Fluids
8. A certain number ofspherical drops of a liquid of radius ‘r’
coalesce to form a single drop ofradius ‘R’ and volume ‘V’.
If ‘T’ is the surface tension of the liquid, then : [2014]
(a) energy =
1 1
4VT
r R
æ ö
-
ç ÷
è ø
is released
(b) energy =
1 1
3VT
r R
æ ö
+
ç ÷
è ø
is absorbed
(c) energy =
1 1
3VT
r R
æ ö
-
ç ÷
è ø
is released
(d) energy is neither released nor absorbed
9. A wind with speed 40 m/s blows parallel to the roof of a
house. The area of the roof is 250 m2. Assuming that the
pressure inside the house is atmospheric pressure, the force
exerted by the wind on the roof and the direction of the
forcewill be (rair = 1.2 kg/m3) [2015]
(a) 4.8 × 105 N, upwards
(b) 2.4 × 105 N, upwards
(c) 2.4 × 105 N, downwards
(d) 4.8 × 105 N, downwards
10. The cylindrical tube ofa spray pump has radius, R, one end
ofwhich has n fine holes, each ofradius r. Ifthe speed of the
liquid in the tube is V, the speed of the ejection of the liquid
through the holes is : [2015 RS]
(a)
2
2
VR
nr
(b)
2
3 2
VR
n r
(c)
2
V R
nr
(d)
2
2 2
VR
n r
11. Water rises to a height 'h' in a capillarytube. If the length of
capaillarytube above the surface of water is made less than
'h' then : [2015 RS]
(a) water rises upto the top of capillary tube and stays
there without overflowing
(b) water rises upto a point a little below the top and stays
there
(c) water does not rise at all.
(d) Water rises uptothe tip of capillarytube and then starts
overflowing like fountain.
12. Two non-mixing liquids of densities r and nr(n > 1) are put
in a container. Theheight ofeach liquid is h.Asolid cylinder
oflength Land densityd is put in thiscontainer. Thecylinder
floats with its axis vertical and length pL(p < 1) in the denser
liquid. The density d is equal to : [2016]
(a) {1 + (n + 1)p}r (b) {2 + (n + 1)p}r
(c) {2 + (n – 1)p}r (d) {1 + (n – 1)p}r
13. A U tube with both ends open tothe atmosphere, is partially
filledwith water.Oil,whichisimmisciblewithwater,ispoured
intoone side until it stands at a distance of10 mm above the
water level on the other side. Meanwhile the water rises by
65 mm from its original level (see diagram). The densityof
the oil is [2017]
A
Oil
B C
Water
Initial water level
Final water level
10 mm
Pa
Pa
F
E
D
65 mm
65 mm
(a) 425 kg m–3 (b) 800 kg m–3
(c) 928 kg m–3 (d) 650 kg m–3](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-279-320.jpg)
![279
Mechanical Properties of Fluids
3. (b) In streamline flow, the speed of liquid of each particle
at a point in a particular cross-section is constant,
between two cross-section of a tube of flow because
AV = constant (law of continuity).
4. (a) As given that,
Diameter at 1st section (d1) = 2.5.
Diameter at 2nd section (d2) = 3.75.
According to equation of continuity,
for cross-sections A1 and A2.
A1v1 = A2v2
2
2
1 2 2 2
2
2 1 1
1
( )
( )
v A r r
v A r
r
æ ö
p
= = = ç ÷
p è ø
2
2
2
2
2
1
1
2
2
d
d
d
d
æ ö
ç ÷ æ ö
è ø
= = ç ÷
è ø
æ ö
ç ÷
è ø
2 1
2 1
,
2 2
d d
r r
é ù
= =
ê ú
ë û
Q
2
3.75 9
2.5 4
æ ö
= =
ç ÷
è ø
1 2
: 9 : 4
v v
=
5. (c) We observed that meniscus of liquid is convex shape
as shown in figure which is possible if only, the angle
of contact is obtuse. Hence, the combination will be of
case of mercury-glass (140°). Hence verifies the option
(c).
140°
convex
mercury
NEET/AIPMT (2013-2017) Questions
6. (c) Wetability of a surface by a liquid primarily depends
on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and
vice-versa.
7. (b) According to Bernoulli’s theorem,
2
1
2
P v
+ r = constant and Av
v = constant
IfA isminimum,v ismaximum,Pisminimum.
8. (c) As surface area decreases so energy is released.
Energy released = 4pR2T[n1/3 – 1]
where R = n1/3r
=
3 1 1
4 R T
r R
é ù
p -
ê ú
ë û
=
1 1
3VT
r R
é ù
-
ê ú
ë û
9. (b) According to Bernoulli’s theorem,
1 2
2 0
P v P 0
+ r = +
So,DP =
1
2
rv2
F = DPA =
1
2
rv2A P0
P
=
1
2
×1.2×40 ×40×250
= 2.4 × 105 N (upwards)
10. (a) Inflow rate of volume of the liquid = Outflow rate of
volume of the liquid
pR2V = npr2(v) Þ v=
2 2
2 2
R V VR
n r nr
p
=
p
11. (a) Water rises upto the top of capillary tube and stays
there without overflowing.
12. (d) As we know,
Pressure P = Vdg
r
d
(1 – p)L
nr pL
Here, LA d g = (pL)A(nr)g + (1 – p)LAr g
Þ d = (1 – p)r + pn r = [1 + (n – 1)p]r
13. (c) Here, oil oil water water
h g h g
´r ´ = ´r ´
r0g × 140 × 10–3 = rwg × 130 × 10–3
roil =
3 3
130
10 928kg / m
140
´ » [Q rw= 1 kgm–3]](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-283-320.jpg)
![282 PHYSICS
Isothermal and adiabatic elasticities:
(i) Isothermal elasticity, .
P
V
dV
dP
Eiso =
÷
ø
ö
ç
è
æ
-
º
(since PV = RT = constant,
V
P
dV
dP
-
= )
(ii) Adiabatic elasticity,
P
V
dV
dP
Eadia g
=
÷
ø
ö
ç
è
æ
-
= ; where
v
p
C
C
=
g
(iii) iso
adia E
E g
=
Expansionof Gases
There are two coefficients of expansion of gases.
(a) Pressure coefficient of expansion of gas (gP)
0
0
-
g = t
P
P P
P t
where P0 = Pressure at 0°C, Pt = Pressure at t°C
(b) Volume coefficient of expansion of gas (gv)
0
0
-
g = t
v
V V
V t
where V0 = Volume at 0°C, Vt =Volume at t°C
HEAT
Heat is the form of energy which is transferred from one body to
another body due to temperature difference between two bodies.
Thermal energyof a bodyis the sum of kinetic energy of random
motion of the molecules/atoms and the potential energy due to
the interatomic forces.
Calorimeter is an instrument used to measure the heat.
The SIunitofheat isjouleand itsdimensionsare[ML2 T–2]. Other
units of heat are :
(i) Calorie: It is the amount of heat required to raise the
temperatureof1gofwater from14.5ºCto15.5ºC.
1 calorie= 4.2joule. Calorieis theunit ofheat in CGSsystem.
(ii) Kilocalorie,1Kcal= 1000cal
(iii) British Thermal Unit (BTU): It is the fps unit of heat
= 252 cal = amount of heat required to raise the temperature
of1 lb ofwater through 1ºF (from 63°F to 64ºF).
SPECIFIC HEAT CAPACITY
It is the amount of heat required to raise the temperature of unit
mass of substance through 1 degree.
Specific heat capacity, =
´ D
Q
c
m t
Its SI unit is J kg–1K–1
CGS unit cal g–1°C–1
It dimensions are [M0L2T–2q–1]
For gases : During the isothermal process
Dt = 0 Þ c = ¥ (infinite)
Thermal Stress
If a rod fixed at two ends is heated or cooled, then stress are
produced in the rod.
Thermal stress = T
Y D
a
´
0
T strain
æ ö
D
= aD =
ç ÷
è ø
l
l
Effectof Temperature on PendulumClock
A Pendulum clock consists of a metal bar attached with a bob at
oneend and fixed at the other end. Let pendulum clock read correct
time when its length is l0. The time period
t0 = l0
g
2π
Suppose that the temperatureis raised byDT, then newtime period
is
g
t 2
= p
l
=
Þ
l
l
0 0
t
t .....(i)
We have = + aD
l l0(1 t)
Þ = + aD
l
l0
(1 t) .....(ii)
From equations (i) and (ii),
= + aD
0
1
2
t
(1 t)
t
= + aD
1
2
1 t (since a is verysmall)
Þ - = aD
0
t 1
1
t 2
t
-
Þ = aD
0
0
t t 1
t 2
t
so, fractional change in time
D
= aD
0
t 1
t 2
T
Loss of time per day = aD ´
1
2
T 86400 second .
(Q no. of seconds in one day = 86400 sec.)
When temperature increases (during summer), the length of the
pendulum increases due to which the time period increases and
the clock loses time. On the other hand, when temperature
decreases (during winter season), the length decreases and the
time period decrease, the clock in this case, gains time.
Error in scale reading :
As the temperature changes, the length ofthe metal scale changes
so is the difference in graduation
a m
R R (1 T)
= + aD
where Ra = actual reading, Rm = measured reading](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-286-320.jpg)
![283
Thermal Properties of Matter
and during the adiabatic process
Q = 0 Þ c = 0 (zero)
Specific heat of gas : The specificheat ofa gas varies from zeroto
infinity. It may have any positive or negative value.
Following of two specific heats of gas are more significant.
(a) Molar specific heat at constant volume (CV) : It is the
amount of heat required to raise the temperature of
1 mole of a gas through 1K or 1°C at constant volume.
1 Q
T
æ ö
= ç ÷
è ø
V
d
C
n d
(b) Molar specific heat at constant pressure (CP) : It is the
amount of heat required toraise the temperature of1 mole of
a gas through 1K or 1°C at constant pressure.
1
C
æ ö
= ç ÷
è ø
P
dQ
n dT
Mayer’s relation : CP – CV = R
[where R is a gas constant. (R = 8.314 Jmol–1K–1)]
Why CP is greater thanCV ?
When heat is supplied to a gas at constant volume entirely used
to raise its temperature. When a gas is heated at constant pressure,
some work is done in expansion of gas which is in addition to
raise its temperature and hence CP > CV.
THERMAL CAPACITY OR HEAT CAPACITY
It is the amount of heat required to raise the temperature of
substance through 1ºC or 1K.
Thermal capacity = mass × specific heat.
Thisis the relationbetween sp. heat capacity and heat capacity.
Its SI unit is JK–1 CGS unit cal/°C. Its dimensions are [ML2T–
2q–1]
Water Equivalent
It is the mass of water in gram which would require the same
amount of heat to raise its temperature through 1ºC as the body
when heated through the same temperature. It is measured in
gram. Hence
Water equivalent of a body = mass of the body × specific heat
Principle of Heat (or Calorimetry)
When two bodies at different temperatures are placed in contact
with each other then heat will flow from the body at higher
temperature to the bodyat lower temperature until both reach toa
common temperature.
i.e. Heat lost by hot body = heat gained by cold body.
It follows the law of conservation of energy.
LATENT HEAT
It is defined as the amount of heat absorbed or given out by a
body during the change of state while its temperature remaining
constant.
It is of two types :
(i) Latent heat of fusion : It is the quantityof heat required to
change unit mass ofthe solid intoliquid at its melting point.
Latent heat of fusion of ice is 80 cal/g.
(ii) Latentheat ofvaporisation:Itisthequantityofheatrequired
to convert unit mass of liquid into vapour at its boiling
point.
Latent heat of vaporisation of water is 536 cal/g.
Latent heat increases the intermolecular potential energy of the
molecules while kinetic energyremains constant.
Different Processes of Phase Change or State Change
(i) Sublimation : It is the process of conversion of solids to
gaseous state on heating. On cooling the vapours get
converted back into solids.
(ii) Condensation : It is the process of conversion of gases/
vapours toliquid state on cooling. On heating, these liquids
are converted into vapours/gases.
(iii) Boiling : It is the process of conversion of liquid to gaseous
state on heating. It is a cooling process.
(iv) Melting : It is the process in which solid gets converted
intoliquid on heating.
(v) Evaporation : Conversion of liquid intogaseous state at all
temperatures is called evaporation. It is a phenomenon that
occurs at the surface of the liquid.
Melting and boiling occur at definite temperature called melting
point and boiling point respectively. The liquid boils at a
temperature, at which its vapour pressure is equal to the
atmospheric pressure.
Howevaporationisdifferentfromboiling?
Evaporation is a slow process occurring at all the temperature at
the surface ofliquid while boiling is a fast process involving whole
ofthe liquid heated toa particular temperature called boiling point
ofthe liquid.
Keep in Memory
1. The heat capacity or thermal capacitydepends on nature of
the substance (specific heat capacity) and mass of quantity
of matter of the body.
2. In case of melting and boiling, the temperature does not
change, only intermolecular potential energyof the system
changes.
3. System are in thermal equilibrium when their temperature
are same or average kinetic energyper molecule is same.
4. Internal energyis the sum of all energies in a system which
includes energy of translational, rotational as well as
vibrational motion of the molecules.
5. The molar specific heat capacity depends on the molecular
weight of the substance.
6. Thetemperature, volume or pressure ofa system mayremain
constant when it absorbs heat.
7. When we press two block of ice together such that after
releasing the pressure two block join this phenomenon is
called regelation.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-287-320.jpg)
![285
Thermal Properties of Matter
Example5.
From what height must a block of ice be dropped into a
well of water so that 5% of it may melt? Given : both ice
and water are at 0°C, L = 80 cal g–1, J = 4.2 J cal–1 and
g = 980 cm s–2.
Solution :
Let m be the mass of ice. Let h be the height from which
block of ice is dropped. Work done,
W = mgh = m× 980 × h erg
Mass of ice to be melted =
5
m
100
´
Heat required,
5
Q
100
= × m × 80 cal or Q = 4 m cal
[Q L = 80 cal g–1] [Q W= JQ]
Now, m × 980 ×h = J× 4m
7
4.2 10 4m
h
m 980
´ ´
=
´
cm= 1714.3m
[Q J = 4.2 × 107 erg cal–1]
SATURATED AND UNSATURATED AIR
The air is said tobe saturatedwhen the maximum possibleamount
of water vapours are present in it. The pressure of the water
vapours in the saturated air is called saturation vapour pressure.
Ifthe air contains vapours less than the maximum possible amount
possible in the air, then it (air) is said to be unsaturated.
Vapour Pressure
A vapour differs from a gas in that the former can be liqufied by
pressure alone, wheras the latter cannot be liqufied unless it is
first cooled.
If a liquid is kept in a closed vessel, the space above the liquid
becomes saturated with vapour. The pressure exerted by this
vapour is called saturated vapour pressure. Its value depends
only on temperature. It is independent of any external pressure.
Humidity
The humidity shows the presence of water vapours in the
atmosphere. The amount ofwater vapours present per unit volume
oftheair is called absolute humidity.
Relative humidity (RH) is defined as the ratio of the mass of
water vapour (m) actually present in the given volume of air to
the mass of water vapour (M) required to saturate the same
volume at the same temperature. Normally RH is expressed in
percentage
i.e., % 100%
= ´
RH
M
m
RH is also defined as the ratio of actual pressure (p) of water
vapour to the saturated vapour pressure (P) of the water at the
same temperature
i.e., % ×100%
=
p
RH
P
Also,
RH =
Saturated vapour pressure of water at dew point
Saturated vapour pressure of water at room temperature
DewPoint :
It is the temperature at which the amount of water vapours
actuallypresent in a certain volume of the air which is sufficient
to saturate that volume of air.
At the dew point the actual vapour pressure becomes the
saturated vapour pressure.
It means that vapour pressure at room temperature = saturated
vapour pressure(S.V.P) at dew point
If room teperature = dewpoint
then Relative humidity(R.H.) = 100%
Keep in Memory
1. Hygrometry is the branch of thermal physics that deals
with the measurement ofamount of water vapours present
in the atmosphere.
2. Hygrometer is an instrument used todetermine dew point
and relative humidity.
3. Relative humidityis low when the air is dry.
4. Saturated vapours do not obey (except Dalton’s law) gas
laws, while unsaturated vapours obey them. The S.V.P.
varies linearlywith temperature (keeping volume constant)
and equal to one atmospheric pressure at boiling
point.(shown in fig.(a)) S.V.P. constant at constant
temperature on increasing or decreasing volume fig. (b).
V = constant
T
Boiling point
1 atm
(S.V.P.)
(a)
P
T = constant
V
(S.V.P.)
(b)
P
5. At 0ºC, the saturation vapour pessure is 4.6 mm of Hg.
Therefore, water vapours are always present around the
ice. It increases with the increase in temperature.
6. The relative humidity is 100% when thetemperatureofthe
atmosphere is equal to the dew point.
7. Cryogenics is the study of low temperature.
8. As RH is expressed in percentage, hence the maximum
value of RH maybe 100%.
9. The saturation vapour pressure of water at 100ºC is 760
mm ofHg at sea level.
10. As the temperature rises, the absolute humidity may
increase and the relative humiditymayremain constant or
may even decrease.
11. Ifthe relative humidityis high then at the same temperatue
one feels hotter.
12. If the air is absolutley dry, the dew point is not observed.
13. The relative humidity decreases with the increase in
temperature.
14. Ifwater is sprinkled in the room, both relative humidityas
well as the dew point increase.
15. Dew appears on the leaves of the trees etc. due to the
condensation of saturated vapours.
16. Themist or fog is thesmall droplets of water formed due to
the condensation of water vapours near the surface of
earth.
17. Mist or fog formed much higher above the surface of earth
is called cloud.
11.1](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-289-320.jpg)
![286 PHYSICS
18. The Relative humidity between 50% and 60% gives
comfortable feeling to human beings.
19. When the relative humidityis 100%, the reading of the dry
and wet bulbs is same.
20. Smaller the difference in the temperature of dry and wet
bulbs, larger is the relative humidity.
21. Water vapours are always seen around the ice, because the
temperature around it is less than the dew point.
HEAT TRANSFER
Three modes of transmission of heat :
(a)Conduction (b) Convection and(c)Radiation
(a) Conduction : In thermal conduction particles of body at
higher temperature transmit heat to the particles at lower
temperature by mutual contact (collision) only and not by
the movement of the particles. All solids are heated by
conduction. Conduction cannot take place in vacuum.
Dx
D
D
Q
t
D
D
Q
t
A
T1
T2
T > T
1 2
Figure shows a solid of cross-section area A and thickness
Dx. Thefaceareatdifferent temperatureT1 and T2 (T1 > T2)
The rate of heat flow ÷
ø
ö
ç
è
æ
D
D
t
Q
as found experimentally is
given by
÷
ø
ö
ç
è
æ
D
D
µ
D
D
x
T
A
t
Q
where 2
1 T
T
T -
=
D
Þ
Δ Δ
Δ Δ
æ ö
= ç ÷
è ø
Q T
KA
t x
…(1)
where K is proportionality constant called thermal
conductivity.
It is a measure of how quickly heat energy can conduct
through the substance.
Coefficientof thermalconductivity(K): Thecoefficientofthermal
conductivity, K, of a material is defined as the amount of heat that
flowing per second through a rod of that material having unit
length and unit area of cross-section in the steady state, when
the difference of temperature between two ends of the rod is 1 ºC
and flow of heat is perpendicular to the end faces of the rod.
Unit of coefficient of thermal conductivity in SI system is
watt/m-K
Dimensions : [M L T–3 q–1]
For a perfect conductor thermal conductivity K is infinite and
for a perfect insulator K is zero.
In general, solids are better conductors than liquids and liquids
are better conductors than gases. (Heat transfer through the
process of conduction is possible in liquids and gases also, if
theyare heated from the top.) Metals are much better conductors
than non-metals.
A good conductor of heat is also a good conductor of electricity
The conduction ofboth heat and electricityis due to the movement
of free electrons.
(b) Convection : It is the process by which heat is transferred
from one place to another in a medium by the movement of
particles of the medium. It occurs due to densitydifference.
This phenomenon occurs in fluids.
(c) Radiation : It is the process by which heat is transferred
from one place to another without any intervening medium.
The light reaches from Sun to Earth byradiation process.
Keep in Memory
1. The equation
Δ Δ
Δ Δ
æ ö
= ç ÷
è ø
Q T
KA
t x
is valid for steady state
condition. The condition is said to occur when no part of
heat is used up in raising the temperature of any part of
cross-section of the solid.
2. On comparing equation (1)with the following equation used
for flowing of charge on account of potential difference.
V1
Dq
V2
s
l
We find :
l
V
A
t
q D
s
=
D
D
The role of resistance (thermal resistance) is played by
Dx
KA
3. Series combination of conductors
l1 l2
K , A
1 1
T1 T T2
K , A
2 2
Equivalent thermal conductivity:
1 2
1 2
2
=
+
eq
K K
K
K K
1 2
1 2
1 2
( – )
A T T
H
l l
K K
=
+
where H = heat flow per second
4. Parallel combination of conductors
l
K , A
1 1
T1 T2
K , A
2 2
Equivalent thermal conductivity:
Keq = 1 1 2 2
2 2
+
+
K A K A
A A
1 2
1 2
1 2
1 2
1 2
K K
T + T
l l
H =
K K
+
l l](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-290-320.jpg)
![287
Thermal Properties of Matter
5. Davy’s safety lamp is based on the conduction. It is used in
mines toknow the ignition temperature of gases. Danger of
explosion can be avoided.
6. (i) Principle of chimney used in a kitchen or a factory is
based on the convection.
(ii) Land and sea breezes are due ot the convection.
(iii) Temperature of theupper part of theflameis more than
the temperature on the sides, because the currents of
air carrythe heat upwards.
(iv) Radiation can be detected by differential air
thermometer, Bolometer, thermopile, etc.
Example6.
Two plates of same area are placed in contact. Their
thickness as well as thermal conductivities are in the ratio
2 : 3. The outer surface of one plate is maintained at 10ºC
and that of other at 0ºC. What is the temp. of the common
surface?
Solution :
Let q be the temp. of common surface.
As
2
1
t
Q
t
Q
÷
÷
ø
ö
ç
ç
è
æ
D
D
=
÷
÷
ø
ö
ç
ç
è
æ
D
D
2
2
2
1
1
1
x
)
0
(
A
K
x
)
10
(
A
K
-
q
=
q
-
or 1 1
2 2
K x
(10 θ) θ
K x
- =
K1
x1 x2
K2
Plate 1 Plate 2
10°C 8°C 0°C
Now,
3
2
K
K
x
x
;
A
A
2
1
2
1
2
1 =
=
= ; .
C
º
5
or
3
2
)
10
(
3
2
=
q
q
=
q
-
Example7.
A slab consists of two parallel layers of two different
materials of same thickness and thermal conductivities
K1 and K2. Find the equivalent thermal conductivity of
the slab.
Solution :
Rate of flow of heat through each layer of slab is same,
because slabs are in series.
x
x
DT1
DT2
l
l
l 2
A
)
T
T
(
K
T
A
K
T
A
K 2
1
2
2
1
1
D
+
D
=
D
=
D
or C
2
)
T
T
(
K
T
K
T
K 2
1
2
2
1
1 =
D
+
D
=
D
=
D , say
K
C
2
T
T
,
K
C
T
,
K
C
T 2
1
2
2
1
1 =
D
+
D
=
D
=
D
or
K
C
2
K
C
K
C
2
1
=
+ or
2
1
2
1
K
K
K
K
2
K
+
=
Example8.
Two walls of thickness d1 and d2 and thermal conductivities
K1 and K2 are in contact. In the steady state, if the
temperatures at the outer surfaces are T1 and T2 then find
the temperature at the common wall.
Solution :
K1
T1 T2
T
d1 d2
K2
Let the temperature of common wall be T. Then
2
2
2
1
1
1
d
t
)
T
T
(
A
K
d
t
)
T
T
(
A
K -
=
-
or,
2
2
2
1
1
1
d
t
)
T
T
(
K
d
t
)
T
T
(
K -
=
-
or, )]
T
T
(
K
[
d
)]
T
T
(
K
[
d 2
2
1
1
1
2 -
=
-
or, 2
2
1
2
1
2
1
2
1
1 T
K
d
T
K
d
T
d
K
d
T
K -
=
-
or, ]
K
d
d
K
[
T
T
d
K
d
T
K 2
1
2
1
2
1
2
2
1
1 +
=
-
ú
û
ù
ê
ë
é
+
-
=
1
2
2
1
2
1
2
2
1
1
d
K
d
K
T
d
K
d
T
K
T
HEAT TRANSFER BY RADIATION AND NEWTON’S LAW
OF COOLING
Stefan’s Law:
This law is also called Stefan Boltzmann law. This law states that
the power radiated for overall wavelength from a black body is
proportional to the fourth power of thermodynamic temperature T.
4 4
E T E T
µ Þ = s for perfectly black body ....(1)
where E is energyemitted per second from unit sruface area of the
black body, T is temperature and s is Stefan’s constant, and
s = 5.67×10–8 Wm–2 K–4.
If body is not perfectly black body, then
E = es T4 ....(2)
where e is emissivity. Then energyradiated per second by a body
of area A is
E1 = es T4 ....(3)
If the body temperature is T and surrounding temperature is T0,
then net rate of loss of energy by body through radiation from
equation (3) is
' 4 4
1 0
( )
E eA T T
= s - ....(4)
Now rate of loss of energy in terms of specific heat c is
'
1
dQ dT
mc E
dt dt
= = ....(5)
and 4 4
0
( )
dT eA
T T
dt mc
s
= - ....(6)
(T- temperature, t - time in second)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-291-320.jpg)
![288 PHYSICS
So body cools by radiation and rate of cooling depends on
e (emissivity of body), A (area of cross-section of body),
m (mass of body) and c (specific heat capacity of body).
A black body is defined as a body which completely absorbs all
the heat radiation falling on it without reflecting and
transmitting any of it.
Weins Law :
lmT = b,
where b is the Weins constant and b = 2.898 × 10–3mK
and T = temperature.
Graphof lm versus T
T
2
1
1
2
l
=
l
m
m
T
T
Hence on increasing temperature of a body, its colour changes
gradually from red to orange ® yellow ® green ® blue ®
violet. Thus the temperature of violet star is maximum and
temperature of red star is minimum. Sun is a medium categorystar
with lm = 4753Aº (yellow colour) and temperature6000K.
Kirchoff’s Law
According to this law, the ratio of emissive power to absorptive
power is same for all surfaces at the same temperature and is
equal to the emissive power of a perfectly black body at the same
temperature.
i.e.
e
E A
a
=
where e = emissive power of a given surface
a = absorptive power of a given surface
E = emissive power of a perfect black body
A = absorptive power of a perfect black body
(i) For a perfect black body, A = 1, =
e
E
a
(ii) If emissive and absorptive power are considered for a
particular wavelenth l then l
l
l
=
e
E
a
(iii) SinceEl is constant at a given temperature, according to
this law if a surface is good absorber of a particular
wavelength then it is also a good emitter of that wavelength.
Fraunhoffer's Line
(i) Fraunhoffer lines are the dark lines in the spectrum of sun
and are explained on the basis ofKirchoff’s law. When white
light emitted from central core of sun (photosphere) passes
through its atmosphere (chromosphere) radiation of those
wavelength will be absorbed by the gases present there,
which theyusuallyemit (as good emitter is a good absorber)
resulting dark lines in the spectrum of sun.
(ii) On the basis ofKirchoff’s law, Fraunhoffer identified some
of the elements present in the chromosphere. They are
hydrogen, helium, sodium, iron, calcium, etc. Fraunhoffer
had observed about 600 darklines in the spectrum of sun.
Newton's Law of Cooling
The rate of loss of heat of a body is directly proportional to the
temperature difference between the body and the surroundings.
i.e. 0
( )
dT
K T T
dt
= - -
where T = temperature of body, T0 = temperature ofsurrounding.
and
mc
eA
K
s
= where, e = emissivity of body, A = area of surface
of body, s = Stefan’s constant, m = mass ofbody, c = specific heat
of body.
It is a special case of Stefan’s lawand above relation is applicable
onlywhen the temperature ofbody is not much different from the
temperature of surroundings.
Solar constant : Solar constant is the solar radiation incident
normally per second on one square metre at the mean distance
of the earth from the sun in free space.
It is given byS = 1.34 × 109 Jm–2s–1.
Temperature of sun is given T4 = S/s(R/r)2, where R is mean
distance of the earth from the sun and r is the radius of the sun.
Keep in Memory
1. Good absorber is a good emitter.
2. Cooking utensils are provided with wooden or ebonite
handles, since wood or ebonite is a bad conductor of heat.
3. Good conductor of heat are good conductors of electricity,
Mica is an exception which being a good conductor ofheat
is a bad conductor of electricity.
Example9.
Compare the rate of loss of heat from a metal sphere of
temperature 827°C, with the rate of loss of heat from the
same sphere at 427 °C, if the temperature of surroundings
is 27°C.
Solution :
Given : T1 = 827 °C = 1100 K, T2 = 427 °C = 700 K
and T0 = 27 °C = 300 K
According to Steafan Boltzmann law ofradiation,
dQ
dt
= s A
Ae (T4 – T0
4)
4 4 4 4
1 0
1
4 4 4 4
2 0
2
dQ
dt (T T ) [(1100) (300) ]
dQ (T T ) [(700) (300) ]
dt
æ ö
ç ÷
è ø - -
= =
æ ö - -
ç ÷
è ø
=6.276
or
1 2
dQ dQ
: 6.276 :1
dt dt
æ ö æ ö
=
ç ÷ ç ÷
è ø è ø
Example10.
The filament of an evacuated light bulb has a length 10 cm,
diameter 0.2 mm and emissivity 0.2, calculate the power it
radiates at 2000 K. (s = 5.67 × 10–8 W/m2 K4)
Solution :
l= 10cm= 0.1 m, d= 0.2mm =2 × 10–4 m
r = 0.1mm=1×10–4 m,
e = 0.2, T = 2000 K, s= 5.67 ×10–8 W/m2 K4
According to stefan's law of radiation, rate of emission of
heat for an ordinary body,
E = sAeT4 = s(2 pr l) eT4 [QA= 2prl]
= 5.67 ×10–8 × 2× 3.14 ×1 ×10–4 × 0.1 × 0.2 × (2000)4
= 11.4 W
Power radiated bythe filament = 11.4 W](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-292-320.jpg)
![295
Thermal Properties of Matter
Exemplar Questions
1. A bimetallic strip is made of aluminium and steel (aAl >
asteel). On heating, the strip will
(a) remain straight
(b) get twisted
(c) will bend with aluminium on concave side
(d) will bend with steel on concave side
2. A uniform metallic rod rotates about its perpendicular
bisector with constant angular speed. Ifit isheated uniformly
to raise its temperature slightly
(a) its speed of rotation increases
(b) its speed of rotation decreases
(c) its speed of rotation remains same
(d) its speed increases because its moment of inertia
increases
3. The graph between two temperature scales A and B isshown
in figure between upper fixed point and lower fixed point
there are 150 equal division on scale A and 100 on scale B.
The relationship for conversion between the two scales is
given by
O 100
180
DtB = 100°
DtA = 150°
Temperature (°B)
Temperature
(°A)
(a)
180
100 150
A B
t t
-
= (b)
30
150 100
A B
t t
-
=
(c)
180
150 100
B A
t t
-
= (d)
40
100 180
B A
t t
-
=
4. An aluminium sphere is dipped into water. Which of the
following is true?
(a) Buoyancywill beless in water at 0°C than that in water
at4°C
(b) Buoyancywill bemorein water at 0°C than that in water
at4°C
(c) Buoyancyin water at 0°C will be same as that in water
at4°C
(d) Buoyancymaybemoreor less in water at 4°Cdepending
on the radius of the sphere
5. As the temperature is increased, the period of a pendulum
(a) increases as its effective length increases even though
its centre of mass still remains at the centre of the bob
(b) decreases as its effective length increases even though
its centre of mass still remains at the centre of the bob
(c) increases as its effectivelength increases due toshifting
to centre of mass below the centre of the bob
(d) decreases as its effective length remains same but the
centre of mass shifts above the centre of the bob
6. Heat is associated with
(a) kinetic energy of random motion of molecules
(b) kinetic energy of orderlymotion of molecules
(c) total kinetic energy of random and orderly motion of
molecules
(d) kinetic energy of random motion in some cases and
kinetic energy of orderlymotion in other
7. The radius ofa metal sphere at room temperature T is R and
the coefficient of linear expansion of the metal is a. The
sphere heated a little by a temperature DT so that its new
temperature is T + DT. The increase in the volume of the
sphere is approximately.
(a) 2 R T
p aD (b) 2
R T
p aD
(c) 3
4 / 3
R T
p aD (d) 3
4 R T
p aD
8. Asphere, a cubeanda thin circular plate, all of samematerial
andsame mass are initiallyheatedtosame high temperature.
(a) Plate will cool fastest and cube the slowest
(b) Sphere will cool fastest and cube the slowest
(c) Plate will cool fastest and sphere the slowest
(d) Cube will cool fastest and plate the slowest
PastYears (2013-2017) NEET/AIPMTQuestions
9. A piece of iron is heated in a flame. It first becomes dull red
then becomes reddish yellow and finallyturns to white hot.
The correct explanation for the above observation is
possible by using [2013]
(a) Wien’s displacement law
(b) Kirchoff’s law
(c) Newton’s law of cooling
(d) Stefan’s law
10. The density of water at 20°C is 998 kg/m3 and at 40°C 992
kg/m3. The coefficient of volume expansion of water is
[NEET Kar. 2013]
(a) 10–4/°C (b) 3 × 10–4/°C
(c) 2 × 10–4/°C (d) 6 × 10–4/°C](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-299-320.jpg)
![296 PHYSICS
11. Two metal rods 1 and 2 of same lengths have same
temperature difference between their ends. Their thermal
conductivitiesareK1 and K2 and cross sectional areas A1
and A2, respectively. If the rate of heat conduction in rod 1
is four times that in rod 2, then [NEET Kar. 2013]
(a) K1A1 = K2A2 (b) K1A1 = 4K2A2
(c) K1A1 = 2K2A2 (d) 4K1A1 = K2A2
12. Certain quantityofwater cools from 70°Cto60°C in the first
5minutes and to54°C in thenext5minutes. Thetemperature
of the surroundings is: [2014]
(a) 45°C (b) 20°C
(c) 42°C (d) 10°C
13. Steam at 100°C is passed into 20 g of water at 10°C. When
water acquires a temperature of 80°C, the mass of water
present will be: [2014]
[Take specific heat ofwater = 1 cal g– 1 °C– 1 and latent heat
of steam = 540 cal g– 1]
(a) 24 g (b) 31.5g
(c) 42.5g (d) 22.5g
14. On observing light from three different stars P, Q and R, it
was found that intensityof violet colour is maximum in the
spectrum of P, the intensity of green colour is maximum in
thespectrum of Rand the intensityof red colour ismaximum
in the spectrum of Q. If TP, TQ and TR are the respective
absolute temperature ofP, Q and R, then it can beconcluded
from the above observations that [2015]
(a) TP > TR > TQ (b) TP < TR < TQ
(c) TP < TQ < TR (d) TP > TQ > TR
15. The two ends of a metal rod are maintained at temperatures
100°C and 110°C. The rate of heat flowin the rod is found to
be 4.0 J/s. If the ends are maintained at temperatures 200°C
and 210°C, the rate of heat flow will be [2015]
(a) 16.8 J/s (b) 8.0 J/s
(c) 4.0 J/s (d) 44.0 J/s
16. The value of coefficient of volume expansion of glycerine is
5 × 10-4 K-1. Thefractional changein thedensityofglycerine
for a rise of 40°C in its temperature, is: [2015 RS]
(a) 0.020 (b) 0.025
(c) 0.010 (d) 0.015
17. Apiece of ice falls from a height h sothat it melts completely.
Only one-quarter of the heat produced is absorbed by the
ice and all energy of ice gets converted into heat during its
fall. The value ofh is : [2016]
[Latent heat ofice is 3.4 × 105 J/kg and g = 10 N/kg]
(a) 34km (b) 544km
(c) 136km (d) 68km
18. Coefficient of linear expansion of brass and steel rods are
a1 and a2. Lengths of brass and steel rods are 1
l and 2
l
respectively. If 2 1
( )
l l
- is maintained same at all
temperatures, which one of the following relations holds
good ? [2016]
(a) 2 2
1 2 2 1
l l
a =a (b) 1 2 2 1
l l
a =a
(c) 1 2 2 1
l l
a =a (d) 1 1 2 2
a = a
l l
19. A black body is at a temperature of 5760 K. The energy of
radiation emitted by the body at wavelength 250 nm is U1,
at wavelength 500 nm is U2 and that at 1000nmisU3.Wien's
constant, b = 2.88 × 106 nmK. Which of the following is
correct ? [2016]
(a) U1 = 0 (b) U3 = 0
(c) U1 > U2 (d) U2 > U1
20. A spherical black body with a radius of 12 cm radiates 450
watt power at 500 K. If the radius were halved and the
temperature doubled, the power radiated in watt would be :
(a) 450 (b) 1000 [2017]
(c) 1800 (d) 225
21. Two rods Aand B of different materials are welded together
as shown in figure. Their thermal conductivities are K1 and
K2. The thermal conductivityof the composite rod will be :
A K1
B K2
T1 T2
d
[2017]
(a)
1 2
3(K K )
2
+
(b) K1 + K2
(c) 2 (K1 + K2) (d) 1 2
K K
2
+](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-300-320.jpg)
![297
Thermal Properties of Matter
EXERCISE - 1
1. (c) 2. (d) 3. (d) 4. (c) 5. (b)
6. (c) 7. (a) 8. (d)
9. (a)
strain
stress
/
A
/
F
E =
D
=
l
l
where Dl=(l'–l) = lat so F = EAat
10. (a) 11. (c)
12. (b) Let the angle subtended by the arc formed be q. Then
2
1
1
2
r
r
r
or
r -
-
=
D
D
=
q
=
q
l
l
l
l
t
T
)
( 1
2 D
a
-
a
=
q
l
t
T
)
(
r
or 1
2 D
a
-
a
=
l
l
So,
T
)
(
t
r
1
2 D
a
-
a
=
13. (a) Initial rate of loss of heat
2
2
2
1
2
4
1
4
R
R
e
A
T
e
A
T
=
´
´
s
´
´
s
=
14. (b) Momentum of photon
E
c
=
Change in momentum
2E
c
=
= momentum transferred to the surface
(the photon will reflect with same magnitude of
momentum in opposite direction).
15. (c) We know
4
eT
E = T
ln
4
e
ln
E
ln +
=
Þ
negative
is
e
ln
1
e
because <
E
ln
T
ln
16. (c) The temperature is measured by the value of the
thermodynamic property of a substance i.e., the
propertywhich varies linearlywith the temperature.
17. (b) Temperature of stars can be determined by Wein’s
displacement law
=
lm constant
18. (c)
19. (b) We know that Q =
T T
R
H L
-
Also Thermal resistance, R=
l l
KA K r
=
p
2
Heat flowwill be maximum when thermal resistance is
minimum. From given option
(i) r = 2r0, l = 2l0
= =
R
K r K r
2
2 2
0
0
2
0
0
2
l l
p p
( )
(ii) r = 2r0, l = l0
= =
R
K r K r
l l
0
0
2
0
0
2
2 4
p p
( )
(iii) r = r0, l = 2l0
= =
R
K r K r
2 2
0
0
2
0
0
2
l l
p p
(iv) r = r0, l = l0
= =
R
K r K r
l l
0
0
2
0
0
2
p p
It is clear that for option (b) resistance is minimum,
hence heat flowwill bemaximum.
20. (c)
21. (a) In condition of weightlessness, convection is not
possible.
22. (a) eff 1 2
r r r
=
2 1 1 2 2 1
KA(T T ) K r r (T T )
dQ
dt L L
- p -
= =
23. (c) When a body is heated then relation between colours
and temperature is according to Prevost’s theory of
radiation which states that everybody emitting radiant
energyin all directions at a rate depending only on the
nature of its surface and its temperature e.g., when a
bodyis placed in an enclosure(furnace) it would acquire
the temperature of furnace and seem white means
radiate white light. So it becomes first dark and then
white.
24. (c) 1 2
kA (T T )
dQ
dt L
-
=
[(T1–T2) is the temperature difference]
25. (b)
EXERCISE - 2
1. (c) )
t
1
(
R
R 0
t a
+
=
)
10
1
(
R
71
.
2 0 ´
a
+
= ...(1)
)
100
1
(
R
70
.
3 0 ´
a
+
= ...(2)
)
t
1
(
R
26
.
3 0 a
+
= ...(3)
Solve these equations to obtain the value of t.
2. (a)
5
C
9
32
140
=
-
or C
9
160
700
=
-
or C = 60ºC
3. (c) t
T
2
1
t ´
D
a
=
D
Hints & Solutions](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-301-320.jpg)
![298 PHYSICS
86400
)
15
T
(
2
1
5 ´
-
a
=
and 86400
)
T
30
(
2
1
10 ´
-
a
=
4. (a) a
µ
a
= F
or
A
t
Y
F
(Q Y t A is same for both copper and iron)
or C C Ι Ι
F α and F α
µ µ
2
3
1
2
/
3
F
FC
=
=
I
5. (b)
Δ
αΔT
=
l
l
5 3
10 100 10
- -
= ´ =
3
Δ
100% 10 100
-
´ = ´
l
l
1
10 0.1%
-
= =
6. (a) Let m be the mass of steam condensed. Then
m×540+m×10/2=22 ×70
m= 2.83gm
Now, total mass = 22 + 2.83 = 24.83 gm
7. (d) Thermal resistance R =l/KA
Where l =20cm. & A(cylindrical rod)
=pr2 =40pcm2
So = =
´ p ´
20 cal
R 0.318
0.5 40 sec º C
8. (c) According to Stefan’s Law, the rate of loss of heat is
e
)
T
T
(
A
t
Q 4
2
4
1 ´
-
s
=
here s = 5.67 × 10–8J/m2 × sec.K2,
T1=527+273=800K,
T2 = 27+ 273 = 300K& A= 200×10–4m2
So. 8 2
4 4
Q
5.67 10 2 10
t
[(800) (300) ] 0.4
- -
= ´ ´ ´
- ´
@ 182 joule
9. (a) According to Stefan’s Law
energy radiated per sec E = sAT4
(here e = 1 for black body)
for first case E = sA(273)4
for second case E1=sA(546)4 so E1=16E
10. (b) According to Wein's displacement law, λ T = b.
m
where b=2.884 × 10–3 mK.
So l = l
m m
1 2
1 2
T T
or 4800 × 6000 = l ´
m2
3000 or l = °
m2
9600A
11. (b) The time of cooling increases as the difference between
the temperature of body& surrounding is reduced. So
T1 < T2 < T3 (according to Newton’s Law of cooling).
12. (c) 5 5
1 2
11.0 10 T 5.5 10 T
- -
´ = ´
2
1
T
T
2
1
=
13. (c)
14. (c)
5
80
9
32
F
=
-
F – 32 = 144 or F = 176ºF.
15. (d) Percentage change in volume = gt × 100
t
300
100
t
3 a
=
´
a
= 2
.
1
200
10
2
300 5
=
´
´
´
= -
16. (b)
17. (c) According to Stefan's Boltzmann law, the energy
radiated per unit time E = sAT4.
5.67 × 104 = 5.67 × 10–8 × 1 × T4 ( A = 1 m2)
or,
1/ 4
4
–8
5.67 10
T 1000K
5.67 10
æ ö
´
= =
ç ÷
´
è ø
or , T = 1000 – 273 = 727°C.
18. (a) When the temperature of cold junction is increased,
the neutral temperature remains constant for a given
thermocouple and it is independent of the temperature
of the cold junction.
qn = 275°C, qi – qn = qn – qc
qi = 2qn – qc = 530°C.
19. (a) According to Newton's law of cooling,
1 2 1 2
0
K
t 2
q - q q + q
é ù
= - q
ê ú
ë û
where q0 is the surrounding temperature.
60 40 60 40
K 10
7 2
- +
æ ö
= -
ç ÷
è ø
Þ
20
40K
7
= Þ
1
K
14
=
40 28 40 28
K 10
t 2
- +
é ù
= -
ê ú
ë û
Þ
12
24K
t
=
or
12 12 14
t 7min
24K 24
´
= = =
20. (a) Let q be temperature of interface.
Q
P
K A( 0)
K A(100 ) q -
- q
=
l l
P
Q
K
K 100
q
=
- q Þ
2
3 100
q
=
-q
or, 200 – 2q = 3q or, 5q= 200 or, q = 40°C
21. (a) Let the final temperature of mixture be T.
Then 100 × 80 + 100 (T – 0) ×½
(as specific heat of ice is 0.5 cal/g C° and specific heat
ofwater is 1 cal/gC°)
= 100 × 1 ×(100 – T)
Solving, we get T = 13.33ºC.
22. (a) 3
2
1 a
+
a
+
a
=
g
7
7
7
10
231
10
231
10
13 -
-
-
´
+
´
+
´
=
7
10
475 -
´
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-302-320.jpg)
![299
Thermal Properties of Matter
23. (a) T
)
(
V
V g
m
0 D
g
-
g
=
D
100
)]
10
10
(
3
10
82
.
1
[
1 6
4 -
-
´
´
-
´
=
100
)]
10
3
.
0
10
82
.
1
[
1 4
4 -
-
´
-
´
= =15.2ml
24. (c) Given Δ / 0.10% 0.001 and ΔT 100º C
= = =
l l
Now
Δ
αΔT
=
l
l
or 0.001 α 100
= ´
or 5
α 10 /º C
-
=
Further 5
γ 3α 3 10 /ºC
-
= = ´
3
Δ V
100 (3 10 ) (100) 0.30%
V
-
´ = ´ =
25. (a) We know that real apparent vessel
γ γ γ
= +
So, app vessel glass app vessel steel
(γ γ ) (γ γ )
+ = +
(Q real
g is same in both cases)
or 6
vessel glass
153 10 (γ )
-
´ +
steel
vessel
6
)
(
10
144 g
+
´
= -
Further vessel steel
(γ )
6 6
3 3 (12 10 ) 36 10 /ºC
- -
= a = ´ ´ = ´
6
vessel glass
153 10 (γ )
-
´ +
6
6
10
36
10
144 -
-
´
+
´
=
Solving we get 6
vessel glass
( ) 27 10 /ºC
-
g = ´
or
glass 6
9 10 /ºC
3
-
g
a = = ´
26. (b) Let the final temperature be T.
Then 200 × 1 × (T – 20)+ 20× (T– 20)
= 440 (92 – T)
Solving it, we get T = 68ºC.
27. (a) )
80
100
(
m
mL -
+
)
15
80
(
02
.
0
)
15
–
80
(
1
1
.
1 -
´
+
´
´
=
30
.
1
5
.
71
m
20
540
m +
=
+
´
80
.
72
m
560 = 130
.
0
m =
28. (d) h
mg
mgh
W
W
W 2
1 ¢
-
=
-
= )
h
h
(
mg ¢
-
=
8
.
19
10
5
)
2
.
0
20
(
10
5 ´
´
=
-
´
=
joule
990
198
5 =
´
=
This energyis converted intoheat when the ball strikes
the earth. Heat produced is
calorie
2
.
4
990
Q =
C
º
32
11
09
.
0
5000
42
100
99
mc
Q
T =
´
´
´
=
=
D
29. (b) 80× 1 × 30= m × 80 Þ m = 30gm.
30. (c)
31. (a) Use 2
2
2
1
2
1
K
K
l
l
=
32. (b) Here flow of heat, area of vessels & temperature
gradient are same so
1 1 2 2
dT dT
Q K A t K A t
dx dx
= =
so
4
7
20
35
t
t
K
K
1
2
2
1
=
=
=
33. (b)
34. (b) 1 1 2 2
λ T λ T
=
2
T
.
9506
4753
6050 =
´
K
3025
T2 =
35. (a)
)
T
T
(
)
T
T
(
E
E
4
0
4
2
4
0
4
1
2
1
-
s
-
s
=
4
4
4
4
)
300
(
)
500
(
)
300
(
)
600
(
-
-
=
36. (d) Thermal capacity= 40 × 0.2 = 8 cal °C–1
37. (b) –13°F is (13 + 32)° below ice point on F scale.
38 (a) Mass ratio 1 : 2; hence DT ratio 2 : 1. Therefore
equilibriumtemp is30°C.
39. (b) Heat capacityof cold : hot =1 : 2
Sofinal temp. is
3
1
× (2 × 92 + 1 × 20) = 68°C.
40. (d) 4
T
area
E ´
´
s
= ; T increases by a factor
2
3
.
Area increases by a factor
4
1
.
41. (d)
2
t , t '
A A / 2
µ µ
l l
42. (a) q
=
q
-
Þ
-
q
=
q
- 4
400
)
0
(
K
)
100
(
K
4
80 C
Þ q = °
43. (a) 540 ×80 +540q = 540 (80–q)
°
=
q
Þ
=
q
Þ
q
-
=
q
+
Þ 0
0
2
80
80
44. (a)
45. (c) The total power radiated by a black body of area A at
temperature TK is given byP = AsT4
Where s = Stefan's constant
= 5.7 × 10–8 W m–2 K–4
Which is best represented in graph. (c)
46. (b) According to Stefan’s law 4
E T ,
= s
T1 = 500K
T2 = 1000 K
4 4
2 2
1 1
E T 1000
16
E T 500
æ ö æ ö
= = =
ç ÷
ç ÷ è ø
è ø
E2 = 16 × 7 = 112 cal / cm2s
47. (a) The rate of heat loss per unit area due to radiation
= Îs (T4–T0
4)
= 0.6 × 5.67 × 10–8 [(400)4–(300)4] = 595 Jm–2s–1.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-303-320.jpg)
![301
Thermal Properties of Matter
5. (a) As the temperature increased the length (L) of the
pendulum increases due to expansion i.e., linear.
Pendulum
L
So, Time period of pendulum
2
L
T
g
= p
T L
µ
Hence on increasing temperature, time period (T) also
increases.
6. (a) As we know that when the temperature increases
vibration of molecules about their mean position
increases hence the kinetic energy associated with
random motion of molecules increases.
7. (d) Let the radius of the sphere is R. As the temperature
increases radius of the sphere increases as shown.
R
dV
Original volume
3
4
3
V R
= p
Coefficient of linear expansion = a
Coefficient of volume expansion = 3a
As we know that
V
Y
V t
D
=
D
By putting the value of V, increase in the volume
3
V V t
Þ D = aD 3
4 R t
= p aD
8. (c) Loss of heat temperature on cooling temperature
increase depend on material of object surface area
exposed to surrounding and temperature difference
between body and surrounding.
Let us consider the diagram where all the three objects
are heated to same temperature T. As we know that
density
mass
volume
r =
where r is same for all the three objects hence, volume
will also be same.
m
T
Plate
Cube
Sphere
T
T
m
m
As thickness of the plate is least so, surface area of the
plateismaximum.
We know that, according to Stefan's law of heat loss
4
H AT
µ
where, A is surface area of object and T is temperature.
So, Hsphere : Hcube : Hplate
= Asphere : Acube : Aplate
Soarea ofcircular plate ismaximum.
For sphere, as the sphere is having minimum surface
area.
Hence, the sphere cools slowest and circular platewill
cool faster.
PastYears (2013-2017) NEET/AIPMTQuestions
9. (a) Wein’s displacement law
According to this law
lmax µ
1
T
or, lmax × T = constant
So, as the temperature increases l decreases.
10. (b) From question,
Dr = (998– 992) kg/m3 = 6 kg/m3
r =
3 3
998 992
kg/m 995 kg/m
2
+
=
r =
m
V
Þ
V
V
Dr D
= -
r
Þ
V
V
Dr D
=
r
Coefficient of volume expansion of water,
1 1
V
V t t
D Dr
=
D r D
=
4
6
3 10 / C
995 20
-
» ´ °
´
11. (b) Q1 = 4Q2 (Given)
Þ 1 1 2 2
4
K A t K A t
L L
D D
= Þ K1A1 = 4K2A2.
12. (a) Let the temperature of surroundings be q0
By Newton's law of cooling
1 2 1 2
0
k
t 2
q - q q + q
é ù
= - q
ê ú
ë û
Þ 0
70 60 70 60
k
5 2
- +
é ù
= - q
ê ú
ë û
Þ 2 = k [65 – q0] ...(i)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-305-320.jpg)
![302 PHYSICS
Similarly, 0
60 54 60 54
k
5 2
- +
é ù
= - q
ê ú
ë û
Þ
6
5
= k [57 – q0] ...(ii)
Bydividing (i) by(ii) we have
0
0
65
10
6 57
- q
=
- q
Þ q0 = 45º
13. (d) According to the principle of calorimetry.
Heat lost = Heat gained
mLv + mswDq = mwswDq
Þ m×540+m×1×(100–80)
=20 ×1×(80 –10)
Þ m= 2.5g
Therefore total mass of water at 80°C
=(20 +2.5) g= 22.5g
14. (a) From Wein’s displacement law
lm × T = constant
P – max. intensityis at violet
Þlm isminimumÞtempmaximum
R – max. intensityis at green
Þ lm is moderate Þ temp moderate
Q – max. intensityis at red Þ lm is maximum Þtemp.
minimum i.e., Tp > TR > TQ
15. (c) As the temperature difference DT = 10°C as well as the
thermal resistance is same for both the cases, sothermal
current or rate of heat flow will also be same for both
the cases.
16. (a) From question,
Rise in temperature Dt = 40°C
Fractional change in the density
0
Dr
r
= ?
Coefficient of volume expansion
g = 5 × 10–4K–1
r = r0 (1 –gDt)
Þ
0
Dr
r
= gDT = (5 ×10–4) (40) = 0.02
17. (c) According to question only one-quarter of the heat
produced by falling piece of ice is absorbed in the
meltingof ice.
i.e.,
mgh
4
=mL
Þ h =
5
4L 4 3.4 10
136 km
g 10
´ ´
= = .
18. (d) From question, (l2 – l1) is maintained same at all
temperatures hence change in length for both rods
should be same
i.e., Dl1 = Dl2
As we know, coefficient of linear expansion,
a =
0 T
D
D
l
l
l1a1DT = l2a2DT
l1a1 = l2a2
19. (d) Accordingtowein'sdisplacementlaw, maximum amount
of emitted radiation corresponding to lm =
b
T
lm =
6
2.88 10 nmK
5760K
´
= 500 nm
U2
U
Emitted
(radiation)
250 nm
500 nm
1000 nm
wave length K
From the graph U1 < U2 > U3
20. (c) Given r1 = 12 cm, r2 =6 cm
T1 = 500 K and T2 = 2 × 500 = 1000 K
P1 = 450 watt
Rate of power loss 2 4
P r T
µ
2 4
1 1 1
2 4
2 2 2
P r T
P r T
=
2 4
2 2
2 1 2 4
1 1
r T
P P
r T
=
Solving we get, P2 = 1800 watt
21. (d) Heat current H = H1 + H2
= 1 1 2 2 1 2
K A(T T ) K A(T T )
d d
- -
+
EQ 1 2 1 2
1 2
K 2A(T T ) A(T T )
[K K ]
d d
- -
= +
Hence equivalent thermal conductivities for two rods
of equal area is given by 1 2
2
EQ
k k
K
+
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-306-320.jpg)
![308 PHYSICS
Example4.
When a system is taken from state a to state b, in fig. along
the path a ® c ® b, 60 J of heat flow into the system, and
30 J of work are done :
(i) How much heat flows into the system along the path
a ® d ® b if the work is 10 J.
(ii) When the system is returned from b to a along the
curved path, the work done by the system is –20 J.
Does the system absorb or liberate heat, and how
much?
(iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the
process a ® d and d ® b.
P
V
c
a
b
d
Solution :
For the path a ® c ® b
dU = dQ – dW = 60 – 30 = 30 J or Ub – Ua = 30 J
(i) Along the path a ® d ® b
dQ = dU+ dW = 30 + 10 = 40 J
(ii) Along the curved path b – a
dQ = (Ua – Ub) + W = (–30) + (–20) = –50 J,
heat flows out the system
(iii) Qad = 32 J; Qdb = 8 J
Example5.
Two samples of a gas initially at same temperature and
pressure are compressed from avolume V to V/2. One sample
is compressed isothermally and the other adiabatically. In
which sample is the pressure greater?
Solution :
Let initial volume, V1 = Vand final volume, V2 = V/2
Initial pressure, P1 = P ; final pressure, P2 = ?
For isothermal compression
P2V2 = P1V1 or 1 1
2
2
P V PV
P 2P
V V / 2
= = =
For adiabatic compression
P2' V2
g = P1V1
g or P2' = P1
1
2
V
V
g
æ ö
ç ÷
è ø
=
V
P
V / 2
g
æ ö
ç ÷
è ø
or P2´ = 2gP
Since g > 1 2g > 2 P2' > P2
Pressure during adiabatic compression is greater than the
pressure during isothermal compression.
Example6.
A Carnot engine working between 300 K and 600 K has a
work output of 800 J per cycle. What is the amount of heat
energy supplied to the engine from source per cycle?
Solution :
W = 800 J, T1 = 600 K, T2 = 300 K
h = 1 – 1
2 1
T W
T Q
= =
1
300 800
1
600 Q
- = or 0.5 =
1
800
Q
Heat energy supplied by source,
800
Q
0.5
= = 1600 joule per cycle
Example7.
The temperatures T1 and T2 of the two heat reservoirs in
an ideal Carnot engine are 1500°C and 500°C respectively.
Which of the following : increasing T1 by 100°C or
decreasing T2 by 100°C would result in a greater
improvement in the efficiency of the engine?
Solution :
The efficiency of a Carnot's engine is given by 2
1
T
1
T
h = -
Given T1 =1500°C =1500+ 273= 1773Kand
T2 = 500°C = 500 + 273 = 773 K.
When the temperature of the source is increased by100°C,
keeping T2 unchanged, the new temperature of the source
is
T´1 =1500+100=1600°C =1873K.Theefficiencybecomes
2
1
T 773
´ 1 1 0.59
T´ 1873
h = - = - =
On the other hand, if the temperature ofthe sink isdecreased
by 100°C, keeping T1 unchanged, the new temperature of
the sink is T´2 = 500 – 100 = 400°C = 673 K. The efficiency
now becomes
2
1
T´ 673
´´ 1 1 0.62
T 1773
h = - = - =
Since h´´ is greater than h´, decreasing the temperature of
the sink by 100°C results in a greater efficiency than
increasing the temperature of the source by100°C.
Example8.
Calculate the work done when 1 mole of a perfect gas is
compressed adiabatically. The initial pressure and volume
of the gas are 105 N/m2 and 6 litre respectively. The final
volume of the gas is 2 litres. Molar specific heat of the gas
at constant volume is 3R/2. [(3)5/3 = 6.19]
Solution :
For an adiabatic change PVg = constant
P1V1
g = P2V2
g](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-312-320.jpg)
![309
Thermodynamics
As molar specific heat of gas at constant volume
v
3
C R
2
=
CP = CV + R =
3 5
R R R
2 2
+ = ;
P
V
C (5 / 2) R 5
C (3/ 2) R 3
g = = =
From eqn. (1) 1
2 1
2
V
P P
V
g
æ ö
= ç ÷
è ø
=
5/3
5 2
6
10 N / m
2
æ ö
´
ç ÷
è ø
= (3)5/3 × 105 = 6.19 ×105 N/m2
Work done =
5 3 5 3
1
[6.19 10 2 10 10 6 10 ]
5
1
3
- - -
´ ´ ´ - ´ ´
æ ö
- ç ÷
è ø
2
2 10 3
(6.19 3)
2
é ù
´ ´
= - -
ê ú
ê ú
ë û
= – 3 × 102 × 3.19 = – 957 joule
[–ve sign shows external work done on the gas]
Example9.
A refrigerator is to maintain eatables kept inside at 90C. If
room temperature is 360C, calculate the coefficient of
performance.
Solution :
Here, T1 =36°C =36 +273= 309K,
T2 =10°C =10 +273 =283K
2
1 2
T 283 283
COP 10.9
T T 309 283 26
= = = =
- -
Example10.
One mole of an ideal gas at pressure P0 and temperature
T0 is expanded isothermally to twice its volume and then
compressed at constant pressure to (V0/2) and the gas is
brought back to original state by a process in which P µ V
(Pressure is directly proportional to volume). The correct
representation of process is
(a)
V
P
V /2
0 V0 2V0
P0 (b)
V
P
V /2
0 V0 2V0
P0
(c)
P0
T
P
P0/2
T0
T /4
0
(d)
T
V
T0
V /2
0
V0
Solution : (c)
ProcessAB is isothermal expansion,
BC is isobaric compression and in process CA
2
nRT
P P T
P
µ Þ µ
P0
A
P
P0/2
T0
T /4
0
B
C
Example11.
A Carnot’s heat engine works with an ideal monatomic
gas, and an adiabatic expansion ratio 2. Determine its
efficiency.
Solution :
Given,
3
2
V
2
V
r = = and g for a monatomicgas = 5/3.
Using,
1
1
1
g -
æ ö
h = -ç ÷
r
è ø
we have, the required efficiency
5
1
3
1
1 1 0.63 0.37 or 37%
2
-
æ ö
h = - = - =
ç ÷
è ø](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-313-320.jpg)
![317
Thermodynamics
6. Three copper blocks of masses M1, M2 and M3 kg
respectivelyare brought intothermal contact till theyreach
equilibrium. Before contact, theywere at T1, T2, T3 (T1 > T2
> T3). Assuming there is no heat loss to the surroundings,
theequilibrium temperature Tis (s is specific heatofcopper)
(a) 1 2 3
3
T T T
T
+ +
=
(b) 1 1 2 2 3 3
1 2 3
M T M T M T
T
M M M
+ +
=
+ +
(c) 1 1 2 2 3 3
1 2 3
3( )
M T M T M T
T
M M M
+ +
=
+ +
(d) 1 1 2 2 3 3
1 2 3
M T s M T s M T s
T
M M M
+ +
=
+ +
NEET/AIPMT (2013-2017) Questions
7. A gas is taken through the cycle A ® B® C ® A, as
shown in figure. What is the net work done by the gas ?
P (10 Pa)
5
V (10 m )
3 3
A
B
C
7
6
5
4
3
2
1
0 2 6
4 8
[2013]
(a) 1000J (b) zero
(c) – 2000 J (d) 2000J
8. During an adiabatic process, the pressure of a gas is found
to be proportional to the cube of its temperature. The ratio
of
p
v
C
C
for the gas is [2013]
(a) 2 (b)
5
3
(c)
3
2
(d)
4
3
9. A system is taken from state a to state c by two paths adc
and abc as shown in the figure. The internal energy at a is
Ua = 10 J. Along the path adc the amount of heat absorbed
dQ1 = 50 J and the work donedW1 = 20 J whereas along the
path abc the heat absorbed dQ2 = 36 J. Theamount ofwork
done along the path abc is [NEET Kar. 2013]
a b
c
d
p
V
(a) 6 J (b) 10J
(c) 12J (d) 36J
10. Which ofthe following relations does not give the equation
of an adiabatic process, where terms have their usual
meaning?
(a) PgT1–g = constant [NEET Kar. 2013]
(b) P1–g Tg = constant
(c) PVg = constant
(d) TVg–1 = constant
11. Two Carnot engines A and B are operated in series. The
engineA receives heat from the source at temperature T1
and rejects the heat to the sink at temperature T. Thesecond
engine B receives the heat at temperature T and rejects to
its sink at temperature T2. For what value of T the
efficienciesofthe two enginesare equal? [NEET Kar. 2013]
(a) 1 2
2
T T
+
(b) 1 2
2
T T
-
(c) T1T2 (d) 1 2
T T
12. Athermodynamic system undergoes cyclic processABCDA
as shown in fig. The work done by the system in the cycle
is : [2014]
(a) P0V0
(b) 2P0V0
(c) 0 0
P V
2
P C B
D
A
V0 2V0 V
P0
2P0
3P0
(d) Zero
13. A monoatomic gas at a pressure P, having a volume V
expands isothermallyto a volume2V and then adiabatically
to a volume 16V. The final pressure of the gas is :(take g =
5
3
) [2014]
(a) 64P (b) 32P
(c)
P
64
(d) 16P
14. Figure below shows two paths that may be taken by a gas
to go from a state A to a state C.
2 × 10 m 4 × 10 m
–3 3 –3 3
2×10 Pa
4
6×10 Pa
4
P B C
A
V](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-321-320.jpg)
![318 PHYSICS
In process AB, 400 J of heat is added to the system and in
process BC, 100 J of heat is added to the system. The heat
absorbed by the system in the process AC will be [2015]
(a) 500J (b) 460J
(c) 300J (d) 380J
15. A carnot engine having an efficiency of
1
10
as heat engine,
is used as a refrigerator. If the work done on the system is
10 J, the amount of energy absorbed from the reservoir at
lower temperature is :- [ 2015, 2017]
(a) 90 J (b) 99 J
(c) 100 J (d) 1 J
16. The coefficient of performance of a refrigerator is 5. If the
insidetemperatureof freezer is–20°C, then thetemperature
of the surroundings to which it rejects heat is [2015 RS]
(a) 41°C (b) 11°C
(c) 21°C (d) 31°C
17. An ideal gas is compressed to half its initial volume by
means of several processes. Which of the process results
in the maximum work done on the gas? [2015 RS]
(a) Isobaric (b) Isochoric
(c) Isothermal (d) Adiabatic
18. A gas is compressed isothermallyto half its initial volume.
Thesame gas is compressedseparatelythrough an adiabatic
process until its volume is again reduced to half. Then :
[2016]
(a) Compressing the gas isothermally will require more
work to be done.
(b) Compressing the gas through adiabatic process will
require more work to be done.
(c) Compressing the gas isothermallyor adiabaticallywill
require the same amount of work.
(d) Which of the case (whether compression through
isothermal or through adiabatic process) requires more
work will depend upon the atomicity of the gas.
19. A refrigerator worksbetween 4°C and 30°C. It is required to
remove 600 calories of heat every second in order to keep
the temperature of the refrigerated space constant. The
power required is: (Take 1 cal = 4.2 joules) [2016]
(a) 2.365W (b) 23.65W
(c) 236.5W (d) 2365W
20. Thermodynamic processes are indicated in the following
diagram : [2017]
P
i
IV
f
III
f
f
f
I
700k
500k
300k V
II
Match the following
Column-1 Column-2
P. Process I A. Adiabatic
Q. Process II B. Isobaric
R. Process III C. Isochoric
S. Process IV D. Isothermal
(a) P ® C, Q ® A, R ® D, S ® B
(b) P ® C, Q ® D, R ® B, S ® A
(c) P ® D, Q ® B, R ® A, S ® C
(d) P ® A, Q ® C, R ® D, S ® B](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-322-320.jpg)
![320 PHYSICS
W
000
,
50
273
300
273
=
-
27 50,000
cal/min
273
W
´
=
4.2 27 50,000
Joule/sec
60 273
W
P
t
´ ´
= =
´
= 346 watt = 0.346 kW
4. (d) )
10
5
.
6
10
8
(
dW
dQ
dU 5
5
´
-
´
=
-
= J
10
5
.
1 5
´
=
5
5
10
5
.
1
10
dU
dQ
dW ´
-
==
-
=
5
0.5 10 J
= - ´
– ve sign indicates that work done on the gas is
J
10
5
.
0 5
´ .
5. (c) v v
dU nC dT or 80 5 C (120 100)
= = ´ -
Cv = 4.0 joule/K
6. (b) 0 0
4/3
V
4/3 4/3
P V = P P = P 8
1 1
8
æ ö Þ
ç ÷
è ø
= 16Po
7. (c)
373
23
100
373
123
373
250
1
T
T
1
1
2 +
=
=
-
=
h
Þ
-
=
h
8. (d)
T
V
)
100
/
90
(
P
T
V
P ¢
=
or
90
10
1
90
100
P
P
+
=
=
¢
or %
1
.
11
90
10
P
P
P
=
=
-
¢
9. (d) PVg =
g
÷
ø
ö
ç
è
æ
32
V
P1
or g
= = =
1.4
1
P (32) , P (32) , P 128P
10. (c) g
-
g
g
-
g
= 1
2
2
1
1
1 P
T
P
T
11. (d) 1
T 18ºC (273 18) 291 K
= = + =
and 8
/
V
V 1
2 =
We know that 1
V
T -
g = constant
or , 1
1
1
1
2
2 V
T
V
T -
g
-
g
=
1
4
.
1
1
2
1
1
2 )
8
(
291
V
V
T
T -
-
g
´
=
÷
÷
ø
ö
ç
ç
è
æ
=
C
º
5
.
395
K
5
.
668 =
=
12. (c) T
Pµ if V is constant, where P is pressure of certain
amount of gas and T is absolute zero temperature.
13. (d) We know that efficiency of carnot engine = I
T
T
L
H
-
Also, Efficiency of Heat engine =
Work output
Heat input
1
1
- =
æ ö
Þ = -
ç ÷
è ø
L
H S
S
H
T W
T Q
TL
W Q
T
4
4
127 273
6 10 1
227 273
1.2 10 cal
+
æ ö
= ´ -
ç ÷
è ø
+
= ´
14. (a) Work done in an isothermal process is given by
2
10
1
V
W 2.3026nRTlog
V
=
Here, n = 3, R = 8.31 J mol–1 degree–1
T = 300 K, V1 =4 litre, V2 = 1litre.
Hence, W = 2.3026 × 3 × 8.31 × 300 × 10
1
log
4
=17221.15(–2log102)
=–17221.15×2×0.3010=– 10368J.
15. (c) The speed of sound in air, v T
µ
1 1
2 2
v T
v T
=
Þ
1
1 2
v (27 273)
2v T
+
=
2
1 300
2 T
=
T2 =1200 K=(1200– 273)°C=927°C
16. (b) Joule
10
]
1
1671
[
10
01
.
1
V
P
dW 6
5 -
´
-
´
=
D
=
1.01 167
cal.
4.2
40cal. nearly
´
=
=
ΔQ mL 1 540,
ΔQ ΔW ΔU
= = ´
= +
or .
cal
500
40
540
U =
-
=
D
17. (b) 2
2
1 2
T 273 13 260,
T
K
T T
= - =
=
-
1
1
260
5
T 260
or T 260 52
=
-
- =
1
2
T 312K,
T 312 273 39ºC
=
= - =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-324-320.jpg)
![321
Thermodynamics
18. (b) p
5
dQ n C dT 1 R 100
2
= = ´ ´
J
10
8
.
20
J
5
.
2077 2
´
=
=
v
dU n C dT
3
1 R 100
2
1246.5J
=
= ´ ´
=
2
dU dQ dU
8.31 10 J
= -
= ´
19. (d) Under adiabatic change
g
g
-
÷
÷
ø
ö
ç
ç
è
æ
=
1
2
1
1
2
P
P
T
T
or g
g
-
=
1
2
1
1
2 )
P
/
P
(
T
T
air
for
5
/
7
4
.
1
;
)
1
/
4
(
300
T )
5
/
7
(
)
5
/
7
(
1
2 =
=
g
=
-
or 7
/
2
2 )
4
(
300
T -
=
20. (d) p
Q nC T
= D and W PΔV nRΔT.
= =
For monatomic gas,
2
R
5
Cp = .
W 2
Q 5
Þ =
21. (c)
1
2
1
2
1
2
1
2
T
T
Q
Q
T
T
1
Q
Q
1 =
Þ
-
=
-
=
h
So K
250
300
500
150
Q
T
Q
T
1
1
2
2 =
´
=
´
=
22. (d) Efficiency,
T2
= 1 –
T
1
η
T1 (source temp.) = 400 K
T2 (sink temp.) = 300 K
300 1
25%
400 4
= 1 – = =
h
23. (c)
)
1
4
.
1
(
2
)
1
4
.
1
(
4
V
.
T
V
)
300
(
-
-
÷
ø
ö
ç
è
æ
=
4
.
0
2 )
4
(
)
300
(
T =
24. (d) From C to D, V is constant. So process is isochoric.
From D to A, the curve represents constant
temperature. So the process is isothermal.
From A to B, pressure is constant . So, the process is
isobaric.
BC represents constant entropy.
25. (d) In cylinder A, heat is supplied at constant pressure
while in cylinder Bheat issupplied at constant volume.
(DQ)A =nCP(DT)A
and (DQ)B = nCV (DT)B
Given : (DQ)A =(DQ)B
(DT)B
P
A
V
C
( T)
C
1.4 30
42K
= D
= ´
=
[Q for diatomic gas P
V
C
1.4
C
= ]
26. (b)
2
1
T
1
T
273 0
1
273 200
200
473
h = -
+
= -
+
=
2
1
T
' 1
T
(273 200)
1
273 0
200
473
h = -
-
= -
+
=
200 273
' 473 200
0.577
h
= ´
h
=
27. (a)
250
300
250
W
750
-
=
Heat rejected = 750 + 150 = 900 cal.
28. (b)
1
2
1
T
T
T -
is maximum in case(b).
29. (c) For adiabatic process, dU = – 100 J
which remains same for other processes also.
Let C be the heat capacity of 2nd process then
–(C) 5 = dU + dW
= – 100+ 25 = – 75
C = 15J/K
30. (c) According to first law of thermodynamics
Q = DU + W
DU = Q – W
= 2 × 4.2 × 1000 – 500
= 8400 –500 = 7900 J
31. (a) PV3/2 = K](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-325-320.jpg)
![323
Thermodynamics
43. (a) Initial and final condition is same for all process
DU1 = DU2 = DU3
from first law of thermodynamics
DQ= DU + DW
Work done
DW1 > DW2 > DW3 (Area of P.V. graph)
So DQ1 > DQ2 > DQ3
44. (d) Efficiency
1
W
h =
q
and W = q1 – q2
1 2
1
–
q q
h =
q
= 2
1
1
q
-
q
45. (b) P1 > P2
P1
P2
T
V
q2
q1
As V = constant Þ Pµ T
Hence from V–T graph P1 > P2
46. (b) Change in internal energy do not depend upon the
path followed by the process. It onlydepends on initial
and final states i.e.,
DU1 = DU2
47. (a) For adiabatic process Q = 0.
Byfirst law of thermodynamics,
Q = DE + W
Þ DEint = – W.
48. (d) The efficiency of carnot's heat engine
2
1
T
1–
T
h =
where T2 is temperature ofsink, and T1 is temperature
of hot reservoir or source.
When efficiencyis 40% i.e. h = 40/100 =
2
1
T
1–
T
or
2
1
T
2
1– [T T(given)]
5 T
= =
T2=
3
T
5
Now, when efficiency is 50%
1
1
3
T
50 6
5
1– T T
100 5
T
h = = =
49. (a) 50. (c)
EXERCISE - 3
Exemplar Questions
1. (c) For the straight line in the graph denoted by 4, that
shows pressure is constant, so curve 4 represents an
isobaric process.
p
V
1
2
3
4
For the straight line in graph denoted by1, that shows
volume is constant, so curve 1 represents isochoric
process. Out of curves 3 and 2, curve 2 is steeper.
Hence, curve 2 is adiabatic and curve 3 is isothermal.
2. (a) As we know that amount of sweat evaporated/minute
Sweat produced/ minute
Number of calories required for evaporation/kg
=
Amount of heat produced per minute in jogging
Latent heat (in cal/kg)
=
580 × 103 calories are needed to convert
1 kg H2O into stream.
1 cal. will produce sweat = 1 kg/ 580 × 103
14.5 × 103 cal will produce (sweat)
3
3
14.5 10 145
kg= kg/m
580
580 10
´
=
´
=0.025kg.
3. (c) According to given P–V diagram that
pV = constant
So we can say that the gas is going through an
isothermal process.
Ifpressure (P)increase at constant temperature volume
V decreases, the graph (iii) shows that pressure (P) is
smaller at point 2 and larger at point 1 point sothe gas
expands and pressure decreases. Hence verifiesoption
(c).
4. (b) According to the given p-V diagram.
Work done in the process ABCD
= (AB) × BC = (3V0 – V0) × (2p0 – p0)
= 2V0 × p0 = 2p0V0
Here the direction of arrow is anti-clockwise, sowork
done is negative.
Hence, work done by the gas = –2p0V0
That shows external work done on the system.
5. (a) Let us consider the p-V diagram for container A
(isothermal) and for container B (adiabatic).](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-327-320.jpg)
![324 PHYSICS
2 2
1 1
p p
p0 p0
V0 V0
2V0 2V0
V V
Container
(Isothermal)
A Container
(Adiabatic)
B
In both process compression of the gas.
For isothermal process (gas A) during 1®2
1 1 2 2
p V p V
= 1 0 2 0
( 2 , )
V V V V
= =
Q
0 0 2 0
(2 ) ( )
p V p V
=
2 0
2
p p
=
For adiabatic process, (gas B)during (1®2)
1 2
1 2
p V p V
g g
= 1 0 2 0
( 2 , )
V V V V
= =
Q
0 0 2 0
(2 ) ( )
p V p V
g g
=
0
2 0 0
0
2
(2)
V
p p p
V
g
g
æ ö
= =
ç ÷
è ø
So, ratio of final pressure
1
0
2
1 0
(2)
( )
2
( ) 2
A
p
p
p p
g
g -
B
æ ö
= = =
ç ÷
è ø
where, g is ratio of specific heat capacities for the gas.
Hence, verifies the option (a).
6. (b) Consider the equilibrium temperature ofthe system is
T.
Let us consider, T1, T2 < T < T3.
As given that, there is no net loss to the surroundings.
Heat lost by M3 = Heat gained by M1
+ Heat gained by M2
3 3 1 1 2 2
( ) ( ) ( )
M s T T M s T T M s T T
- = - + -
3 3 3 1 1 1
M sT M sT M sT M sT
- = -
2 2 2
M sT M sT
+ -
(where, s is specified heat ofthe copper material)
1 2 3 3 3 1 1 2 2
[ ]
T M M M M T M T M T
+ + = + +
1 1 2 2 3 3
1 2 3
M T M T M T
T
M M M
+ +
=
+ +
NEET/AIPMT (2013-2017) Questions
7. (a) Wnet = Area of triangleABC
=
1
2
AC × BC
=
1
2
× 5 × 10–3 × 4 × 105 = 1000 J
8. (c) According to question P µ T3
But as we know for an adiabatic process the
pressure P µ 1
T
g
g -
.
So,
1
g
g -
= 3 Þ g =
3
2
or,
,
p
v
C
C
=
3
2
9. (a) From first law ofthermodynamics
Qadc = DUadc + Wadc
50J = DUadc + 20 J
DUadc = 30J
Again, Qabc = DUabc + Wabc
Wabc = Qabc – DUabc
= Qabc – DUadc
= 36 J – 30 J
= 6 J
10. (a) Adiabatic equations of state are
PVg = constant
TVg–1 = constant
P1–gTg = constant.
11. (d) Efficiency of engine A, 1
1
1 ,
T
T
h = -
Efficiencyof engine B, 2
2 1
T
T
h = -
Here, h1 = h2
2
1
T
T
T T
= Þ 1 2
T T T
=
12. (d) Work done by the system in the cycle
=Area under P-V curve andV-axis
= 0 0 0 0
1
(2P P )(2V V )
2
- - +
0 0 0 0
1
(3P 2P )(2V V )
2
é ù
æ ö
- - -
ç ÷
ê ú
è ø
ë û
=
0 0 0 0
P V P V
0
2 2
- =
13. (c) For isothermal process P1V1 = P2V2
Þ PV= P2(2V) Þ P2 =
P
2
For adiabatic process
2 3
2 3
P V P V
g g
=
Þ 3
P
(2v) P 16v)
2
g g
æ ö
=
ç ÷
è ø
Þ P3 =
5/3
3 1 P
2 8 64
æ ö
=
ç ÷
è ø](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-328-320.jpg)
![338 PHYSICS
NEET/AIPMT (2013-2017) Questions
9. In the given (V – T) diagram, what is the relation between
pressure P1 and P2 ? [2013]
(a) P2 > P1
(b) P2 < P1
T
V
P2
P1
q1
q2
(c) Cannot be predicted
(d) P2 = P1
10. The amount of heat energyrequired to raisethe temperature
of 1g ofHelium at NTP, from T1K toT2K is [2013]
(a)
3
2
NakB(T2 – T1)
(b)
3
4
NakB(T2 – T1)
(c)
3
4
NakB
2
1
T
T
(d)
3
8
NakB(T2 – T1)
11. In a vessel, the gas is at a pressure P. If the mass of all the
molecules is halved and their speed is doubled, then the
resultant pressure will be [NEET Kar. 2013]
(a) 4P (b) 2P
(c) P (d) P/2
12. The mean free path of molecules of a gas, (radius ‘r’) is
inversely proportional to : [2014]
(a) r3 (b) r2
(c) r (d) r
13. The ratio of the specific heats
p
v
C
C
= g in terms of degrees
of freedom (n) is given by [2015]
(a)
n
1
3
æ ö
+
ç ÷
è ø
(b)
2
1
n
æ ö
+
ç ÷
è ø
(c)
n
1
2
æ ö
+
ç ÷
è ø
(d)
1
1
n
æ ö
+
ç ÷
è ø
14. One mole of an ideal diatomic gas undergoes a transition
from Ato B along a path AB as shown in the figure.
P(in kPa)
V(in m )
3
2
5
A
B
4 6
The change in internal energy of the gas during the
transition is: [2015]
(a) – 20 kJ (b) 20J
(c) –12kJ (d) 20kJ
15. Two vessels separately contain two ideal gases A and B at
the same temperature. The pressure ofAbeing twice that of
B. Under such conditions, the density of A is found to be
1.5 times the densityofB. The ratio of molecular weight of
A and B is : [2015 RS]
(a)
3
4
(b) 2
(c)
1
2
(d)
2
3
16. 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat
capacity of the gas at constant volume is 5.0JK–1. If the
speed of sound in this gas at NTP is 952 ms–1, then the
heat capacity at constant pressure is (Take gas constant R
= 8.3 JK–1 mol–1) [2015 RS]
(a) 7.5 JK–1 mol–1
(b) 7.0 JK–1 mol–1
(c) 8.5 JK–1 mol–1
(d) 8.0 JK–1 mol–1
17. The molecules of a given mass of a gas have r.m.s. velocity
of 200 ms–1 at 27°C and 1.0 × 105 Nm–2 pressure. When the
temperature and pressure ofthe gas are respectively, 127°C
and 0.05 × 105 Nm–2, the r.m.s. velocity ofits molecules in
ms–1 is : [2016]
(a) 100 2 (b)
400
3
(c)
100 2
3
(d)
100
3
18. Agas mixture consists of 2 moles ofO2 and 4 moles ofAr at
temperature T. Neglecting all vibrational modes, the total
internal energy of the system is :- [2017]
(a) 15RT (b) 9 RT
(c) 11 RT (d) 4 RT](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-342-320.jpg)
![341
Kinetic Theory
20. (c) According to Dalton’s Law
P= P1 +P2.....................
here 1 litre of orygen at a pressure of1atm, & 2 litre of
nitrogen at a pressure of 5 atm are introduced in a
vessel of 1 litre capacity.
so 1 1 2 2
V P P V 1 1 2 .5
P 2atm
volumeof vessel 1
+ ´ + ´
= = =
21. (b) Relative humidity %
100
M
m
´
=
where m is the mass of vapour actually present in
certain volume of air & M is the mass required to
saturate the air fully of that volume
so %
75
100
20
15
H
.
R =
´
=
22. (c) Bydefinition
Þ
100
60
;
100
.
temp
room
at
P
.
V
.
S
.
temp
room
at
P
.
V
H
.
R ´
=
15
(V.P.)
12.67
=
(V.P)15 = 7.6mm of Hg.
Now since unsaturated vapour obeys gas equation &
mass of water remains constant so
nRT
P PαT
V
æ ö
= Þ
ç ÷
è ø
15
20
20
(V.P) 273 15
(V.P)
(V.P) 273 20
+
= Þ
+
=7.73 mm ofHg
So
20
20
20
(V.P)
(R.H) 100 44.5%
(S.V.P)
= ´ =
23. (a)
273
327
273
t
)
v
(
)
v
(
1
rms
2
rms
+
+
= or
1 273
2 600
t +
=
or t = – 123ºC
24. (a)
1
1
1
T
V
P
= constant for one mole
&
1
1
T
P
= constant for constant volume
so 1 1
1
1 1
P 1.004P
T 250K
T (T 1)
= Þ =
+
25. (d) Total K.E. ofAtype molecules 2
m
2
3
w
=
Total K.E. ofAtype of molecule is
[ ]
2
z
ms
.
r
2
y
ms
.
r
2
x
ms
.
r
A )
v
(
)
v
(
)
v
(
m
2
1
E
.
K +
+
=
but w
=
x
ms
.
r )
v
(
so w
=
= z
ms
.
r
y
ms
.
r )
v
(
)
v
(
Total K.E. of B type molecules
2 2
1
2 m v m v
2
= ´ =
Now, 2 2
3
m ω m v
2
´ = or
2 2 2
( ω / v )
3
=
26. (d) rms
v T
µ
At 0ºC, rms
v 273
µ
At temp. T, rms
v
T
2
µ
K
69
~
K
2
.
68
4
273
T
or
273
T
2
1
-
=
=
=
C
º
204
273
69
T -
=
-
=
27. (c) For a monatomic gas
R
2
3
Cv =
So correct graph is
v
C
R
2
/
3
®
T
28. (d) Molar volume = 22400 cm3
Bubble volume =
3
3
4 d 4
10
3 8 3
-
p = p
No. of molecules
23 3
17
6 10 4 10
10
22400 3
-
´ ´ p
= »
´
29. (d) Molecule number ratio is
3
1
:
3
2
O
:
H 2
2 = .
That gives
2 1
2
(c ) 16 1
rms
3 3
= +
æ ö æ ö
è ø è ø
times the
value for O2.
30. (b) The Vander waal’s equation of state is
RT
)
b
V
(
V
a
P
2
=
-
÷
÷
ø
ö
ç
ç
è
æ
+ or 2
V
a
b
V
RT
P -
-
=
At the critical point, P = Pc, V = Vc, and T = Tc
2
c
c
c
c
V
a
b
V
RT
P -
-
= ....(i)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-345-320.jpg)
![342 PHYSICS
At the critical point on the isothermal, 0
dV
dP
c
c
=
c
2 3
c c
RT 2a
0
(V b) V
-
= +
-
or 3
c
2
c
c
V
a
2
)
b
V
(
RT
=
-
....(ii)
Also at critical point, 0
dV
P
d
2
c
c
2
=
4
c
3
c
c
V
a
6
)
b
V
(
RT
2
0 -
-
= or
c
3 4
c c
2RT 6a
(V b) V
=
-
..(iii)
Dividing (ii) by (iii), weget
c
c V
3
1
)
b
V
(
2
1
=
- or Vc = 3 b ....(iv)
Putting this value in (ii), we get
2 3
RTc 2a
4b 27 b
= or
R
b
27
a
8
Tc = ...(v)
Putting the values of Vc and Tc in (i), we get
2
2
c
b
27
a
b
9
a
R
b
27
a
8
b
2
R
P =
-
÷
÷
ø
ö
ç
ç
è
æ
=
31. (c) g for resulting mixture should be in between 7/5 and
5/3.
32 (b) For 1 molecule of a gas, vrms =
3KT
m
where m is the mass of one molecule
For N molecule of a gas, v1= 3KT N
m
´
For 2N molecule of a gas v2 =
3KI N
N
(2m)
´
´
1
2
v
1
v
=
33. (a) Since VP2 = constant,
2 '2
VP 2VP
=
P
P'
2
=
As
P
constant or Tα P,
T
= thusT becomes T/Ö2
34. (a) We know that, ÷
ø
ö
ç
è
æ g
=
M
RT
Crms
Here 1
2
rms 2
rms
rms 1
C m
1
C ;
M C m
æ ö
µ = ç ÷
è ø
35. (b) 3 12 2 13
dU d M N 6M 12N
F
dr dr r R r r
-
é ù é ù
= = - - = - +
ê ú ê ú
ë û ë û
In equilibrium position, F = 0
2 13
6M 12N
0
r r
- = or,
6 2N
r
M
=
Potential energyat equilibrium position
2 2 2
2
M N M M M
U
(2N / M) 2N 4N 4N
(2N / M)
= = = - =
36. (b) 1 1 2 2
1 2
P V P V
T T
= Here, P1 =200kPa
T1 =22°C =295K T2 =42°C =315K
V2 1 1 1
2
V V 1.02V
100
= + =
1
2
1
200 315V
P
295 1.02V
´
=
´
209.37kPa
=
37. (a) According to given Vander Waal's equation
2
2
nRT n
P
V n V
a
= -
- b
Workdone,
2 2 2
1 1 1
2
2
V V V
V V V
dV dV
W PdV nRT n
V n V
= = - a
- b
ò ò ò
[ ]
2
2
1
1
2 1
log ( )
V
V
e V
V
nRT V n n
V
é ù
= - b + a ê ú
ë û
2
2 1 2
1 1 2
loge
V n V V
nRT n
V n V V
æ ö é ù
- b -
= + a
ç ÷ ê ú
- b
è ø ë û
38. (d) Internal energyof 2 moles of oxygen
2
5 5
Uo µ RT 2. RT 5RT
2 2
æ ö
= = =
ç ÷
è ø
Internal energyof 4 moles ofArgon.
Ar
3 3
U µ RT 4. RT 6RT
2 2
æ ö
= = =
ç ÷
è ø
Total internal energy
2
O Ar
U U U 11RT
= + =
39. (c)
1 v 2 v
1 2
vmix
1 2
n C n C
C
n n
+
=
+
Þ
1 v 1 v
1 2
1 1
n C 2n C
13R
6 n 2n
+
=
+
1
2
n 1
n 2
é ù
=
ê ú
ë û
Q
Þ v v
1 2
13R
C 2C
2
= +](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-346-320.jpg)
![344 PHYSICS
EXERCISE - 3
Exemplar Questions
1. (b) As the relative velocity of molecule with respect to
the walls of container does not change in rocket, due
to the mass ofa molecule is negligible with respect to
the mass of whole system and system of gas moves
as a whole and (g = 0) on molecule energy where.
Hence pressure of the gas inside the vessel, as
observed by us, on the ground remain the same.
2. (d) Pressure on the wall due to force exerted bymolecule
on walls due to its rate of transfer of momentum to
wall.
In an ideal gas, when a molecule collides elastically
with a wall, themomentum transferredto each molecule
will betwice themagnitude ofits normal momentum is
2 mv. For the wall EFGH, absorbs those molecules.
Which strike toit so rate of change in momentum toit
become only mv so the pressure of EFGH would half
ofABCD.
3. (b) Boyle's law is applicable at constant temperature and
temperature remains constant in isothermal process.
For ideal gas, pV = nRT = constant
So, pV = constant (at constant temperature)
1
p
V
µ
So, this process can be called as isothermal process.
4. (c) Let us consider the given diagram where an ideal gas
is contained in a cylinder, having a piston of mass M.
The pressure on gas does not change.
M
p
pa
p
pa Mg/A
A
The pressure inside the gas will be
p = pa + Mg /A
where, pa = atmospheric pressure
A = area of cross-section of the piston.
Mg = weight of piston
Hence, p = constant.
As the piston and cylinder is frictionless so the
equation for ideal gas
pV = nRT, volume (V) increases at constant pressure.
as p, R, n are constant so,
V T
µ
so on increasing temperature of system its volume
increased but pwill remain constant.
5. (a) As we know that an ideal gas equation,
as the pressure and quantity of gas in system are
constant
nR
pV nRT V T
p
æ ö
= Þ = ç ÷
è ø
T
V
p
µ as n, R are constant
1
V
T p
æ ö
µ ç ÷
è ø
Slope of the V – T graph,
dV nR
m
dT p
= = [m = slope of V – t graph]
1
dV
dT p
µ or
1
m
p
µ [Q nR = constant]
So,
1
p
m
µ
hence, 1 2
2 1
=
p m
p m
where, m1 is slope of the graph corresponding to p1
and similarlym2 is slope corresponding to p2. Soslope
of p1
is smaller than p2
. Hence, (p1
> p2
).
6. (d) Pressure exerted by gas is due to rate of change of
momentum (p) imparted byparticles to wall.
When the molecules breaks into atoms, the number of
moles would become twice.
From ideal gas equation,
pV = nRT
where, p = Pressure of gas, n = Number of moles
R = Gas constant, T = temperature
As gases breaks number of moles becomes twice of
initial, so n2 = 2n1
So, p nT
µ
or
2 2 2 1
1 1 1 1
(2 )(3000)
20
(300)
p n T n
p n T n
= = =
So, p2 = 20p1
Hence, final pressure of the gas would be 20 times to
the initial pressure.
7. (b) For a function f1(v) thenumber ofmolecules (n) which
will have their speeds between v and v + dv.
For each function f1(v) and f2(v) number of molecules
remain same 1 mole each but due to mass difference
their speed will be different.
Hence both gases of each function f1(v) and f2(v) will
obeythe Maxwell's distribution law separately.
8. (d) As we know that an ideal gas equation,
pV = nRT
where, n = Number of moles, p = Pressure,
V = Volume, R = Gas constant,
T = Temperature
pV
n
RT
=
If n, R are constant for the system or as number of
moles of the gas remains fixed, hence, we can write
pV
T
= constant
or
1 1 2 2
1 2
p V p V
RT RT
=](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-348-320.jpg)
![345
Kinetic Theory
2
2 1 1
2 1
( )
T
p p V
V T
æ ö
= ç ÷
è ø
( )( )(1.1 )
(1.05) ( )
p V T
V T
= [p1 = p, V2 = 1.05 V
and T2 = 1.1 T]
1.1
1.05
p
æ ö
= ´ç ÷
è ø
= p(1.0476)
So, final pressure p2
will lies between p and 1.1p.
NEET/AIPMT (2013-2017) Questions
9. (b) P1 > P2
P1
P2
T
V
q2
q1
As V = constant Þ Pµ T
Hence from V–T graph P1 > P2
10. (d) From first law ofthermodynamics
DQ = DU+ DW =
3
2
.
1
4
R (T2 – T1) + 0
=
3
8
NaKB (T2 – T1) [Q K =
R
N
]
11. (b) Q 2
rms
1
3
mn
P V
V
=
When mass is halved and speed is doubled then
Resultant pressure, Pt =
2
rms
1
(2 )
3 2
m n
v
V
´ ´
= 2 P.
12. (b) Mean free path lm = 2
1
2 d n
p
where d = diameter of molecule and d = 2r
lm µ 2
1
r
13. (b) Let ‘n’ be the degree of freedom
p
v
n
1 R
C 2
2
1
n
C n
R
2
æ ö
+
ç ÷
è ø æ ö
g = = = +
ç ÷
è ø
æ ö
ç ÷
è ø
14. (a) Change in internal energyfrom A® B
DU=
f
2
nRDT =
f
2
nR (Tf – Ti)
=
5
2
{PfVf –PiVi}
(As gas is diatomic f = 5)
=
5
2
{2 × 103 × 6 – 5 × 103 × 4}
=
5
2
{12 – 20} × 103 J= 5 × (–4) × 103 J
DU= –20 KJ
15. (a) FromPV=nRT
PA = A A
M
RT
r
and B B
B
M
P
RT
r
=
From question,
A A A A
B B B B
P M M 3
2
P M M 2
r
= = =
r
So,
A
B
M 3
M 4
=
16. (d) Molar mass of the gas = 4g/mol
Speed of sound
V =
RT
m
g
Þ 952= 3
3.3 273
4 10-
g´ ´
´
Þ g = 1.6 =
6 8
10 5
1
=
Also, g =
P
V
C 8
C 5
=
So, CP =
8 5
5
´
= 8JK–1mol–1 [CV = 5.0 JK–1 given]
17. (b) Here v1 = 200 m/s;
temperature T1 = 27°C = 27 + 273 = 300 k
temperature T2 =127°C=127+273=400 k,V=?
R.M.S. Velocity, Vµ T
Þ
v 400
200 300
=
Þ v =
200 2
m / s
3
´
Þ v =
400
m / s
3
18. (c) Internal energy of the system is given by
U=
f
nRT
2
Degree of freedom
Fdiatomic = 5
fmonoatomic = 3
and, number of moles
n(O2)= 2
n(Ar) = 4
Utotal =
5 3
(2)RT (4)RT
2 2
+ = 11RT
T](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-349-320.jpg)
![348 PHYSICS
SOME SYSTEMS EXECUTING S.H.M.
Case 1 Spring mass system :
(i) When two springs having force constants k1 and k2
connected in parallel, then
k2
k1
M
The force constant of the combination is
k = k1 + k2 and hence T = 2p[M/(k1 + k2)]1/2
(ii) When two springs of force constants k1 and k2 are
connected in series, then
k2
k1
M
The force constant of the combination is
1/k = 1/k1 + 1/k2. i.e.,
k = k1k2/(k1 + k2) Hence 2
m
T
K
= p
(iii) If two mass M1 and M2 are connected at the two ends of
the spring, then their period of oscillation is given by
M2
M1
T = 2p[m/k)]1/2 where 1 2
1 2
M M
M M
m =
+
is the reduced mass.
(iv) When the length of spring increases, spring constant
decreases. If thelength ofspring becomes n times, its spring
constant becomes
n
1
times and therefore time period will
be increased by n times.
(v) If we divide the spring into n equal parts, the spring
constant of each part becomes n k. Hence time period when
the same mass is suspended from each part is:
1/2
2
M
T
nk
é ù
= pê ú
ë û
Case 2 Simple pendulum : A simple pendulum consists of a
point mass suspended by a weightless inextensible cord from a
rigid support.
q
x =lq
mg sinq mg mg cosq
l l
Let a bob of mass m is displaced from its, equilibrium position
and released, then it oscillates in a vertical plane under gravity.
Let q be the angular displacement at any time t, then
corresponding linear displacement along the arc is
x = lq.
It is clear from the diagram that mg sinq, is the restoring force
acting on m tending to return it to mean position. So from
Newton’s second law
q
-
=
= sin
mg
dt
x
md
F
2
2
...(i)
where negative sign indicates that restoring force mg sin q (= F)
is opposite to displacement q. If q is very small, then
sin q » q, so from equation (i)
l
x
mg
sin
mg
dt
x
md
2
2
-
=
q
-
= or ...(ii)
where w2 = g/l.
This is the equation of S.H.M. of the bob with time period
2
2
T
g
æ ö
p
= = p ç ÷
w è ø
l
OM
How to find the time period of a body undergoing S.H.M.?
Step 1 : First, find the equilibrium position. Equilibrium position
will be one for which 0
F =
S and 0
St =
Step 2 :Displacethebody, from the equilibrium position byx.
Find therestoring force acting on the bodyF = –kx (for translation)
Find the restoring torque acting on the body q
-
=
t k (for
rotational)
Step 3 : Since 2
2
dt
x
md
ma
F =
=
Use kx
dt
x
md
2
2
-
= ... (i) for translational
q
-
= k
dt
x
md
2
2
… (ii) for rotational
Step 4 :
k
m
2
T p
= (for translational)
k
I
2
T p
= (for rotational) where, I = moment ofinertia
COMMON DEFAULT
Incorrect. The time period of spring mass-system is
dependent on the value of g.
Correct. Time period ofspring-mass system shifts onlythe
equilibrium position. It does not change the time period.
Because of this reason, time period of spring mass system
remains same on plains / mountains / in satellites.
Incorrect. The time taken tocover halfthe amplitude form
equilibrium position is
8
T
.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-352-320.jpg)
![349
Oscillations
Correct. The actual time taken is
12
T
.
Incorrect. In a spring mass system, mass oscillate about
the end of a spring when the spring is in unstretched
condition.
Correct. Themass oscillates about the equilibrium position
which may or may not be at the unstretched length.
Case 3 Liquid in U-tube :AU-tube ofuniform cross-sectional
area A has been set up vertically with open ends facing up.
x
x
[Restoring force = –2Adg x]
If m gm of a liquid of densityd is poured into it then time period
of oscillation.
Adg
2
m
2
T p
=
Case 4 Rectangular block in liquid :
d
h'
h
d'
Rectangular block floating in a liquid,
gd
'
d
'
h
2
T p
= ; where
d = density of liquid, d¢ = density of block, h = height of block
Case 5 : Vibration of gas system in a cylinder with frictionless
piston.
A
m
p
h
Time period,
AP
mh
2
T p
=
where m = mass of gas, A = cross sectional area of piston
P = pressure exerted by gas on the piston, h = height of piston
Case 6: If a tunnel is dig in the earth diametrically or along a
chordthen timeperiod, sec
36
min
84
g
R
2
T =
p
= for a particle
released in the tunnel.
Case 7: The time period of a ball oscillating in the neck of a
chamber
B
mV
A
2
T
p
=
Case 8 : Ifa dipole of dipole moment p is suspended in a uniform
electric field E then time period of oscillation
PE
I
T p
2
=
Keep in Memory
1. In S.H.M. the phase relationship between displacement,
velocity and accleration, is as follows :
(a) The velocity is leading the displacement by a phase
2
p
radian
(b) The acceleration is leading the displacement by a
phase p radian
(c) The acceleration is leading the velocity bya phase
2
p
radian.
2. (a) When
2
A
x = , then velocityV = 0.86Vmax.
(b) When V = Vmax/2, the displacement x = 0.87A.
(c) When
2
A
x = , the kineticenergyof S.H.M. is 75% of
the total energy and potential energy25% of the total
energy.
(d) When the kinetic energyof S.H.M. is 50% of the total
energy, the displacement is 71% of the amplitude.
3. The time period of a simple pendulum of length l which is
comparable with radius of earth.
1/2
1/2
R R
T 2 2
R ( R)g
1 g
é ù
é ù
= p = p ê ú
ê ú +
æ ö ë û
ê ú
+
ç ÷
è ø
ê ú
ë û
l
l
l
where R = radius of the earth and g is the acceleration due
to gravity on the surface of the earth.
(a) When l << R, then
g
2
T
l
p
=
(b) When l = R, we find hour
1
g
2
R
2
T
2
/
1
»
ú
ú
û
ù
ê
ê
ë
é
p
=
(c) When l = ¥ , then
2
/
1
g
R
2
T
ú
ú
û
ù
ê
ê
ë
é
p
= = 84.6 minutes. Thus maximum ofT
is 84.6 minutes.
(d) Under weightlessness or in the freely falling lift
¥
=
ú
û
ù
ê
ë
é
p
=
2
/
1
0
2
T
l
. This means, the pendulum does
not oscillate at all.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-353-320.jpg)
![353
Oscillations
1
2
1
2
1 k
1
k
2
1
k
or
k
1
k
2
1
k
1
-
ú
û
ù
ê
ë
é
+
=
ú
û
ù
ê
ë
é
+
=
Example3.
A particle starts with S.H.M. from the mean position as
shown in figure below. Its amplitude isA and its time period
is T. At one time, its speed is half that of the maximum
speed. What is this displacement at that time ?
(a)
2 A
3
(b)
3 A
2
(c)
2A
3
(d)
3A
2
Solution : (b)
We know that 2 2 1/ 2
v ω[A x ]
= -
1/ 2
2
2
2
v
x A
ω
é ù
= -
ê ú
ê ú
ë û
Given that
max
v Aω
v
2 2
= = .
so, A
.
2
3
4
A
A
2
/
1
2
2
2
2
=
ú
ú
û
ù
ê
ê
ë
é
w
w
-
Example4.
A cylindrical piston of mass M slides smoothly inside a
long cylinder closed at one end, enclosing a certain mass
of gas. The cylinder is kept with its axis horizontal. If the
piston is disturbed from its equilibrium position, it
oscillates simple harmonically. The period of oscillation
will be
P A
M
h
(a)
Mh
T 2π
PA
æ ö
= ç ÷
è ø
(b)
MA
T 2π
Ph
æ ö
= ç ÷
è ø
(c)
M
T 2π
PAh
æ ö
= ç ÷
è ø
(d) T 2π (MPhA)
=
Solution : (a)
We know that PV = R T
)
V
/
V
(
P
P
or
0
V
P
P
V D
=
D
=
D
+
D
or ÷
ø
ö
ç
è
æ
-
=
÷
÷
ø
ö
ç
ç
è
æ
-
=
h
x
P
h
A
x
A
P
A
F
÷
ø
ö
ç
è
æ
-
=
h
x
A
P
F x
h
M
A
P
M
1
h
x
A
P
a ÷
÷
ø
ö
ç
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
-
=
This gives, 2 2
2
æ ö
= = = ç ÷
è ø
P A M h
or T
M h P A
p
w p
w
Example5.
A simple harmonic oscillator has an amplitude A and time
period T. Determine the time required by it to travel from
x = A to
2
A
x = .
Solution :
For S.H.M., ÷
ø
ö
ç
è
æ p
= t
T
2
sin
A
x
When x =A, ÷
ø
ö
ç
è
æ p
= t
.
T
2
sin
A
A
1
t
.
T
2
sin =
÷
ø
ö
ç
è
æ p
Þ ÷
ø
ö
ç
è
æ p
=
÷
ø
ö
ç
è
æ p
2
sin
t
.
T
2
sin Þ t = (T/4)
When
2
A
x = and ÷
ø
ö
ç
è
æ p
= t
.
T
2
sin
A
2
A
or ÷
ø
ö
ç
è
æ p
=
p
t
T
2
sin
6
sin or t = (T/12)
Now, time taken to travel from
x =A to x =A/2 = T/4 –T/12 = T/6
Example6.
Calculate the increase in velocity of sound for 1ºC rise of
temperature, if the velocity of sound at 0ºC is 332 m/sec.
Solution :
t
1
0 0
1/2
1 0 0
0 0 0
t 0 0
T
v 273 t
;
v T 273
t t
v v 1 v 1
273 2 273
t t
v 1 v v
546 546
t
v v v = 0.61 m/sec.
546
æ ö +
æ ö
= = ç ÷
ç ÷ è ø
è ø
æ ö æ ö
= + = +
ç ÷ ç ÷
è ø è ø
´
æ ö æ ö
= + = +
ç ÷ ç ÷
è ø è ø
æ ö
- = ç ÷
è ø
(As v0 = 332 m/sec and Dt = 1ºC)
Example7.
Which one of the following equations does not represent
S.H.M.; where x = displacement and t = time. Parameters
a, b and c are the constants of motion
(a) x = a sin bt
(b) x = a cos bt + c
(c) x = a sin bt + c cos bt
(d) x = a sec bt + c cosec bt
Solution : (d)
Sec bt is not defined for bt = p/2.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-357-320.jpg)
![354 PHYSICS
x = a sec bt + c cosec bt =
bt
cos
bt
sin
bt
cos
c
bt
sin
a +
This equation cannot be modified in the form of simple
equation of S.H.M.
i.e. x = a sin (wt + f).
so, it cannot represent S.H.M.
Example8.
If length of pendulum is increased, time period T increases.
So, frequency decreases or angular velocity decreases. As
maximum velocity max
v Aω
= , so max. velocity will also
decrease. (where A is amplitude of bob).
A pendulum bob has a speed of 3 m/s at its lowest position.
The pendulum is 0.5 m long. What will be the speed of the
bob, when the length makes an angle of 60º to the vertical?
(g = 10 m/s2)
Solution :
Here, r w = 3 m/s; l = 0.5 m; K.E. at the lowest point
m
2
9
)
3
(
m
2
1
r
m
2
1 2
2
2
=
=
w
=
Letv bethevelocityofthe bobatB when ÐOAB= 60° then,
A
O
C B
60º
l
OC = h = (l – l cos q) = (l – l cos 60º) = l /2
If v is the velocity of bob at position B, then using law of
conservation of energy, we have,
2
1 1
m 9 m v mg (1 cos θ)
2 2
´ = + -
l
or
2
1 1
m v m 9 mg (1 cos θ)
2 2
= ´ - -
l
1/2
1/2
1/2
[9 2 (1 cos )]
[9 2 10 0.5 (1 1/ 2)]
(9 – 5) 2 /
= - -
= - ´ ´ -
= =
l
v g
m s
q
Example9.
A small spherical steel ball is placed a little away from the
centre of a large concave mirror whose radius of curvature
R = 2.5 cm. When the ball is released, it begins to oscillate
about the centre. Show that the motion of the ball is simple
harmonic and find the period of motion. Neglect friction
and take g = 10 m/sec2.
Solution :
F = – mg sin q = – mg q (Q q issmall)
= – mg (x/R)
F= – kx, wherek = (mg/R)
Now, a = F/m = – kx/m = – w 2x
Hence, motion is simpleharmonic motion and
T =
2p
w
=
R
2
g
æ ö
p ç ÷
è ø
= 2 × 3.14 ×
2.5
10
æ ö
ç ÷
è ø = 3.142 sec.
Example10.
A spring of stiffness constant k and natural lengthl is cut
into two parts of length 3l / 4 and l / 4 respectively, and an
arrangement is made as shown in the figure. If the mass is
slightly displaced, find the time period of oscillation.
m
(3/4)l l/4
Solution :
The stiffness of a spring is inversely proportional to its
length. Therefore the stiffness of each part is
1
4
k k
3
= and k2 = 4k
m
k k
1 = (4/3) k k
2 = 4
Timeperiod,
1 2
m
T 2
k k
= p
+
or
3m 3m
T 2
16k 2 k
p
= p =](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-358-320.jpg)
![359
Oscillations
19. A boy is executing simple Harmonic Motion. At a
displacement x its potential energy is E1 and at a
displacement y its potential energy is E2. The potential
energyE at displacement (x + y) is
(a) 1 2
E E E
= - (b) 1 2
E E E
= +
(c) 1 2
E E E
= + (d) 1 2
E E E
= -
20. A particle undergoes simpleharmonic motion having time
period T. The time taken in 3/8th oscillation is
(a) T
8
3
(b) T
8
5
(c) T
12
5
(d) T
12
7
21. Twoparticles are executing S.H.M. of same amplitude and
frequency along the same straight line path. They pass
each other when going in opposite directions, each time
their displacement is half of their amplitude. What is the
phase difference between them ?
(a) 5p/6 (b) 2p/3
(c) p/3 (d) p/6
22. Lissajous figure obtained by combining x = a sin wt and
y= a sin (wt + p/4) will be a/an
(a) ellipse (b) straight line
(c) circle (d) parabola
23. A simple spring has length l and force constant K. It is cut
into two springs of lengths l1 and l2 such that l1 = n l2
(n = an integer). The force constant ofspring of length l1 is
(a) K(1 + n) (b) (K/n)(1 + n)
(c) K (d) K/(n + 1)
24. The spring constant from the adjoining combination of
springs is
m
2K
K K
(a) K (b) 2K
(c) 4K (d) 5K/2
25. A Second’s pendulum is placed in a space laboratory
orbiting around the earth at a height 3 R from the earth’s
surface where R is earth’s radius. The time period of the
pendulum will be
(a) zero (b) 2 3
(c) 4 sec (d) infinite
26. A simple pendulum attached to the roof of a lift has a time
period of 2s in a stationary lift. If the lift is allowed to fall
freely the frequency of oscillations of pendulum will be
(a) zero (b) 2Hz
(c) 0.5Hz (d) infinity
27. Twosimple pendulumsoflength 0.5 m and 20 mrespectively
are given small linear displacement in one direction at the
same time. They will again be in the phase when the
pendulum of shorter length has completed oscillations
[nT1=(n–1)T2, where T1 is timeperiod ofshorter length &
T2 be time period of longer wavelength and n are no. of
oscillations completed]
(a) 5 (b) 1
(c) 2 (d) 3
28. A clock which keeps correct time at 20ºC, is subjected to
40ºC. If coefficient oflinear expansion of the pendulum is
12×10–6 per ºC. Howmuch will it gain or loose in time ?
(a) 10.3 seconds/day (b) 20.6 seconds/day
(c) 5 seconds/day (d) 20 minutes/day
29. A mass is suspended separately by two different springs
in successive order then time periods is t1 and t2
respectively. It is connected by both springs as shown in
fig. then time period is t0, thecorrect relation is
k1 k2
m
(a) 2
2
2
1
2
0 t
t
t +
= (b) 2
2
2
1
2
0 t
t
t -
-
-
+
=
(c) 1
2
1
1
1
0 t
t
t -
-
-
+
= (d) 2
1
0 t
t
t +
=
30. When an oscillator completes 100 oscillations itsamplitude
reduces to
3
1
ofits initial value. What will be its amplitude,
when it completes 200 oscillations ?
(a)
8
1
(b)
3
2
(c)
6
1
(d)
9
1
31. A particle of mass m is fixed to one end of a light spring of
force constant k and unstretched length l. The system is
rotated about the other end of the spring with an
angular velocity w, in gravity free space. The increase in
length of the spring will be
k
m
w
(a)
k
m 2
l
w
(b) 2
2
m
k
m
w
-
w l
(c) 2
2
m
k
m
w
+
w l
(d) None of these
32. Frequency of oscillation is proportional to
2k
k
m
(a)
m
k
3
(b)
m
k
(c)
m
k
2
(d)
k
3
m](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-363-320.jpg)
![362 PHYSICS
Exemplar Questions
1. The displacement of a particle is represented by the
equation 3cos 2 .
4
y t
p
æ ö
= - w
ç ÷
è ø
The motion of the particle
is
(a) simple harmonic with period 2p/w
(b) simple harmonic with period p/w
(c) periodicbut not simpleharmonic
(d) non-periodic
2. The displacement of a particle is represented by the
equation y = sin3 wt. The motion is
(a) non-periodic
(b) periodicbut not simpleharmonic
(c) simple harmonic with period 2p/w
(d) simple harmonic with period p/w
3. The relation between acceleration and displacement offour
particles are given below
(a) ax = +2x (b) ax = +2x2
(c) ax = –2x2 (d) ax = –2x
Which, one of the particle is exempting simple harmonic
motion?
4. Motion of an oscillating liquid column in a U-tube is
(a) periodicbut not simpleharmonic
(b) non-periodic
(c) simple harmonic and time period is independent ofthe
density of the liquid
(d) simpleharmonicand timeperiod isdirectlyproportional
to the density of the liquid
5. Aparticle isacted simultaneouslybymutuallyperpendicular
simple harmonic motion x = a cos wt and y = a sin wt. The
trajectoryof motion of the particle will be
(a) an ellipse (b) a parabola
(c) a circle (d) a straight line
6. The displacement of a particle varies with time according to
the relation sin cos .
y a t b t
= w + w
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude a2 + b2
(d) The motion is SHM with amplitude 2 2
a b
+
7. Four pendulums A, B, C and D are suspended from the
same
G G
D
C B A
elastic support as shown in figure. A and C are of the same
length, while B is smaller than A and D is larger than A. IfA
is given a transverse displacement,
(a) Dwill vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude
8. Figure shows the circular motion of a particle. The radius of
the circle, the period, sense of revolution and the initial
position are indicated on the figure. The simple harmonic
motion of the x-projection ofthe radius vector of the rotating
particle P is
(a)
2
( ) sin
30
t
x t B
p
æ ö
= ç ÷
è ø
y
x
B
p t
T
( = 0)
= 30s
(b) ( ) cos
15
t
x t B
p
æ ö
= ç ÷
è ø
(c) ( ) sin
15 2
t
x t B
p p
æ ö
= +
ç ÷
è ø
(d) ( ) cos
15 2
t
x t B
p p
æ ö
= +
ç ÷
è ø
9. The equation of motion of a particle is x = a cos(at)2. The
motion is
(a) periodic but not oscillatory
(b) periodic and oscillatory
(c) oscillatory but not periodic
(d) neither periodic nor oscillatory
10. A particle executing SHMmaximum speed of30 cm/s and a
maximum acceleration of 60cm/s2. Theperiod ofoscillation
is
(a) p sec (b)
2
p
sec
(c) 2p sec (d)
t
p
sec
PastYears (2013-2017) NEET/AIPMTQuestions
11. A particle of mass m oscillates along x-axis according to
equation x = a sin wt. The nature of the graph between
momentum and displacement ofthe particle is
[NEET Kar. 2013]
(a) straight line passing through origin
(b) circle
(c) hyperbola
(d) ellipse](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-366-320.jpg)
![363
Oscillations
12. The oscillation of a body on a smooth horizontal surface is
represented by the equation,
X = A cos (wt)
where X = displacement at time t
w = frequency of oscillation
Which one of the following graphs shows correctly the
variation of‘a’ with ‘t’? [2014]
(a)
a
O
T t
(b)
a
O
T t
(c)
a
O
T t
(d)
a
O
T t
13. A particle is executing SHM along a straight line. Its
velocities at distances x1 and x2 from the mean position
are V1 and V2, respectively. Its time period is [2015]
(a)
2 2
2 1
2 2
1 2
x – x
2
V – V
p (b)
2 2
1 2
2 2
1 2
V V
2
x x
+
p
+
(c)
2 2
1 2
2 2
1 2
V – V
2
x – x
p (d)
2 2
1 2
2 2
1 2
x – x
2
V – V
p
14. When two displacements represented by y1 = asin(wt) and
y2 = b cos(wt) are superimposed the motion is: [2015]
(a) simple harmonic with amplitude
a
b
(b) simple harmonic with amplitude 2 2
a b
+
(c) simple harmonic with amplitude
(a b)
2
+
(d) not a simpleharmonic
15. Aparticleisexecutingasimpleharmonicmotion.Itsmaximum
acceleration is a and maximum velocityis b.
Then its time period of vibration will be : [2015 RS]
(a)
a
b
(b)
2
b
a
(c)
2pb
a
(d)
2
2
b
a
16. A particleexecutes linear simple harmonic motion with an
amplitude of 3 cm. When the particle is at 2 cm from the
mean position, the magnitude of its velocityis equal to that
ofits acceleration. Then its time period in seconds is [2017]
(a)
5
2p
(b)
4
5
p
(c)
2
3
p
(d)
5
p
17. A spring of force constant k is cut into lengths of ratio 1 : 2
: 3. Theyare connected in series and the new force constant
is k'. Then theyare connected in parallel and force constant
is k¢¢ . Then k' : k¢¢ is [2017]
(a) 1 :9 (b) 1 : 11
(c) 1 : 14 (d) 1 : 6](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-367-320.jpg)
![366 PHYSICS
16. (a) )
2
(
T
D
1
'
,
T
2
)
D
3
(
1
r
p
=
n
pr
=
n
l
l
3
2
2
3
2
'
=
´
=
n
n
17. (b)
18. (c) Kinetic energy
2 2 2
1
( )
2
K m a y
w
= -
2
2 2 2
1 2
10 [10 5 ] 375
2 2
æ ö
= ´ ´ - =
ç ÷
è ø
erg
p
p
19. (b)
2 2
1 2
1 2
2 2
1 1
,
2 2
= Þ = = Þ =
E E
E kx x E ky y
k k
and
2 2
( )
2
1
= + Þ + =
E
E k x y x y
k
1 2
1 2
2 2 2
E E E
E E E
k k k
Þ + = Þ + =
20. (c) Time to complete 1/4th oscillation is
4
T
s. Time to
complete
8
1
th vibration from extreme position is
obtained from
t
T
2
cos
a
t
cos
a
2
a
y
p
=
w
=
= or s
6
T
t =
Sotime to complete 3/8th oscillation
=
12
T
5
6
T
4
T
=
+
21. (b) )
t
(
sin
a
y f
+
w
= ; when y = a/2,
then )
t
(
sin
a
2
a
f
+
w
=
or
6
sin
2
1
)
t
(
sin
p
=
=
f
+
w or
6
5
sin
p
So phase of two particles is p/6 and 5 p/6 radians
Hence phase difference = (5 p/6) – p/6 = 2 p/3
22. (a) sin wt = x/a and
2
2
2
a
/
x
1
t
sin
1
t
cos -
=
w
-
=
w
y= sin (wt + p/4)
= sin wt cos p/4 + cos wt sin p/4
=
2
1
a
/
x
1
2
1
a
x 2
2
´
÷
ø
ö
ç
è
æ
-
+
´
or 2
2
x
a
x
y
2 -
+
=
or 2
2
2
2
2
2
x
a
x
2
)
x
a
(
x
y
2 -
+
-
+
=
=
2
2
2
x
a
x
2
a -
+
It is an equation of an ellipse.
23. (b) Let k be the force constant of spring of length l2.
Since l1 = n l2, where n is an integer, so the spring is
made of (n + 1) equal parts in length each of length l2.
k
)
1
n
(
K
1 +
= or k = (n + 1) K
The spring of length l1 (= n l2) will be equivalent to n
springs connected in series where spring constant
n
/
K
)
1
n
(
n
k
k +
=
=
¢ & spring constant of length l2
isK(n+1).
24. (c) Here all the three springs are connected in parallel to
mass m. Hence equivalent spring constant
k = K + K+ 2 K= 4 K.
25. (d) The second pendulum placed in a space laboratory
orbiting around the earth is in a weightlessness state.
Hence g = 0 so T = ¥
26. (a) When lift is falling freely, the effective acceleration
due to gravity inside the lift is zero i.e. g' = g – g = 0.
Therefore time period will be infinityand frequencyis
zero
27. (b) Let T1, T2 be the time period of shorter length and
longer length pendulums respectively.As per question,
n T1 = (n – 1) T2 ;
so
g
20
2
)
1
n
(
g
5
.
0
2
n p
-
=
p
or 6
)
1
n
(
40
)
1
n
(
n -
»
-
=
Hence n = 6/5 » 1
28. (a)
g
2
T
l
p
= ; 1/ 2
T 1
2 ( / g) /
T 2
-
D
= p ´ ´ D
l l l
)
g
2
T
T
(
l
l
Q
D
+
p
=
D
+
q
D
a
=
D
=
D
2
1
2
1
T
T
l
l
=
5
6
10
12
)
20
40
(
10
12
2
1 -
-
´
=
-
´
´
´
5
10
12
T
T -
´
´
=
D = 24 ×60× 60× 12 ×10–5
= 10.3 s/day
29. (b)
1
1
k
m
2
t p
= or
1
2
2
1
k
m
4
t
p
= or 2
1
2
1
t
m
4
k
p
=
Similarly, 2
2
2
2
t
m
4
k
p
= and
2
0
2
2
1
t
m
4
)
k
k
(
p
=
+
2
2
2
2
1
2
2
0
2
t
m
4
t
m
4
t
m
4 p
+
p
=
p
or 2
2
2
1
2
0 t
1
t
1
t
1
+
=
30. (d) It is a damped oscillation, where amplitude of
oscillation at time t is given by γ t
0
A a e-
=
where a0 = initial amplitude of oscillation
g = damping constant.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-370-320.jpg)
![367
Oscillations
As per question, γ 100 / ν
0
0
a
a e
3
-
= ……(i)
(where n is the frequency of oscillation)
and γ 200 / ν
0
A a e-
= ……(ii)
From (i) ; γ 100/ ν
o
0
a
a e
3
- ´
= ……(iii)
Dividing equation (ii) by(iii), wehave
γ 200 / ν
γ 100 / ν
0
A e
a (1/ 3) e
-
- ´
= =
γ 100 / ν 1
e
3
- ´
=
or 0
0 a
9
1
3
1
3
1
a
A =
´
´
=
31. (b) Let x = the increase in the length of spring. Then the
particle moves along a circular path of radius (l + x),
and the spring force = kx = centripetal force
mw2 (l+ x) =kx or
2
2
m
k
m
x
w
-
w
=
l
32. (a)
k k
2
x
Let mass is displaced towards left by x then force on
mass = – kx –2kx = – 3kx
[negative sign is taken because force is opposite to
the direction of motion]
F 3kx
Þ = - x
m 2
w
-
=
3k
.
m
Þ w =
1 3k
f .
2 2 m
w
= =
p p
Thus it is propotional to m
/
k
3
33. (a) Maximum velocity can be found out by energy
conservation.
2
mv
2
1
mgh = gh
2
v
or max = ;
2
gh
'
v
2
vmax
=
=
'
mgh
)
'
v
(
m
2
1
mgh 2
+
= ;
'
mgh
2
mgh
2
1
mgh +
=
Þ '
mgh
4
mgh
3
=
4
h
3
'
h
or =
34. (b) Loss in PE of mass = gain in PE of the spring
mg(h + x) =
2
kx
2
1
2
kx – 2mgx – 2mgh = 0
Þ x=
k
2
)
mgh
2
(
k
4
g
m
4
mg
2 2
2
-
-
±
or x=
k
mgh
2
k
g
m
k
mg
2
2
2
+
±
x = A–
k
mg
=
k
mg
k
mgh
2
k
g
m
2
2
2
-
+
=
k
mg
ú
ú
û
ù
ê
ê
ë
é
-
+ 1
mg
kh
2
1
35. (a) t = 0, v maximum. The motion begins from mean
position. So it represents S.H.M.
36. (a) The both of the pendulum will more along a straight
lineAB.
q q
Re
mg
x
A B
The direction ofthe Earth's gravitational field is radial
Now, mg
R
m
GM
F 2
e
e
=
=
x
R
m
GM
R
x
F
cos
F
F 3
e
e
e
x -
=
-
=
q
-
= =–kx
where
e
3
e
GM m
k
R
=
Time period ofa simple harmonic oscillator,
3
e
e R
/
m
GM
m
2
k
m
2
T p
=
p
=
or
g
R
2
R
GM
R
2
T e
2
e
e
e
p
=
p
=
37. (a) For the block is about to slip, a
g 2
w
=
m
or
g
a
m
Þ w =
g 1 g
2 ν ν
a 2 a
m m
Þ p = Þ =
p
38. (d) Oscillations along spring length are independent of
gravitation.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-371-320.jpg)
![368 PHYSICS
39. (c )
2 2
1 4
I M(2L) ML
3 3
= =
Force applied by the spring is F = – kx
Þ F = – k (2Lq)
(q is the angular displacement from the equilibrium
position). Further
2 2
| F | 4L ksin 4L k
t = ´ = q = - q
r r
l
Also,
..
2
I I 4L k
t = a = q = - q
Þ
.. 3k
0
M
q+ q =
Þ 0
3k
M
w =
40. (a) V(x)=k |x|3
since, 2
dV(x)
F 3k | x |
dx
= - = - ……(i)
x = a sin (wt)
This equation always fits to the differential equation
2
2
2
d x
x
dt
= -w or
2
2
d x
m m x
dt
= - w
Þ F = – mw2x ……(ii)
Equations (i) and (ii) give
2 2
3k | x | m x
- = - w
Þ 1/ 2
3kx 3ka
[sin( t)]
m m
w = = w
Þ a
wµ
Þ
1
T
a
µ
41. (b)
2R
T 2 2
g g
= p = p
l
42. (d) Let the extension in the spring be x0 at equilibrium. If
F0 be the tension in the string then F0 = kx0. Further if
T0 is the tension in the thread then T0 = mg and
2T0 =kx0.
Let the mass m be displaced through a slight
displacement x downwards. Let the the new tension in
the string and spring be T and F respectively.
Þ 0
x
F x
2
æ ö
= ç ÷
è ø
and F = 2T
Þ 0
x
2T k x
2
æ ö
= ç ÷
è ø
Þ 0
kx
(T T )
4
- =
Þ Time period
m m
2 4
k k
4
= p = p
43. (b)
2
1
mg(h y) ky
2
+ =
Þ
mg mg 2kh
y 1
k k mg
= ± +
At equilibrium
mg = ky0
Þ 0
mg
y
k
=
Þ Amplitude A = y – y0
mg 2kh
1
k mg
= +
Energyof oscillation is
2
2
1 mg
E kA mgh
2 2k
æ ö
= = + ç ÷
è ø
44. (b) For block Ato move in S.H.M.
mg x
N
A
mean
position
mg–N =mw2x
where x is the distance from mean position
For block to leave contact N = 0
2
2 g
x
x
m
mg
w
=
Þ
w
=
Þ
45. (a) The displacement of a particle in S.H.M. is given by
y= a sin (wt + f)
velocity=
dy
dt
= wa cos (wt + f)
The velocity is maximum when the particle passes
through the mean position i.e.,
max
dy
dt
æ ö
ç ÷
è ø = w a](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-372-320.jpg)
![369
Oscillations
The kinetic energy at this instant is given by
1
2
m
2
max
dy
dt
æ ö
ç ÷
è ø =
1
2
mw2 a2 = 8 × 10–3 joule
or
1
2
× (0.1) w2 × (0.1)2 = 8 × 10–3
Solving we get w= ±4
Substituting the values of a, w and f in the equation of
S.H.M., we get
y= 0.1 sin (± 4t + p /4) metre.
46. (d) Slope of F - x curve = – k =
80
0.2
- Þ k = 400 N/m,
Timeperiod, T = 2p
m
k
= 0.0314 sec.
47. (b) At resonance, amplitude ofoscillation is maximum
Þ 2w2 –36w+ 9 isminimum
Þ 4w– 36 = 0 (derivative is zero)
Þ w= 9
48. (b) Torque about hinge
2.5 g × 0.40 cos q – 1g × 1 cos q = 0
49. (c) 50. (b)
EXERCISE - 3
Exemplar Questions
1. (b) As given that, 3cos 2
4
y t
p
æ ö
= - w
ç ÷
è ø
Velocityof the particle
3cos 2
4
dy d
v t
dt dt
é p ù
æ ö
= = - w
ç ÷
ê ú
è ø
ë û
3( 2 ) sin 2
4
t
é p ù
æ ö
= - w - - w
ç ÷
ê ú
è ø
ë û
6 sin 2
4
t
p
æ ö
= w - w
ç ÷
è ø
So, acceleration,
6 sin 2
4
dv d
a t
dt dt
é p ù
æ ö
= = w - w
ç ÷
ê ú
è ø
ë û
(6 ) ( 2 )cos 2
4
t
p
æ ö
= w ´ - w - w
ç ÷
è ø
2
4 3cos 2
4
t
é p ù
æ ö
= - w - w
ç ÷
ê ú
è ø
ë û
2
4
a y
= - w
In simple harmonic motion acceleration (or force) is
directly proportional to the negative of displacement
ofparticle
Þ as acceleration, a y
µ -
Hence, due to negativesign motion is simple harmonic
motion (SHM.)
Asimpleharmonic motion isalwaysperiodic. Somotion
is periodic simpleharmonic.
From the given equation,
3cos 2
4
y t
p
æ ö
= - w
ç ÷
è ø
Compare it by standard equation
cos( )
y a t
= w + f
So, ' 2
w = w
2 2
2 '
' 2
T
T
p p p
= w Þ = =
w w
Hence, the motion is SHM with period
p
w
.
2. (b) A motion will be harmonic if a µ displacement and a
simple harmonic motion is always periodic but all
simple harmonic motion are periodic but all periodic
are not harmonic.
As given equation of motion is
3
sin
y t
= w
(3sin 4sin3 )/ 4
t t
= w - w
3
[ sin3 3sin 4sin ]
q = q- q
Q
Differentiating both side w.r.t. t
So, (3sin ) (4sin3 ) / 4
dy d d
v t t
dt dt dt
é ù
= = w - w
ê ú
ë û
or
3 (3sin sin3 )
sin
4
q - q
é ù
q = ê ú
ë û
4 3 cos 4 [3 cos3 ]
dy
t t
dt
= w w - ´ w w
Again, differentiating both side w.r.t. t
2
2
2
4 3 sin 36 sin3
d y
t t
dt
´ = - w w + w w
2 2
2
3 sin 36 sin3
4
d y t t
a
dt
- w w + w w
= =
So,
2
2
d y
dt
æ ö
ç ÷
ç ÷
è ø
is not directly proportional to y.
Somotion is not harmonic.
3
( ) sin ,
y t t
= w
3
( ) sin ( )
y t T t T
+ = w +
3 2
sin ( )
t T
T
p
é ù
= +
ê ú
ë û
3 3
sin (2 ) sin
t t
= p + w = w](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-373-320.jpg)
![370 PHYSICS
3. (d) For motion to be SHM acceleration of the particle must
be opposite of restoring force and proportional to
negative of displacement. So, F = ma = m(–2x)
i.e., F = –2mx, so F x
µ -
Hence, ax = –2x
We should be clear that x has to be linear.
4. (c) Consider a U-tube filled in which a liquid column
oscillates. When liquid column lifted upto height y
from A toB in arm Q. The liquidlevel in arm Pbecomes
at C', so the difference between the height of two
columns are = AB +A'C = y + y = 2y.
In this case, restoring force acts on the liquid due to
gravity. Acceleration of the liquid column, can be
calculated in terms of restoring force.
P Q
B
A
C
B'
A'
C'
h
y
Equilibrium level
Restoring force,
f = Weight of liquid column of height 2y
( 2 ) 2
f A y g A gy
= - ´ ´r ´ = - r
[ ]
v
h =r
Q
As restoring force at A opposite to gravitational force
as liquid is lifted against ( ) 2
t mg k A g
w m = r
Then f y
µ - so motion is SHM.
Timeperiod
(2 )
2 2 2
2
m A h h
T
k A g g
´ ´r
= p = p = p
r
2
h
T
g
= p
Time period is independent ofthe density of the liquid
and motion is harmonic.
5. (c) The resultant-displacement can be find by adding x
and y-components.
According to variation of x and y, trajectory will be
predicted, so resultant displacement is y' = (x + y)
As given that,
cos
x a t
= w ...(i)
sin
y a t
= w ...(ii)
So, ' ( cos sin )
y a t a t
= w + w
' (cos sin )
y a t t
Þ = w + w
cos sin
' 2
2 2
t t
y a
w w
é ù
= +
ê ú
ë û
' 2(cos cos 45 sin sin 45 )
y a t t
= w °+ w °
2 cos( 45 )
a t
= w - °
So, the displacement is not straight line and not
parabola also.
Now, squaring and adding eqs. (i) and (ii),
2 2 2
x y a
+ = 2 2
[ cos sin 1]
t t
w + w =
Q
This is the equation of a circle, so motion in circular
(independent of time).
Clearly, the locus is a circle of constant radius a.
6. (d) As given that, the displacement is
sin cos
y a t b t
= w + w
Let sin
a A
= f ...(i)
and cos
b A
= f ...(ii)
Squaring and adding (i) and (ii)
2 2 2 2 2 2
sin cos
a b A A
+ = f+ f
2 2
A a b
= + (amplitude)
sin sin cos cos
y A t A t
= f× w + f w
cos( )
y A t
= w - f
sin( )
dy
A t
dt
= - w w - f
2
2 2
2
cos( )
d y
A t y
dt
= - w w - f = -w
Thus,
2
2
( )
d y
y
dt
µ - so, motion is SHM.
Hence, it is an equation of SHM with amplitude
2 2
.
A a b
= +
7. (b) When pendulum vibrate with transverse vibration then
2
l
T
g
= p
l = length of pendulum.
G G
Elastic support
A
B
C
D
e
Through the elastic rigid support the disturbance is
transferred to all the pendulum A and C are having
length and hence same frequency. They will be in
resonance, because their time period of oscillation.
So, a periodic force of period (T) produces resonance
in AandCandtheywillvibratewith maximumamplitude
as in resonance.](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-374-320.jpg)
![371
Oscillations
8. (a) As the particle (P) is executing circular motion with
radius B.
Let particle P is at Q at instant any (t), foot of
perpendicular on x axis is at R vector OQ makes Ðq ,
with its zero position not P displacement of particles
for O to R.
Consider angular velocity of the particle executing
circular motion is w and when it is at Q makes and
angle q as shown in the diagram.
BR
x
y
p t
( =0)
Q
q
q
r
90–q
O
Clearly, t
q = w
Now, we can write
cos(90 – )
OR OQ
= q ( )
OR X
=
Q
sin sin
x OQ OQ t
= q = w
sin
r t
= w [ ]
OQ r
=
Q
2 2
sin sin
30
x B t B t
T
p p
æ ö
= = ç ÷
è ø
2
sin
30
x B t
p
æ ö
= ç ÷
è ø
Hence, this equation represents SHM.
9. (c) According to the question,
2
cos( )
x a t
= a
is a cosine function, so it is an oscillatory motion.
Now, at, t = (t + T). The equation of motion of particle
2
( ) cos[ ( )]
x t T a t T
+ = a + 2
[ ( ) cos( ) ]
x t a t
= a
Q
2 2
cos[ 2 ] ( )
a t T tT x t
= a + a + a ¹
where, T is supposed as period of the function w(t).
Hence, it is oscillatory but not periodic.
10. (a) Let us consider a equation of an SHM is represented
by y = a sin wt
cos
dy
v a t
dt
= = w w
max
( ) 30 cm/sec
v a
= w = (given)...(i)
Acceleration 2
( ) sin
= = - w w
dv
a a t
dt
2
max 60
= w =
a a (given)...(ii)
Eqs. (i) and (ii), we get
( ) 60 (30) 60
a
w w = Þ w =
w= 2 rad/s
2
2 rad/s
T
p
=
sec
= p
T
PastYears (2013-2017) NEET/AIPMTQuestions
11. (d) As
2 2
2 2 2
1
v y
a a
+ =
w
This is the equation of ellipse.
Hence the graph is an ellipse. P versus x graph is
similar to V versus x graph.
12. (c) Displacement, x = Acos (wt) (given)
Velocity, v =
dx
A sin ( t)
dt
= - w w
Acceleration, a = 2
dv
A cos t
dt
= - w w
Hence graph (c) correctly dipicts the variation of a
with t.
13. (a) As we know, for particle undergoing SHM,
2 2
V A – X
= w
2 2 2 2
1 1
V (A – x )
=w
2 2 2 2
2 2
V (A – x )
=w
Substracting we get,
2 2
2 2
1 2
1 2
2 2
V V
x x
+ = +
w w
Þ
2 2
2 2
1 2
2 1
2
V – V
x – x
=
w
Þ
2 2
1 2
2 2
2 1
V – V
w
x – x
=
Þ
2 2
2 1
2 2
1 2
x – x
T 2
V – V
= p
14. (b) The two displacements equations are y1 = a sin(wt)
and y2 = b cos(wt) = b sin t
2
p
æ ö
w +
ç ÷
è ø
yeq = y1 + y2
= a sinwt + b coswt = a sinwt + b sin t
2
p
æ ö
w +
ç ÷
è ø
Since the frequencies for both SHMs are same,
resultant motion will be SHM.
NowAeq =
2 2
a b 2ab cos
2
p
+ +
b
a
2 2
eq
A a b
Þ = +](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-375-320.jpg)
![376 PHYSICS
By comparison of the standard equation to above given
equation,
we get .
sec
10
1000
1
T 3
-
=
=
Particle velocity
÷
ø
ö
ç
è
æ p
-
p
p
´
p
=
=
17
x
t
2000
cos
2000
10
dt
dy
s
/
m
200
s
/
cm
000
,
20
dt
dy
max
=
=
÷
ø
ö
ç
è
æ
Þ (as vmax = A
Aw)
Example2.
A progressive wave of frequency 500 Hz is travelling with
a velocity of 360 m/s. Howfar apart are two points 60o out
of phase?
Solution :
We known that for a wave v = f l
so l=
v 360
f 500
= = 0.72 m
Now as in a wave path difference is related to phase
difference by the relation.
Phase difference Df= 60o = (p/180) x 60 = (p/3) rad
Path difference Dx=
2
l
p
(Df)=
0.72
2 3
p
p
= 0.12 m
Example3.
Determine the change in volume of 6 litres of alcohol if
the pressure is decreased from 200 cm of Hg to 75 cm.
[velocity of sound in alcohol is 1280 m/s, density of
alcohol = 0.81 gm/cc, density of Hg = 13.6 gm/cc and
g = 9.81 m/s2].
Solution :
For propagation of sound in liquid
v = ( )
B / r , i.e., B = v2r
But by definition B= – V
P
V
D
D
so –V
P
V
D
D
= v2r, i.e., DV = 2
V( P)
v
-D
r
Here DP = H2rg – H1rg = (75 – 200) ´ 13.6 ´ 981
= –1.667 ´ 106 dyne/cm2
so DV =
( )( )
( )
3 6
2
5
6 10 1.667 10
0.81 1.280 10
´ ´
´ ´
= 0.75 cc
Example4.
(a) Speed of sound in air is 332 m/s at NTP. What will be
the speed of sound in hydrogen at NTP if the density
of hydrogen at NTP is (1/16) that of air?
(b) Calculate the ratio of the speed of sound in neon to that
in water vapour at any temperature. [Molecular weight of
neon = 2.02 ´ 10–2 kg/mol and for water vapours = 1.8 ´
10–2 kg/mol]
Solution :
The velocity of sound in air is given by
v =
E
r
=
P
g
r
=
RT
M
g
(a) In terms of density and pressure
H
air
v
v =
air
H
H air
P
P
r
´
r =
air
H
r
r [as Pair = PH]
or vH = vair ´
air
H
r
r = 332 ´
16
1
= 1328 m/s
(b) In terms of temperature and molecular weight
Ne
W
v
v =
Ne W
Ne W
M
M
g
´
g [as TN = TW]
Now as neon is monatomic (g = 5/3) while water vapours
polyatomic (g = 4/3) so
Ne
W
v
v =
( )
( )
2
2
5 / 3 1.8 10
4 / 3 2.02 10
-
-
´ ´
´ ´
=
5 1.8
4 2.02
´ = 1.05
BEATS
When two wave trains slightly differing in frequencies travel
along the same straight line in the same direction, then the
resultant amplitude is alternately maximum and minimum at a
point in the medium. This phenomenon of waxing and waning
of sound is called beats.
Let two sound waves of frequencies n1 and n2 are propagating
simultaneouslyand in same direction. Then at x=0
y1 = A sin 2p n1t, and y2 = A sin 2p n2t,
For simplicity we take amplitude of both waves to be same.
By principle of superposition, the resultant displacement at any
instant is
y = y1 + y2 = 2A cos 2p nAt sin 2p navt
where
2
n
n
n 2
1
av
+
= ,
2
n
n
n 2
1
A
-
=
Þ y = Abeat sin 2p navt ..................(i)
It is clear from the above expression (i) that](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-380-320.jpg)
![378 PHYSICS
Solution :
In this case, we can assume that both the source and
observer are moving towards each other with velocityv. If
c be the velocity of signal, then
n
v
c
v
c
n ÷
ø
ö
ç
è
æ
-
+
=
¢ or n
)
v
c
(
)
v
c
(
)
v
c
(
n
2 ÷
÷
ø
ö
ç
ç
è
æ
-
-
+
=
¢
or
2 2 2
2 2 2
(c v ) c
n n n; (as c v)
c v 2c v c 2cv
-
¢ ¢
= Þ = >>
+ - -
n
v
2
c
c
n ÷
÷
ø
ö
ç
ç
è
æ
-
=
¢
STATIONARY OR STANDING WAVES
When two progressive waves having the same amplitude, velocity
and time period but travelling in opposite directions
superimpose, then stationary wave is produced.
Let two waves ofsame amplitude and frequencytravel in opposite
direction at same speed, then
y1 = Asin (wt –kx) and
y2 = Asin (wt + kx)
By principle of superposition
y = y1 + y2 = (2A cos kx) sin wt ...(i)
s
y A sin ωt
=
It is clear that amplitude ofstationary waveAs varywith position
(a) As = 0, when cos kx = 0 i.e., kx = p/2, 3p/2............
i.e.,x =l/4, 3l/4...................[ask=2p/l]
These points are called nodes and spacing between two
nodes is l/2.
(b) As is maximum, when cos kx is max
i.e., kx= 0,p,2p,3pi.e.,x=0,ll/2,2l/2....
It is clear that antinode (where As is maximum) are also
equally spaced with spacing l/2.
(c) The distance between node and antinode is l/4 (see figure)
Antinode Antinode
segment 1 segment 2 segment 3
Node
x
l
l
/2
/4
o
2A
Keep in Memory
1. When a string vibrates in one segment, the sound produced
is called fundamental note. The string is said to vibrate in
fundamental mode.
2. The fundamental note is called first harmonic, and is given
by
l
2
v
0 =
n , where v = speed of wave.
3. If the fundamental frequencybe 0
n then 2 0
n , 3 0
n , 4 0
n
... are respectively called second third, fourth ... harmonics
respectively.
4. If an instrument produces notes of frequencies
4
3
2
1 ,
,
, n
n
n
n .... where 1 2 3 4
ν ν ν ν
< < < ....., then 2
n is
(ii) If source and observer both move towards each other.
0
3 0
s
V V
V V
æ ö
+
¢
n = n
ç ÷
-
è ø
;
Whenthewindblows inthedirection of sound, then in all above
formulaeVisreplacedby(V+ W) whereWis thevelocityofwind.
Ifthe wind blowsin theoppositedirectiontosoundthenVisreplaced
by(V– W).
Keep in Memory
1. The motion of the listener causes change in number of
waves receivedbythe listener andthis producesan apparent
change in frequency.
2. The motion of the source of sound causes change in
wavelength of the sound waves, which produces apparent
change in frequency.
3. If a star goes away from the earth with velocity v, then the
frequencyof the light emitted from it changes from n ton'.
n' = n (1–v/c), where c is the velocity of light and
or
v v
c c
Dn Dl
= =
n l
where l
D is called Doppler’s shift.
If wavelength of the observed waves decreases then the
object from which the waves are coming is moving towards
the listener and vice versa.
Example7.
Two engines cross each other travelling in opposite
direction at 72 km/hour. One engine sounds a whistle of
frequency 1088 cps. What are the frequencies as heard by
an observer on the other engine before and after crossing.
Take the speed of sound as 340 m/s.
Solution :
The apparent frequency before crossing,
÷
÷
ø
ö
ç
ç
è
æ
-
+
=
¢
s
0
v
v
v
v
n
n
HereVo = 72 km/hour = 20 m/s
vs = 72 km/hour = 20 m/s
÷
ø
ö
ç
è
æ
-
+
´
=
¢
20
340
20
340
1088
n Hz
1224
320
360
1088 =
´
=
The apparent frequency after crossing
0
s
v v
n'' n
v v
æ ö
-
= ç ÷
+
è ø
÷
ø
ö
ç
è
æ
+
-
´
=
20
340
20
340
1088
Hz
11
.
967
36
32
1088 =
´
=
Example8.
A rocket is going towards moon with a speed v. The
astronaut in the rocket sends signals of frequency n
towards the moon and receives them back on reflection
from the moon. What will be the frequency of the signal
received by the astronaut? (Take v << c)](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-382-320.jpg)
![379
Waves
called first overtone, 3
n is called second overtone, 4
n is
called third overtone ... so on.
5. Harmonics are the integral multiples of the fundamental
frequency.If n0 be the fundamental frequency, then 0
nn is
the frequencyof nth harmonic.
6. Overtones are the notes of frequency higher than the
fundamental frequencyactuallyproduced bythe instrument.
7. In the strings all harmonics are produced.
StationaryWaves in an Organ Pipe :
In the open organ pipe all the harmonics are produced.
In an open organ pipe, the fundamental frequency or first
harmonic is 0
2
v
n =
l
, where v is velocity of sound and l is the
length of air column [see fig. (a)]
(a)
l
(b)
l
2
l
=
l ,
1
2l
=
l
2
2l
=
l ,
2
2l
=
l
(c)
l
2
3l
=
l ,
3
L
3
=
l
Similarly the frequency of second harmonic or first overtone is
[see fig (b)], 01
2
2
v
n =
l
Similarly the frequency of third harmonic and second overtone
is, [(see fig. (c)] 02
3
2
v
n =
l
Similarly 03 04
4 5
,
2 2
v v
n n
= =
l l
..................
In the closed organ pipeonlythe odd harmonics are produced. In
a closed organ pipe, the fundamental frequency (or first
harmonic) is (see fig. a)
4
c
v
n =
l
(a)
l
(b)
l
(c)
l
4
l
=
l
4
3l
=
l
4
5l
=
l
Similarly the frequencyof third harmonic or first overtone (IInd
harmonic absent) is (see fig. b)
l
4
v
3
n 2
c =
Similarly
l
l 4
v
7
n
,
4
v
5
n 4
c
3
c =
= ........
End Correction
It is observed that the antinode actually occurs a little above the
open end. A correction is applied for this which is known as end
correction and is denoted bye.
(i) For closed organ pipe : l is replaced by l+ e where
e = 0.3D, D is the diameter of the tube.
(ii) For open organ pipe: l is replaced by l + 2e where
e=0.3D
In resonance tube, the velocity of sound in air given by
( )
2 1
2ν
v l l
= -
where n = frequency of tuning fork, ll = 1st resonating
length, l2 = 2nd resonating length.
ResonanceTube :
It is used to determine velocity of sound in air with the help of a
tuning fork of known frequency.
n n
A
e
l
1
e
l
2
Let l1 and l2 are lengths of first and second resonances then
1 e
4
l
+ =
l and 2
3
e
4
l
+ =
l
2 1
2
l
Þ - =
l l 2 1
2( )
Þ l = -
l l
Speed of sound in air is
v = ul where u is the frequency
2 1
2 ( )
v = u -
l l](https://image.slidesharecdn.com/dishaneetphysicsguide-240427161812-19b55928/85/Disha-NEET-Physics-Guide-for-classes-11-and-12-pdf-383-320.jpg)
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Disha NEET Physics Guide for classes 11 and 12.pdf
- 2. • Corporate Office: 45, 2nd Floor, Maharishi Dayanand Marg, Corner Market, Malviya Nagar, New Delhi-110017 Tel. : 011-26692293 / 26692294 DISHA PUBLICATION All Rights Reserved © Copyright Publisher No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried and made our best efforts to provide accurate up-to-date information in this book. For further information about the books from DISHA PUBLICATION, Log on to www.dishapublication.com or www.aiets.co.in or email to info@aiets.co.in Typeset by Disha DTP Team (ii) By Sanjeev Kumar Jha
- 3. 1. Physical World1-7 2. Units and Measurements 8-34 3. Motion in a Straight Line 35-66 4. Motion in a Plane 67-98 5. Laws of Motion 99-138 6. Work, Energy and Power 139-172 7. System of Particles and Rotational Motion 173-210 8. Gravitation 211-240 9. Mechanical Properties of Solids 241-259 10. Mechanical Properties of Fluids 260-279 11. Thermal Properties of Matter 280-302 12. Thermodynamics 303-325 13. Kinetic Theory 326-345 14. Oscillations 346-372 15. Waves 373-400 Contents (iii)
- 4. 16. Electric Chargesand Fields 401-429 17. Electrostatic Potential and Capacitance 430-463 18. Current Electricity 464-511 19. Moving Charges and Magnetism 512-542 20. Magnetism and Matter 543-566 21. Electromagnetic Induction 567-587 22. Alternating Current 588-609 23. Electromagnetic Waves 610-623 24. Ray Optics and Optical Instruments 624-660 25. Wave Optics 661-684 26. Dual Nature of Radiation and Matter 685-710 27. Atoms 711-727 28. Nuclei 728-748 29. Semiconductor Electronics : Materials, Devices and Simple Circuits 749-784 (iv)
- 5. WHAT IS SCIENCE? Science refers to a system of acquiring knowledge. This system uses observation and experimentation to describe and explain natural phenomena. The term science also refers to the organized body of knowledge that people have gained using that system. Less formally, the word science often describes any systematic field of study or the knowledge gained from it. The purpose of science is to produce useful models of reality. Most scientific investigations use some form of the scientific method. The scientific method is a logical and rational order of steps by which scientists come to conclusions about the world around them. The Scientific method helps to organize thoughts and procedures so that scientists can be confident in the answers theyfind. Scientists use observations, hypotheses, and deductions to make these conclusions. Science as defined above, is sometimes called pure science to differentiate it from applied science, which is the application of research tohuman needs. Fields of science are commonlydivided into two major categories. 1. Natural science : The science in which we study about the natural world. Natural science includes physics, chemistry, biology, etc. 2. Social science : It is the systematic studyof human behavior and society. WHAT IS PHYSICS ? The word physics originates from a Greek word which means nature. Physics is the branch of science that deals with the study of basic laws of nature and their manifestation of various natural phenomena. There are two main thrusts in physics : 1. Unification : In physics, attempt is made to explain various physical phenomena in terms of just few concepts and laws. Attempts are being made to unify fundamental forces of nature in the persuit of unification. 2. Reductionism.Another attempt madein physics is to explain a macroscopic system in terms of its microsocopic constituents. This persuit is called reductionism. Keep in Memory 1. Informationreceivedthrough thesensesiscalled observation. 2. An idea that mayexplain a number ofobservations is called hypothesis. 3. A hypothesis that has been tested many times is called scientific theory. 4. A scientific theory that has been tested and has always proved true is called scientific law. SCOPE AND EXCITEMENT OF PHYSICS The scope of physics is very vast. It covers a tremendous range ofmagnitude of physical quantities like length, mass, time, energy, etc. Basically, there are twodomains ofinterest : macroscopic and microscopic. The macroscopic domain includes phenomena at the laboratory, terrestrial andastronomical scales. Themicroscopic domain includes atomic, molecular and nuclear phenomena. Classical physics deals mainly with macroscopic phenomena consisting ofthe studyofheat, light, electricity, magnetism, optics, acoustics, and mechanics. Since the turn of the 20th century, the study of quantum mechanics and relativity has become more important. Today, physics is often divided into branches such as nuclear physics, particle physics, quantum physics, theoretical physics, and solid-state physics. The study of the planets, stars, and their interactions is known as astrophysics, the physics of the Earth is called geophysics, and the study of the physical laws relating to living organisms is called biophysics. Physics isexciting in many ways. Application and exploitation of physical laws to make useful devices is the most interesting and exciting part and requires great ingenuityand persistence ofeffort. PHYSICS, TECHNOLOGY AND SOCIETY Physics generates new technologies and these technologies have made our lives comfortable and materially prosperous. We hear terms like science, engineering , technology all in same context though they are not exactly same. 1 Physical World
- 6. 2 PHYSICS Scientists Investigate the naturalworld Technology The products and processes created by engineers Scientific knowledge What scientists have learned about the natural world Engineers Create the designed world Values Needs Environment Economy Society Science: • A body of knowledge • Seeks to describe and understand the natural world and its physical properties • Scientific knowledge can be used to make predictions • Science uses a process--the scientific method--to generate knowledge Engineering: • Design under constraint • Seeks solutions for societal problems and needs • Aims toproduce the best solution from given resources and constraints • Engineering uses a process–the engineering design process–to produce solutions and technologies Technology: • The body of knowledge, processes and artifacts that result from engineering • Almost everything made by humans to solve a need is a technology • Examples of technologyinclude pencils, shoes, cell phones, and processes to treat water In the real world, these disciplines are closely connected. Scientists often use technologies created by engineers to conduct their research. In turn, engineers often use knowledge developed by scientists to inform the design of the technologies they create. Link between technology and physics Technology Scientific principle(s) Aeroplane Bernoulli’s principle in fluid dynamics Bose-Einstein condensate Trapping and cooling of atoms by laser beams and magnetic fields Computers Digital logic Electric generator Faraday’s laws of electromagnetic induction Electron microscope Wave nature of electrons Fusion test reactor (Tokamak) Magnetic confinement of plasma Giant Metrewave Radio Telescope (GMRT) Detection of cosmic radio waves Hydroelectric power Conversion of gravitational potential energy into electrical energy Lasers Light amplification by stimulated emission of radiation Non-reflecting coatings Thin filmoptical interference Nuclear reactor Controlled nuclear fission Optical fibres Total internal reflection of light Particle accelerators Motion of charged particles in electromagnetic fields Photocell Photoelectric effect Production of ultra high magnetic fields Superconductivity Radio and television Generation, propagation and detection of electromagnetic waves Rocket propulsion Newton’s laws of motion Sonar Reflection of ultrasonic waves Steam engine Laws of thermodynamics
- 7. 3 PhysicalWorld Some physicists fromdifferent countries of the world and their major contributions Name of physicists Major contribution/discovery Country of origin Abdus Salam Unification of weak and electromagnetic interactions Pakistan Albert Einstein Explanation of photoelectric effect; Theory of relativity Germany Archimedes Principle of buoyancy; Principle of the lever Greece C.H. Townes Maser; Laser U.S.A. Christiaan Huygens Wave theory of light Holland C.V. Raman Inelastic scattering of light by molecules India Edwin Hubble Expanding universe U.S.A. Enrico Fermi Controlled nuclear fission Italy Ernest Orlando Lawrence Cyclotron U.S.A. Ernest Rutherford Nuclear model of atom New Zealand Galileo Galilei Law of inertia Italy Heinrich Rudolf Hertz Generation of electromagnetic waves Germany Hideki Yukawa Theory of nuclear forces Japan Homi Jehangir Bhabha Cascade process of cosmic radiation India Issac Newton Universal law of gravitation; Laws of motion; Reflecting telescope U.K. James Clark Maxwell Electromagnetic theory; Light - an electromagnetic wave U.K. James Chadwick Neutron U.K. J.C. Bose Ultra short radio waves India J.J. Thomson Electron U.K. John Bardeen Transistors; Theory of super conductivity U.S.A. Lev Davidovich Landau Theory of condensed matter; Liquid helium Russia Louis Victor de Broglie Wave nature of matter France Marie Sklodowska Curie Discovery of radium and polonium; Studies on natural radioactivity Poland Michael Faraday Laws of electromagnetic induction U.K. M.N. Saha Thermal ionisation India Niels Bohr Quantum model of hydrogen atom Denmark Paul Dirac Relativistic theory of electron; Quantum statistics U.K. R.A. Millikan Measurement of electronic charge U.S.A. S. Chandrashekhar Chandrashekhar limit, structure and evolution of stars India S.N. Bose Quantum statistics India Victor Francis Hess Cosmic radiation Austria Werner Heisenberg Quantum mechanics; Uncertainty principle Germany W.K. Roentgen X-rays Germany Wolfgang Pauli Exclusion principle Austria Example 1. Does imagination play any role inphysics? Solution : Yes, imagination has clayed an important role in the development of physics. Huygen’s principle, Bohr’s theory, Maxwell equations, Heisenberg’s uncertaintyprinciple, etc. were the imaginations of the scientists which successfully explained the various natural phenomena. Example 2. How isscience different fromtechnology? Solution : Science is the studyofnatural laws while technologyis the practical application of these laws to the daily life problems. FUNDAMENTAL FORCES IN NATURE We come across several forces in our day-to-day lives eg., frictional force, muscular force, forces exerted by springs and strings etc. These forces actuallyarisefrom four fundamental forces of nature. Following are the four fundamental forces in nature. 1. Gravitational force 2. Weak nuclear force 3. Electromagnetic force 4. Strong nuclear force Among these forces gravitational force is the weakest and strong nuclear force is the strongest force in nature.
- 8. 4 PHYSICS Fundamental forcesof nature Name Relative strength Range Exchange particles Major role Important properties Gravitational force (Force of attraction between any two boies by virtue of their masses) 10–39 Infinite Gravitons Large-scale structure Universal attractive, weakest, long range, central, conservative force. Weak nuclear force (Changes quark types as in Beta-decay of nucleus 10–13 Very short, sub-nuclear size (on 10–16 m) Weak bosons Nuclear reactions Govern process involving neutrino and antineutrino, operates only through a range of nuclear size. Electromagnetic force (Force between particles with charge/magnetism) 10–2 Infinite Photons Chemistry and Biology Either attractive or repulsive, long range, central, conservative force. Strong nuclear force (Strong attractive force which binds together the protons and neutrons in nucleus) 1 Very Short, nuclear size (on 10–15 m) Gluons Holding nuclei together Basically an attractive becomes repulsive (when distance between nucleons < 0.5 fermi) strongest, short range, non-central, non- conservative force. Physicists are trying toderive a unified theorythat would describe all the forces in nature as a single fundamental law. So far, theyhave succeeded in producing a unified description of the weak and electromagnetic forces, but a deeper understanding of the strong and gravitational forces has not yet been achieved. Theories that postulate the unification of the strong, weak, and electromagnetic forces are called Grand Unified Theories (often known bythe acronym GUTs). Theories that add gravitytothe mix and tryto unify all four fundamental forces into a single force are called Superunified Theories. The theory that describes the unified electromagnetic and weak interactions is called the StandardElectroweak Theory, or sometimes just theStandardModel. Progress in unification of different forces/domains in nature Name of the physicist Year Achievement in unification Issac Newton 1687 Unified celestial and terrestrial mechanics; showed that the same laws of motion and the law of gravitation apply to both the domains. Hans Christian Oersted 1820 Michael Faraday 1830 James Clark Maxwell 1873 Unified electricity, magnetism and optics; showed that light is an electromagnetic wave. Sheldon Glashow, Abdus Salam, Steven Weinberg 1979 Showed that the 'weak' nuclear force and the electromagnetic force could be viewed as different aspects of a single electro-weak force. Carlo Rubia, Simon Vander Meer 1984 Verified experimentally the predictions of the theory of electro-weak force. Showed that electric and magnetic phenomena are inseparable aspects of a unified domain : electromagnetism. Example 3 .What is electromagnetic force? Solution : It is the force due to interaction between two moving charges. It is caused by exchange of photons (g) between two charged particles. NATURE OF PHYSICAL LAWS A physical law, scientific law, or a law of nature is a scientific generalization based on empirical observations of physical behavior. Empirical laws are typically conclusions based on repeatedscientific experimentsand simpleobservations over many years, and which have become accepted universally within the scientific community. The production ofa summarydescription of nature in the form of such laws is a fundamental aim of science. Physical laws are distinguished from scientific theories by their simplicity. Scientifictheories are generallymorecomplex than laws. They have many component parts, and are more likely to be changed as the body of available experimental data and analysis develops. This is because a physical law is a summaryobservation of strictly empirical matters, whereas a theory is a model that accounts for the observation, explains it, relates it to other observations,and makes testable predictions based upon it. Simply stated, while a lawnotes that something happens, a theoryexplains why and how something happens.
- 9.
- 10. 6 PHYSICS 1. Theman who has won Nobel Prize twice in physics is (a) Einstein (b) Bardeen (c) Heisenberg (d) Faraday 2. Prof. Albert Einstein got nobel prize in physics for his work on (a) special theory of relativity (b) general theoryof relativity (c) photoelectric effect (d) theory of specific heats 3. Which of the following is wronglymatched ? (a) Barometer-Pressure (b) Lactometer-Milk (c) Coulomb’s law-charges (d) Humidity-Calorimeter 4. C.V. Raman got Nobel Prize for his experiment on (a) dispersion of light (b) reflection of light (c) deflection of light (d) scattering of light 5. A scientific way of doing things involve (a) identifying the problem (b) collecting data (c) hypothesising a possible theory (d) All of the above 6. The scientific principle involves in production of ultra high magnetic fields is (a) super conductivity (b) digital logic (c) photoelectric effect (d) laws of thermodynamics 7. Which of the following has infinite range? (a) Gravitational force (b) Electromagnetic force (c) Strong nuclear force (d) Both (a) and (b) 8. Which of the following is the correct decreasing order of the strengths of four fundamental forces of nature ? (a) Electromagnetic force > weak nuclear force > gravitational force > strong nuclear force (b) Strong nuclear force > weak nuclear force > electromagnetic force > gravitational force (c) Gravitational force > electromagnetic force > strong nuclear force > weak nuclear force (d) Strong nuclear force > electromagnetic force > weak nuclear force > gravitational force 9. The exchange particles for the electromagnetic force are (a) gravitons (b) gluons (c) photons (d) mesons 10. Louis de-Broglie is credited for his work on (a) theory of relativity (b) electromagnetic theory (c) matter waves (d) law of distribution of velocities 11. Madam Marie Curiewon Nobel Prize twicewhich were in the field of (a) Physics and chemistry(b) Chemistryonly (c) Physics only (d) Biology only 12. The man whois known as theFather of Experimental Physics is (a) Newton (b) Albert Einstein (c) Galileo (d) Rutherford 13. The person who has been awarded the title of the Father of Physics of 20th century is (a) Madame Curie (b) SirC.V.Raman (c) Neils Bohar (d) Albert Einstein 14. Which ofthe following istrueregarding the physical science? (a) Theydeal with non-living things (b) The study of matter are conducted at atomic or ionic levels (c) Both (a) and (b) (d) None of these 15. The branch of science which deals with nature and natural phenomena is called (a) Sociology (b) Biology (c) Civics (d) Physics 16. Who gave general theory of relativity? (a) Einstein (b) Marconi (c) Ampere (d) Newton 17. Who discovered X-rays? (a) Chadwick (b) Roentgen (c) Thomson (d) Madam Curie 18. Which of the following is the weakest force? (a) Nuclear force (b) Gravitational force (c) Electromagnetic force (d) None of these 19. The field of work ofS. Chandrashekar is (a) theory of black hole (b) Cosmic rays (c) theory of relativity (d) X-rays 20. Two Indian born physicists who have been awarded Nobel Prize in Physics are (a) H. J. Bhabha andAPJ Kalam (b) C.V. Raman and S. Chandrasekhar (c) J.C. Bose and M.N. Saha (d) S. N. Bose and H. J. Bhabha 21. Science is exploring, ...x... and ...y... from what we see around us. Here, x and y refer to (a) qualitative, modify (b) experiment, predict (c) verification, predict (d) reasoning, quantitative 22. Macroscopic domain includes (a) phenomena at the laboratory (b) terrestrial scales (c) astronomical scales (d) All of the above
- 11. 7 PhysicalWorld EXERCISE - 1 1.(b) 2. (c) 3. (d) 4. (d) 5. (d) 6. (a) 7. (d) 8. (d) 9. (c) 10. (c) 11. (a) 12. (c) 13. (d) 14. (c) 15. (d) 16. (a) 17. (b) 18. (b) 19. (a) 20. (b) 21. (b) Science is exploring, experimenting and predicting from what we see around us. 22. (d) The macroscopic domain includes phenomena at the laboratory, terrestrial and astronomical scales. 23. (b) The alpha particle scattering experiment ofRutherford gave the nuclear model of the atom as shown in figure q Flash of light a Lead block Fluorescent screen Scattering angle B A Gold foil Polonium sample 23. In Rutherford, alpha particle scattering experiment asshown in given figure,Aand B refer to q Flash of light a Lead block Fluorescent screen Scattering angle B A Microscope (a) polonium sample and aluminium foil (b) polonium sample and gold foil (c) uranium sampleand gold foil (d) uraniumsampleand aluminium foil DIRECTIONS for Qs. (24-25) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethe correctanswer (ONLYONE option iscorrect)fromthe following (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 istrue, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 24. Statement-1 : The concept of energy is central to Physics and expression for energy can be written for every physical system. Statement-2 : Law of conservation of energy is not valid for all forces and for any kind of transformation between different forms of energy. 25. Statement-1 : Electromagneticforce is much stronger than the gravitational farce. Statement-2 : Electromagnetic force dominates all phenomena at atomic and molecular scales. 24. (d) The concept of energy is central to Physics and the expressions for energycan bewritten for everyphysical system. When all formsofenergye.g., Heat, mechanical energy, electrical energy etc., are counted, it turns out that energy is conserved. The general law of conservation of energyis true for all forces and for any kind of transformation between different forms of energy. 25. (b) It is mainly the electromagnetic force that governs the structure of atoms and molecules, the dynamics of chemical reactions and the mechanical, thermal and other properties of materials. Hints & Solutions
- 12. 8 PHYSICS PHYSICAL QUANTITY Allthe quantities in terms of which laws of physics are described and which can be measured directly or indirectly are called physical quantities. For example mass, length, time, speed, force etc. Types of Physical Quantity 1. Fundamental quantities : Thephysical quantities which do not depend upon other physical quantities are called fundamental or base physical quantities. e.g. mass, length, time temperature electric current, luminous intensity and amount of substance. 2. Derived quantities :The physical quantities which depend on fundamental quantities are called derived quantities e.g. speed, acceleration, force, etc. UNIT The process of measurement is a comparison process. Unit is the standard quantity used for comparision. The chosen standard for measurement of a physical quantity, which has the same nature as that of the quantity is called the unit of that quantity. Choice of a unit (Characteristics of a unit): (1) It should be suitable in size (suitable to use) (2) It should be accurately defined (so that everybody understands the unit in same way) (3) It should be easily reproducible. (4) It should not change with time. (5) It should not change with change in physical conditions i.e., temperature, pressure, moisture etc. (6) It should be universally acceptable. Every measured quantity (its magnitude) comprises of a number and a unit. Ex: In the measurement of time, say 15 s Number (n) Unit (u) If Q is the magnitude of the quantity (which does not depend on the selection of unit) then Q = n u = n1 u1 = n2 u2 Þ 1 µ n u Where u1 and u2 are the units and n1 and n2 are the numerical values in two different system of units. Fundamental (or Base) and Derived Units Fundamental units are those, which are independent of unit of other physical quantity and cannot be further resolved into any other units or the units of fundamental physical quantities are called fundamental or base units. e.g., kilogram, metre, second etc, All units other than fundamental are derived units (which are dependent on fundamental units)e.g., unit ofspeed(ms–1)which depends on unit of length (metre) and unit of time (second), unit ofmomentum (Kgms–1) depends on unit ofmass, length and time etc. SYSTEM OF UNITS A system of units is a complete set of fundamental and derived units for all physical quantities. Different types of system of units F.P.S. (Foot - Pound - Second)system. (British engineeringsystem of units.): In this system the unit of length is foot, mass is pound and time is second. C.G.S. (Centimetre - Gram- Second) system. (Gaussian system of units): In this system the unit of length is centimetre, mass is gram and time is second. M.K.S (Metre - Kilogram-Second) system. Thissystem isrelated tomechanics only. In this system the unit oflength is metre, mass is kilogram and time is second. S.I. (International system) units: (Introduced in 1971) Different countries use different set of units. To avoid complexity, by international agreement, seven physical quantities have been chosen as fundamental or base physical quantities and two as supplementary. These quantities are S.No Base physical quantity Fundamental unit Symbol 1 Mass kilogram kg 2 Length metre m 3 Time second s 4 Temperature kelvin K 5 Electric current ampere A 6 Luminous intensity candela cd 7 Amount of substance mole mol 2 Units and Measurements
- 13. 9 Units and Measurements S.NoSupplementary physical Supplementary unit Symbol quantity 1 Plane angle radian rad 2 Solid angle steradian sr Merits of S.I. Units : (1) SI is a coherent system ofunits: This means that all derived units are obtained by multiplication and division without introducing anynumerical factor. (2) SI is a rational system of units: This is because it assigns only one unit to a particular physical quantity. (3) SI is an absolute system of units: There is no gravitational unit in this system. (4) SI system is applicable to all branches of science. Conventions of writing of Units and their Symbols · Unit is never writtenwith capital initial letter. · For a unit named after scientist the symbol is a capital letter otherwise not. · The unit or symbol is never written in plural form. · Punctuationsmarks are not written after the symbol. Definitions of Fundamental Units (i) Metre : One metre is equal to 1650763.73 wavelength in vacuum of the radiation corresponding to transition between the levels 2p10 and 5d5 of the krypton – 86 atom Or The distance travelled by light in vacuum in 458 , 792 , 299 1 second is called 1 metre. (ii) Kilogram : The mass of cylinder (of height and diameter 39 cm) made ofPlatinum-iridium alloykept at International Bureau of weights and measures in paris is defined as 1kg. (iii) Second : It is the duration of 9,192,631,770 periods of radiation corresponding to the transition between the two hyperfine levels of the ground state of Caesium (133) atom. (iv) Ampere : It is the current which when flows through two infinitely long straight conductors of negligible cross- section placed at a distance of one metre in air or vacuum produces a force of 2 × 10–7 N/m between them. (v) Candela : It is the luminous intensity in a perpendicular direction, of a surface of 1/600,000 square metre of a black bodyat the temperatureoffreezingplatinum under a pressure of1.013 × 105 N/m2. (vi) Kelvin: It is the1/273.16part ofthermodynamictemperature of triple point of water. (vii) Mole : It is the amount ofsubstance which contains as many elementaryentities as there are in 0.012 kgofCarbon-12. S.I. Prefixes : The magnitudes of physical quantities vary over a wide range. For example, the atomicradius, is equal to10–10m, radius ofearth is 6.4×106 m and the mass of electron is 9.1×10–31 kg. The internationallyrecommended standard prefixes for certain powers of 10 are given in the table: Prefix Power of 10 Symbol exa 18 E peta 15 P tera 12 T giga 9 G mega 6 M kilo 3 k hecto 2 h deca 1 da deci –1 d centi –2 c milli –3 m micro –6 m nano –9 n pico –12 p femto –15 f atto –18 a Some Important Practical Units : (1) For large distance (macro-cosm) (a) Astronomical unit: It is the average distance of the centre of the sun from the centre of the earth. 1A.U. = 1.496 ×1011m (b) Light year: It is the distance travelled by the light in vacuum in one year. 1 ly = 9.46 × 1015m (c) Parsec: One parsec is the distance at which an arc 1A.U. long subtends an angle of one second. 1 parsec = 3.1 × 1016m (2) For small distance (micro-cosm) 1 micron = 10–6m 1nanometre = 10–9m 1angstorm = 10–10m 1fermi = 10–15m (3) For small area 1 barn = 10–28m2 (4) For heavy mass 1 ton = 1000kg 1quintal = 100kg 1slug= 14.57kg 1C.S.L (chandrasekhar limit)= 1.4 times the mass of the sun (5) For small mass 1 amu = 1.67 x 10–27kg 1 pound = 453.6g = 0.4536 kg (6) For small time 1 shake = 10–8s (7) For large time Lunar month: It is the time taken by the earth to complete one rotation about its axis with respect to sun. 1L.M. = 27.3 days. Solar day: It is the time taken by the earth to complete one rotation about its axis with respect to sun. Sedrial day: It is the time taken by earth to complete one rotation on its axis with respect to distant star. (8) For measuring pressure 1 bar =1atm pressure= 105N/m2 = 760mmHg 1torr = 1mmHg 1 poiseuille = 10 Poise. DIMENSIONS The powers to which the fundamental units of mass, length and time must be raised to represent the physical quantity are called the dimensions of that physical quantity.
- 14. 10 PHYSICS For example: Force = mass × acceleration = mass × 1 v–u [LT ] [M] t [T] - = = [MLT T–2] Hence the dimensions of force are 1 in mass 1 in length and (– 2) in time. Dimensional Formula : Unit of a physical quantity expressed in terms of M, L and T is called dimensional formula. It shows how and which of the fundamental quantities represent the dimensions. For example, the dimensional formula of work is [ML2T–2] Dimensional Equation : When we equate the dimensional formula with the physical quantity, we get the dimensional equation. For example Work=[ML2T–2] Classification of Physical Quantities (On the basis of dimensions) : Physical Quantity Dimensional physical quantity Dimensional constant For e.g. c (velocity of light in vaccum) For e.g. distance, displacement, force, mass, time etc. For e.g. 0, 1, 2, ....., e, , sin , cos , tan , etc. q p q q For e.g. plane angle, solid angle, strain, refractive index, dielectric constant, relative density, specific gravity, poisson's ratio etc. R(universal gas constant), stefan's constant), h (Planck's constant), k (Boltzmann constant), G (universal gravitational constant) etc. s( Dimensionless constant Dimensional variable Dimensionless variable Dimensionless physical quantity Dimensional Formula of Some Important Physical Quantities : S.No. Physical quantity Relation with other quantities Dimensional formula 1. Velocity(v) Length Time [M0LT–1] 2. Acceleration (a) Velocity Time [M0LT–2] 3. Momentum (p) Mass × velocity [MLT–1] 4. Force (F) Mass × acceleration [MLT–2] 5. Work Force × displacement [ML2T–2] 6. Power (P) Work Time [ML2T–3] 7. Universal gravitational constant 2 1 2 m m Fr G = [M–1L3T–2] 8. Torque F r ´ = t [ML2T–2] 9. Surface tension F S = l [MT–2] 10. Gravitational potential G W V m = [M0L2T–2] 11. Coefficient of viscosity dx dv A F = h [ML–1T–1] 12. Impulse Force×time(F×t) [MLT–1]
- 15. 11 Units and Measurements 13.Strain Changein length L Originallength L D æ ö ç ÷ è ø [M0L0T0] 14. Pressure gradient Pressure P Distance æ ö ç ÷ è ø l [ML–2T–2] 15. Plane angle Arc s Radius of circle r æ ö ç ÷ è ø [M0L0T0] 16. Angular velocity Angle Time t q æ ö ç ÷ è ø [M0L0T–1] 17. Radius of gyration i Moment of inertia of body I Total mass of the body m æ ö ç ÷ ç ÷ å è ø [M0L1T0] 18. Moment of force, moment of couple Force × distance (F × s) [ML2T–2] 19. Angular frequency 2p×frequency(2pn) [M0L0T–1] 20. Pressure Force Area [ML–1T–2] 21. Efficiency Output work or energy W Input work or energy Q æ ö ç ÷ è ø [M0L0T0] 22. Angular impulse forque×time (t × t) [ML2T–1] 23. Planck’s constant Energy E Frequency æ ö ç ÷ n è ø [ML2T–1] 24. Heat capacity, Entropy Heat energy Q Temperature T æ ö ç ÷ è ø [ML2T–2K–1] 25. Specific heat capacity Heat energy Q Mass temperature m T æ ö ç ÷ ´ ´D è ø [M0L2T–2K–1] 26. Thermal conductivity, K Heat energy thickness Q T KA Area temperature time t x ´ D æ ö = - ç ÷ ´ ´ D è ø [MLT–3K–1] 27. Thermal Resistance, R Length Thermal conductivity × area [M–1L–2T3K] 28. Bulk modulus (B) or (compressibility)–1 Volume (changein pressure) P V Change in volume V ´ D æ ö - ç ÷ D è ø [ML–1T–2] 29. Stefan’s constant (s) 4 4 4 Q AtT (Energy/area) Time (temperature) E Q/ A.t T æ ö = s ç ÷ ç ÷ ´ = = s è ø [ML0T–3K–4] 30. Universal gas constant R Pressure volume PV Mole temperature nT ´ æ ö ç ÷ ´ è ø ] mol K T ML [ 1 1 2 2 - - - 31. Voltage, electric potential (V) or Work W Charge q æ ö ç ÷ è ø [ 1 3 2 A T ML - - ] electromotive force (e)
- 16. 12 PHYSICS 32. Capacitance(C) Charge q Potentialdifference V æ ö ç ÷ è ø [ 2 4 2 1 A T L M - - ] 33. Electricfield ( ) E r Electric force F Charge q æ ö ç ÷ è ø [ 1 3 A MLT - - ] 34. Magnetic field(B) r , magnetic induction, Force [F I Bsin ] Current length = q ´ l [ML0T–2A–1] magnetic flux density 35. Magnetic flux (fm) f = BAcosq [ML2T–2A–1] 36. Inductance m Magneticflux Current I f æ ö ç ÷ è ø [ 2 2 2 A T ML - - ] coefficient of self inductance (L) or coefficient of mutual inductance (M) 37. Magnetic field strength or magnetic moment density(I) Magneticmoment M Volume V æ ö ç ÷ è ø [M0L–1T0A] 38. Permittivity constant in free space eo 2 2 2 2 (Charge) q 4π electrostaticforce(distance) 4 F r æ ö ç ÷ ç ÷ ´ p´ ´ è ø 1 3 4 2 M L T A - - é ù ë û 39. Faraday constant (F), charge Avagadro constant × elementry charge ] mol TA L M [ 1 0 0 - 40. Mass defect, (Dm) (Sum of masses of nucleons 0 0 [ML T ] – mass of nucleus)(MP+MN –Mnucleus) 41. Resonant frequency (fr) r 1 T ] T L M [ 1 0 0 - 42. Power of lens -1 1 (Focal length) f æ ö ç ÷ è ø 0 1 0 M L T - é ù ë û 43. Refractive index Speedof lightin vacuum Speedof lightin medium 0 0 0 [M L T ] 44. Wave number 2 Wavelength p 0 1 0 M L T - é ù ë û 45. Binding energy of nucleus Mass defect × (speed of light in vacuum)2 2 2 ML T- é ù ë û 46. Conductance (c) 1 Resistance 1 2 2 3 M L A T - - é ù ë û 47. Fluid flow rate 4 (Pressure) (radius) 8 (Viscosity coefficient) (length) p ´ æ ö ç ÷ è ø ´ 0 3 –1 M L T é ù ë û 48. Inductive reactance (Angular frequency× inductance) 2 3 2 ML T A - - é ù ë û 49. Capacitive reactance (Angular frequency × capacitance)–1 2 3 2 ML T A - - é ù ë û 50. Magnetic dipole moment Torque or Current area Magneticfield ´ 0 2 0 M L T A é ù ë û
- 17. 13 Units and Measurements Shortcuts / Time saving techniques 1. To find dimensions of a typical physical quantity which is involved in a number of formulae, try to use that formula which is easiest for you. For example if you want to find the dimensional formula of magnetic induction then you can use the following formulae q = t = m = q p m = sin MB , qvB F , nI B , r sin Idl 4 dB 0 2 0 Out of these the easiest is probably the third one. 2. Ifyou have tofind the dimensional formula of a combination of physical quantities, then instead of finding the dimensional formula ofeach, tryto correlate the combination ofphysical quantities with a standard formula. For example, if you have to find the dimension of CV2, then try to use formula 2 CV 2 1 E = where E is energyofa capacitor. . 3. = = e m c 1 0 0 velocityof light in vacuum • Dimensions ofthe following are same 2 2 2 2 LI t R V c q CV qV nRT PV Work = ´ = = = = = = [ML2T–2] • Dimensions ofthe following are same Force = Impulse / time = q v B = q E = Thrust = weight = energygradient [MLT–2] • The dimension of RC = R L is same as that of time • Dimensions ofthe following are same velocity = 0 0 T 1 f = = ´l m m e [M°LT T–1] • Dimensions ofthe following are same Frequency R k MB 1 L m I LC = = = = [M°L°T–1] • Dimensions ofthe following are same (E) Modulus of elasticity =Y (Young's modulus) = B (Bulk modulus) = h (Modulus of rigidity) = Stress = Pressure = ility Compressib 1 [ML–1T–2] • Dimensions ofthe following are same Acceleration, retardation, centripetal acceleration, centrifugal acceleration, gravitational intensity/strength. [M°LT–2] • Dimensions ofthe following are same Water equivalent, thermal capacity, entropy, Boltzmann's constant. [ML2T–2K–1] Keep in Memory The dimensional formula of • all trigonometricratio is [M0L0T0] • x in ex is [M0L0T0] • ex is [M0L0T0] • x in log x is [M0L0T0] • log x is [M0L0T0] Example1. Find out the unit and dimensions of permittivity of free space. Solution : According to Coulomb’s law 2 2 2 1 3 4 2 0 2 2 2 A T [q ] [M L T A ] 4 [F][r ] [MLT ][L ] - - - é ù ë û e = = = p Its unit = 2 2 2 coulomb (coulomb) coulomb joule metre volt metre newton metre = = ´ ´ ´ CV–1m–1 Example2. Find out the unit and dimensions of coefficient of self or mutual inductance of. Solution : e = ÷ ø ö ç è æ dt d L I or ÷ ø ö ç è æ dt d M I , where e is induced electromotive force(e.m.f.) dt W t L e d q æ ö æ ö = = ç ÷ ç ÷ è ø è ø I I or [L] = ] A T ML [ ] A ][ AT [ ] T ][ T ML [ 2 2 2 2 2 - - - = Its unit is volt × sec/amp or ohm × sec or henry. Example3. Find out the unit and dimensions of magnetic field intensity . Solution : As B = mH, hence 2 r sin d 4 1 B H q p = m = l I ; 1 2 [A][L] H [M L T A] [L ] - = = ° ° Its unit is ampere /metre in SI system. In c.g.s. system, the unit is oersted. Example4. Find out the unit and dimensions of magnetic permeability of free space or medium. Solution : According to Biot-Savart’s law 2 0 r sin d 4 B q p m = l I and F = BI l sinq or 0 2 d sin F B sin 4 r m q = = q p l l I I ;
- 18. 14 PHYSICS 2 0 22 Fr or m m = l I (dimensionally) Hence [m] or ] A MLT [ ] L ][ A [ ] L ][ MLT [ ] [ 2 2 2 2 2 2 0 - - - = = m Its units are 2 2 2 2 2 N m newton joule/ metre volt coulomb amp amp metre amp m amp amp - ´ = = = ´ ´ - ohm sec henry tesla metre metre metre amp. ´ ´ = = = Example5. The dimensions of physical quantity X in the equation Force = X Density is given by (a) [ML4T–2] (b) [M2L–2T–1] (c) [M2L–2 T–2] (d) [ML–2T–1] Solution : (c) Q Force = X denisty X = Force × density Hence X has dimensions [MLT–2] 3 [M] [L ] = [M2L–2T–2] Example6. If force, acceleration and time are taken as fundamental quantities, then the dimensions of length will be- (a) [FT2] (b) [F–1A2T–1] (c) [FA2T] (d) [AT2] Solution : (d) x y z L F A T = 0 1 0 2 x 2 y z M L T [MLT ] [LT ] T - - = = x x y 2x 2y z M L T + - - + x = 0, x + y= 1, – 2x – 2y+z = 0 x =0, y= 1, z = 2 Hence, 2 L AT = DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Principle of Homogeneity: Only those physical quantities can be added /subtracted/equated /compared which have the same dimensions. Uses of Dimensions : (1) Conversion of one systemof unit intoanother Example : Convert a pressure of 106 dyne/cm2 in S.I units. Sol. We know that 1N = 105 dyne Þ 1 dyne = 10–5 N Also1m = 100 cm Þ 1cm = 10–2 m Now, the pressure 106 dyne/cm2 in SI unit is 106 2 5 2 2 – 5 6 m / N 10 m 10 m 10 N 10 10 cm cm dyne = ´ ´ = ´ - - (2) Checking the accuracy of various formulae Example : Check the correctness of the following equation dimensionally dv F = ηA sinθ dx where F=force, h =coefficientofviscosity, , A = area, dv = velocity dx gradient w.r.t distance, q = angle of contact Sol. L.H.S = force = [MLT–2] R.H.S= 1 1 1 1 2 2 dv LT A (sin ) M L T L MLT dx L - - - - é ù h q = ´ = ë û The equation is dimensionallycorrect. CAUTION : Please note that the above equation is not correct numerically. The conclusion is that an equation, if correct dimensionally, may or may not be numerically correct. Also remember that if an equation is dimensionallyincorrect, we can conclude with surety that the equation is incorrect. (3) Derivationof formula Example : The air bubble formed by explosion inside water performed oscillation with time period T which is directly proportional to Pa db Ec where P is pressure, d is density and E is the energy due to explosion. Find the values of a, b and c. Sol. Let us assume that the required expression for time period is T = K Pa db Ec where K is a dimensionless constant. Writing dimensions on both sides, c 2 2 b 3 a 2 1 1 0 0 ] T ML [ ] ML [ ] T ML [ ] T L M [ - - - - = ] T [ ] T [ ] L [ ] M [ 1 c 2 a 2 c 2 b 3 a c b a = = - - + - - + + Equating the powers, a + b + c = 0 ....(1) 0 c 2 b 3 a = + - - ....(2) – 2a – 2c = 1 ....(3) Solving these equations, we get, a = 6 5 - , b = 2 1 , c = 3 1 . Limitationsof DimensionalAnalysis: (1) Noinformation aboutthedimensionlessconstant isobtained during dimensional analysis (2) Formula cannot be found if a physical quantityis dependent on more than three physical quantities. (3) Formula containing trigonometrical /exponential function cannot be found. (4) Ifan equation is dimensionally correct it mayor may not be absolutely correct. Example7. Find the dimensions of a and b in the Van der waal's equation æ ö ç ÷ è ø 2 a P+ (V - b) = RT V where P is pressure and V is volume of gas.
- 19. 15 Units and Measurements Solution: Dimensionally 2 a P [Byprincipleof homogeneity] V = ] T [ML a V a T ML 2 5 2 2 1 - - - = Þ = Þ AlsodimensionallyV=b [Byprinciple ofhomogeneity] b= [L3] Example8. The formula 2 2 3mg (1 cos ) v M (1 sin ) - q = + q l is obtained as the solution of a problem. Use dimensions to find whether this is a reasonable solution (v is a velocity, m and M are masses, l is a length and g is gravitational acceleration). Solution : Dimensions of L.H.S. = [LT–1]2 = [L2T–2] Dimensions of R.H.S. = 2 2 2 [M][LT ][L] [L T ] [M] - - = Hence formula is reasonable. Example9. In the formula; 2 1 2 1 n n N D x x é ù - = - ê ú - ë û , D = diffusion coefficient, n1 and n2 isnumber of molecules in unit volume along x1 and x2 which represents distances where N is number of molecules passing through per unit area per unit time calculate the dimensions of D. Solution : By homogeneity theory of dimension Dimension of (N) = Dimension of D × 2 1 2 1 dimension of (n n ) dimension of (x x ) - - 2 1 L T = Dimension of D × 3 L L - Þ Dimensions of 'D' = 3 2 L L L T - ´ = 2 L T = [L2T–1] Example10. Let us consider an equation 2 1 mv mgh 2 = where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct. Solution : The dimensions of LHS are [M] [L T–1 ]2 = [M] [ L2 T–2] = [M L2 T–2] The dimensions of RHS are [M][L T–2] [L] = [M][L2 T–2] = [M L2 T–2] The dimensions of LHS and RHS are the same and hence the equation is dimensionallycorrect. Example11. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals b m, the unit of time isg s. Show that a calorie has a magnitude 4.2 a–1 b –2 g2 in terms of the new units. Solution : 1 cal = 4.2 kg m2 s–2. SI system New system n1 = 4.2 n2 = ? M1 = 1 kg M2 = a kg L1 =1m L2 = b metre T1 = 1s T2 = g second Dimensional formula of energyis [ML2T–2] Comparing with [MaLbTc], we find that a = 1, b = 2, c = – 2 Now, a b c 1 1 1 2 1 2 2 2 M L T n n M L T é ù é ù é ù = ê ú ê ú ê ú ë û ë û ë û = 1 2 2 1kg 1m 1s 4.2 kg m s - é ù é ù é ù ê ú ê ú ê ú a b g ë û ë û ë û = 4.2 a–1 b –2 g2 SIGNIFICANT FIGURES The number of digits, which are known reliably in our measurement, and one digit that is uncertain are termed as significant figures. Rules to determine the numbers of significant figures: 1. All non-zerodigits are significant. 235.75has fivesignificant figures. 2. All zeroes between two non-zero digits are significant. 2016.008 has seven significant figures. 3. All zeroes occurring between the decimal point and the non- zero digits are not significant. provided there is only a zero to left of the decimal point. 0.00652 has three significant figures. 4. All zeroes written to the right of a non-zero digit in a number written without a decimal point are not significant. This rule does not work if zero is a result of measurement. 54000 has two significant figures whereas 54000m has five significant figures. 5. All zeroes occurring to the right of a non-zero digit in a number written with a decimal point are significant. 32.2000 has six significant figures. 6. When a number is written in the exponential form, the exponential term does not contribute towards the significant figures. 2.465 × 105 has four significant figures. Keep in Memory 1. The significant figures depend upon the least count of the instrument. 2. The number of significant figure does not depend on the units chosen.
- 20. 16 PHYSICS ROUNDING OFF 1.If digit to be dropped is less than 5 then preceding digit should be left unchanged. 2. If digit to be dropped is more than 5 then one should raise preceding digit by one. 3. If the digit to be dropped is 5 followed bya digit other than zero then the preceding digit is increased by one. 4. If the digit to be dropped is 5 then the preceding digit is not changed if it is even. 5. Ifdigit to be droppedis5 then the precedingdigit isincreased by one if it is odd. Arithmetical Operations with Significant Figures and Rounding off : (1) For addition or subtraction, write the numbers one below the other with all the decimal points in one line. Nowlocate the first column from the left that has a doubtful digit. All digits right tothis column are dropped from all the numbers and rounding is done to this column. Addition subtraction is then done. Example : Find the sum of 23.623 and 8.7 to correct significant figures. Sol. Step-1:- 23.623+ 8.7 Step-2:- 23.6+ 8.7=32.3 (2) In multiplication and division oftwoor more quantities, the number of significant digits in the answer is equal to the number of significant digits in the quantity, which has minimum number ofsignificant digits. The insignificant digits are dropped from the result if they appear after the decimal point. Theyare replaced byzeroes if they appear to the left of the decimal point. The least significant digit is rounded off. Example:107.88(5.S.F.) × 0.610 (3 S. F.) = 65.8068 @ 65.8 ACCURACY, PRECISION OF INSTRUMENTS ANDERRORS IN MEASUREMENTS: Accuracy and Precision are two terms that have very different meanings in experimental physics. We need to be able to distinguish between an accurate measurement and a precise measurement. An accurate measurement is one in which the results of the experiment are in agreement with the ‘accepted’ value. This onlyapplies to experiments where this is the goal like measuring the speed of light. A precise measurement is one that we can make to a large number of decimal places. Thefollowingdiagrams illustratethemeaningof termsaccuracy and precision : In the above figure : The centre of the target represents the accepted value. The closer to the centre, the more accurate the experiment. The extent of the scatter of the data is a measure of the precision. A- Precise and accurate, B-Accurate but imprecise, C- Precisebut not accurate, D- Not accurate nor precise When successive measurements ofthe same quantity are repeated there are different values obtained. In experimental physics it is vital to be able to measure and quantify this uncertainty. (The words "error" and "uncertainty" are often used interchangeably by physicists - this is not ideal - but get used to it!) Error in measurements is the difference of actual or true value and measured value. Error = True value – Measured value Keep in Memory 1. Accuracy depends on the least count of the instrument used for measurement. 2. In the addition and subtraction operation, the result contains the minimum number of decimal places of the figures being used 3. In the multiplication and division operation, the result contains the minimum number of significant figures. 4. Least count (L.C.) ofvernier callipers = one MSD– oneVSD where MSD = mains scale division VSD = vernier scale division 5. Least count of screw gauge (or spherometer) pitch = no of divisions on circular scale where pitch is the ratio of number of divisions moved on linear scale and number of rotations given to circular scale. 6. Pure number or unmeasured value do not have significant numbers 7. Change in the position of decimal does not change the number of significant figures. Similarlythe change in the units ofmeasured value does not change the significant figures.
- 21. 17 Units and Measurements Methodsof Expressing Error: Absolute error: It is the difference between the mean value and the measured value of the physical quantity. |DX1| = |Xmean–X1| .................................. .................................. |DXn| = |Xmean–Xn| Meanabsolute error: DXmean or DX = 1 2 | | | | ......... | | D + D + + D n X X X n Relative error: It is the ratio of the mean absolute error and the value of the quantity being measured. ( ) D d = mean mean X Relative error a X Percentage error: It is the relative error expressed in percent Percentage error 100% D = ´ X X Tofind the maximum error in compoundquantitieswe proceed as : (i) Sum and difference : We have to find the sum or difference of two values given as (a ± Da) and (b ± Db), we do it as follows X ± DX = (a ± Da) + (b± Db) = (a + b) ± (Da + Db) Þ X = a + b and DX = Da + Db in case of sum And X = (a – b) and DX = Da + Db in case of difference. (ii) Product and quotient : We add the fractional or percentage errors in case of finding product or quotient. If P = ab then P a b P a b D D D æ ö = ± + ç ÷ è ø If b a Q = then Q a b Q a b D D D æ ö = ± + ç ÷ è ø (iii) Power of a quantity : If x = an then X a n X a D D æ ö = ç ÷ è ø Example : For y x Q 2 = , If % 3 100 x x = ´ D and % 4 100 y y = ´ D then Q Q D × 100 = (2 × 3 + 4)% = 10% Similarly : If p q r m n A m n c A then p q r A m n c c D D D D é ù = = ± + + ê ú ë û Keep in Memory 1. More the accuracy, smaller is the error. 2. Absolute error |DX| is always positive. 3. |DX| has the same dimensions as that of X. 4. If the least count of measuring instrument is not given and the measured valueisgiven the least error in themeasurement can be found by taking the last digit to be 1 and rest digit to be zero. For e.g. if the measured value of mass m = 2.03 kg then 01 . 0 m ± = D kg. 5. If a number of physical quantities are involved in an expression then the one with higher power contributes more in errors and therefore should be measured more accurately. 6. Relative error is a dimensionless quantity. Example12. Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures? Solution : The number of significant figures in the measured length 7.203 m is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures. Surface area of the cube =6(7.203)2 m2 =311.299254m2 =311.3m2 Volume of the cube= (7.203)3 m3=373.714754m3 =373.7m3 Example13. 5.74 g of a substance occupies volume 1.2 cm3. Express its density by keeping the significant figures in view. Solution : There are 3 significant figures in the measured masswhereas there are only2 significant figures in the measured volume. Hence the density should be expressed to only2 significant figures. Density= mass volume = 3 5.74 gcm 1.2 - = 4.8 g cm–3 . Example14. The mass of a box measured by a grocer’s balance is 2.300 kg. Twogold pieces ofmasses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? Solution : (a) Total mass = 2.3403 kg = 2.3 kg (upto 2 S. F.) (b) Difference = 20.17 g – 20.15 g (upto4 S. F.) COMMON ERRORS IN MEASUREMENTS It is not possible tomeasure the 100% correct valueofanyphysical quantity, even after measuring it so many times. There always exists someuncertainty, which isusuallyreferred toasexperimental error. Experimental errors : (i) Random error : It is the error that has an equal chance of being positive or negative. It occurs irregularly and at random in magnitude and direction. It can be caused (a) by the lack of perfection of observer (b) if the measuring instrument is not perfectlysensitive. (ii) Systematic error : It tends to occur in one direction either positive or negative. It occurs due to (a) measuring instrument having a zeroerror. (b) an instrument being incorrectly calibrated (such as slow- running-stop clock) (c) the observer persistentlycarrying out a mistimed action (e.g., in starting and stopping a clock) For measuring a particular physical quantity, we takea number of readings. Let the readings be X1, X2............,Xn. Then the mean value is found as follows ..... 1 2 ( ) + + + = = X X Xn X or true value X mean n
- 22. 18 PHYSICS 7. Wearealways interestedin calculating the maximum possible error. Example15. In an experiment, the refractive index of water was observed as1.29, 1.33,1.34,1.35,1.32, 1.36,1.30and1.33. Calculatethe mean value, mean absolute error and percentage error in the measurement. Solution : Mean value of refractive index, 1.29 1.33 1.34 1.35 1.32 1.36 1.30 1.33 8 + + + + + + + m = = 1.33 Absoluteerror in measurement, 1 1.33 1.29 0.04 Dm = - = + , 2 1.33 1.33 0.00 Dm = - = , 3 1.33 1.34 0.01 Dm = - = - , 4 1.33 1.35 0.02 Dm = - = - , 5 1.33 1.32 0.01 Dm = - = + , 6 1.33 1.36 0.03 Dm = - = - , 7 1.33 1.30 0.03 Dm = - = + , 8 1.33 1.33 0.00 Dm = - = So, mean absolute error, | 0.04 | | 0.01| | 0.02| | 0.01| | 0.03| | 0.03| | 0 | ( ) 8 + + + + + + Dm = =0.0175 0.02 » Relative error = 0.02 0.015 0.02 1.33 Dm ± = + = ± = ± m Percentage error = 0.02 100% 1.5% 1.33 ± ´ = ± Example16. The length and breadth ofa rectangle are (5.7 ± 0.1) cm and (3.4 ± 0.2) cm. Calculatearea oftherectanglewith error limits. Solution : Here, l= (5.7 ± 0.1) cm, b= (3.4 ± 0.2) cm Area A = l × b = 5.7× 3.4= 19.38cm² = 19.0 cm² (Rounding off totwo significant figures) A A D = ± b b D D æ ö + ç ÷ è ø l l = ± 0.1 0.2 5.7 3.4 æ ö + ç ÷ è ø = ± 0.34 1.14 5.7 3.4 + æ ö ç ÷ è ø ´ A A D = ± 1.48 19.38 Þ DA = ± 1.48 A 19.38 ´ = ± 1.48 19.38 1.48 19.38 ´ = ± DA = ± 1.5 (Rounding off to two significant figures) Area = (19.0 ± 1.5) sq.cm. Example17. A body travels uniformly a distance of (13.8 ± 0.2) m in a time (4.0 ± 0.3) s. Calculate its velocity with error limits. What is percentage error in velocity? Solution : Here, s= (13.8 ±0.2) m, t = (4.0 ± 0.3) s velocity, s 13.8 v t 4.0 = = = 3.45 ms–1 = 3.5 ms–1 (rounding off to two significant figures) v s t v s t D D D æ ö = ± + ç ÷ è ø ( ) 0.8 4.14 0.2 0.3 13.8 4.0 13.8 4.0 + æ ö = ± + = ± ç ÷ è ø ´ Þ v 4.94 0.0895 v 13.8 4.0 D = ± = ± ´ D v = ± 0.0895 × v = ± 0.0895 × 3.45 = ± 0.3087 = ± 0.31 (Rounding off to two significant fig.) Hence, v = (3.5 ± 0.31) ms–1 % age error in velocity = v 100 v D ´ = ± 0.0895 × 100 = ± 8.95 % = ± 9% Example18. Two resistances are expressed as R1 = (4 ± 0.5) W and R2 = (12 ± 0.5) W. What is the net resistance when theyare connected (i) in series and (ii) in parallel, with percentage error ? (a) 16W ± 23%, 3W ± 6.25% (b) 3W ± 2.3%, 3W ± 6.25% (c) 3W ± 23%, 16W ±6.25% (d) 16W ± 6.25%, 3W ± 23% Solution : (d) S 1 2 R R R 16 = + = W ; 1 2 1 2 P 1 2 S R R R R R 3 R R R = = = W + S 1 2 R R R 1 D = D + D = W Þ S S R 1 100 100% R 16 D ´ = ´ = 6.25% Þ S R 16 6.25% = W ± Similarly, 1 2 P S R R R R = S P 1 2 P 1 2 S R R R R R R R R D D D D = + + Þ P P R 0.5 0.5 1 0.23 R 4 12 16 D = + + = Þ P P R 100 23% R D ´ = Þ P R 3 23% = W ±
- 23.
- 24. 20 PHYSICS 1. Temperaturecan be expressed as derived quantityin terms of (a) length and mass (b) mass and time (c) length, mass and time (d) None of these 2. What is the unit of “a” in Vander Waal’s gas equation? (a) Atm litre–2 mol2 (b) Atm litre2 per mol (c) Atm litre–1 mol2 (d) Atm litre2 mol–2 3. Random error can be eliminated by (a) careful observation (b) eliminating the cause (c) measuring the quantitywith more than one instrument (d) taking large number of observations and then their mean. 4. Ifeisthecharge, V thepotentialdifference, T thetemperature, then the units of eV T are the same as that of (a) Planck’s constant (b) Stefan’s constant (c) Boltzmann constant (d) gravitational constant 5. If f = x2, then the relative error in f is (a) x x D 2 (b) x x 2 ) (D (c) x x D (d) (Dx)2 6. Two quantities A and B have different dimensions which mathematical operation given below is physically meaningful? (a) A/B (b) A+ B (c) A – B (d) A= B 7. Which of the following systems of units is not based on units of mass, length and time alone (a) SI (b) MKS (b) CGS (d) FPS 8. Unit of latent heat is (a) J Kg–1 (b) J mol–1 (c) N Kg–1 (d) N mol–1 9. Dyne-sec is the unit of (a) momentum (b) force (c) work (d) angular momentum 10. Illuminance of a surface is measured in (a) Lumen (b) candela (c) lux (d) luxm–2 11. SI unit of electric polarisation is (a) Cm–2 (b) coulomb (c) ampere (d) volt 12. The SI unit of coefficient of mutual inductance of a coil is (a) henry (b) volt (c) farad (c) weber 13. Light year is (a) light emitted by the sun in one year. (b) time taken by light to travel from sun to earth. (c) the distance travelled bylight in free space in one year. (d) time taken by earth to go once around the sun. 14. The SI unit of pressure is (a) atmosphere (b) bar (c) pascal (d) mmofHg 15. Electron volt is a unit of (a) potential difference (b) charge (c) energy (d) capacity 16. Dimensions of impulse are (a) ] T ML [ 1 - (b) ] MLT [ 2 (c) ] MT [ 2 - (d) ] T ML [ 3 1 - - 17. The S.I. unit of pole strength is (a) Am2 (b) Am (c) A m–1 (d) Am–2 18. Which is dimensionless? (a) Force/acceleration (b) Velocity/acceleration (c) Volume/area (d) Energy/work 19. Potential is measured in (a) joule/coulomb (b) watt/coulomb (c) newton-second (d) None of these 20. Maxwell is the unit of (a) magnetic susceptibility (b) intensity of Magnetisation (c) magneticFlux (d) magnetic Permeability 21. The mass of the liquid flowing per second per unit area of cross-section of the tube is proportional to (pressure difference across the ends)n and (average velocity of the liquid)m. Which of the following relations between m and n is correct? (a) m= n (b) m = – n (c) m2 = n (d) m = – n2 22. Which of the following is a derived physical quantity? (a) Mass (b) Velocity (c) Length (d) Time 23. N kg–1 is the unit of (a) velocity (b) force (c) acceleration (d) None of these 24. Which physical quantities have same dimensions? (a) Moment of couple and work (b) Force and power (c) Latent heat and specific heat (d) Work and power 25. The expression [ML–1 T–2] does not represent (a) pressure (b) power (c) stress (d) Young’s modulus
- 25. 21 Units and Measurements 1.What are the units of magnetic permeability? (a) Wb A–1 m–1 (b) Wb–1 Am (c) Wb A m–1 (d) Wb A–1 m 2. The dimensions of pressure gradient are (a) [ML–2 T–2] (b) [ML–2 T–1] (c) [ML–1 T–1] (d) [ML–1 T–2] 3. The dimensions of Rydberg’s constant are (a) [M0 L–1 T] (b) [MLT–1] (c) [M0 L–1 T0] (d) [ML0 T2] 4. The dimensions of universal gas constant are (a) [L2 M1 T–2 K–1] (b) [L1 M2 T–2 K–1] (c) [L1 M1 T–2 K–1] (d) [L2 M2 T–2 K–1] 5. The dimensions of magnetic moment are (a) [L2 A1] (b) [L2 A–1] (c) [L2 / A3] (d) [LA2] 6. The dimensions of Wien’s constant are (a) [ML0 T K] (b) [M0 LT0 K] (c) [M0 L0 T K] (d) [MLTK] 7. The unit and dimensions ofimpedance in terms of charge Q are (a) mho, [ML2 T–2 Q–2] (b) ohm, [ML2 T–1 Q–2] (c) ohm, [ML2 T–2 Q–1] (d) ohm, [MLT–1 Q–1] 8. If L denotes the inductance of an inductor through which a current i is flowing, the dimensions of L i2 are (a) [ML2 T–2] (b) [MLT–2] (c) [M2 L2 T–2] (d) Not expressiblein M, L, T 9. The dimensional formula ofwave number is (a) [M0 L0 T–1] (b) [M0 L–1 T0] (c) [M–1 L–1 T0] (d) [M0 L0 T0] 10. The period of a bodyunder S.H.M. is represented by: T = Pa Db Sc where P is pressure, D is density and S is surface tension, then values of a, b and c are (a) 1 , 2 1 , 2 3 - (b) 3 , 2 , 1 - - (c) 2 1 , 2 3 , 2 1 - - (d) 1, 2 1/3 11. The time of oscillation T of a small drop of liquid depends on radius r, density r and surface tension S. The relation between them is given by (a) 3 r S T r µ (b) S r T 3 r µ (c) r µ 3 2 r S T (d) S r T 3 r µ 12. Thepotential energyofaparticlevarieswith distance x from a fixed origin as B x x A V + = where A and B are constants. The dimensions ofAB are (a) 1 5/2 2 [M L T ] - (b) 1 2 2 [M L T ] - (c) 3/ 2 5/2 2 [M L T ] - (d) 1 7 /2 2 [M L T ] - 13. Distance travelled by a particle at any instant ‘t’ can be represented as S =A(t + B) + Ct2. The dimensions of B are (a) 0 1 1 [M L T ] - (b) 0 0 1 [M L T ] (c) 0 1 2 [M L T ] - - (d) 0 2 2 [M L T ] - 14. The deBroglie wavelength associated with a particle of mass m and energy E is h / 2m E . The dimensional formula of Planck’s constant h is (a) ] T L M [ 2 2 2 - (b) ] T L M [ 1 2 - (c) ] LT M [ 2 - (d) ] T L M [ 2 2 - 15. The velocity of a body which falls under gravity varies as ga hb, where g is acc. due to gravity and h is the height. The values of a and b are (a) a = 1, b = 1/2 (b) a = b = 1 (c) a = 1/2, b= 1 (d) a = 1/2; b= 1/2 16. The velocityvofa particle at time t is given by b v a t t c = + + The dimensions of a, b c are respectively (a) [LT–2], [L], [T] (b) [L2], [T] and [LT2] (c) [LT2], [LT] and [L] (d) [L], [LT] and [T2] 17. The dimensions of Hubble’s constant are (a) ] T [ 1 - (b) ] T L M [ 2 0 0 - (c) ] MLT [ 4 (d) 1 [MT ] - 18. Error in the measurement of radius ofa sphere is 1%. Then error in the measurement of volumeis (a) 1% (b) 5% (c) 3% (d) 8% 19. Subtract 0.2 J from 7.26 J and express the result with correct number of significant figures. (a) 7.1J (b) 7.06J (c) 7.0J (d) 7 J 20. Multiply107.88 by0.610 and express the result with correct number of significant figures. (a) 65.8068 (b) 65.807 (c) 65.81 (d) 65.8
- 26. 22 PHYSICS 21. When97.52 is divided by2.54, the correct result is (a) 38.3937 (b) 38.394 (c) 38.39 (d) 38.4 22. Relative density of a metal may be found with the help of spring balance. In air the spring balancereads (5.00 ± 0.05) N and in water it reads (4.00 ± 0.05) N. Relative density would be (a) (5.00±0.05)N (b) (5.00±11%) (c) (5.00±0.10) (d) (5.00± 6%) 23. Area of a square is (100 ± 2) m2. Its side is (a) (10±1)m (b) (10±0.1)m (c) m ) 2 10 ( ± (d) % 2 10 ± 24. Let Q denote the charge on the plate of a capacitor of capacitance C. The dimensional formula for C Q2 is (a) ] T M L [ 2 2 (b) ] LMT [ 2 (c) ] MT L [ 2 2 - (d) ] T M L [ 2 2 2 25. IfL and Rdenote inductance and resistance then dimension of L/R is (a) ] T L M [ 0 0 0 (b) ] T L M [ 0 0 (c) ] T L M [ 2 0 2 (d) ] MLT [ 2 26. The dimensional formula of current densityis (a) 0 2 1 [M L T Q] - - (b) ] Q T L M [ 1 1 2 0 - (c) ] Q MLT [ 1 - (d) ] Q T ML [ 2 1 2 - - 27. The least count ofa stop watch is 0.2 second. The time of20 oscillations of a pendulum is measured to be 25 second. The percentage error in the measurement of time will be (a) 8% (b) 1.8% (c) 0.8% (d) 0.1% 28. The dimensional formula for relative densityis (a) [M L–3] (b) [M0 L–3] (c) [M0 L0 T–1] (d) [M0 L0 T0] 29. The solid angle sustended by the total surface area of a sphere, at the centre is (a) 4p (b) 2p (c) p (d) 3p 30. If C and L denote the capacitance and inductance, the dimensions of LC are (a) [M0 L0 T–1] (b) [M0 L–1 T0] (c) [M–1 L–1 T0] (d) [M0 L0 T2] 31. The dimensions of solar constant is (a) [M0 L0 T0] (b) [MLT–2] (c) [ML2 T–2] (d) MT–3 32. The dimensions of hc e 1 2 o Î are (a) M–1 L–3 T4 A2 (b) ML3 T–4 A–2 (c) M0 L0 T0 A0 (d) M–1 L–3 T2 A 33. The dimensional formula for entropyis (a) [MLT–2 K1] (b) [ML2 T–2] (c) [ML2 T–2 K–1] (d) [ML2 T–2 K] 34. Dimensions of specific heat are (a) [ML2 T–2 K] (b) [ML2 T–2 K–1] (c) [ML2 T2 K–1] (d) [L2 T–2 K–1] 35. L, C, Rrepresent physical quantities inductance, capacitance and resistance respectively. The combinations which have the dimensions of frequency are (a) 1/RC (b) R/L (c) LC / 1 (d) C/L 36. The physical quantity which has the dimensional formula [M1T–3] is (a) surface tension (b) solar constant (c) density (d) compressibility 37. Which of the following is the most accurate? (a) 200.0m (b) 20 ×101 m (c) 2 × 102 m (d) data is inadequate 38. The velocity of water waves (v) may depend on their wavelength l, the density of water r and the acceleration due to gravity, g. The method of dimensions gives the relation between these quantities is (a) v (b) 2 v g µ l (c) 2 2 v g µ l (d) 2 1 2 v g- µ l 39. The time dependence of a physical quantity p is given by p = p0 exp. (– a t2), where a is a constant and t is the time. The constant a (a) is dimensionless (b) has dimensions T–2 (c) has dimensions T2 (d) has dimensions of p. 40. In the eqn. 2 a P (V b) constant, V æ ö + - = ç ÷ è ø the unit of a is (a) dyne × cm5 (b) dyne × cm4 (c) dyne/cm3 (d) dyne × cm2 41. Dimensions of ‘ohm’ are same as (where h is Planck’s constant and e is charge) (a) e h (b) e h2 (c) 2 e h (d) 2 2 e h 42. The Richardson equation is given by I = AT2e–B/kT. The dimensional formula for AB2 is same as that for (a) I T2 (b) kT (c) I k2 (d) I k2/T 43. The unit of current in C.G..S. system is (a) 10A (b) 1/10A (c) 1/100A (d) 1/1000A
- 27. 23 Units and Measurements 44.Which of the following do not have the same dimensional formula as the velocity? Given that m0 = permeabilityof free space, e0 = permittivity offree space, n = frequency, l = wavelength, P= pressure, r = density, w = angular frequency, k = wave number, (a) 0 o 1 m e (b) nl (c) P/r (d) wk 45. A cubehas numericallyequal volume and surface area. The volume of such a cube is (a) 1000unit (b) 200unit (c) 216unit (d) 300unit 46. A spherical body ofmass m and radius r is allowed tofall in a medium of viscosity h. The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity(v) is called time constant t. Dimensionallytcan be represented by (a) h p 6 mr2 (b) ÷ ÷ ø ö ç ç è æ h p 2 g r m 6 (c) m 6 r v ph (d) None of these 47. A quantity is represented by X = Ma Lb Tc. The % error in measurement of M, L and T are a%, b% and g% respectively. The % error in X would be (a) % ) c b a ( g + b + a (b) % ) c b a ( g + b - a (c) % 100 ) c b a ( ´ g - b - a (d) None of these 48. In a Vernier calliper, N divisions of vernier scale coincide with (N–1) divisions of main scale (in which one division represents 1 mm). the least count of the instrument in cm. should be (a) N (b) N – 1 (c) N 10 1 (d) 1 N 1 - 49. What is the fractional error in g calculated from g / 2 T l p = ? Given fraction errors in T and l are ± x and ± y respectively. . (a) x+ y (b) x – y (c) 2x + y (d) 2x – y 50. Conversion of 1 MW power in a New system of units having basic units of mass, length and time as 10 kg, 1 dm and 1 minute respectively is (a) 2.16 × 1010 unit (b) 2 × 104 unit (c) 2.16 × 1012 unit (d) 1.26 × 1012 unit 51. A resistor of 10 k W having tolerance 10% is connected in series with another resistor of20 k W having tolerance 20%. The tolerance of the combination will be (a) 10% (b) 13% (c) 30% (d) 20% 52. Using mass (M), length(L), time (T) and electric current (A) as fundamental quantities the dimensions of permittivity will be (a) [MLT–1A–1] (b) [MLT–2A–2] (c) [M–1L–3T+4A2] (d) [M2L–2T–2A2] 53. Thepercentageerrors in the measurement ofmass and speed are 2% and 3% respectively. The error, in kinetic energy obtained by measuring mass and speed, will be (a) 12% (b) 10% (c) 8 % (d) 2 % 54. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement ofmass and length are4% and3% respectively, themaximum error in themeasurement of densitywill be (a) 7% (b) 9% (c) 12% (d) 13% 55. The speed of sound in a gas is given by RT v M g = R = universal gas constant, T = temperature M = molar mass of gas The dimensional formula of g is (a) ] T L M [ 0 0 0 (b) ] LT M [ 1 0 - (c) ] MLT [ 2 - (d) ] T L M [ 1 0 0 - 56. Specific gravity has ............ dimensions in mass, ............ dimensions in length and ............ dimensions in time. (a) 0,0, 0 (b) 0,1, 0 (c) 1,0, 0 (d) 1,1, 3 57. If I is the moment of inertia and w the angular velocity, , what is the dimensional formula ofrotational kinetic energy 2 I 2 1 w ? (a) ] T ML [ 1 2 - (b) ] T L M [ 2 1 2 - - (c) ] T ML [ 2 2 - (d) ] T L M [ 2 1 2 - - 58. Given that r = m2 sin pt , where t represents time. If the unit of m is N, then the unit of r is (a) N (b) 2 N (c) Ns (d) s N2 59. The dimensional formula of farad is (a) ] TQ L M [ 2 1 - - (b) 1 2 2 2 [M L T Q ] - - (c) ] TQ L M [ 2 2 1 - - (d) ] Q T L M [ 2 2 1 - - 60. If time T, acceleration A and force F are regarded as base units, then the dimensional formula of work is (a) [FA] (b) ] FAT [ (c) ] FAT [ 2 (d) ] T FA [ 2
- 28. 24 PHYSICS 61. Turpentineoil is flowing through a capillary tube of length l and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given by : 2 2 p(r x ) 4v - h = l . Here vis velocityof oil at a distance x from the axis of the tube. From this relation, the dimensional formula of h is (a) ] T ML [ 1 1 - - (b) ] MLT [ 1 - (c) ] T ML [ 2 2 - (d) ] T L M [ 0 0 0 62. The dimensional formula of velocity gradient is (a) ] T L M [ 1 0 0 - (b) ] MLT [ 1 (c) 0 1 [ML T ] - (d) ] LT M [ 2 0 - 63. The thrust developed by a rocket-motor is given by ) P P ( A mv F 2 1 - + = where m is the mass of the gas ejected per unit time, v is velocity of the gas, A is area of cross- section of the nozzle, 1 P and 2 P are the pressures of the exhaust gas and surrounding atmosphere. The formula is dimensionally (a) correct (b) wrong (c) sometimes wrong, sometimes correct (d) Data is not adequate 64. IfE, m, J and G represent energy, mass, angular momentum and gravitational constant respectively, then the dimensional formula of 2 5 2 G m / EJ is (a) angle (b) length (c) mass (d) time 65. In a vernier callipers, ten smallest divisions of the vernier scale are equal tonine smallest division on the main scale. If the smallest division on the main scale is half millimeter, then the vernier constant is (a) 0.5mm (b) 0.1mm (c) 0.05mm (d) 0.005mm 66. Avernier calliper has20 divisions on thevernier scale, which coincide with 19 on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are of (a) 0.5mm (b) 1mm (c) 2mm (d) 4 1 mm 67. The pitch of the screw gauge is 0.5 mm. Its circular scale contains 50 divisions. The least count ofthe screw gauge is (a) 0.001mm (b) 0.01mm (c) 0.02mm (d) 0.025mm 68. If x = a – b, then the maximum percentage error in the measurement ofx will be (a) % 100 b b a a ´ ÷ ø ö ç è æ D + D (b) % 100 b b a a ´ ÷ ø ö ç è æ D - D (c) % 100 b a b b a a ´ ÷ ø ö ç è æ - D + - D (d) % 100 b a b b a a ´ ÷ ø ö ç è æ - D - - D 69. The heat generated in a circuit is given by 2 I Q = Rt, where I is current, R is resistance and t is time. If the percentage errorsin measuring I,Randt are2%,1%and1%respectively, then themaximum error in measuring heat will be (a) 2% (b) 3% (c) 4% (d) 6% 70. The pressure on a square plate is measured by measuring the force on the plate and length of the sides of the plate by using the formula 2 F P = l . If the maximum errors in the measurement of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is (a) 1% (b) 2% (c) 8% (d) 10% 71. In a simple pendulum experiment for the determination of acceleration due to gravity, time period is measured with an accuracy of 0.2% while length was measured with an accuracy of 0.5%. The percentage accuracy in the value of g so obtained is (a) 0.25% (b) 0.7% (c) 0.9% (d) 1.0% 72. The dimensions of a rectangular block measured with callipershaving least count of0.01 cm are5 mm × 10mm × 5 mm. The maximum percentageerror in the measurement of the volume of the block is (a) 5% (b) 10% (c) 15% (d) 20% 73. Consider the following pairs of quantities 1. Young's modulus; pressure 2. Torque; energy 3. Linear momentum; work 4. Solar day; light year. In which cases are the dimensions, within a pair, same? (a) 1 and 3 (b) 1 and 4 (c) 1 and 2 (d) 2 and 4 74. Which one of the following has the same dimension as that oftime, ifRisresistance, Linductance and C is capacitance? (a) RC (b) LC (c) R L (d) All of the above 75. The equation of a wave is given by x y asin k v æ ö = w - ç ÷ è ø , where w is angular velocity and v is linear velocity. The dimensions of K will be (a) [T2] (b) [T–1] (c) [T] (d) [LT]
- 29. 25 Units and Measurements 76.In the equation X = 3YZ2, X and Z are dimensions of capacitance and magnetic induction respectively. In MKSQ system, the dimensional formula for Y is (a) [M–3 L–2 T–2 Q–4] (b) [M L–2] (c) [M–3 L–2 Q4 T8] (d) [M–3 L–2 Q4 T4] 77. A force is given by F = at + bt2, where t is time, the dimensions of a and b are (a) [M L T–4] and [M L T–1] (b) [M L T–1] and [M L T0] (c) [M L T–3] and [M L T–4] (d) [M L T–3] and [M L T0] 78. The frequency of vibration of a string is given by f = L 2 n T m , where T is tension in the string, L is the length, n is number ofharmonics. The dimensional formula for m is (a) [M0 L T] (b) [M1 L–1 T–1] (c) [M1 L–1 T0] (d) [M0 L T–1] 79. The dimensions of voltage in terms of mass (M), length (L) and time (T) and ampere (A) are (a) [ML2T–2A–2] (b) [ML2T3A–1] (c) [ML2T–3A1] (d) [ML2T–3A–1] 80. Suppose the kinetic energy of a body oscillating with amplitude A and at a distance x is given by 2 2 Bx K x A = + The dimensions of B are the same as that of (a) work/time (b) work × distance (c) work/distance (d) work× time 81. Thedimensions ofmagneticfield in M, L, T and C (coulomb) are given as (a) [MLT–1 C–1] (b) [MT2 C–2] (c) [MT–1 C–1] (d) [MT–2 C–1] 82. Two full turns of the circular scale of a screw gauge cover a distance of 1mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circulr scale divisions in line with the main scale as 35. The diameter of the wire is (a) 3.32mm (b) 3.73mm (c) 3.67mm (d) 3.38mm 83. If the dimensions of a physical quantity are given by Ma Lb Tc, then the physical quantity will be (a) velocity if a = 1, b = 0, c = – 1 (b) acceleration if a = 1, b = 1, c = – 2 (c) force if a = 0, b = – 1, c = – 2 (d) pressure if a = 1, b = – 1, c = – 2 84. The density ofa material in CGS system ofunits is 4g/cm3. In a system ofunits in which unit of length is 10 cm and unit of mass is 100 g, the value of densityof material will be (a) 0.4 (b) 40 (c) 400 (d) 0.04 85. In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows P = 3 2 a b cd % error in P is (a) 10% (b) 7% (c) 4% (d) 14% 86. What is the fractional error in g calculated from g / 2 T l p = ? Given fractional errors in T and l are ± x and ± y respectively. . (a) x+ y (b) x – y (c) 2x + y (d) 2x – y 87. The resistance R ofa wire is given bythe relation R = 2 . r r p l Percentage error in the measurement ofr, l and r is1%, 2% and 3% respectively. Then the percentage error in the measurement ofR is : (a) 6 (b) 9 (c) 8 (d) 10 88. What are the dimensions of permeability? (a) [M1L1T1A–2] (b) [M1L1T–2A–2] (c) [M2L2T1A0] (d) [M1L2T2A–2] 89. The physical quantity having the dimensions [M–1L–3T3A2] is (a) resistance (b) resistivity (c) electrical conductivity (d) electromotive force 90. The time of reverberation of a room Ais one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room A? (a) 2 (b) 4 (c) 1 2 (d) 1 91. Which of the following is the unit of molar gas constant? (a) JK–1 mol–1 (b) Joule (c) JK–1 (d) J mol–1 92. Densityof liquid is 16.8 gcm–3. Its valuein the International System of Units is (a) 3 16.8 kgm- (b) 3 168 kgm- (c) 3 1680 kgm- (d) 3 16800 kgm- 93. The dimensional formula of couple is (a) ] T ML [ 2 2 - (b) ] MLT [ 2 (c) 1 3 [ML T ] - - (d) ] T ML [ 2 2 - -
- 30. 26 PHYSICS 94. Therefractive index of water measured by the relation m = real depth apparent depth is found to have values of 1.34, 1.38, 1.32 and 1.36; the mean value of refractive index with percentage error is (a) 1.35± 1.48 % (b) 1.35 ± 0 % (c) 1.36 ± 6 % (d) 1.36 ± 0 % 95. Diameter of a steel ball is measured using a Vernier callipers which has divisions of0.1 cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given below: S.No. MS(cm) VS divisions 1. 0.5 8 2. 0.5 4 3. 0.5 6 Ifthezero error is–0.03 cm, then mean corrected diameter is (a) 0.52cm (b) 0.59cm (c) 0.56cm (d) 0.53cm 96. Theperiod ofoscillation ofa simplependulumisT = L 2 g p . Measured valueofL is20.0 cm known to1 mm accuracyand time for 100 oscillations of the pendulum is found tobe 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is (a) 1% (b) 5% (c) 2% (d) 3% Exemplar Questions 1. The number of significant figures in 0.06900 is (a) 5 (b) 4 (c) 2 (d) 3 2. The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is (a) 663.821 (b) 664 (c) 663.8 (d) 663.82 3. The mass and volume of a body are 4.237 g and 2.5 cm3, respectively. The density of the material of the body in correct significant figures is (a) 1.6048g cm–3 (b) 1.69gcm–3 (c) 1.7 g cm–3 (d) 1.695gcm–3 4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75and2.74 (b) 2.74and2.73 (c) 2.75and2.73 (d) 2.74and2.74 5. The length and breadth of a rectangular sheet are 16.2 ± 0.1 cm and 10.1 ± 0.1 cm, respectively. The area of the sheet in appropriate significant figures and error is (a) 164 ±3 cm2 (b) 163.62±2.6cm2 (c) 163.6± 2.6 cm2 (d) 163.62± 3cm2 6. Which of the following pairs of physical quantities does not have same dimensional forrmula? (a) Work and torque (b) Angular momentum and Planck'sconstant (c) Tension and surface tension (d) Impulseand linear momentum 7. Measure of two quantities along with the precision of respective measuring instrument is A = 2.5 ms–1 ± 0.5ms–1, B = 0.10 s ± 0.01 s. Thevalue of AB will be (a) (0.25±0.08)m (b) (0.25±0.5)m (c) (0.25±0.05)m (d) (0.25±0.135)m 8. You measure twoquantities asA =1.0m ± 0.2m, B =2.0 m± 0.2 m. We should report correct value for AB as (a) 1.4m±0.4m (b) 1.41m±0.15m (c) 1.4m±0.3m (d) 1.4m±0.2m 9. Which of the following measurement is most precise? (a) 5.00mm (b) 5.00cm (c) 5.00m (d) 5.00km DIRECTIONS for Qs. (97 to 100) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethe correctanswer (ONLYONE option iscorrect)fromthe following (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 97. Statement 1 : The number of significant figures depends on the least count of measuring instrument. Statement 2 : Significant figures define the accuracy of measuring instrument. 98. Statement 1 : Absolute error maybe negative or positive. Statement 2 : Absolute error is the difference between the real value and the measured value of a physical quantity. 99. Statement 1 : In the measurement of physical quantities direct and indirect methods are used. Statement 2 : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result. 100. Statement 1 : Energy cannot be divided by volume. Statement 2 : Because dimensions for energy and volume are different.
- 31. 27 Units and Measurements 10.The mean length ofan object is 5 cm. Which is the following measurement is most accurate? (a) 4.9cm (b) 4.805cm (c) 5.25cm (d) 5.4cm 11. Young's modulus ofsteel is1.9 × 1011 N/m2.When expressed in CGS units ofdyne/cm2, it will beequal to(1N = 105 dyne, 1 m2 =104 cm2) (a) 1.9× 1010 (b) 1.9× 1011 (c) 1.9× 1012 (d) 1.9× 1013 12. If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula (a) [pA–1T1] (b) [p2AT] (c) [pA–1/2T] (d) [pA1/2T] NEET/AIPMT (2013-2017) Questions 13. In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. QuantityP is calculated as follows P = 3 2 a b cd % error in P is: (a) 10% (b) 7% [2013] (c) 4% (d) 14% 14. The pair of quantities having same dimensions is (a) Young’s modulus and energy [NEET Kar. 2013] (b) impulse and surface tension (c) angular momentum and work (d) work and torque 15. Ifforce(F),velocity(V)and time(T)aretaken asfundamental units, then the dimensions of mass are : [2014] (a) [F V T– 1] (b) [F V T– 2] (c) [F V– 1 T– 1] (d) [F V– 1 T] 16. If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be : [2015] (a) [EV–1T–2] (b) [EV–2T–2] (c) [E–2V–1T–3] (d) [EV–2T–1] 17. If dimensions of critical velocity ucof a liquid flowing through a tube are expressed as x y x [ r ] h r , where h, r and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by : [2015 RS] (a) –1, –1, 1 (b) –1, –1, –1 (c) 1,1, 1 (d) 1, –1, –1 18. A physical quantityof the dimensions of length that can be formed out of c, G and 2 0 e 4pe is [c is velocity of light, G is universal constant of gravitation and e is charge] [2017] (a) 1/ 2 2 2 0 e c G 4 é ù ê ú pe ê ú ë û (b) 1/2 2 2 0 1 e G4 c é ù ê ú pe ê ú ë û (c) 2 0 1 e G c 4pe (d) 1/ 2 2 2 0 1 e G 4 c é ù ê ú pe ê ú ë û
- 32. 28 PHYSICS EXERCISE -1 1. (d) Temperature is one of the basic physical quantities. 2. (b) The vander Waal’s gas equation is RT ) b V )( V a P ( 2 = - + for one mole. & = m - ÷ ÷ ø ö ç ç è æ m + ) b V ( V a P 2 mRT for m mole Dimensionallyin first bracket on L.H.S m = Þ ú û ù ê ë é m = 2 2 V ] P [ ] a [ V a ] P [ Dimension of ú ú û ù ê ê ë é = mole litre . atm ] a [ 2 3. (d) 4. (c) R T PV T W T eV = = = and N R = Boltzmann constant. 5. (a) 6. (a) 7. (a) SI is based on seven fundamental units. 8. (a) 1 kg J kg J m Q L - = = = 9. (a) As force = change in momentum/time. force × time= change in momentum 10. (c) Illuminanceis intensityofillumination measured in lux. 11. (a) 2 2 i Cm m C A Q P - = = = 12. (a) 13. (c) 1 light year = speed oflight in vacuum × no. ofseconds in one year = (3 × 108) × (365 × 24 × 60 × 60) = 9.467 × 1015 m. 14. (c) 1 pascal = 1 N / m2. 15. (c) Electron volt is a unit of energy & 1eV = 1.6×10–19 joule 16. (a) Impulse = force × time ] LT M [ T MLT 1 1 2 - - = ´ = . 17. (b) Pole strength, . m A m m A 2 M m 2 = = = l 18. (d) Both energy and work have same unit. energy/work is a pure number. 19. (a) Potential is work done per unit charge. 20. (c) Maxwell is the unit ofmagnetic flux in C.G.S system. 1 Wb(S.I unit) = 108 maxwell 21. (b) Let M = pnvm m 1 n 2 1 1 2 ) LT ( ) T ML ( T ML - - - - - = m n 2 m n n T L M - - + - = 2 m n ; 1 n - = + - = 1 1 2 n 2 m - = + - = + - = n m - = 22. (b) 23. (c) N kg–1 = force/mass = acceleration 24. (a) Moment of couple = force × distance ] T L M [ 2 2 1 - = work = force × distance ] T L M [ 2 2 1 - = . 25. (b) ] T ML [ T T ML time energy Power 3 2 2 2 - - = = = . EXERCISE - 2 1. (a) From Biot Savart’s law 2 0 r sin dl i 4 B q p m = 2 2 2 1 1 0 4 Br Wb m m Wb A m idl sin Am - - - p m = = = q 2. (a) Pressure gradient Pressure difference distance = . [Pressure gradient] 1 2 ML T L - - = 2 2 ML T - - é ù = ë û 3. (c) From , n 1 n 1 R 1 2 2 2 1 ÷ ÷ ø ö ç ç è æ - = l dimensions of ] T L M [ L L 1 R 0 1 0 1 - - = = = 4. (a) 2 2 PV W ML T R T T molK - = = = m m where m is number of mole of the gas = [M1L2T–2K–1mol–1] 5. (a) M = current × area =AL2 = [L2 A1] 6. (b) 0 1 0 1 m b T LK [M L T K ] = l = = 7. (b) 2 2 1 V W ML T Impedance Q Q QT - - = = = I I ] Q T ML [ 2 1 2 - - = . Hints & Solutions
- 33. 29 Units and Measurements 8.(a) Energy stored in an inductor ] T ML [ i L 2 1 2 2 2 - = = 9. (b) Wave number ] T L M [ L 1 1 0 1 0 - = = l = n . 10. (a) T = Pa Db Sc c 2 b 3 a 2 1 1 0 0 ) MT ( ) ML ( ) T ML ( T L M - - - - = c 2 a 2 b 3 a c b a T L M - - - - + + = Applying principle of homogeneity a + b + c = 0; – a – 3b = 0; – 2a – 2c = 1 on solving, we get a = – 3/2, b = 1/2, c = 1 11. (b) T MT L ML S r 2 3 3 3 = = r - - 12. (d) x V A ; Vx x A ]; L [ x B = = = = 2 2 / 5 2 / 1 2 2 T L M L T L M - - = = ] T L M [ ) L ( ) T ML ( AB 2 2 / 7 1 2 2 / 5 - - = = 13. (b) In S = A(t + B) + Ct2 ; B is added to time t. Therefore, dimensions of B are those of time. 14. (b) ) T ML ( M L E m 2 h 2 2 - = l = ] T L M [ 1 2 - = 15. (d) a b v g h ; = a 2 b a b a 2 1 0 T L L ) LT ( ] LT M [ - + - - = = 2 / 1 a 1 a 2 ; 1 b a = - = - = + 2 / 1 b = 16. (a) As c is added to t, c = [T] 1 2 v LT a [LT ] t T - - = = = , 1 b v(t c) LT T [L] - = + = ´ = 17. (a) Hubble’s constant, ] L [ ] LT [ ce tan dis velocity H 1 - = = ] T [ 1 - = . m / N 10 70 3 - ´ = 18. (c) ; r 3 4 V 3 p = % 3 % 1 3 100 r r 3 100 V V = ´ = ´ ÷ ø ö ç è æ D = ´ D 19. (a) Subtraction is correct upto one place of decimal, corresponding to the least number of decimal places. 7.26–0.2=7.06=7.1J. 20. (d) Number of significant figures in multiplication is three, corresponding tothe minimum number 107.88× 0.610=65.8068=65.8 21. (d) 4 . 38 393 . 38 54 . 2 52 . 97 = = (with least number of significant figures, 3). 22. (d) Relative density = Weight of body in air Loss of weight in water 5.00 5.00 5.00 – 4.00 1.00 = = 0.05 0.05 100 100 5.00 1.00 Dr æ ö ´ = + ´ ç ÷ è ø r = (0.01 + 0.05) × 100 = 0.06 × 100 = 6% Relative density = ± 5.00 6% 23. (a) Area = (Length)2 or length 1 2 (Area) = 1 2 (100 2) = ± 1 2 1 (100) 2 2 = ± ´ (10 1)m = ± 24. (c) We know that C 2 Q2 is energyofcapacitor so it represent the dimension of energy ] T ML [ 2 2 - = . 25. (b) Volt/amp. sec/amp. Volt L/R ´ = = sec. ] T L M [ 0 0 = 26. (a) Current Q Current density area area t = = ´ 27. (c) 8 . 0 100 25 2 . 0 = ´ 28. (d) 29. (a) 30. (d) From 1 2 LC n = p 2 0 0 2 2 1 2 1 1 LC T [M L T ] (2 ) (T ) - = = = = p n 31. (d) Solar constant = energy/sec/area ] MT [ TL T L M 3 2 2 2 - - = = 32. (c) From 2 2 o r e 4 1 F e p = 2 o 2 r F 4 e p = e (dimensionally) 2 2 2 2 2 1 1 o e 4 Fr (MLT )L hc hc ML T [LT ] - - - p = = e ] A T L M [ 0 0 0 0 = , hc e o 2 e is called fine structure constant & has value 137 1 .
- 34. 30 PHYSICS 33. (c) K T ML T Q Entropy 2 2- = = ] K T ML [ 1 2 2 - - = 34. (d) 2 2 2 2 1 Q ML T s [L T K ] m MK - - - = = = q 35. (c) 2 2 2 1 2 4 2 1 1 LC (ML T A ) (M L T A ) - - - - = ´ = 1 2 1 T T - = 36. (b) Solar constant = energy/area/time ] T M [ T L T L M 3 1 2 2 2 - - = = . 37. (a) 38. (b) a b c v k g = l r c 2 b 3 a 1 0 ) LT ( ) ML ( L ] LT M [ - - - = c 2 c b 3 a b T L M - + - = 1 c b 3 a ; 0 b = + - = 1 c 2 - = - Þ 2 / 1 c = 2 1 a = 1/ 2 0 1/ 2 2 v g or v g µ l r µ l 39. (b) In p = p0 exp. (– a t2) is dimensionless ] T [ T 1 t 1 2 2 2 - = = = a 40. (b) As P V a 2 = 2 a PV = 3 2 4 2 dyne (cm ) dyne cm cm = = ´ 41. (c) 2 3 2 2 1 2 2 A T ML ) AT ( T ML e h - - - = = = Resistance (ohm) 42. (c) I = AT2 e–B/kT Dimensions ofA = I /T2; Dimensions of B = kT (Q power of exponential is dimensionless) 2 2 2 2 AB (kT) k T = = I I 43. (a) The C.G.S unit ofcurrent is called biot (Bi) i.e., 1C (1/10)e.m.u of charge 1 1A Bi 1sec sec 10 = = = or 1Bi = 10A 44. (d) ] T L [ L 1 T 1 k 1 1 - - = ´ = w The dimensions of the quantities in a, b, c are of velocity [LT–1] 45. (c) Volume (L3) = surfacearea (6L2) L= 6, volume = 63 = 216 46. (d) None of the expressions has the dimensions of time. 47. (a) ; T L M X c b a = X a M b L c T 100 100 X M L T D D D D æ ö ´ = + + ´ ç ÷ è ø )% c b a ( g + b + a = 48. (c) NVD = (N – 1) MD MD N 1 N VD 1 ÷ ø ö ç è æ - = L.C. = Least count = 1MD – 1VD MD N 1 N 1 . C . L ÷ ø ö ç è æ - - = cm N 10 1 cm N 1 . 0 . D . M N 1 = = = scale on vernier parts of number scale main on part 1 of value = whereV.D. = vernier division, M.D. Main scaledivision. 49. (c) From 2 2 T 4 g ; g 2 T l l p = p = ) x 2 y ( T T 2 g g + = D + D = D l l 50. (c) We have, n1u1 = n2u2 n2 1 1 2 æ ö = ç ÷ è ø u n u 2 3 6 1 1 1 2 2 2 10 - æ ö æ ö æ ö = ´ ç ÷ ç ÷ ç ÷ è ø è ø è ø M L T M L T 3 2 6 1 1 1 1 10 10 60 1 10 kg m s kg s m - - æ ö æ ö æ ö = ´ ç ÷ ç ÷ ç ÷ è ø è ø è ø ´ ( ) ( ) 2 3 6 1 10 10 60 10 æ ö = ´ ç ÷ è ø ( )3 7 10 60 = ´ = 2.16 × 1012 units. 51. (c) Effective resistance S R (10k 10%) (20k 20%) = W ± + W ± Tolerance of the combination = 30 30 W ± ( k %) 52. (c) Force, F 2 2 1 0 r q q 4 1 pe = 2 2 1 0 Fr 4 q . q p = e Þ
- 35. 31 Units and Measurements Sodimension of e0 ] A T L M [ ] L ][ MLT [ ] AT [ 2 4 3 1 2 2 2 - - - = = 53. (c) Percentage error in mass % 2 100 m m = ÷ ø ö ç è æ ´ D and percentage error in speed % 3 100 v v = ÷ ø ö ç è æ ´ D . 2 1 E mv 2 = E m V 100 100 2 100 E m V D D D ´ = ´ + ´ = 2% + 2 × 3% = 8%. 54. (d) Density = Mass Volume 3 L M = r , L L 3 M M D + D = r r D % error in density= % error in Mass + 3 (% error in length) = 4 + 3(3) = 13% 55. (a) Ratio of specific heat, p v C C g = 56. (a) Specific gravity is the ratio of density of substance and densityofwater at 4°C. The ratioof likequantities is dimensionless. 57. (c) DimensionallyK.E = Work 58. (b) Trigonometric ratio are a number and hence demensionless 59. (b) 1 2 2 2 Q Q [C] [M L T Q ] V W 2 - - é ù é ù = = = ê ú ê ú ë û ê ú ë û 60. (c) ] LT [ ] A [ 2 - = or ] AT [ ] L [ 2 = [Work] = [Force ×Distance] = ] FAT [ ] FL [ 2 = 61. (a) h is the coefficient of viscosity. . 62. (a) Velocity gradient is velocity per unit distance. 63. (a) Use principle of homogeneity. 64. (a) ] T L M [ ] T L M ][ M [ ] T ML ][ T ML [ 0 0 0 2 2 3 1 5 2 1 2 2 2 = - - - - = angle. 65. (c) 10VD= 9MD, 1VD = 9 MD 10 Vernier constant = 1 MD – 1VD = 9 1 1 1 1 MD MD 10 10 10 2 æ ö - = = ´ ç ÷ è ø =0.05mm 66. (c) 0.1 19 1 MSD 10 20 æ ö = - ç ÷ è ø Þ 1 1 1MSD 100 20 = ´ Þ 1 MSD = 1 10 2 5 ´ = 67. (b) Least count 0.5 0.01mm 50 = = 68. (c) Maximum absolute error is b a D + D . Therefore the percentage error = absolute error 100 actual value ´ 69. (d) 100 t t 100 R R 100 I I 2 100 Q Q ´ D + ´ D + ´ D = ´ D 2 2% 1% 1% = ´ + + = 6%. 70. (c) P F 100 100 2 100 P F D D D ´ = ´ + ´ = l l 4% + 2 × 2% =8% 71. (c) 2 T g , g 2 T l l µ p = 100 g g ´ D = 0.5% + 2 × 0.2% = 0.9% 72. (a) % error = 100 5 . 0 01 . 0 100 0 . 1 01 . 0 100 5 . 0 01 . 0 ´ + ´ + ´ = 2 + 1 + 2 = 4 + 1 = 5 73. (c) 74. (b) 2 2 2 1 2 4 2 [ LC] ML T A .M L T A T - - - - = = 75. (c) The quantity ( v wx – wk) has dimension of angle and hence wk is dimensionless being angle. 76. (d) 1 2 4 2 3 2 4 4 2 2 4 2 [X] M L T A [Y] M L Q T [Z ] M T A - - - - - - = = = Q A T æ ö = ç ÷ è ø Q 77. (c) [at] = [F] amd [bt2] = [F] Þ [a] = MLT–3 and [b] = MLT–4 78. (c) Clearly, 2 2 2 n T m 4f L = ; 2 2 2 MLT [m] T .L - - = 79. (d) 2 2 2 1 3 W ML T [V] ML A T Q AT - - - é ù = = = ê ú ë û 80. (b) From 2 2 2 Bx Bx B K x x A x = = = + B = K × x = K.E. × distance = work × distance
- 36. 32 PHYSICS 81. (c)We know that F = q v B 2 1 1 1 F MLT B MT C qv C LT - - - - = = = ´ 82. (d) Least count of screw gauge = 0.5 mm 0.01mm 50 = Reading = [Main scale reading + circular scale reading × L.C] – (zeroerror) =[3 +35×0.01]–(–0.03)=3.38mm 83. (d) Pressure = [ ] –2 –1 –2 2 MLT ML T L = Þ a = 1, b = – 1, c = – 2. 84. (b) In CGS system, 3 g 4 cm d = The unit of mass is 100g and unit of length is 10 cm, so density = 3 100g 4 100 10 cm 10 æ ö ç ÷ è ø æ ö ç ÷ è ø = 3 3 4 (100g) 100 (10cm) 1 10 æ ö ç ÷ è ø æ ö ç ÷ è ø 3 3 4 100g (10) · 100 (10cm) = ´ = 40 unit 85. (d) P = 3 2 a b cd , P P D × 100% = a 3 a D × 100% + b 2 b D × 100%+ c c D × 100%+ d d D ×100%. = 3 × 1% + 2 × 2% + 3% + 4% = 14% 86. (c) From 2 2 T 4 g ; g 2 T l l p = p = ) x 2 y ( T T 2 g g + = D + D = D l l 87. (b) Given = 2 R , r r = p l then R 100 R D ´ = r 100 100 2 100 r Dr D D ´ + ´ + ´ r l l = 1% + 2% + 2 × 3% = 9% 88. (b) The magnetic field at a point near a long straight conductor is given by B = 0 2 . 4 I r m p Þ 0 m = 4 2 rB I p [m0] = [ ][ ] [ ] r B I = -2 -1 L.MT A I F B qv æ ö = ç ÷ è ø Q = MLT–2A–2 89. (c) Resistivity, m ne2t r = 3 –3 –2 [ML T A ] r = So, electrical conductivity 1 s = r –1 –3 3 2 [M L T A ] s = 90. (a) Reverberation time, t = forward backward d d V V æ ö æ ö + ç ÷ ç ÷ è ø è ø when dimensions double then, d¢ = 2d d d 2d 2d d d t 2 V V V V V V ¢ ¢ æ ö ¢ = + = + = + ç ÷ è ø Þ t¢ = 2t d d t V V æ ö + = ç ÷ è ø Q 91. (a) 1 1 mol K J K mol J nT PV R - - = = = . 92. (d) 16.8 gcm–3 =16800Kgm–3. 93. (a) Dimensionallycouple = Torque =Work 94. (a) The mean value ofrefractiveindex, 1.34 1.38 1.32 1.36 1.35 4 + + + m = = and | (1.35 1.34) | | (1.35 1.38) | | (1.35 1.32) | | (1.35 1.36) | 4 - + - + - + - Dm = = 0.02 Thus 100 Dm ´ m = 0.02 100 1.48 1.35 ´ = 95. (b) Least count = 0.1 10 = 0.01 cm d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm Mean diameter = 0.61 0.57 0.59 3 + + = 0.59 cm 96. (d) As, 2 2 g 4 T = p l So, g T 100 100 2 100 g T D D D ´ = ´ + ´ l l 0.1 1 100 2 100 20 90 = ´ + ´ ´ = 2.72 ; 3% 97. (c) 98. (b) 99. (b) 100. (a)
- 37. 33 Units and Measurements EXERCISE- 3 Exemplar Questions 1. (b) In 0.06900, the twozeroes before six are not siginificant and two zeroes on right side of9 are significant figures. Hence, number of significant figures are four (6900). 2. (b) In addition the result will be in least number of places after decimal andminimum number of significant figure. The sum of the given numbers can be calculated as 663.821. Thenumber with least decimal places is 227.2 is correct to only one decimal place but in addition of numbers, the final result should be rounded off to one decimal place i.e., 664. 3. (c) As we know that in multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. Thesignificant figure in given numbers4.237 g and2.5 cm3 are four and two respectivey so, density should be reported to two significant figures. 3 3 4.237 g Density 1.6948 1.7 g.cm 2.5 cm - = = = As rounding off the number upto 2 significant figures, we get density = 1.7. 4. (d) Rounding off2.745 upto 3 significant figures here IVth digit is 5 and its preceding is even, so no change in 4. Thus answer would be 2.74. Rounding off 2.735 upto3 significant figures, here IV digit is 5 and its preceding digit is 3 (odd). So 3 is increased by 1 answer become would be 2.74. 5. (a) If Dx is error in a physical quantity, then relativeerror is calculated as Dx x . Given that, Length l = (16.2 ± 0.1) cm Breadth b = (10.1 ± 0.1) cm Q Dl = 0.1 cm, Db= 0.1 cm Area (A)= l × b= 16.2×10.1 = 163.62cm2 In significant figure rounding off to three significant digits, area A = 164 cm2 0.1 0.1 16.2 10.1 D D D = + = + A l b A l b 1.01 1.62 2.63 16.2 10.1 163.62 + = = ´ So, 2.63 2.63 164 163.62 163.62 D = ´ = ´ A A = 2.636 cm2 Now rounding off up to one significant figure DA =3cm2. So,Area A = A ± DA =(164 ±3) cm2. 6. (c) a. Work = force × distance = [MLT–2][L] = [ML2T–2] Torque = F × d = [ML2T–2] LHS and RHS has same dimensions. b. Angular momentum (L) = mvr=[M][LT–1][L] = [ML2T–1] Planck's constant h = n E 2 2 2 1 1 [ML T ] [ML T ] [T ] - - - = = 1 and æ ö = n n = ç ÷ è ø E h T Q So, dimensions of h and L are equal. c. Tension = force = [MLT–2] Surface tension 2 force [MLT ] length [L] - = = =[ML0T–2] d. Impulse= force ×time = [MLT–2][T] = [MLT–1] Momentum = mass × velocity = [M][LT–1] = [MLT–1] LHS and RHS has same dimensions. Hence only (c) options of Physical quantities does not have same dimensions. 7. (a) By applying the Rule of significant figure in multiplication and addition. As given that, A = 2.5ms–1 ± 0.5 ms–1, B = 0.10s ± 0.01 s x = AB = (2.5)(0.10) = 0.25 m (consider onlysignificant figure value) D D D = + x A B x A B 0.5 0.01 0.05 0.025 0.075 2.5 0.10 0.25 0.25 D + = + = = x x Dx =0.075=0.08m (rounding off to two significant figures.) AB =(0.25±0.08)m 8. (d) As given that, A = 1.0m ±0.2 m, B =2.0m ±0.2m So, X = (1.0)(2.0) = AB = 1.414 m Rounding off upto two significant digit X =1.4 m=(r) (Let) 1 1 0.2 0.2 2 2 1.0 2.0 D D D é ù é ù = + = + ê ú ê ú ë û ë û x A B x A B 0.6 2 2.0 = ´ 0.6 0.6 1.4 0.212 2 2.0 2 2.0 ´ Þ D = = = ´ ´ x x Rounding off upto one significant digit Dx =0.2 m = Dr (Let) So, correct value of = + D AB r r =(1.4 ±0.2)m
- 38. 34 PHYSICS 9. (a)For the most precise measurement, the unit must be least and number of digits including zeroes after decimal must be zero. Now, take first option, As here 5.00 mm has the smallest unit and the error in 5.00 mm is least (commonly taken as 0.01 mm if not specified), hence, 5.00 mm is most precise. 10. (a) Now, checking the errors with each options one by one, |Dl1|= |5 –4.9|= 0.1cm |Dl2|=|5–4.805|= 0.195cm |Dl3|=|5.25–5|=0.25cm |Dl4|= |5.4–5|= 0.4cm Error Dl1 isleast or minimum. So, 4.9 cm is most precise. 11. (c) It is given that Young's modulus (Y) is, Y =1.9 ×1011 N/m2 1N = 105 dyne So, Y = 1.9 × 1011 × 105 dyne/m2 Convert meter tocentimeter Q 1m=100cm Y = 1.9 ×1011 × 105 dyne/(100)2 cm2 = 1.9 × 1016 – 4 dyne/cm2 Y = 1.9 × 1012 dyne/cm2 12. (d) Given that fundamental quantities are momentum (p), area (A) andtime (T). Let us consider the dimensional formula for [ ] a b c E p A T µ ] = a b c E kp A T where k is dimensionless constant of proportionality. Dimensions ofenergy[E] = [ML2T–2] and Dimension of momentum p= mv = [MLT–1] Dimension ofArea [A] = [L2] Dimension ofTime [T] = [T] Dimension of energy[E] = [K] [p]a[A]b[T]c Putting all the dimensions, value ML2T–2 = [MLT–1]a [L2]b [T]c = MaL2b + aT–a + c Byprinciple of homogeneity of dimensions, a = 1, 2b +a = 2 Þ 2b + 1 = 2 Þ 1 2 = b 2 - + = - a c c = –2 + a = –2 + 1 = –1 So, Dimensional formula (ofenergy) E = [pA1/2T–1] 1/2 1 [ ] - = E pA T NEET/AIPMT (2013-2017) Questions 13. (d) P = 3 2 a b cd , P P D × 100% = a 3 a D × 100% + b 2 b D × 100%+ c c D × 100% + d d D ×100%. = 3 × 1% + 2 × 2% + 3% + 4% = 14% 14. (d) Work = Force × displacement Torque = Force × force arm = mass × acceleration × length = [M] × [LT–2] × [L] = [M L2T–2] 15. (d) Force = mass × acceleration Þ [Mass] = force acceleration é ù ê ú ë û = force velocity / time é ù ê ú ë û = [F V– 1 T] 16. (b) Let surface tension s = Ea VbTc b –2 2 –2 a C MLT L (ML T ) (T) L T æ ö = ç ÷ è ø Equating the dimension ofLHS and RHS ML0T–2 = MaL2a + b T–2a – b + c Þ a = 1, 2a + b = 0, –2a – b + c = –2 Þ a = 1, b = – 2, c = – 2 Hence, the dimensions of surface tension are [EV–2 T–2] 17. (d) Applying dimensional method : vc = hxryrz [M0LT–1]= [ML–1T–1]x [ML–3T0]y [M0LT0]z Equating powers both sides x + y= 0; –x = –1 x =1 1 + y = 0 y = –1 –x – 3y + z = 1 –1 – 3(–1) + z = 1 –1 + 3+ z = 1 z = –1 18. (d) Let dimensions of length is related as, L z 2 x y 0 e [c] [G] 4 é ù = ê ú pe ê ú ë û 2 0 e 4pe = ML3T–2 L= [LT–1]x [M–1L3T–2]y[ML3T–2]z [L] = [Lx + 3y + 3z M –y + z T–x – 2y – 2z] Comparing both sides –y+ z = 0 Þ y= z ...(i) x+ 3y+ 3z =1 ...(ii) –x – 4z = 0 (Q y= z) ...(iii) From(i),(ii)&(iii) z = y= 1 , 2 x = –2 Hence, L = 1/2 2 2 0 e c G 4 - é ù × ê ú pe ê ú ë û
- 39. BASIC DEFINITIONS Mechanics :Branch of physics, which deals with the study of objects in rest and in motion. Statics : Studyof objects at rest or in equilibrium. Kinematics : Studyof motion of objects without considering the cause of motion. Dynamics : Study of motion of objects considering the cause of motion. Rest : An object is said to be at rest if it does not change its position with time, with respect to its surrounding (a reference point which is generallytaken as origin in numerical problems) Motion : An object is said tobe in motion if it changes its position with time, with respect to its surroundings. Rest and motion are relative terms. Point mass/Point object : An object is said to be a point mass if during its motion it covers distance much greater than its own size. One dimensional motion : An object travels in a straight line. It is also called rectilinear or linear motion. The position change of the object with time in one dimension can be described by only one co-ordinate. Ex. Astone falling freelyunder gravity. Two dimensional motion or motion in a plane : For an object travelling in a plane two coordinates sayX and Y are required to describe its motion. Ex. An insect crawling over the floor. Three dimensional motion : An objecttravels in space.Todescribe motion of objects in three dimension require all three coordinates x, yand z. Ex. Akite flying in the sky. DISTANCE AND DISPLACEMENT Distance or Path length : The length of the actual path travelled by an object during motion in a given interval of time is called the distance travelled by that object or path length. It is a scalar quantity. Displacement : It is the shortest distance between the initial and final position of an object and is directed from the initial position to the final position. It is a vector quantity. Keep in Memory 1. Displacement maybe positive, negative or zero but distance is always positive. 2. Displacement is not affected by the shift of the coordinate axes. 3. Displacement of an object is independent of the path followed by the object but distance depends upon path. 4. Displacement and distance both have same unit as that of length i.e. metre. 5. Distance 1 |Displacement| ³ 6. For a moving body distance always increases with time 7. For a bodyundergoing one dimensional motion, in the same direction distance = | displacement |. For all other motion distance > | displacement |. SPEED It is the distance travelled per unit time by an object. It is a scalar quantity. It cannot be negative. Uniform speed : An object is said to be moving with a uniform speed, if it covers equal distances in equal intervals of time, howsoever small the time intervals maybe. Non-uniform speed : If an object covers unequal distances in equal interval of time or equal distances in unequal interval of time. Instantaneousspeed : Thespeed ofan object at a particular instant of time is called the instantaneous speed. Instantaneous speed, Vinst = 0 lim D ® D = D t s ds t dt . Average speed : It is ratio of the total distance travelled by the object to the total time taken. Average speed Vav x t D = D Dimensions : [M0LT-1]; Unit: In SI systems. VELOCITY It is the displacement of an object per unit time. It is a vector quantity. It can be positive negative or zero. Uniform velocity : An object is said tobemoving with a uniform velocity, if it covers equal displacements in equal intervals of time, howsoever small the time intervals maybe. Non-uniform velocity : If an object covers unequal displacements in equal interval oftime or equal displacements in unequal interval of time. Instantaneous velocity : The velocity of an object at a particular instant of time is called the instantaneous velocity. 3 Motion in a Straight Line
- 40. 36 PHYSICS Variable ornon-uniform acceleration : If the velocity of body changes in different amounts during same time interval, then the acceleration of the body is known as variable acceleration. Acceleration is variable if either its direction or magnitude or both changes with respect to time. A good example of variable acceleration is the acceleration in uniform circular motion. EQUATIONS FOR UNIFORMLY ACCELERATED MOTION When the motion is uniformlyaccelerated i.e., when acceleration is constant in magnitude and direction : (i) v = u + at (ii) 2 at 2 1 ut s + = (iii) v2 – u2 = 2as where u = initial velocity; v = final velocity; a = uniform acceleration and s = distance travelled in time t, (iv) u v s t 2 + æ ö = ç ÷ è ø (v) Distance travelled in nth second sn = 1) (2n 2 a u - + ; sn = distance covered in nth second Above equations in vector form , t a 2 1 t u s , t a u v 2 + = + = ) s . a 2 u . u v . v or ( s . a 2 u v 2 2 r r r r r r r r r r = + = - t ) v u ( 2 1 s r r r + = When displacement (s) is given as a function of time t [s = f(t)] then 0 t s µ Body at rest v = 0 a = 0 1 t s µ Uniform velocity; acceleration zero 0 t v µ a = 0 2 t s µ Uniform acceleration 1 t v µ 0 t a µ more or 3 t s µ Non uniform acceleration more or 2 t v µ more or 1 t a µ We use calculus method (integration and differentiation) for displacement, velocity, acceleration as a function of time. We know that ò = Þ = dt v s dt ds v ; ò = Þ = dt a v dt dv a ; when a = f(s) ò ò = Þ = dv v ds a ds dv v a , where s = displacement, v = instantaneous velocity, a = instantaneous acceleration Instantaneous velocity, 0 lim D ® D = = D r r r inst t r dr V t dt . Average Velocity : It is ratio of the total displacement to the total timetaken. Average velocity, D = D r av r v t Dimensions : [M0LT–1] ; Unit: In SI system, m/s Keep in Memory 1. |Average velocity| can be zero but average speed cannot be zero for a moving object. 2. |Average velocity| 1 Average speed £ 3. | Instantaneous velocity | = Instantaneous speed. 4. A particle may have constant speed but variable velocity. It happens when particle travels in curvilinear path. 5. If the body covers first half distance with speed v1 and next half with speed v2 then Average speed 1 2 1 2 2 = + v v v v v 6. If a bodycovers first one-third distance at a speed v1, next one-third at speed v2 and last one-third at speed v3, then Average speed 1 2 3 1 2 2 3 1 3 3 = + + v v v v v v v v v v 7. If a bodytravels with uniform speed v1 for time t1 and with uniform speed v2 for time t2, then Average speed 1 1 2 2 1 2 + = + v t v t v t t ACCELERATION The rate of change of velocity with respect to time is called acceleration. It is a vector quantity. Let velocity changes by v r D during some interval of time t D . Average acceleration av a r is given by D = D r r av v a t Instantaneous acceleration a r is given by 0 lim D ® D = = D r r r t v dv a t dt Its SI unit is meter/sec2 (ms–2). A body moving with uniform velocity has zero acceleration. It means that neither its speed nor its direction ofmotion is changing withtime. Uniform acceleration : If the velocity of the body changes in equal amount during same time interval, then the acceleration of the body is said to be uniform. Acceleration is uniform when neither its direction nor magnitude changes with respect to time.
- 41. 37 Motionina Straight Line VERTICALMOTION UNDER GRAVITY (i) For a bodythrown downward with initial velocityu from a height h, the equations of motion are v = u +gt ; 2 2 v u 2gh = + h = ut + 2 1 gt 2 ; th n h = g u (2n 1) 2 + - (ii) If initial velocity is zero, then the equations are v = gt ; v 2gh = h = 2 1 gt 2 ; th n g h (2n 1) 2 = - (iii) When a bodyis thrown upwards with initial velocity u, the equations of motion are v = u – gt h = ut – 2 1 gt 2 v2 = u2 – 2gh. To summarise : Motion of a body Displacement Velocity Acceleration Velocity constant Uniformly accelerated motion Non-uniformly accelerated motion Accelerated motion Velocity = Displacement Time t ) v u ( 1 s ) 1 n 2 ( 2 a u s as 2 u v at 2 1 ut s at u v nth 2 2 2 ´ + = - + = = - + = + = dt ds v = dt dv a = Differentiation Integration Differentiation Integration ò ò = dt v ds ò ò = dt a dv dv v ds = 2 NOTE: Calculus method asshown in non-uniformlyaccelerated motion may also be used for uniformly accelerated motion. Uniformly Accelerated Motion : A Discussion While using equations of motion we can have two approaches. Approach1: Take a = +ve when velocity increases and a = –ve when velocity decreases. Take rest of physical quantities such as u, v, t and s as positive. Approach 2 : (Vector method) Assume one direction to be positive and other negative. Assign sign to all the vectors (u, v, a, s), +ve sign is given to a vector which is directed to the positive direction and vice-versa – – – + + + Normallythedirection taken is asdrawn above. But it is important tonote that you can takeanydirection ofyour choice tobepositive and the opposite direction to be negative. NOTE: The second method (or approach) is useful only when there is reversal of motion during the activity concerned. Keep in Memory 1. The direction of average acceleration vector is the direction of the change in velocity vector. t v v a i f r r r - = a r has a direction of ) v (– v v v i f i f r r r r + = - i.e., the resultant of f v r and i v – r 2. There is no definite relationship between velocity vector and acceleration vector. 3. For a body starting from rest and moving with uniform acceleration, the ratio of distances covered in t1 sec., t2 sec, t3 sec, etc. are in the ratio t1 2 : t2 2 : t3 2 etc. 4. A body moving with a velocity v is stopped by application ofbrakes after covering a distance s. Ifthe same bodymoves with a velocity nv, it stops after covering a distance n2s by the application of same retardation. Example1. The displacement of particle is zero at t = 0 and displacement is ‘x’ at t = t. It starts moving in the positive x-direction with a velocity which varies as v = k x where k is constant. Show that the velocity varies with time.
- 42. 38 PHYSICS Solution : dt k x dx or x k dt dx or x k v= = = Given that when t = 0, x = 0 and when t = t, x = x, Hence ò ò = t 0 x 0 dt k x dx ; ò ò = - t 0 x 0 ½ dt k dx x or [ ] x 1 1 2 t 0 0 x k t 1 1 2 - + é ù ê ú = ê ú ê ú - + ê ú ë û or 2 t k x or t k x 2 = = Now, 2 t k 2 t k k v 2 = ú û ù ê ë é ´ = Thus the velocity varies with time. Example2. A particle covers each 1/3 of the total distance with speed vl, v2 and v3 respectively. Find the average speed of the particle. Solution : Average speed 1 2 3 Total distance travelled s v s s s Total time taken 3v 3v 3v = = + + Þ 1 2 3 1 2 2 3 3 1 3v v v v v v v v v v = + + Example3. A cheetah can accelerate from0 to 96 km/h in 2 sec., whereas a cat requires 6 sec. Compute the average accelerations for the cheetah and cat. Solution : For cheetah | r a av| = f i | v v | t - D r r = 96km / h 0 2sec - = 1000m 96 3600sec 2sec ´ = 15 m/s2 For cat | r a av| = 10 96 36 6 ´ = 5 m/s2. Example4. A particle is moving in east direction with speed 5 m/s. After 10 sec it starts moving in north direction with same speed. Find average acceleration. Solution : | r v f| = | r v i| = 5 m/s Acceleration ¹ 0 E S N W (due to change in direction of velocityAv. acceleration, D r v = r v f – r v i = r v f + (– r v i) r a = f i v v t - D r r = ˆ ˆ 5j 5i 10 - Þ r a = – 1 2 $ i + 1 2 $ j | r a | = 1 2 1 2 2 2 F HG I KJ + F HG I KJ = 1 2 m/s2. Keep in Memory 1. An object moving under the influence of earth's gravity in which air resistance and small changes in g are neglected is called a freely falling body. 2. In the absence of air resistance, the velocity of projection is equal to the velocity with which the body strikes the ground. 3. Distance travelled by a freely falling body in 1st second is always half of the numerical value of g or 4.9 m, irrespective of height h. 4. For a freely falling body with initial velocityzero (i) Velocity µ time (v = gt) (ii) Velocity Distance fallen µ (v2 = 2gs) (iii) Distance fallen a (time)2 ÷ ø ö ç è æ = 2 gt 2 1 s , where g is the acceleration due to gravity. 5. If maximum height attained bya body projected vertically upwards is equal to the magnitude of velocityof projection, then velocityof projection is 2g ms–1 and time of flight is 4 sec. 6. If maximum height attained bya body projected upward is equal to magnitude of acceleration due to gravity i.e., 'g', the time of ascent is 2 sec. and velocity of projection is 2 g . 7. Ratio of maximum heights reached by different bodies projected with velocities u1, u2, u3 etc. are in the ratio of 3 3 2 2 2 1 u : u : u etc. and ratio of times of ascent are in ratio of u1 : u2 : u3 etc. 8. During free fall velocity increases by equal amount every decend and distance covered during 1st, 2nd, 3rd seconds offall,are4.9m,14.7m,24.5m. t = 3 4.9 m 14.7 m 24.5 m t = 2 t = 1 t = 0 u = 0 9.8 m/s 19.6 m/s 29.4 m/s 9. If a body is projected horizontally from top of a tower, the time taken by it to reach the ground does not depend on the velocity of projection, but depends on the height of tower and is equal to t = g h 2 .
- 43. 39 Motionina Straight Line 10If velocity v of a body changes its direction by q without change in magnitude then the change in velocity will be 2 sin v 2 q . 11. From the top of a tower a body is projected upward with a certain speed, 2nd body is thrown downward with same speed and 3rd is let to fall freely from same point then 2 1 3 t t t = where t1 = time taken by the body projected upward, t2 = time taken by the body thrown downward and t3 = time taken by the body falling freely. 12. If a body falls freelyfrom a height h on a sandysurface and it buries into sand upto a depth of x, then the retardation produced by sand is given by h x a g x + æ ö = ç ÷ è ø . 13. In case of air resistance, the time of ascent is less than time of descent of a body projected vertically upward i.e. ta < td. 14. When atmosphere is effective, then buoyancy force always acts in upward direction whether bodyis moving in upward or downward direction and it depends on volume of the body. The viscous drag force acts against the motion. 15. If bodies have same volume but different densities, the buoyant force remains the same. CAUTION : Please note that dropping bodygets the velocity of the object but if the object is in acceleration, the body dropped will not acquire the acceleration ofthe object. COMMON DEFAULT Incorrect. In the question, if it is given that a body is dropped, taking its initial velocity zero. Correct. The initial velocity is zero if the object dropping the body is also at rest (zero velocity). But if the object dropping the bodyis having a velocity, then the bodybeing dropped will alsohaveinitial velocitywhich will be same as that of the object. For example : (a) When an aeroplane flying horizontallydrops a bomb. (b) An ascending helicopter dropping a food packet. (c) A stone dropped from a moving train etc. Incorrect. Applying equations of motion in case of non- uniform acceleration of the body. Correct. The equations of motion are for uniformly accelerated motion of the body. Pleasenote that when thecaseis ofnon-uniform acceleration we use calculus (differentiation and integration). dt ds v ; ds vdv dt dv a = = = . In fact calculus method is a universal method which can be used both in case of uniform as well as non-uniform acceleration. Incorrect. Taking average velocity same as that of instantaneous velocity. Correct. Average velocity i f i f av t – t r r t r V - = D D = = (where i r is position vector at time ti and f r is position vector at time tf). Whereas instantaneous velocity inst. t 0 r dr V lim t dt D ® D = = D u r u r uu r It is important to note that average velocity is equal to instantaneous velocity only when the case is of uniform velocity. Incorrect. Taking acceleration as negative (– a)even when acceleration is an unknown. Correct. Take acceleration as (a) when it is unknown even if we know that the motion is a case of deceleration or retardation. On solving, we will find the value of (a) to be negative . Incorrect. Magnitude of instantaneous velocity is different from instantaneous speed . Correct. Magnitude of instantaneous velocity is equal to the instantaneous speed in any case. Example5. The numerical ratio of average velocity to average speed is (a) always less than one (b) always equal to one (c) always more than one (d) equal to or less than one Solution : (d) It is equal to or less than one as average velocity depends upon displacement whereas average speed depends upon path length. Example6. The distance travelled by a body is directly proportional to the time taken. Its speed (a) increases (b) decreases (c) becomes zero (d) remains constant Solution : (d) When s µ t, so s v constant. t = = Example7. A particle is projected vertically upwards. Prove that it will be at 3/4 of its greatest height at time which are in the ratio 1 : 3. Solution : Ifu is the initial velocity of a particle while going vertically upwards, then the maximum height attained is g 2 u h 2 = . If t is the time when particle reaches at a height 4 3 h, then using the relation 2 at 2 1 ut s + = ; we have 2 t ) g ( 2 1 ut h 4 3 - + = or, 2 2 gt 2 1 ut g 2 u 4 3 - = ÷ ÷ ø ö ç ç è æ or 0 g u 4 3 t g u 2 t 2 2 2 = + -
- 44. 40 PHYSICS Solving itfor t, we have g 2 u g u 2 g u 4 3 1 4 g u 4 g u 2 t 2 2 2 2 ± = ´ ´ - ± = t = u/g t = 3u/2g 2 3h/4 t1 h Taking negative sign, g 2 u g 2 u g u t1 = - = ; Taking positive sign, . g 2 / u 3 g 2 u g u t2 = + = 3 1 g 2 / u 3 g 2 / u t t 2 1 = ÷ ÷ ø ö ç ç è æ = . Example8. A police party in a jeep is chasing a dacoit on a straight road. The jeep is moving with a maximum uniform speed v. The dacoit rides on a motorcycle of his waiting friend when the jeep is at a distance d from him and the motorcycle starts with constant acceleration a. Show that the dacoit will be caught if ³ v 2ad . Solution : Suppose the dacoit is caught at a time t sec after the motorcycle starts. The distance travelled by the motorcycle during this interval is 2 at 2 1 s = ...(1) During the interval, the jeep travels a distance, s + d = vt ...(2) By(1) and (2), 0 d vt at 2 1 2 = + - or a ad 2 v v t 2 - ± = The dacoit will becaught if t is real andpositive. This will be possible if ad 2 v or 0 ad 2 v2 > > - . Example9. Two trains, each of length 100 m, are running on parallel tracks. One overtakes the other in 20 second and one crosses the other in 10 second. Calculate the velocities of each train. Solution : Let u and v be the velocities of the trains. The relative velocityofovertaking is u – vwhile the relative velocity of crossing is u + v. Total distance = 100 + 100 = 200 m. 10 v u or v u 200 20 = - - = and 20 v u or v u 200 10 = + + = By solving, we get, u = 15 m/sec and v = 5 m/sec. Example10. A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth sec to the distance covered in n sec is (a) 2 2 1 – n n (b) 2 1 1 – n n (c) 2 2 2 1 – n n (d) 2 2 1 + n n Solution : (a) The distance covered in nth second ) 1 n 2 ( 2 a u sn - + = or ) 1 n 2 ( 2 a 0 sn - + = ...(1) Further distance covered in n second 2 2 n a 2 1 0 t a 2 1 t u s + = + = ...(2) 2 2 n n 1 n 2 ) 2 / an ( ) 1 n 2 ( 2 a s s - = - = Example11. The water drop falls at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant? (a) 1.25 m (b) 2.50 m (c) 3.75 m (d) 4.00 m Solution : (c) See fig. Let t be the time interval between any two drops. 5 m x h Ist drop 3rd drop ground 2nd drop Tap For third drop 2 / 5 t g or ) t 2 ( g 2 1 5 2 2 = = For second drop 2 t g 2 1 x = or m 25 . 1 4 5 2 5 2 1 x = = ´ = Thereforeh = 5– x = 5– 1.25 = 3.75 m Hence option (c)
- 45. 41 Motionina Straight Line Example12. Theheight of a tower is h metre. A body is thrown from the top of tower vertically upward with some speed, it takes t1, second to reach the ground. Another body thrown from the top of tower with same speed downwards and takes t2 seconds to reach the ground. If third body, released from same place takes ‘t’ second to reach the ground, then (a) 1 2 t +t t = 2 (b) 1 2 t t = t (c) 1 2 2 1 1 = + t t t (d) 1 2 t = (t t ) Solution : (d) Let u be the initial velocity of the body. Then at time t1 2 1 1 t g 2 1 t u h + - = ...(1) and at time t2, h = 2 2 2 t g 2 1 t u + ...(2) From eqs. (1) and (2), we get 2 2 2 2 1 1 t g 2 1 t u t g 2 1 t u + = + - or ) t t ( u ) t t ( g 2 1 2 1 2 2 2 1 + = - or ) t t ( u ) t t ( ) t t ( g 2 1 2 1 2 1 2 1 + = - + or ) t t ( g 2 1 u 2 1 - = ...(3) Substituting the value of u in eqn. (2) we get 2 2 2 2 1 t g 2 1 t ) t t ( g 2 1 h + - = or 2 1 t t g 2 1 h = ...(4) For third body, 2 t g 2 1 h = ...(5) From eqs. (4) and (5) we get, 1 2 t t t = Example13. A bullet moving with a speed 10 m/s hits the wooden plank and is stopped when it penetrates the plank 20 cm deep. Calculate retardation of the bullet. Solution : v0 = 10 m/s, v = 0 and s = 20 cm. = 2 0.02m 100 = Using v2 – v0 2 = 2ax 2 100 0 (10) 2a (0.02) a 2 0.02 - - = Þ = ´ or a = – 2500 m/s2 Retardation = 2500 m/s2 Example14. A body covers a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it cover in 15th sec? Solution : From formula, th n S u a(2n –1) 2 1 = + 7 a S th u (2 7 1) 2 = + ´ - but 7 S th 20m = a 13a 20 u 13 20 u 2 2 = + ´ Þ = + ...(1) also 9 s th 24m = 17a 24 u 2 = + ...(2) From eqn. (1) 13a u 20 2 = - ...(3) Substitute this value of in eqn. (2) 13a 17a 24 20 2 2 = - + 17a 13a 24 20 2 2 - = - 4a 4 2 = Þ 4 = 2a Þ 2 4 a 2 m /s 2 = = Substituting this value of a in eqn. (3) 13a u 20 2 = - Þ 13 2 u 20 u 20 13 2 ´ = - Þ = - u = 7 m/s Now, s15th = a u (2 15 1) 2 + ´ - = 2 7 (29) 7 29 36m 2 + = + = Example15. A train starts from rest and for the first kilometer moves with constant acceleration, for the next 3 kilometers it has constant velocity and for another 2 kilometers it moves with constant retardation to come to rest after 10 min. Find the maximum velocity and the three time intervals in the three types of motion. Solution : Let the three time intervals be t1 min, t2 min, and t3 min. respectively. Let the maximum velocityattained be vm/min. A B C D 1000m 3000m 2000m t1 v–t2 t3 For Ato B 1 0 v 1000 t 2 + æ ö = ç ÷ è ø Þ 1 vt 2000 = ...........(1) For B to C 2 vt 3000 = ...........(2) [Using in both equations disp. = mean vel. × time] and for C to D 3 v 0 2000 t 2 + æ ö = ç ÷ è ø Þ 3 vt 4000 = ...... (3) Adding eqs. (1), (2) and (3), we get ) t t t ( v 9000 3 2 1 + + =
- 46. 42 PHYSICS 3 9000 90010 v 900 m / min. km 54 km / hr. 10 1/ 60 hr. - ´ = = = = Now, from eqs. (1), (2) and (3) we get 1 2000 20 2 t 2 min. 900 9 9 = = = , 2 3000 10 1 t 3 min. 900 9 3 = = = and 3 4000 40 4 t 4 min. 900 9 9 = = = Example16. A falling stone takes 0.2 seconds to fall past a window which is 1m high. From howfar above the top of the window was the stone dropped ? Solution : 2 1 h gt 2 = ; 2 1 h 1 g(t 0.2) 2 + = + 2 2 2 1 1 1 1 gt 1 gt g(0.2) g 2 0.2t 2 2 2 2 + = + + ´ ´ u=0 1m h 1 1 0.2gt 5 = + ; 4 2 2t t 5 5 = Þ = 1 4 4 h g m 2 25 5 = = Example17. From the top of a multi-storeyed building 40m tall, a boy projects a stone vertically upwards with an initial velocity of 10 ms–1 such that it eventually falls to the ground. (i) After how long will the stone strike the ground ? (ii) After how long will it pass through the point from where it was projected ? (iii) What will be its velocity when it strikes the ground ? Take g = 10 ms–2. Solution : Using, 2 1 S ut at 2 r r r = + (i) 2 1 40 10t gt 2 - = - or 2 40 10t 5t - = - [Q taking g = 10m/s2] or 2 5t 10t 40 0 - - = h v u u=10m/s 40m or 2 10 10 4 5( 40) 10 100 800 t 2 5 10 + - ´ - + + = = ´ 10 30 4 sec. 10 + = = (ii) It will pass from where it was projected after 2 10 t 2 sec. g ´ = = (iii) Velocity with which stone strikes the ground V =10 + g× 2 =30m/s VARIOUS GRAPHS RELATED TO MOTION (a) Displacement-time graph - In this graph time is plotted on x-axis and displacement on y-axis. (i) For a stationary body (v = 0) the time-displacement graph is a straight line parallel to time axis. Displ Velocity = 0 Time Body is at rest (ii) When the velocity of a body is constant then time- displacement graph will be an oblique straight line. Greater the slope ÷ ø ö ç è æ q = tan dt dx of the straight line, higher will be the velocity. (iii) If the velocity of a body is not constant then the time- displacement curve is a zig-zag curve. (iv) For an accelerated motion the slope of time- displacement curve increases with time while for decelerated motion it decreases with time.
- 47. 43 Motionina Straight Line a> 0 t Displ. a < 0 t Displ. (v) When the particle returns towards the point of reference then the time-displacement line makes an angle q > 90° with the time axis. t Displ q > 90º (b) Velocity-time graph - In this curve time is plotted along x-axis and velocity is plotted along y-axis. (i) When the velocity of the particle is constant or acceleration is zero. t Velocity Vel. const, a = 0 (ii) When the particle is moving with a constant acceleration and its initial velocity is zero. (iii) When the particle is moving with constant retardation. t V Retardation (iv) When the particlemoveswith non-uniform acceleration and its initial velocity is zero. (v) When the acceleration decreases and increases. t V a d e c r e a s i n g a increasing (vi) The total area enclosed by the time - velocity curve represents the distance travelled by a body. NOTE: While finding displacement through v – t graph, keeping sign under consideration. (c) Acceleration-time graph - In this curve the time is plotted along X-axis and acceleration is plotted alongY-axis. (i) When the acceleration of the particle is zero. t a Acc = 0 (ii) When acceleration is constant t a = const a
- 48. 44 PHYSICS (iii) Whenacceleration is increasing and is positive. t a a increasing (iv) When acceleration is decreasing and is negative t a decreas i n g a (v) When initial acceleration is zero and rate of change of acceleration is non-uniform t a (vi) The change in velocityof the particle = area enclosed bythe time-acceleration curve. t a Example18. The velocity-time graph of a body moving in a straight line is shown in fig. Find the displacement and distance travelled by the body in 10 seconds. 20 10 0 –10 –20 A 4 2 6 8 10 C D t (s) u (m/s) Solution : The area enclosed by velocity-time graph with time axis measuresthedistance travelled in a given time. Displacement covered from 0 to 6 seconds is positive; from 6 to 8 seconds is negative and from 8 to 10 seconds is positive; whereas distance covered is always positive. Total distance covered in 10 s 10 2 2 1 20 2 2 1 20 6 2 1 ´ ´ + ´ ´ + ´ ´ = =90m Total displacement in 10s m 50 10 2 2 1 20 2 2 1 20 6 2 1 = ´ ´ + ´ ´ - ´ ´ = Example19. Two trains, which are moving along different tracks in opposite directions, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down, The separation, between the trains when both have stopped, is 10 20 40 V(m/s) t (s) Train I 8 –20 V(m/s) t (s) Train II (a) 120m (b) 280m (c) 60m (d) 20m. Solution : (d) Initial distance between trains is 300m. Displacement of1st train is calculated by area under v – t. Curve of train 1 I displacement 10 40 200m 2 = = ´ ´ = Displacement of train 1 II 8 ( 20) 80m 2 = ´ ´ - = - Distance between the two trains is 20m. Example20. Figure given below shows the variation of velocity of a particle with time. 1 2 3 4 5 6 7 velocity (m/s) 2 4 time(sec) Y X 6 8 Find the following : (i) Displacement during the time intervals (a) 0 to 2 sec, (b) 2 to 4 sec. and (c) 4 to 7 sec. (ii) Accelerations at – (a) t = 1 sec, (b) t = 3 sec. and (c) t = 7 sec. (iii) Average acceleration – (a) between t = 0 to t = 4 sec. (b) between t = 0 to t = 7 sec. (iv) Average velocity during the motion.
- 49. 45 Motionina Straight Line Solution: (i) (a) Displacement between t = 0 sec. to t = 2 sec. Þ m 8 s / m 8 sec 2 2 1 = ´ ´ (b) Between t = 2 sec. to t = 4 sec. Þ m 16 s / m 8 sec 2 = ´ (c) Between t = 4 sec. to t = 7 sec. Þ m 12 s / m 8 sec 3 2 1 = ´ ´ (ii) Acceleration = slope of v – t curve (a) At t = 1 sec, slope = 2 s / m 4 sec / m sec 2 sec / m 8 = (b) At t = 3 sec, slope = 0 (c) At t = 7 sec, slope = 2 s / m 3 2 2 3 8 - = - (iii) Average acceleration = Total change in velocity Total change in time (a) Between t = 0 to t = 4 sec., Average acceleration = 2 s / m 2 4 8m/s = (b) Between t = 0 to t = 7 sec., Average acceleration = 0 7 0 = (iv) Average velocity = Total displacement Total time = s / m 7 1 5 7 36 7 12 16 8 = = + + Example21. The velocity-time graph of a particle moving along a straight line is shown below. The acceleration and deceleration are same and it is equal to 4 m/s². If the average velocity during the motion is 15 m/ s and total time of motion is 20 second then find t t 20 20–t 20 – 2t v (a) the value of t (b) the maximum velocity of the particle during the journey. (c) the distance travelled with uniform velocity. Solution : v = 0 + at Total displacement = 1 (20 2t 20) 4t 2t (40 2t) 2 - + ´ = - Average velocity = Total displacement Total time 2t (40 2t) 15 20 - = Solving quadratic equation, 150 = 40t – 2t2 Þ t = 5 sec. (another solution not acceptable think why!) Maximum velocity= 4t = 4 × 5= 20 m/s Distance travelled with uniform velocity = (20 – 2t) V = (20 – 2 × 5) × 20 = 200 m RELATIVE VELOCITY (In one dimension) ThevelocityofArelative toB isthe velocitywith whichAappears tobe movingw.r.t.an observer who is moving with the velocityofB Relative velocity ofAw.r.t. B AB A B v v v = - r r r Similarly, relative velocityofB w.r.t.A BA B A v v v = - r r r Case 1: Bodies moving in same direction : vA vB AB A B v v v = - r r r Þ vAB = vA – vB Case 2:Bodiesmovingin opposite direction: – + vA vB ( ) = - - r r r AB A B v v v Þ vAB = vA + vB Example22. Two trains, one travelling at 54kmph and the other at 72kmph, are headed towards each other on a level track. When they are two kilometers apart, both drivers simultaneously apply their brakes. If their brakes produces equal retardation in both the trains at a rate of 0.15 m/s2, determine whether there is a collision or not. Solution : Speed offirst train = 54 kmph = 15m/s. Speed ofsecond train = 72kmph = 20 m/s As both the trains are headed towards each other, relative velocity of one train with respect to other is given as vr=15+ 20 =35m/s Both trains are retarded byacceleration of 0.15 m/s2. Relative retardation ar = 0.15 + 0.15 = 0.3 m/s2. Now, we assume one train is at rest and other is coming at 35m/s retarded by 0.3 m/s2 at a distance of two kilometer. The maximum distance travelled bythe moving train while retarding is smax = v a r r 2 2 = ( ) . 35 2 0 3 2 ´ = 2041.66m It is more than 2km, which shows that it will hit the second train. Example23. Two cars started simultaneously towards each other from towns A and B which are 480 km apart. It took first car travelling from A to B 8 hours to cover the distance and second car travelling from B to A 12 hours. Determine when the car meet after starting and at what distance from town A. Assuming that both the cars travelled with constant speed. Solution : Velocity of car from A = hour / km 60 8 480 = velocityof car from B = hour / km 40 12 480 = Let the two cars meet at t hour total distance 480 t 4.8 relative velocity of meeting 60 40 = = = + hour The distance s = vA × t = 60 × 4.8 = 288 km.
- 50.
- 51. 47 Motionina Straight Line 1.The study of motion, without consideration of its cause is studied in (a) statistics (b) kinematics (c) mechanics (d) modern physics 2. The ratio of the numerical values of the average velocity and average speed of a body is always: (a) unity (b) unity or less (c) unityor more (d) less than unity 3. A particlehas moved from one position to another position (a) its distance is zero (b) its displacement is zero (c) neither distance nor displacement is zero (d) average velocity is zero 4. The displacement of a body is zero. The distance covered (a) is zero (b) is not zero (c) may or may not be zero (d) depends upon the acceleration 5. Which of the following changes when a particle is moving with uniform velocity? (a) Speed (b) Velocity (c) Acceleration (d) Position vector 6. The slope of the velocity time graph for retarded motion is (a) positive (b) negative (c) zero (d) can be +ve, –ve or zero 7. The area of the acceleration-displacement curve of a body gives (a) impulse (b) change in momentum per unit mass (c) change in KE per unit mass (d) total change in energy 8. If the displacement of a particle varies with time as 7 t x + = , the (a) velocityofthe particle is inverselyproportional to t (b) velocity of the particle is proportional to t (c) velocity of the particle is proportional to t (d) the particle moves with a constant acceleration 9. The initial velocity of a particle is u (at t = 0) and the acceleration a is given by f t. Which of the following relation is valid? (a) v = u + f t2 (b) v = u + f t2/2 (c) v = u + f t (d) v = u 10. The displacement xofa particle movingalong a straight line at time t is given by x = a0 + a1 t + a2 t2 What is the acceleration of the particle (a) a1 (b) a2 (c) 2 a2 (d) 3 a2 11. The displacement-time graphs of two particlesA and B are straight lines making angles ofrespectively30º and 60º with thetime axis. Ifthe velocityofAis vA and that of Bis vB, the value of vA/vB is (a) 1/2 (b) 3 / 1 (c) 3 (d) 1/3 12. A person travels along a straight road for the first half time with a velocity v1 and the second half time with a velocity v2. Then the mean velocity v is given by (a) 2 v v v 2 1 + = (b) 2 1 v 1 v 1 v 2 + = (c) 2 1 v v v = (d) 1 2 v v v = 13. A particle covers half of the circle of radius r. Then the displacement and distance of the particle are respectively (a) 2pr,0 (b) 2r,pr (c) r 2 , 2 r p (d) pr,r 14. The distance through which a body falls in the nth second is h. The distancethrough which it falls in the next second is (a) h (b) 2 g h + (c) h – g (d) h + g 15. It is given that t = px2 + qx, where x is displacement and t is time. The acceleration ofparticle at origin is (a) 3 q p 2 - (b) 3 p q 2 - (c) 3 q p 2 (d) 3 p q 2 16. Figure shows the v-t graph for twoparticles P and Q. Which of the following statements regarding their relative motion is true ? Their relative velocityis O V Q T P (a) is zero (b) is non-zero but constant (c) continuously decreases (d) continuously increases 17. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocityofsound, the time T after which the splash is heard is given by (a) T = 2h/v (b) 2h h T g v æ ö = + ç ÷ è ø (c) g h v h 2 T + ÷ ø ö ç è æ = (d) v h 2 g 2 h T + ÷ ÷ ø ö ç ç è æ =
- 52. 48 PHYSICS 18. Apoint traversed half of the distance with a velocity v0. The half ofremaining part of the distance was covered with velocity v1 & second half of remaining part by v2 velocity. The mean velocity of the point, averaged over the whole time of motion is (a) 3 v v v 2 1 0 + + (b) 3 v v v 2 2 1 0 + + (c) 3 v 2 v 2 . v 2 1 0 + + (d) ) v v v 2 ( ) v v ( v 2 2 1 0 2 1 0 + + + 19. The acceleration ofa particle is increasing linearlywith time t as bt. The particle starts from the origin with an initial velocity v0. The distance travelled by the particle in time t will be (a) 2 0 bt 3 1 t v + (b) 3 0 bt 3 1 t v + (c) 3 0 bt 6 1 t v + (d) 2 0 bt 2 1 t v + 20. The deceleration experienced by a moving motorboat after its engine is cut off, is given by dv/dt = – kv3 where k is constant. If v0 is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut-off is (a) ) 1 kt v 2 ( v 2 0 0 + (c) kt 0 e v - (c) 2 / v0 (d) 0 v 21. The displacement x ofa particle varies with timeaccording to the relation ). e 1 ( b a x bt - - = Then select the false alternatives. (a) At b 1 t = , the displacement of the particle is nearly ÷ ø ö ç è æ b a 3 2 (b) the velocity and acceleration of the particle at t = 0 are a and –ab respectively (c) the particle cannot go beyond b a x = (d) the particle will not come back to its starting point at t ® ¥ 22. Thedisplacementof a particleisgiven by x t 1 = + . Which of the following statements about its velocity is true ? (a) It is zero (b) It is constant but not zero (c) It increases with time (d) It decreases with time 23. Two bodies of masses m1 and m2 fall from heights h1 andh2 respectively. The ratio of their velocities, when theyhit the ground is (a) 2 1 h h (b) 2 1 h h (c) 1 1 1 2 m h m h (d) 2 2 2 1 h h 24. Two cars A and B are travelling in the same direction with velocities vA and vB (vA>vB).When the car Ais at a distance d behind the car B the driver of the car A applies brakes producing a uniform retardation a. Therewill benocollision when (a) a 2 ) v v ( d 2 B A - < (b) a 2 v v d 2 B 2 A - < (c) a 2 ) v v ( d 2 B A - > (d) a 2 v v d 2 B 2 A - > 25. A bodyis thrown upwards and reaches its maximum height. At that position (a) itsacceleration isminimum (b) its velocity is zero and its acceleration is also zero (c) its velocityis zerobut its acceleration is maximum (d) its velocityis zeroanditsacceleration is the acceleration due to gravity. 1. The position x of a particle varies with time (t) as x =At2 – Bt3.The acceleration at timet ofthe particle will be equal to zero. What is the value of t? (a) B 3 A 2 (b) B A (c) B 3 A (d) zero 2. The acceleration of a particle, starting from rest, varies with time according to the relation t sin s a 2 w w - = The displacement of this particle at a timet will be (a) s sin w t (b) s w cos w t (c) s w sin w t (d) 2 2 t ) t sin s ( 2 1 w w - 3. The displacement of a particle is given by y = a + b t + c t2 – d t4 The initial velocityand acceleration are respectively (a) b, – 4 d (b) – b, 2 c (c) b, 2 c (d) 2 c, – 4 d 4. A passenger travels along the straight road for half the distance with velocity v1 and the remaining half distance with velocity v2. Then average velocity is given by (a) v1v2 (b) v2 2/ v1 2 (c) (v1 + v2 )/2 (d) 2v1v2 / (v1 + v2) 5. A point moves with uniform acceleration and v1, v2 and v3 denote the average velocities in t1, t2 and t3 sec. Which of the following relation is correct? (a) ) t t ( : ) t t ( ) v v ( : ) v v ( 3 2 2 1 3 2 2 1 + - = - - (b) ) t t ( : ) t t ( ) v v ( : ) v v ( 3 2 2 1 3 2 2 1 + + = - - (c) ) t t ( : ) t t ( ) v v ( : ) v v ( 3 1 2 1 3 2 2 1 - - = - - (d) ) t t ( : ) t t ( ) v v ( : ) v v ( 3 2 2 1 3 2 2 1 - - = - -
- 53. 49 Motionina Straight Line 6.Abus starts moving with acceleration 2 m/s2.Acyclist 96 m behind the bus starts simultaneously towards the bus at 20 m/s. After what time will he be able to overtake the bus? (a) 4 sec (b) 8 sec (c) 12 sec (d) 16 sec 7. When the speed of a car is v, the minimum distance over which it can be stopped is s. If the speed becomes n v, what will be the minimum distance over which it can be stopped during same retardation (a) s/n (b) n s (c) s/n2 (d) n2 s 8. The two ends of a train moving with constant acceleration pass a certain point with velocities u and v. The velocitywith which the middle point of the train passes the same point is (a) (u + v)/2 (b) (u2 + v2)/2 (c) 2 2 (u v )/ 2 + u (d) 2 2 u v + 9. A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is (a) 32m (b) 16m (c) 4m (d) None of the above 10. The velocityof a particle at an instant is 10 m/s. After 5 sec, the velocity of the particle is 20 m/s. Find the velocity at 3 seconds before from the instant when velocity of a particle is 10m/s. (a) 8 m/s (b) 4 m/s (c) 6 m/s (d) 7 m/s 11. A particle experiences constant acceleration for 20 seconds after startingfromrest.If it travelsadistances1 in the first 10 seconds and distance s2 in the next 10 seconds, then (a) s2 = s1 (b) s2 = 2 s1 (c) s2 = 3 s1 (d) s2 = 4 s1 12. A train of 150 m length is going towards north direction at a speed of 10 ms–1. Aparrot flies at a speed of 5 ms–1 towards south direction parallel to the railwaytrack. The time taken by the parrot to cross the train is equal to (a) 12 s (b) 8 s (c) 15 s (d) 10 s 13. A particle is moving along a straight line path according to the relation s2 = at2 + 2bt + c s represents the distance travelled in t seconds and a, b, c are constants. Then the acceleration of the particle varies as (a) s–3 (b) s3/2 (c) s–2/3 (d) s2 14. A stone thrown vertically upwards with a speed of 5 m/sec attains a height H1. Another stone thrown upwards from the same point with a speed of 10 m/sec attains a height H2. The correct relation between H1 and H2 is (a) H2 = 4H1 (b) H2 = 3H1 (c) H1 =2H2 (d) H1 = H2 15. A body covers 26, 28, 30, 32 meters in 10th, 11th, 12th and 13th seconds respectively. The body starts (a) from rest and moves with uniform velocity (b) from rest and moves with uniform acceleration (c) with an initial velocity and moves with uniform acceleration (d) with an initial velocityandmoveswith uniform velocity 16. A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is (a) 3u2/g (b) 4u2/g (c) 6u2/g (d) 9u2/g 17. A smooth inclined plane is inclined at an angle q with horizontal. A body starts from rest and slides down the inclined surface. h q Then the time taken by it to reach the bottom is (a) ÷ ÷ ø ö ç ç è æ g h 2 (b) ÷ ÷ ø ö ç ç è æ g 2l (c) 1 2h sin g q (d) g ) h 2 ( sin q 18. A ball is dropped downwards, after 1 sec another ball is dropped downwards from the same point. What is the distance between them after 3 sec? (a) 25m (b) 20m (c) 50m (d) 9.8m 19. Twotrains are each 50 m long moving parallel towards each other at speeds 10 m/s and 15 m/s respectively. After what time will theypass each other? (a) sec 3 2 5 (b) 4 sec (c) 2 sec (d) 6 sec 20. A ball is projected verticallyupwards with kinetic energyE. The kinetic energy of the ball at the highest point of its flight will be (a) E (b) 2 / E (c) E/2 (d) zero 21. A particle is moving in a straight line with initial velocity and uniform acceleration a. Ifthe sum ofthe distancetravelled in tth and (t + 1)th seconds is 100 cm, then its velocity after t seconds, in cm/s, is (a) 80 (b) 50 (c) 20 (d) 30 22. Similar balls are thrown vertically each with a velocity 20 ms–1, one on the surface of earth and the other on the surface of moon. What will be ratioof the maximum heights attained bythem? (Acceleration on moon = 1.7ms–2 approx) (a) 6 (b) 6 1 (c) 5 1 (d) None of these
- 54. 50 PHYSICS 23. Therelative velocityVAB or VBA of two bodies A& B may be (1) greater than velocity of body A (2) greater than velocity of body B (3) less than the velocity of body A (4) less than the velocity of body B (a) (1) and (2) only (b) (3) and (4) only (c) (1), (2) and (3) only (d) (1),(2),(3) and(4) 24. A stone is thrown vertically upwards. When the particle is at a height halfofits maximum height, its speed is 10m/sec, then maximum height attained byparticle is (g = 10m/sec2) (a) 8m (b) 10m (c) 15m (d) 20m 25. From a 10m high building a stone ‘A’ is dropped & simultaneouslyanother stone ‘B’ is thrown horizontallywith an initial speed of 5 m/sec–1. Which one of the following statements is true? (a) It is not possible to calculate which one of two stones will reach ground first (b) Both stones ‘A’ & ‘B’ will reach the ground simultaneously. (c) ‘A’stones reach the ground earlier than ‘B’ (d) ‘B’ stones reach the ground earlier than ‘A’ 26. An automobiletravelling with a speed of60 km/h, can apply brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be (a) 60m (b) 40m (c) 20m (d) 80m 27. The motion of a particle is described bythe equation u = at. The distance travelled by particle in first 4 sec is (a) 4a (b) 12a (c) 6a (d) 8a 28. If you were tothrow a ball verticallyupward with an initial velocityof 50 m/s, approximatelyhow long would it takefor the ball to return to your hand? Assume air resistance is negligible. (a) 2.5 s (b) 5.0 s (c) 7.5 s (d) 10 s 29. A body travels 2 m in the first two second and 2.20 m in the next 4 second with uniform deceleration. The velocityofthe body at the end of 9 second is (a) – 10 1 s m - (b) –0.20 1 s m - (c) –0.40 1 s m - (d) –0.80 1 s m - 30. From a 200 m high tower, one ball is thrown upwards with speed of 10 1 s m - and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to (a) 12 s (b) 6 s (c) 2 s (d) 1 s 31. Arocket is fired upward from the earth’s surface such that it creates an acceleration of 19.6 2 s m - . If after 5 s, its engine isswitched off, themaximumheight oftherocket from earth’s surface would be (a) 980m (b) 735m (c) 490m (d) 245m 32. A particle travels halfthe distance with a velocityof6 1 s m - . The remaining half distance is covered with a velocityof 4 1 s m - for halfthe time and with a velocityof 8 1 s m - for the rest of the half time. What is the velocity of the particle averaged over the whole time of motion ? (a) 9 ms–1 (b) 6 ms–1 (c) 5.35ms–1 (d) 5 ms–1 33. A ball released from a height falls 5 m in one second. In 4 seconds it falls through (a) 20m (b) 1.25m (c) 40m (d) 80m 34. A food packet is released from a helicopter rising steadilyat the speed of 2 m/sec. After 2 seconds the velocity of the packet is (g = 10 m/sec2) (a) 22 m/sec (b) 20 m/sec (c) 18 m/sec (d) none of these 35. The displacement x of a particle along a straight line at time t is given by : x = a0 + 2 t a1 + 3 a2 t2. The acceleration of the particle is (a) 3 a2 (b) 3 a 2 2 (c) 2 a1 (d) a0 + 3 a2 36. A rubber ball is dropped from a height of 5 metre on a plane where the acceleration due to gravity is same as that onto the surface of the earth. On bouncing, it rises to a height of 1.8 m. On bouncing, the ball loses its velocitybya factor of (a) 5 3 (b) 25 9 (c) 5 2 (d) 25 16 37. A boy moving with a velocityof 20 km h–1 along a straight line joining two stationary objects. According to him both objects (a) move in the same direction with the same speed of 20 km h–1 (b) move in different direction with the same speed of 20 km h–1 (c) move towards him (d) remain stationary 38. A man leaves his house for a cycle ride. He comes back to his house after half-an-hour after covering a distance of one km. What is his average velocity for the ride ? (a) zero (b) 2 km h–1 (c) 10 km s–1 (d) -1 1 km s 2 39. Acar travels fromAtoBat a speed of20 km h–1 andreturns at a speed of 30 km h–1. The average speed of the car for the whole journey is (a) 5 km h–1 (b) 24 km h–1 (c) 25 km h–1 (d) 50 km h–1
- 55. 51 Motionina Straight Line 40.A body dropped from a height ‘h’ with an initial speed zero, strikes the ground with a velocity3 km/hour. Another body of same mass dropped from the same height ‘h’ with an initial speed u' = 4 km/hour. Find the final velocityofsecond mass, with which it strikes the ground (a) 3km/hr (b) 4km/hr (c) 5km/hr (d) 6km/hr 41. An electron starting from rest has a velocity that increases linearlywith timei.e. v= kt where 2 s m 2 k - = . Thedistance covered in the first 3 second is (a) 9m (b) 16m (c) 27m (d) 36m 42. A body released from the top of a tower falls through half theheight ofthetower in 2 s. In what time shall the bodyfall through the height of the tower ? (a) 4 s (b) 3.26s (c) 3.48s (d) 2.828s 43. The displacement x of a particle at the instant when its velocity is v is given by 16 x 3 v + = . Its acceleration and initial velocity are (a) 1.5 units, 4 units (b) 3 units, 4 units (b) 16 units, 1.6 units (d) 16 units, 3 units 44. Let A, B, C, D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the times of descent through AB, BC and CD are in the ratio. (a) 2 3 : 2 3 : 1 + - (b) 2 3 : 1 2 : 1 - - (c) 3 : 1 2 : 1 - (d) 1 3 : 2 : 1 - 45. A body moves in a straight line along Y-axis. Its distance y (in metre) from theorigin isgiven byy= 8t – 3t2. Theaverage speed in the time interval from t = 0 second to t = 1 second is (a) – 4 ms–1 (b) zero (c) 5 ms–1 (d) 6 ms–1 46. The acceleration duetogravityon planet Ais nine times the acceleration due to gravity on planet B. A man jumps to a height 2m on the surface of A. What is height of jump by same person on planet B? (a) 2/3m (b) 2/9m (c) 18m (d) 6m 47. In the given figure the distance PQ is constant. SQ is a vertical line passing through point R. Aparticle is kept at R and the plane PR is such that angle q can be varied such that R lies on line SQ. The time taken by particle to come down varies, as the q increases q S P Q R (a) decreases continuously (b) increases (c) increases then decreases (d) decreases then increases 48. The displacement x of a particle varies with time t as x = ae-at + bebt, where a, b, a and b are positive constants. The velocity of the particle will (a) be independent of a and b (b) drop to zero when a = b (c) go on decreasing with time (d) go on increasing with time 49. Which oneof the following equations represents the motion of a body with finite constant acceleration ? In these equations, y denotes the displacement of the body at time t and a, b and c are constants of motion. (a) y = at (b) 2 bt at y + = (c) 3 2 ct bt at y + + = (d) bt t a y + = 50. The dependence of velocity of a body with time is given by the equation 2 v 20 0.1t . = + The body is in (a) uniform retardation (b) uniform acceleration (c) non-uniform acceleration (d) zero acceleration. 51. Two stones are thrown from the top of a tower, one straight down with an initial speed u and the second straight up with the same speed u. When the twostones hit the ground, they will have speeds in the ratio (a) 2: 3 (b) 2: 1 (c) 1: 2 (d) 1: 1 52. A graph of acceleration versus time of a particle starting from rest at t = 0 is as shown in Fig. The speed of the particle at t = 14 second is .) sec in ( t ) ms in ( a 2 - 0 2 2 4 4 6 8 10 12 14 2 - 4 - (a) 2 ms–1 (b) 34 ms–1 (c) 20 ms–1 (d) 42 ms–1 53. In the displacement d versus time t graph given below, the value of average velocityin the time interval 0 to 20 s is (in m/s) 0 0 10 10 30 30 40 50 20 20 s / t m / d (a) 1.5 (b) 4 (c) 1 (d) 2
- 56. 52 PHYSICS 54. Figureshows the position of a particle moving along the X-axis as a function of time. 2 4 6 10 20 ) m ( x ) s ( t Which of the following is correct? (a) The particle has come to the rest 6 times (b) The maximum speed is at t = 6 s. (c) The velocity remains positive for t = 0 to t = 6 s. (d) The average velocity for the total period shown is negative. 55. A steel ball is bouncing up and down on a steel plate with a period of oscillation of 1 second. If g = 10 ms–2, then it bounces up to a height of (a) 5m (b) 10m (c) 2.5m (d) 1.25m 56. A bodystarts from rest and travels a distance x with uniform acceleration, then it travelsa distance2x with uniform speed, finallyit travels a distance 3x with uniform retardation and comes to rest. Ifthe complete motion ofthe particle is along a straight line, then the ratio of its average velocity to maximumvelocityis (a) 2/5 (b) 3/5 (c) 4/5 (d) 6/7 57. When two bodies move uniformly towards each other, the distance decreases by 6 ms–1. If both bodies move in the same directions with the same speeds (as above), the distance between them increases by 4 ms–1. Then the speeds of the two bodies are (a) 3 ms–1 and 3 ms–1 (b) 4 ms–1 and 2 ms–1 (c) 5 ms–1 and 1 ms–1 (d) 7 ms–1 and 3 ms–1 58. A ball is thrown verticallyupward with a velocity ‘u’ from the balloon descending with velocity v. The ball will pass by the balloon after time (a) g 2 v u - (b) g 2 v u + (c) 2(u v) g - (d) g ) v u ( 2 + 59. Two bodies begin a free fall from the same height at a time interval of N s. If vertical separation between the two bodies is 1 after n second from the start of the first body, then n is equal to (a) nN (b) gN 1 (c) 2 N gN 1 + (d) 4 N gN 1 - 60. A particle starting from rest falls from a certain height. Assuming that the acceleration due to gravity remain the same throughout the motion, its displacements in three successive half second intervals are S1, S2 and S3 then (a) S1 : S2 : S3 = 1 : 5 : 9 (b) S1 : S2 : S3 = 1 : 3 : 5 (c) S1 : S2 : S3 = 9 : 2 : 3 (d) S1 : S2 : S3 = 1 : 1 : 1 61. For the velocity time graph shown in the figure below the distance covered by the body in the last two seconds of its motion is what fraction ofthe total distance travelled by it in all the seven seconds? 10 8 6 4 2 0 1 2 3 4 5 6 7 8 B C D A ms velocity –1 (a) 2 1 (b) 4 1 (c) 3 2 (d) 3 1 62. Which of the following graph cannot possibly represent one dimensional motion ofa particle? (a) t x (b) x t (c) t speed (d) All of the above 63. The distance time graph of a particle at time t makes angles 45° with the time axis.After one second, it makes angle 60° with the time axis. What is the acceleration of the particle? (a) 3 1 - (b) 3 1 + (c) 3 (d) 1 64. TwoparticlesA and B are connected bya rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity ofA to the left is 10 m/s. What is the velocity of B when angle a = 60°? B A a (a) 9.8m/s (b) 10m/s (c) 5.8m/s (d) 17.3m/s 65. A balloon starts rising from the ground with an acceleration of 1.25 ms–2. After 8 s, a stone is released from the balloon. The stone will (Taking g = 10 m s–2) (a) begin to move down after being released (b) reach the ground in 4 s (c) cover a distance of 40 m in reaching the ground (d) will have a displacement of 50 m. 66. A point initiallyat rest moves along x-axis. Its acceleration varies with time as 2 a (6t 5)m /s = + . Ifit starts from origin, the distance covered in 2 s is (a) 20m (b) 18m (c) 16m (d) 25m
- 57. 53 Motionina Straight Line 67.The relation between time t and distance x is 2 t x x = a +b where a and b are constants. The retardation is (a) 3 2 v a (b) 3 2 v b (c) 3 2 v ab (d) 2 3 2 v b 68. A stone is just released from the window of a train moving alonga horizontal straight track. The stonewill hit theground following a (a) straight line path (b) circular path (c) parabolic path (d) hyperbolic path 69. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2 . He reaches the ground with a speed of3 m/s. At what height, did he bail out ? (a) 182m (b) 91m (c) 111m (d) 293m 70. A car accelerates from rest at a constant rate a for some time after which it decelerates at a constant rate b to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by (a) t 2 2 ÷ ÷ ø ö ç ç è æ b a b + a (b) t 2 2 ÷ ÷ ø ö ç ç è æ b a b - a (c) t ÷ ÷ ø ö ç ç è æ b a b + a (d) t ÷ ÷ ø ö ç ç è æ b + a ab 71. A car moving with a speed of 40 km/hour can be stopped by applying brakes after at least 2m. If the same car is moving with a speed of 80km/hour, what is the minimum stopping distance. (a) 8m (b) 6m (c) 4m (d) 2m 72. A man throws balls with same speed verticallyupwards one after the other at an interval of 2 sec. What should be the speed of throw so that more than two balls are in air at any time (a) onlywith speed 19.6 m/s (b) morethan 19.6 m/s (c) at least 9.8 m/s (d) any speed less then 19.6 m/s. 73. A ball is dropped from a high rise platform at t = 0 starting from rest.After 6 seconds another ball is thrown downwards from the same platform with a speed v. The twoballs meet at t = 18s. What is the value of v? (take g = 10 m/s2) (a) 75m/s (b) 55m/s (c) 40m/s (d) 60m/s 74. A particle moves a distance x in time t according toequation x = (t + 5)–1. The acceleration of particle is proportional to (a) (velocity) 3/2 (b) (distance)2 (c) (distance)–2 (d) (velocity)2/3 75. A particle has initial velocity (2 3 ) i j + r r and acceleration (0.3 0.2 ) i j + r r . The magnitude of velocity after 10 seconds will be (a) 9 2 units (b) 5 2 units (c) 5 units (d) 9 units 76. Astone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is (a) h1 = 2 h 3 = 3 h 5 (b) h2 = 3h1 and h3 = 3h2 (c) h1 = h2 = h3 (d) h1 = 2h2 = 3h3 77. A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball at 3 T second (a) 9 h 8 meters from the ground (b) 9 h 7 meters from the ground (c) 9 h meters from the ground (d) 18 h 17 meters from the ground 78. The motion of particle is described by the equation x = a + bt2, where a = 15 cm and b= 3 cm/sec2. Its instant velocityat time 3 sec will be (a) 36 cm/sec (b) 9 cm/sec (c) 4.5 cm/sec (d) 18 cm/sec 79. Which one of the following equation represents the motion ofa body moving with constant finite acceleration? In these equation, y denotes the displacement in time t and p, q and r are constant: (a) y = (p + qt )(t + pt) (b) y = p + t/r (c) y= (p + t) (q + t ) (r + t) (d) (p qt) y rt + = 80. A ball is thrown up with velocity 19.6 m/s. The maximum height attained by the ball is (a) 29.2m (b) 9.8m (c) 19.6m (d) 15.8m 81. A train A which is 120 m long is runningwith velocity20 m/ s while train B which is 130 m long is running in opposite direction with velocity 30 m/s. What is the time taken by train B to cross the train A ? (a) 5 sec (b) 25 sec (c) 10 sec (d) 100 sec 82. A car travelling ata speed of30 km h–1 is brought to a halt in 8 m byapplying brakes. Ifthe same car is travelling at 60 km h–1, it can be brought to a halt with the same braking power in (a) 32m (b) 24m (c) 16m (d) 8cm 83. Velocitytime curve for a bodyprojected verticallyupwards is (a) parabola (b) ellipse (c) hyperbola (d) straight line 84. A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is (a) east-north direction (b) west-north direction (c) south-east direction (d) None of the above
- 58. 54 PHYSICS 85. Ametro train starts from rest and in 5 s achieves 108 km/h. After that it moves with constant velocityand comes torest after travelling 45 m with uniform retardation. Iftotaldistance travelled is 395 m, find total time oftravelling. (a) 12.2s (b) 15.3s (c) 9 s (d) 17.2s 86. A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O is given by x= 40 + 12t – t3. Howlong would the particle travel before coming to rest? (a) 24m (b) 40m (c) 56m (d) 16m 87. The displacement of particle is given by x = 2 1 2 0 . 2 3 a t a t a + - What is its acceleration? (a) 2 2 3 a (b) 2 2 3 a - (c) a2 (d) zero 88. Figure here gives the speed-time graph for a body. The displacement travelled between t = 1.0 second and t = 7.0 second is nearest to t .) sec in ( ) ms in ( v 1 - 4 - 4 4 6 8 2 0 (a) 1.5m (b) 2m (c) 3m (d) 4m 89. A boat takes 2 hours to travel 8 km and back in still water lake. With water velocity of 4 km h–1, the time taken for going upstream of 8 km and coming back is (a) 160 minutes (b) 80 minutes (c) 100 minutes (d) 120 minutes 90. A lift in which a man is standing, is moving upwards with a speed of 10 ms–1. The man drops a coin from a height of 4.9 metre and if g = 9.8 ms–2, then the coin reaches the floor of thelift after a time (a) 2 s (b) 1 s (c) 2 1 (d) 2 1 91. If a ball is thrown vertically upwards with a velocity of 40m/s, then velocity of the ball after two seconds is : (g = 10 m/sec2) (a) 15m/s (b) 20m/s (c) 25m/s (d) 28m/s 92. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 sec., it covers a distance of (a) 20cm (b) 400m (c) 1440cm (d) 2980cm 93. Thewater drops fall at regular intervals from a tap5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant ? (Take g = 10 m/s2) (a) 1.25m (b) 2.50m (c) 3.75m (d) 5.00m 94. The displacement time graph of a moving particle is shown below C D E F DISPLACEMENT S Time The instantaneous velocityof the particle is negative at the point (a) D (b) F (c) C (d) E 95. A particle moves along a straight line such that its displacement at any time t is given by s = (t3 – 6t2 + 3t + 4) metres The velocity when the acceleration is zero is (a) 3 ms–1 (b) – 12 ms –1 (c) 42 ms–2 (d) – 9 ms–1 96. A body starts from rest, what is the ratio of the distance travelled by the body during the 4th and 3rd seconds ? (a) 5 7 (b) 7 5 (c) 3 7 (d) 7 3 97. Which of the following curve does not represent motion in one dimension? (a) v t (b) v t (c) v t (d) v t DIRECTIONS for Qs. (98 to 100) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethe correctanswer (ONLYONE option is correct ) from the following. (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 98. Statement 1 : Velocity-time graph for an object in uniform motion along a straight path is a straight line parallel to the time axis. Statement 2 : In uniform motion of an object velocity increases as the square of time elapsed. 99. Statement 1 : Apositiveacceleration can be associated with a ‘slowing down’ of the body. Statement 2 : The origin and the positive direction of an axis are a matter of choice. 100. Statement 1 :In a free fall, weight of a body becomes effectivelyzero. Statement 2 : Acceleration due to gravity acting on a body having free fall is zero.
- 59. 55 Motionina Straight Line ExemplarQuestions 1. Among the four graph shown in the figure there is onlyone graph for which average velocityover the time interval (O, T) can vanish for a suitably chosen T. Which one is it? (a) t x (b) t x (c) x t (d) t x 2. A lift is coming from 8th floor and isjust about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct? (a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0 (c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a < 0 3. In one dimensional motion, instantaneous speed v satisfies 0 0 v v . £ < (a) The displacement in time T must always take non- negative values (b) The displacement x in time T satisfies 0 0 – < < v T x v T (c) The acceleration is always a non-negative number (d) The motion has no turning points 4. A vehicle travels half the distance l with speed v1 and the other half with speed v2, then its average speed is (a) 1 2 2 + v v (b) 1 2 1 2 2 + + v v v v (c) 1 2 1 2 2 + v v v v (d) 1 2 1 2 ( ) + L v v v v 5. The displacement of a particle is given byx = (t – 2)2 where x is in metre and t in second. The distance covered by the particle in first 4 seconds is (a) 4m (b) 8m (c) 12m (d) 16m 6. At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be (a) 1 2 2 + t t (b) 1 2 2 1 - t t t t (c) 1 2 2 1 + t t t t (d) t1 – t2 NEET/AIPMT (2013-2017) Questions 7. Astone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is [2013] (a) h1 = 2 h 3 = 3 h 5 (b) h2 = 3h1 and h3 = 3h2 (c) h1 = h2 = h3 (d) h1 = 2h2 = 3h3 8. The displacement ‘x’ (in meter) ofa particle of mass ‘m’ (in kg) moving in one dimension under the action of a force, is related to time ‘t’ (in sec) by t = 3 x + . The displacement of the particle when its velocityis zero, will be [NEET Kar. 2013] (a) 2m (b) 4m (c) zero (d) 6m 9. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx–2n where b and n are constants and x is the position of the particle. The acceleration ofthe particle as d function of x, is given by: [2015] (a) –2nb2x–4n–1 (b) –2b2x–2n+1 (c) –2nb2e–4n+1 (d) –2nb2x–2n–1 10. Ifthe velocityof a particle is v=At + Bt2, whereA and B are constants, then the distance travelled by it between 1s and 2s is : [2016] (a) 3 A 4B 2 + (b) 3A+7B (c) 3 7 A B 2 3 + (d) A B 2 3 + 11. Preeti reached the metrostation and found that the escalator was not working. She walked up the stationaryescalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be: [2017] (a) 1 2 2 1 t t t t - (b) 1 2 2 1 t t t t + (c) t1 – t2 (d) 1 2 t t 2 +
- 60. 56 PHYSICS EXERCISE -1 1. (b) 2. (b) 3. (c) 4. (c) 5. (d) 6. (b) 7. (c) 8. (b) 2 ) 7 t ( x or ) 7 t ( x + = + = time velocity ), 7 t ( 2 dt dx µ + = 9. (b) a = f t, t f dt dv a = = at t = 0, velocity = u v t u 0 dv f t dt = ò ò , v – u = 2 t f 2 Þ v = u+ 2 t f 2 NOTE : Do not use v = u + at directly because the acceleration is not constant. 10. (c) 1 2 dx v a 2a t dt = = + 2 dv a 2a dt = = 11. (d) A B v tan30º and v tan 60º = = A B v tan30º 1/ 3 1 v tan 60º 3 3 = = = 12. (a) Let for the first halftime t, the person travels a distance s1. Hence 1 1 1 1 s v or s v t t = = For second half time, 2 2 2 2 s v or s v t t = = Now, 1 2 s s Total displacement v Total time 2t + = = 1 2 1 2 v t v t v v 2t 2 + + = = 13. (b) When a particle cover half of circle of radius r, then displacement isAB = 2r & distance = half ofcircumference of circle = pr A B r r pr 14. (d) 2 2 gn 2 1 ) 1 n ( g 2 1 y - + = ) 1 n 2 ( 2 g ] n ) 1 n [( 2 g 2 2 + = - + = ......(i) Also, ) 1 n 2 ( 2 g h - = ......(ii) From (i)and(ii) y = h + g 15. (a) Differentiate two times and put x = 0. 16. (d) The difference in velocities is increasing with time as both of them have more constant but different acceleration. 17. (b) Time taken by the stone to reach the water level g h 2 t1 = Time taken bysound to come to the mouth of the well, v h t2 = Total time v h g h 2 t t 2 1 + = + 18. (d) Let the total distance be d. Then for first half distance, time = 0 v 2 d , next distance. = v1t and last halfdistance = v2t 1 2 d v t v t 2 + = ; 1 2 d t 2(v v ) = + Now average speed ) v v ( 2 d ) v v ( 2 d v 2 d d t 2 1 2 1 0 + + + + = 0 1 2 1 2 0 2v (v v ) (v v ) 2v + = + + 19. (c) a = bt or bt dt dv = . Integrating, we get c 2 bt v 2 + = , where c is a constant of integration. At t = 0, v = v0. Thus v0 = c. Now, o 2 v 2 bt dt ds v + = = dt v 2 bt ds o 2 ÷ ÷ ø ö ç ç è æ + = Integrating we get, t v 6 bt s o 3 + = 20. (a) 3 kv dt dv - = or dt k v dv 3 - = Integrating we get, c kt v 2 1 2 + - = - ...(1) Hints & Solutions
- 61. 57 Motionina Straight Line Att = 0, v = v0 2 o 1 c 2v - = Putting in (1) 2 2 2 2 0 0 1 1 1 1 kt or kt 2v 2v 2v 2v - = - - - = - or 2 2 0 v 2 1 kt v 2 1 = ú ú û ù ê ê ë é + or [ ] 2 2 0 2 0 v v kt v 2 1 = + or kt v 2 1 v v or kt v 2 1 v v 2 0 0 2 0 2 0 2 + = + = 21. (d) ) e 1 1 ( b a ) e 1 ( b a ) e 1 ( b a x 1 b 1 b - = - = - = - ´ - a (e 1) b e - = b / a 3 2 ~ b a 637 . 0 718 . 2 ) 718 . 1 ( b a 718 . 2 ) 1 718 . 2 ( b a - = = - = velocity bt ae dt dx v - = = , v0 = a accleration ab a & abe dt dv a 0 bt - = - = = - At t = 0, 0 ) 1 1 ( b a x = - = and At b 1 t = , b / a 3 2 ) e 1 1 ( b a ) e 1 ( b a x 1 = - = - = - At a t , x b = ¥ = It cannot go beyond this, so point a x b > is not reached by the particle. At t = 0, x = 0, at a t , x b = ¥ = , therefore the particle does not come back to its starting point at ¥ = t . 22. (c) 2 x t 2t 1 = + + Hence dx v 2t 2 dt = = + . It increases with time. 23. (b) When a body falls through a height h, it acquires a velocity gh 2 . 24. (c) Initial relative velocity vA – vB, is reduced to 0 in distance d¢ (<d) with retardation a. d a 2 ) v v ( 0 2 B A 2 ¢ - = - - 2 2 A B A B (v v ) (v v ) d d 2a 2a - - = > ¢ 25. (d) EXERCISE -2 1. (c) Given that x = At2 – B t3 velocity 2 dx 2At 3Bt dt = = - and acceleration Bt 6 A 2 dt dx dt d - = ÷ ø ö ç è æ = For acceleration to be zero 2A – 6Bt = 0. B 3 A B 6 A 2 t = = 2. (a) t sin s dt x d a 2 2 2 w w - = = On integrating, t cos s t cos s dt dx 2 w w = w w w = Again on integrating, we get t sin s t sin s x w = w w w = 3. (c) 3 dy v b 2ct 4d t dt = = + - 3 0 v b 2c(0) 4d(0) b = + - = (Q for initial velocity, t = 0) Now 2 dv a 2c 12d t dt = = - 2 0 a 2c 12d(0) 2c = - = , (at t = 0) 4. (d) 1 2 1 2 2 1 1 2 1 2 x x 2v v 1 2 2 x x v v v v 2v 2v 2v v + = = + æ ö + + ç ÷ è ø 5. (b) Let u be the initial velocity 1 1 2 1 2 v u a t , v u a (t t ) ¢ ¢ = + = + + and 3 1 2 3 v u a (t t t ) ¢ = + + + Now 1 1 1 u v u (u a t ) v 2 2 ¢ + + + = = 1 t a 2 1 u + = 1 2 2 1 2 v v 1 v u a t a t 2 2 ¢ ¢ + = = + + 2 3 3 1 2 3 v v 1 v u a t a t a t 2 2 ¢ ¢ + = = + + + So, 1 2 1 2 1 v v a (t t ) 2 - = - + and 2 3 2 3 1 v v a (t t ) 2 - = - + 1 2 2 3 1 2 2 3 (v v ) :(v v ) (t t ) :(t t ) - - = + +
- 62. 58 PHYSICS 6. (b)Let after a time t, the cyclist overtake the bus. Then t 20 t 2 2 1 96 2 ´ = ´ ´ + or t2 – 20 t + 96 = 0 1 2 96 4 400 20 t ´ ´ - ± = . sec 8 2 4 20 = ± = . sec 12 and 7. (d) v2 = u2 + 2 a s or v2 – u2 = 2 a s Maximum retardation, a = v2/2 s When the initial velocityis n v, then the distance over which it can be stopped is given by 2 2 2 0 n 2 u (n v) s n s 2a 2(v / 2s) = = = 8. (c) Let the length of train is s, then by third equation of motion, 2 2 v u 2a s = + ´ ....(1) Where v is final velocity after travelling a distance s with an acceleration a & u is initial velocity as per question Let velocityof middle point of train at same point is v', then ) 2 / s ( a 2 u ) v ( 2 2 ´ + = ¢ ....(2) By equations (1) and (2), we get 2 u v v 2 2 + = ¢ 9. (b) 8 = a t1 and 0 = 8 – a (4 – t1) or ÷ ø ö ç è æ - = = a 8 4 a 8 a 8 t1 8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec Now, m 8 s or ) 2 ( 4 2 1 2 0 s 1 2 1 = ´ + ´ = m 8 s or ) 2 ( 4 2 1 2 8 s 2 2 2 = ´ ´ - ´ = m 16 s s 2 1 = + 10. (b) u = 10 m/s, t = 5 sec, v = 20 m/s, a = ? 2 20 10 a 2 ms 5 - - = = From the formula v1 = u1 + a t, we have 10 = u1 + 2 × 3 or u1 = 4 m/sec. 11. (c) Let a be the constant acceleration of the particle. Then 2 t a 2 1 t u s + = or a 50 ) 10 ( a 2 1 0 s 2 1 = ´ ´ + = and 2 2 1 s 0 a(20) 50a 150a 2 é ù = + - = ê ú ë û 2 1 s 3s = Alternatively: Let a be constant acceleration and 2 1 s ut at 2 = + , then 1 1 s 0 a 100 50a 2 = + ´ ´ = Velocityafter 10 sec. is v = 0 + 10a So, 2 2 1 1 s 10a 10 a 100 150a s 3s 2 = ´ + ´ = Þ = 12. (d) So by figure the velocity of parrot w.r. t. train is = 5–(–10) = 15m/sec so time taken to cross the train is velocity relative train of length = sec 10 15 150 = = North East West South train 10m/sec parrot 5m/sec 13. (a) s2 = at2 + 2bt + c b 2 at 2 dt ds s 2 + = or s b at dt ds + = , again differentiating 2 2 2 2 s s b at ) b at ( as dt ds . s ) b at ( s . a dt s d ÷ ø ö ç è æ + + - = + - = 3 2 2 2 2 s ) b at ( as dt s d + - = 2 3 2 d s a s . dt - = ¥ 14. (a) From third equation of motion 2 2 v u 2ah = + In first case initial velocityu1 = 5 m/sec final velocity v1 = 0, a = – g and max. height obtained is H1, then, g 2 25 H1 = In second case u2 = 10 m/sec, v2 = 0, a = –g and max. height is H2 then, g 2 100 H2 = . It implies that H2 = 4H1 15. (c) The distance covered in nth second is a ) 1 n 2 ( 2 1 u Sn - + = where u is initial velocity& a is acceleration then 2 a 19 u 26 + = ....(1) 2 a 21 u 28 + = ....(2) 2 a 23 u 30 + = ....(3) 2 a 25 u 32 + = ....(4) From eqs. (1) and (2) we get u = 7m/sec, a=2m/sec2 The body starts with initial velocity u =7m/sec and moves with uniform acceleration a = 2m/sec2
- 63. 59 Motionina Straight Line 16.(b) The stone rises up till its vertical velocity is zero and again reached the top of the tower with a speed u (downward). The speed of the stone at the base is 3u. – + v, g, h u Hence 2 2 2 4u (3u) ( u) 2gh or h g = - + = 17. (c) Sobysecond equation ofmotion, we get 2 1 S ut at 2 = + here S = l, u = 0, a = g sinq 2 2 2h 1 2h t a sin g gsin = = = q q l h sin æ ö q = ç ÷ è ø Q l h g cosq g sinq g q 18. (a) S = ut + ½at2 here a = g For first body u1 =0 Þ S1=½g × 9 For second body u2=0 Þ S2= ½g × 4 So difference between them after 3 sec. = S1 – S2 = ½ g×5 If g = 10m/sec2 then S1–S2 = 25 m. 19. (b) Relative speed of each train with respect to each other be, n = 10 + 15 = 25 m/s Here distance covered by each train = sum of their lengths= 50 + 50 = 100 m Required time . sec 4 25 100 = = 20. (d) At highest point of the trajectory velocity becomes zero and all kinetic energychanges to potential energy so at highest point, K.E. = 0 21. (b) The distance travel in nth second is Sn = u + ½ (2n–1)a ....(1) so distance travel in tth & (t+1)th second are St = u +½ (2t–1)a ....(2) St+1= u+½ (2t+1)a ....(3) As per question, St+St+1 = 100 = 2(u + at) ....(4) Now from first equation of motion the velocity, of particle after time t, ifit moves with an accleration a is v = u + a t ....(5) where u is initial velocity Sofrom eq(4) and (5), we get v = 50cm./sec. 22. (b) Since v2 = u2 + 2as For first case u1=20m/sec, v1= 0, a1= g = 10, s1= ? So m 20 10 2 400 s1 = ´ = For second case (at moon) u2 =20m/sec, v2=0, ? s , sec / m 7 . 1 6 g a 2 2 2 = = = 6 / 10 2 400 7 . 1 2 400 s2 ´ = ´ = so 6 1 s s 2 1 = 23. (d) All options are correct : (i) When two bodies A & B move in opposite directions then relative velocity between A & B either VAB or VBA both are greater than VA&VB. (ii) When twobodiesA& Bmove in parallel direction then A AB B A AB V V V V V < Þ - = B BA A B BA V V V V V < Þ - = 24. (b) From third equation of motion v2 = u2 – 2gh ( a g) = - Q Given, v = 10 m/sec at h/2. But v = 0, when particle attainedmaximum height h. Therefore (10)2 = u2 – 2gh/2 or 100 = 2gh –2gh/2 (Q0 = u2 – 2gh) Þ h = 10m 25. (b) Since in both case the height of building and down ward acceleration ‘g’ is same. So both stones reach simultaneouslyi.e., 2 2 t 10 2 1 10 gt 2 1 S ´ = Þ = or t 2 sec, = for both stone. 26. (d) Speed 1 5 v 60 18 = ´ m/s 50 3 = m/s = = ´ = 1 1 5 100 d 20m, v' 120 18 3 m/s Let dceleration be a 2 1 1 0 v 2ad = - ....(1) or 2 1 1 v 2ad = 2 1 2 (2v ) 2ad = ...(2) (2) divided by (1) gives, m 80 20 4 d d d 4 2 1 2 = ´ = Þ = 27. (d) Equation of motion is at u = we know that ds u dt = at dt ds = Þ atdt ds or = integrating it we get, ò = ò 4 0 s 0 tdt a ds a 8 ] t [ 2 a s 4 0 2 = = 28. (d) The onlyforce acting on the ball is the force ofgravity. The ball will ascend until gravity reduces its velocity to zero and then it will descend. Find the time it takes for the ball toreach its maximumheight and then double the time to cover the round trip.
- 64. 60 PHYSICS Using vatmaximum height = v0 + at = v0 – gt, we get: 0 m/s = 50 m/s – (9.8 m/s2) t Therefore, t = (50 m/s)/(9.8 m/s2) ~ (50 m/s)/ (10 m/s2) ~ 5s This is the time it takes the ball to reach its maximum height. The total round trip time is 2t ~ 10s. 29. (b) A B C AtoB 1 2 u 2 a 2 2 1 u a, 2 = ´ + ´ ´ ´ Þ = + AtoC 6 6 a 2 1 6 u 20 . 4 ´ ´ + ´ = Þ a 3 u 7 . 0 + = , 2 s m 15 . 0 a or 3 . 0 a 2 - - = - = , 1 1 s m 15 . 1 s m ) 15 . 0 1 ( a 1 u - - = + = - = Velocity at t = 9 sec. 1 s m 2 . 0 35 . 1 15 . 1 9 15 . 0 15 . 1 v - - = - = ´ - = 30. (c) The ball thrown upward will losevelocityin 1s. It return back to thrown point in another 1 s with the same velocity as second. Thus the difference will be 2 s. 31. (b) Velocity when the engine is switched off 1 v 19.6 5 98ms- = ´ = 2 1 max h h h + = where 2 2 1 2 1 v h at & h 2 2a = = max 1 98 98 h 19.6 5 5 2 2 9.8 ´ = ´ ´ ´ + ´ =245 +490= 735m 32. (b) Average velocity for the second half of the distance is = 1 1 2 v v 4 8 6ms 2 2 - + + = = Given that first half distance is covered with a velocity of 1 s m 6 - . Therefore, the average velocity for the whole time of motion is 1 s m 6 - 33. (d) Since S = ut + ½ gt2 where u is initial velocity& a is acceleration. In this case u = 0 & a = g so distance travelled in 4 sec is, S=½ ×10×16= 80m 34. (c) The food packet has an initial velocity of 2 m/sec in upward direction, therefore v = – u + gt or v = –2 + 10 × 2 = 18 m /sec. 35. (b) Differentiated twice. 36. (c) Downward motion 2 v 0 2 9.8 5 2 - = ´ ´ Þ v 98 9.9 = = Also for upward motion 2 2 0 u 2 ( 9.8) 1.8 - = ´ - ´ Þ 94 . 5 3528 u = = Fractional loss 4 . 0 9 . 9 94 . 5 9 . 9 = - = 37. (a) Use AB A B v v v = - r r r . 38. (a) Since displacement is zero. 39. (b) Average velocity 1 h km 24 30 20 30 20 2 - = + ´ ´ = . 40. (c) From third equation of motion, v2 = u2 + 2as where v& uarefinal & initial velocity, a is acceleration, s is distance. For first casev1= 3km/hour, u1= 0, a1=g & s1=? 20 36 36 100 9 s1 ´ ´ ´ = metre For second case v2=?, u2=4km/hour , a2= g = 10m/sec & 20 36 36 100 9 s s 2 1 ´ ´ ´ = = so 36 36 20 100 9 10 2 3600 3600 1000 1000 16 v2 2 ´ ´ ´ ´ ´ + ´ ´ ´ = or v2 = 5 km/hour 41. (a) kt dt ds = Þ 2 1 1 s kt 2 3 3 2 2 = = ´ ´ ´ = 9 m. 42. (d) For constant acceleration and zero initial velocity 2 t h µ 2 1 1 2 2 2 h t h t = 2 2 1 1 1 h t t 2 t 2 2s h Þ = = ´ = ´ 43. (a) 16 x 3 v + = Þ 16 x 3 v2 + = Þ x 3 16 v2 = - Comparing with , aS 2 u v 2 2 = - we get, u= 4 units, 2a = 3 or a = 1.5 units 44. (b) 2 1 t g 2 1 AB S = = Þ 2 1 2 1 2S AC g (t t ) 2 = = + and 2 3 2 1 ) t t t ( g 2 1 AD S 3 + + = = g S 2 t1 = g S 2 g S 4 t , g S 4 t t 2 2 1 - = = + 2S A S B S C S 3S D g S 6 t t t 3 2 1 = + + g S 4 g S 6 t3 - = ) 2 3 ( : ) 1 2 ( : 1 : : t : t : t 3 2 1 - -
- 65. 61 Motionina Straight Line 45.(c) 1 (8 1 3 1 1) 0 v 5ms 1 - ´ - ´ ´ - = = 46. (c) Sincethe initial velocityofjump issame on both planets So 0 = u2– 2gAhA 0 = u2–2gBhB or m 18 2 1 9 h h g h g B B B A A = ´ = Þ = ´ 47. (d) q S R P Q mg q sin m g q cos mg Let distance (PR) iscovered by the particle in time ‘t’. Þ PR = q = q + sin gt 2 1 t . sin g 2 1 0 2 2 Further PR = q cos PQ (Given PQ = constant) Þ PQ = q qcos sin gt 2 1 2 = q 2 sin gt 4 1 2 Þ t = 2 q 2 sin g PQ q µ 2 sin 1 t So as q increases, q 2 sin first increases and then decreases. Hence ‘t’ first decreases and then increases. 48. (d) Given x = ae–at + bebt Velocity, v = dx dt = –aae–at + bbebt = - + a e b e t t a b a b i.e., goon increasing with time. 49. (b) 2 y t ; v- t';a t° µ µ µ 50. (c) On differentiating, acceleration = 0.2t ) t ( f a = Þ 51. (d) Use v2 – u2 = 2aS. In both the cases, (u positive or negative) u2 is positive. 52. (b) Area under a-t graph is change in velocity. Area 1 1 1 (4 4) 6 4 2 4 2 2 2 2 2 = ´ + ´ + ´ ´ - ´ ´ 1 36 2 34 ms- = - = As initial velocity is zero therefore, the velocity at 14 second is 1 s m 34 - . 53. (c) At t = 20s, d = 20 m 54. (a) At six points in the graph the tangents have zero slope i.e. velocity is zero. 55. (d) Time fall is 1 second. 2 2 1 1 10 h g 2 2 8 æ ö = = ç ÷ è ø =1.25m 56. (b) 3 2 1 av t t t x 3 x 2 x v + + + + = 1 max 2x t v = , 2 max 2x t v = , 3 max 6x t v = max av 6x v v 10x = av max v 3 v 5 = 57. (c) Let A v and B v are the velocities of two bodies. In first case, A v + B v = 6m/s ........(1) In second case, A v – B v = 4m/s .........(2) From (1) &(2) we get, A v = 5 m/s and B v =1 m/s. 58. (d) BB v r = Relative velocityofballw.r.t. balloon = v u + gt ) v u ( 0 + + - = of g v u t + = Total time g ) v u ( 2 + = 59. (c) 2 2 2 1 ) N n ( g 2 1 y , gn 2 1 y - = = ] ) N n ( n [ g 2 1 y y 2 2 2 1 - - = - Þ g 1 (2n N)N 2 = - ] 1 y y [ 2 1 = - Q Þ 2 N gN 1 n + = 60. (b) S1 = 2 1 g 2 2 1 ÷ ø ö ç è æ , S1 + S2 = 1 2 g(1)2 2 1 g 2 2 3 ÷ ø ö ç è æ = S1 + S2 + S3 By solving we get S1 : S2 : S3 = 1 : 3 : 5 61. (b) Distance in last two second = 2 1 ×10× 2 =10m. Total distance = 2 1 × 10 × (6 + 2) = 40 m. 62. (d) In (a), at the same time particlehas twopositionswhich is not possible. In (b), particle has two velocities at the same time. In (c), speed is negative which is not possible. 63. (a) Velocityat timet is tan 45° = 1.Velocityat time(t = 1) is tan 60 3 ° = . Acceleration is change in velocity in one second = 3 1 - . 64. (c) Here, the particle B moves upwards. Let the upward velocity of B be v then v tan 60 10 = °
- 66. 62 PHYSICS 65. (b)v= 1.25 × 8 ms–1 = 10 ms–1 1 s 1.25 8 8m 40m 2 = ´ ´ ´ = Now, 2 1 40 10t 10 t 2 = - + ´ ´ or 2 5t 10t 40 0 - - = or 2 t 2t 8 0 - - = or t = 4 s. 66. (b) Given acceleration a 6t 5 = + dv a 6t 5 dt = = + , dv (6t 5)dt = + Integrating it, we have v t 0 0 dv (6t 5)dt = + ò ò 2 v 3t 5t C, = + + where C is constant of integration. When t = 0 , v = 0 so C = 0 2 ds v 3t 5t dt = = + or 2 ds (3t 5t)dt = + Integrating it within the conditions of motion, i.e., as t changes from 0 to 2 s, s changes from 0 to s, we have s 2 2 0 0 ds (3t 5t)dt = + ò ò 2 3 2 0 5 s t t 8 10 18m 2 = + = + = 67. (a) 2 t x x = a +b Differentiating w.r.t. time on both sides, we get dx dx 1 2 .x dt dt = a + b 3 2 dx 1 dv 2 v v ; 2 v dt 2 x dt ( 2 x) - a = = = = - a b + a b + a Negative sign shows retardation. 68. (c) The horizontal velocityofthe stone will be the same as that of the train. In this way, the horizontal motion will beuniform motion. Theverticalmotion will becontrolled bythe forceofgravity, i. e., verticalmotion isaccelerated motion. Thus the resultant motion will be along a parabolic trajectory. 69. (d) Initial velocity of parachute after bailing out, u = gh 2 u = 50 8 . 9 2 ´ ´ = 5 14 s / m 2 a - = s / m 3 v m 50 2 The velocity at ground, v= 3m/s S = 2 2 u v 2 2 ´ - = 4 980 32 - » 243m Initiallyhe has fallen 50 m. Total height from where he bailed out =243+50=293m 70. (d) As per question, Let max. velocity is v then v = a t1 & v – b t2 = 0, where t = t1 + t2 Now t1 + t2 = t or v v t + = a b t v t 1 1 æ ö ab = = ç ÷ a +b æ ö è ø + ç ÷ a b è ø and 1 2 s s s = + 2 2 2 v v v 1 1 2 2 2 æ ö = + = + ç ÷ a b a b è ø 71. (a) From third equation of motion : 2 2 v u 2as = + for first case sec / m 36 10 40 u ´ = , v=0, a = ?, s = 2 m so, 2 2 sec / m 4 1 36 10 40 a ÷ ø ö ç è æ ´ = for second case sec / m 36 10 80 u ´ = , v = 0, So 2 2 2 80 10 1 40 10 s / 2 8meter 36 4 36 ´ ´ æ ö æ ö = ´ ´ = ç ÷ ç ÷ è ø è ø 72. (b) Height attained by balls in 2 sec is m 6 . 19 4 8 . 9 2 1 = ´ ´ = the same distance will be covered in 2 second (for descent) Time interval of throwing balls, remaining same. So, for twoballs remaining in air, the time ofascent or descent must be greater than 2 seconds. Hence speed of balls must be greater than 19.6 m/sec. 73. (a) Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s. Now, distance moved in 18 s by 1st ball = 1 2 ×10 × 182 =90× 18=1620m Distance moved in 12 s by 2nd ball = ut + 1 2 gt2 1620 = 12 v + 5 × 144 Þ v= 135 – 60 = 75 ms –1 74. (a) 1 5 x t = + v = dx dt 2 1 ( 5) t - = + a = 2 2 d x dt = 3 2 ( 5) t + = 2x3 Now 1 2 1 ( 5) v t µ + 3 2 3 1 ( 5) v a t µ µ +
- 67. 63 Motionina Straight Line 75.(b) v u at = + r r r ˆ ˆ ˆ ˆ (2 3 ) (0.3 0.2 ) 10 v i j i j = + + + ´ = ˆ ˆ 5 5 i j + | | v r = 2 2 5 5 + ; | | 5 2 v = r 76. (a) Q h = 1 2 gt2 h1 = 1 2 g(5)2 = 125 h1 + h2 = 1 2 g(10)2 = 500 Þ h2 = 375 h1 + h2 + h3 = 1 2 g(15)2 = 1125 Þ h3 = 625 h2 = 3h1 , h3 = 5h1 or h1 = 2 h 3 = 3 h 5 77. (a) h = 2 gT 2 1 now for t = T/3 second vertical distance moved is given by h¢ 2 1 T g 2 3 æ ö = ç ÷ è ø Þ h¢ 2 1 gT h 2 9 9 = ´ = position of ball from ground 9 h h - = 9 h 8 = 78. (d) x = a + bt2 = 15 + 3t2 v = dx dt = 3 × 2t = 6t Þ t 3s v = = 6 × 3 = 18cm/s 79. (a) Motion with constant acceleration is represented by a quadratic equation of t Y = (p + qt) (r + pt) = pr + qrt + p2t + pqt2 80. (c) Let the maximum height attained bythe ball be h. At maximum height , velocityof ball, v = 0 Given, initial velocity, u = 19.6 m/s Using the equation of motion, v2 = u2 + 2gh We get 0= (19.6)2 + 2 (–9.8) ×h Þ h = 2 (19.6) 2 9.8 ´ = 19.6m 81. (a) Here, length of train A, LA = 120 m length of train B, LB = 130 m velocity of train A, vA = 20 m/s velocity of train B, vB = 30 m/s Train B is running in opposite direction to train B, velocityof train B relative to train A, vBA = vB + vA = (30 + 20) m/s = 50 m/s Total distance to be covered by train B = LA +LB=(120+130)m =250m Hence, time required by train B to cross train A t = 250 50 sec = 5sec 82. (a) Retardation, 2 2 2 0 (25/3) 2 2 8 v u a s - - = = ´ 2 1 25 16 3 a æ ö = - ´ ç ÷ è ø For u = 60 km h–1 = 50 3 ms–1 2 2 0 (50/ 3) 2 [ (25/3) 1/16] s - = ´ - ´ =32m 83. (d) Velocitytime curve will be a straight line as shown: o t v At the highest point v = 0. 84. (b) Let O be the origin, then O Q N Car E Train P N W S E passenger in the train at P observes the car at Q along the direction PQ; i.e. west north direction. 85. (d) Given : u = 0, t = 5 sec, v= 108 km/hr = 30m/s By eqn of motion v = u + at or 2 v 30 a 6 m / s [ u 0] t 5 = = = = Q 2 1 1 S at 2 = 2 1 6 5 75 m 2 = ´ ´ = Distance travelled in first 5 sec is 75m. Distance travelled with uniform speed of 30 m/s is S2 395 = S1 + S2 + S3 395 = 75+ S2 +45 S2 =395–120= 275m Timetaken to travel 275 m = 275 9.2sec 30 = For retarding motion, we have 02 – 302 = 2 (– a) × 45
- 68. 64 PHYSICS We get,a = 10 m/s2 Now by, S = ut + 2 1 at 2 45 = 30t + 2 1 (–10)t 2 45 = 30t – 5t2 Pn solving we get, t = 3 sec Total time taken = 5 + 9.2 +3 = 17.2 sec. 86. (c) When particle comes to rest, V = 0 = dx d dt dt = (40 + 12t – t3) Þ 12 – 3t2 = 0 Þ t2 = 12 4 3 = t = 2 sec Therefore distance travelled byparticle before coming to rest, x = 40+ 12t – t3 = 40 + 12 × 2 – (2)3 = 56m 87. (b) We get acceleration by double differentiation of displacement. V= 1 2 0 a t a dx d a t dt dt 2 3 2 æ ö = + - ç ÷ è ø = 1 2 a 2 a t 2 3 - a = 1 2 2 a 2 d a t dv 2 2 3 a dt dt 3 æ ö - ç ÷ è ø - = = 88. (b) 1 1 1 (2 4) 2 1 4 3 4 2m 2 2 2 + ´ + ´ ´ - ´ ´ = 89. (a) Velocity of boat 1 h km 8 2 8 8 - = + = Velocityof water 1 h km 4 - = 160 h 3 8 4 8 8 4 8 8 t = = + + - = minute 90. (b) Using relative terms s / m 0 u . rel = ? t , m 9 . 4 S , s m 8 . 9 a 2 = = = - 2 t 8 . 9 2 1 t 0 9 . 4 ´ ´ + ´ = Þ 9 . 4 t 9 . 4 2 = Þ t = 1 s 91. (b) From first equation of motion v = u + a t here u = 40, a = g = – 10, t = 2 so v = 40 – 10 × 2 = 20 m/sec 92. (b) v= [144 × 1000/(60 × 60)] m/sec. v = u + at or (144 ×1000)/(60 ×60) =0+ a ×20 2 144 1000 a 2m / sec 60 60 20 ´ = = ´ ´ Now 2 1 s u t at 2 = + m 400 ) 20 ( 2 2 1 0 2 = ´ ´ + = 93. (c) Height of tap = 5m and (g) = 10 m/sec2. For the first drop, 5 = ut 2 1 2 + gt 2 1 (0 ) 10 2 = ´ + ´ t t = 5t2 or t2 = 1 or t = 1 sec. It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec. Distance covered by the second drop in 0.5 sec 2 1 1 (0 0.5) 10 2 2 = + = ´ + ´ ut gt ×(0.5)2 =1.25m. Therefore, distance of the second drop above the ground = 5 – 1.25 = 3.75 m. 94. (d) At E, the slope of the curve is negative. 95. (d) Velocity, 2 3 –12 3 = = + ds v t t dt Acceleration, 6 –12; = = dv a t dt For a = 0, we have, 0 = 6 t – 12 or t = 2s. Hence, at t = 2 s the velocitywill be 2 –1 3 2 –12 2 3 9 ms = ´ ´ + = - v 96. (a) 4 3 0 (2 4 1) 7 2 5 0 (2 3 1) 2 + ´ - = = + ´ - a D a D 97. (b) In one dimensional motion, thebodycan have at a time one velocity but not two values of velocities. 98. (d) Inuniformmotion theobject moveswithuniformvelocity, the magnitude of its velocity at different instane i.e., at t = 0, t =1, sec, t =2sec..... will alwaysbeconstant. Thus velocity-timegraphfor an objectinuniform motion along a straight path isa straight lineparallelto timeaxis. 99. (b) 100. (d) EXERCISE -3 Exemplar Questions 1. (b) Ifwe draw a line parallel to time axis from the point (A) on graph at t = 0 sec. This line can intersect graph at B. In graph (b) for one value of displacement there are twodifferent pointsoftime. so, for onetime, the average velocity is positive and for other time is equivalent negative. As there are opposite velocities in the inteval 0 to T hence average velocity can vanish in (b). This can be seen in the figure given below. t x O A B T Here, OA = BT (same displacement) for two different points of time.
- 69. 65 Motionina Straight Line 2.(a) As the lift is moving downward directions so displacement isnegative (zero).Wehavetoseewhether the motion is accelerating or retarding. Dueto downward motion displacement is negative the lift reaches 4th floor is about to stop hence, motion is retarding (–a) downward in nature hence, x < 0; a > 0. 8th floor x < 0 4th floor Ground floor x < 0 O 6th floor As displacement is in negative direction, x < 0 velocity will also be negative i.e., v < 0 but net acceleration is +ve a > 0, that can be shown in the graph. 3. (b) In one dimensional motion, for the maximum and minimum displacement we must have the magnitude and direction of maximum velocity. As maximum velocityin positive direction is v0, hence maximum velocity in opposite direction is also –v0. Maximum displacement in one direction = v0T Maximum displacement in opposite directions = –v0T. Hence, 0 0 - < < v T x v T 4. (c) Time taken totravel first half distance, 1 1 1 / 2 2 = = l l t v v 1 ( / 2) = QL l Time taken to travel second half distance, 2 2 2 = l t v 2 ( / 2) = QL l So, total time taken to travel full distance = t1 + t2 1 2 1 2 1 1 2 2 2 é ù = + = + ê ú ë û l l l v v v v Total time 2 1 1 2 2 é ù + = ê ú ë û v v l v v So, average speed, vav. = Total distance Total time = 1 2 av. 1 2 1 2 2 1 1 2 Þ = = + é ù + ê ú ë û v v l v v v l v v 1 2 av. 1 2 2 = + v v v v v 5. (b) As given that, x = (t – 2)2 Now, velocity 2 ( 2) = = - dx d v t dt dt = 2 (t – 2) m/s Acceleration, [2( 2)] = = - dv d a t dt dt = 2[1–0]=2m/s2 =2ms–2 at t = 0; v0 = 2 (0 – 2) = –4 m/s t = 2 s; v2 = 2 (2 – 2) = 0 m/s t = 4 s; v4 = 2 (4 – 2) = 4 m/s B D O A V t 4 2 C –4 m/s 4 m/s (Time) v-t graph is shown in diagram. Distance travelled = area between time axis of the graph = area OAC + are ABD 1 1 2 2 = ´ + ´ OA OC AD BD 4 2 1 2 4 8 m 2 2 ´ = + ´ ´ = If displacement occurs 1 1 2 2 = ´ ´ + ´ ´ OA OC AD BD 1 1 2( 4) 2 4 0 2 2 = ´ - + ´ ´ = 6. (c) Let us consider, displacement is L, then velocity of girl with respect to ground, 1 = g L v t Velocity of escalator with respect to ground, 2 = e L v t Net velocityof the girl on moving escalator with respect to ground 1 2 = + = + g e L L v v t t Þ vge 1 2 1 2 é ù + = ê ú ë û t t L t t
- 70. 66 PHYSICS Now, ift is total time taken bygirl on moving escalator in covering distance L, then distance speed = t 1 2 1 2 1 2 1 2 = = + æ ö + ç ÷ è ø t t L t t t t L t t NEET/AIPMT (2013-2017) Questions 7. (a) Q h = 1 2 gt2 h1 = 1 2 g(5)2 = 125 h1 + h2 = 1 2 g(10)2 = 500 Þ h2 = 375 h1 + h2 + h3 = 1 2 g(15)2 = 1125 Þ h3 = 625 h2 = 3h1 , h3 = 5h1 or h1 = 2 h 3 = 3 h 5 8. (c) Q t = 3 x + Þ x = t – 3 Þ x = (t – 3)2 v = dx dt = 2(t – 3) = 0 Þ t = 3 x = (3 – 3)2 Þ x = 0. 9. (a) According to question, V(x) = bx–2n So, dv dx = – 2 nb x– 2n – 1 Acceleration of the particle as function of x, a = v dv dx = bx–2n { } –2n–1 b (–2n)x = – 2nb2x–4n–1 10. (c) Given : Velocity V =At + Bt2 Þ dx dt =At + Bt2 By integrating we get distance travelled Þ ( ) x 2 2 0 1 dx At Bt dt = + ò ò Distance travelled by the particle between 1s and 2s x= ( ) ( ) 2 2 3 3 A B 3A 7B 2 1 2 1 2 3 2 3 - + - = + 11. (b) Velocity of preeti w.r.t. elevator v1= 1 d t Velocityof elevator w.r.t. ground 2 2 d v t = then velocityof preeti w.r.t. ground v = v1 + v2 1 2 d d d t t t = + 1 2 1 1 1 t t t = + t = 1 2 1 2 t t (t t ) + (time taken by preeti to walk up on the moving escalator)
- 71. SCALARS AND VECTORS Scalars: The physical quantities which have only magnitude but no direction, are called scalar quantities. For example - distance, speed, work, temperature, mass, etc. · Scalars are added, subtracted, multiplied and divided by ordinarylaws of algebra. Vectors: For any quantity to be a vector, (i) it must have magnitude. (ii) it must have direction. (iii) it must satisfy parallelogram law of vector addition. For example – displacement, velocity, force, etc. Electric current has magnitude as well as direction but still it is not treated as a vector quantity because it is added by ordinarylaw of algebra. Types of Vectors (i) Like vectors : Vectors having same direction are called like vectors. The magnitude may or may not be equal. A B A r and B r are like vectors. These are also called parallel vectors or collinear vectors. (ii) Equal vectors : Vectors having same magnitude and same direction are called equal vectors. A B Here A r and B r are equal vectors A B = r r Thus, equal vector is a special case of like vector. (iii) Unlike vectors : Vectors having exactlyopposite directions are called unlike vectors. The magnitude mayor maynot be equal. A B A r and B r are unlike vectors. (iv) Negative vectors :Vectors having exactlyopposite direction and equal magnitudes are called negative vectors. A B Here A r and B r are negative vectors, A B = - r r Thus negative vectors is a special case of unlike vectors. (v) Unit vector : Vector which hasunit magnitude. It represents direction only. For example take a vector B r . Unit vector in the direction of B r is | B | B , which is denoted as B̂. B̂, is read as “B cap” or "B caret". (vi) Orthogonal unit vector : A set of unit vectors, having the directions of thepositivex, yand z axes of three dimensional rectangular coordinate system are denoted by ˆ ˆ ˆ i, j and k . They are called orthogonal unit vectors because angle between any of the two unit vectors is 90º. o X Y Z j ˆ k̂ î The coordinate system which has shown in fig. is called right handed coordinate system. Such a system derives its name from the fact that right threaded screwrotated through 90º from OX to OY will advance in positive Z direction as shown in the figure. (vii) Null vector (zero vector) : A vector of zero magnitude is called a zero or null vector. Its direction is not defined. It is denoted by 0. Properties of Null or Zero Vector : (a) The sum of a finite vector A ur and the zero vector is equal to the finite vector i.e., A A + = 0 ur ur (b) The multiplication ofa zerovector bya finite number n is equal to the zero vector i.e., 0 n = 0 (c) The multiplication of a finite A ur by a zero is equal to zero vector i.e., A 0 = 0 ur 4 Motion in a Plane
- 72. 68 PHYSICS (viii) Axialvector : Vector associated with rotation about an axis i.e., produce rotation effect is called axial vector. Examples are angular velocity, angular momentum, torque etc. (ix) Coplanar vectors : Vectors in the same plane are called coplanar vectors. (x) Position vectors and displacement vectors : The vector drawn from the origin of the co-ordinate axes to the position ofa particle is called position vector ofthe particle. IfA(x1, y1, z1) and B (x2, y2, z2) be the positions of the particle at two different times of its motion w.r.t. the origin O, then position vector ofA and B are X Y O rA rB A B P a t h o f p a r t i c l e D isplacem ent vector A 1 1 1 ˆ ˆ ˆ r OA x i y j z k = = + + uu r uuur B 2 2 2 ˆ ˆ ˆ r OB x i y j z k = = + + uu r uuu r . The displacement vector is AB = OA OB - . 2 2 2 1 2 1 ˆ ˆ ˆ (x x )i (y y )j (z z )k = - + - + - Laws of Vector Algebra 1. A B B A r r r r + = + (Commutative law ofaddition) 2. C ) B A ( ) C B ( A r r r r r r + + = + + (Associative law of addition) 3. m A A m r r = 4. A ) mn ( ) A n ( m r r = 5. A n A m A ) n m ( r r r + = + 6. B m A m ) B A ( m r r r r + = + ADDITION OF VECTORS Triangle Law of Vector Addition It states that if two vectors acting on a particle at the same time are represented in magnitude and direction by the two sides of a triangle taken in one order, their resultant vector is represented in magnitude and direction by the third side of the triangle taken in opposite order. O P B Q q A R N b Magnitude of R ur is given by 2 2 2 cos = + + q R A B AB whereq is the angle between A ur and B u r . Direction of R ur : Let the resultant R ur makes an angle b with the direction of A ur . Then from right angle triangle QNO, tan sin cos QN QN B ON OP PN A B q b = = = + + q (i) | R ur | is maximum, if cosq = 1, q = 0° (parallel vector) A B ¾¾ ® ¾¾ ® ur u r P Rmax 2 2 A B 2AB = + + = A + B (ii) | R ur | is minimum, if cosq = –1, q = 180° (opposite vector) B A ¬¾ ¾ ¾¾ ® uuu r uur Rmin 2 2 A B 2AB A B = + - = - (iii) If the vectors A and B are orthogonal, i.e., o 2 2 90 ,R A B q = = + Parallelogram Law of Vector Addition It states that if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram then their resultant is represented in magnitude and direction by the diagonal of the parallelogram. Let the twovectors A ur and B u r , inclined at angle q are represented bysides OPandOS uuu r uuu r of parallelogram OPQS, then resultant vector R ur is represented by diagonal OQ uuur of the parallelogram. S Q P O q b R B A 2 2 sin 2 cos ; tan cos q = + + q b = + q B R A B AB A B If q < 90° , (acute angle) R ur = A ur + B u r , R ur is called main (major)diagonal ofparallelogram If q > 90° , (obtuse angle) R ur = A ur + B u r , R ur is called minor diagonal. Polygon Law of Vector Addition If a number of non zero vectors are represented by the (n–1) sides of an n-sided polygon then the resultant is given by the closing side or the nth side of the polygon taken in opposite order. B D E C O E B A C D A R So, R A B C D E = + + + + ur ur u r u r ur u r or, OA AB BC CD DE OE + + + + = uuur uuu r uuu r uuur uuu r uuu r
- 73. 69 Motion in aPlane 15. Magnitude of a vector is independent of co-ordinate axes system. 16. Component of a vector perpendicular to itself is zero. 17. (a) Resultant of two vectors is maximum when angle between the vectors is zero i.e., q = 0° Rmax = A + B (b) Resultant of two vectors is minimum when q =180° Rmin = A – B (c) The magnitude of resultant of A and B can vary between (A + B) and (A – B) SUBTRACTION OF VECTORS We convert vector subtraction into vector addition. q 180° – q A – B B ) B ( A B A - + = - If the angle between A and B is q then the angle between 1 8 0 ° – q A R – B a A and B - is (180° – q). q - + = - cos AB 2 B A | B A | 2 2 o o sin(180 ) tan A Bcos(180 ) B - q a = + - q sin A Bcos B q = - q RESOLUTION OF A VECTOR Rectangular Components of a Vector in Plane X Y q A A j ˆ y j ˆ Ay î Ax The vector A may be written as ˆ ˆ x y A A i A j = + uu r where î Ax is the component of vector A in X-direction and ĵ Ay is the component of vector A in the Y-direction. 1. Resultant of two unequal vectors cannot be zero. 2. Resultant of three co-planar vectors may or may not be zero. 3. Minimum no. ofcoplanar vectors for zeroresultant is 2 (for equal magnitude) and 3 (for unequal magnitude). 4. Resultant of three non coplanar vectors cannot be zero. Minimum number of non coplanar vectors whose sum can be zero is four. 5. Polygon law should be used only for diagram purpose for calculation of resultant vector (For addition of more than 2 vectors) we use components of vector. Keep in Memory 1. If B A r r = , then 0 A B - = r r is a null vector. . 2. Null vector or zero vector is defined as a vector whose magnitude is zero and direction indeterminate. Null vector differs from ordinaryzero in the sense that ordinary zero is not associated with direction. 3. | A | A Â r r = is called a unit vector. It is unitless and dimensionless vector. Its magnitude is 1. It represents direction only. 4. If B A r r = , then | B | | A | r r = and B̂ Â = , where ˆ ˆ A and B are unit vectors of A and B respectively. 5. A vector can be divided or multiplied bya scalar. 6. Vectors of the same kind can only be added or subtracted. It is not possible to add or subtract the vectors of different kind. This rule is also valid for scalars. 7. Vectors of same as well as different kinds can be multiplied. 8. A vector can have any number of components. But it can have only three rectangular components in space and two rectangular components in a plane. Rectangular components are mutuallyperpendicular. 9. The minimum number of unequal non-coplanar whose vector sum is zero is 4. 10. When x y z ˆ ˆ ˆ A A i A j A k = + + ur 2 2 2 x y z | A | A A A = + + r , where | A | r is modulus or magnitude of vector A r . 11. ˆ ˆ i j + makes 45° with both X andY-axes. It makes angle 90° with Z-axis. 12. k̂ j ˆ î + + makes angle 54.74° with each of the X, Y and Z-axes. 13. A B B A - ¹ - 14. If | B A | | B A | - = + then angle between A and B is 2 p .
- 74. 70 PHYSICS Also Ax= A cos q and Ay = A sin q ĵ ) sin A ( î ) cos A ( A q + q = Þ A cos q and A sin q are the magnitudes of the components of A in X andY-direction respectively. . Also 2 2 | | x y A A A = + uu r Rectangular components of a vector in 3D : Three rectangular components along X, Y and Z direction are given by x y z ˆ ˆ ˆ A i,A j,A k. Therefore, vector A r may be written as x y z ˆ ˆ ˆ A A i A j A k = + + ur and 2 2 2 x y z A A A A = + + If a , b and g are the angles subtended by the rectangular components of vector then cos a = x A , A cos b = y A A and cos g = z A A Also, cos2 a + cos2 b + cos2 g = 1 CAUTION : Do not resolve the vector at its head. The vector is always resolved at its tail. B cos q B sin q Wrong q B B cos q B sin q Correct q B Example1. X and Y component of vector A are 4 and 6 m respectively. The X and Y component of A + B uur uu r are 10 m and 9 m respectively. Calculate the length of vector B and its angle with respect to X-axis Solution : j ˆ 6 î 4 A + = and j ˆ 9 î 10 B A + = + ĵ 3 î 6 ) j ˆ 6 î 4 ( – ) ĵ 9 î 10 ( A – ) B A ( B + = + + = + = length of B is m 5 3 3 6 | B | 2 2 = + = Also ÷ ø ö ç è æ = q Þ = = = q - 2 1 tan 2 1 6 3 B B tan 1 x y where q is the angle which (A B) + r r is making with X-axis. Example2. Find the resultant of vectors given in figure X O Y P 4 cm M Q 3 cm 60° 6cm Solution : Here ĵ 2 . 5 î 3 ĵ 60 sin 6 î 60 cos 6 OP + = ° + ° = ` j ˆ 3 QM and î 4 PQ = = j ˆ 2 . 8 î 7 QM PQ OP OM + = + + = Example3. The resultant of two forces F1 and F2 is P. If F2 is reversed, the resultant is Q. Show that 2 2 2 2 1 2 2 + = + P Q ( F F ) . a q q 2 F 2 F P 1 F 1 F Q Solution : Suppose q be the angle between the forces 1 F r and 2 F r , then q + + = cos F F 2 F F P 2 1 2 2 2 1 2 ......(i) 2 F 2 F P 1 F 1 F q Q When the force F2 is reversed, then the magnitude of their resultant is 2 2 2 1 2 1 2 2 cos(180 ) Q F F F F = + + ° - q = q - + cos F F 2 F F 2 1 2 2 2 1 ......(ii) Adding equations (i) and (ii), ) F F ( 2 F 2 F 2 Q P 2 2 2 1 2 2 2 1 2 2 + = + = + Example4. Find the components of vector ˆ ˆ A = 2i +3j r along the directions of ˆ ˆ i + j and ˆ ˆ i – j. Solution : Here j ˆ 3 î 2 A + = r In order to find the component of A r along the direction of ĵ î + , let us find out the unit vector along ĵ î + . If â is the unit vector along ĵ î + , then 2 j ˆ î | j ˆ î | j ˆ î â + = + + = Hence, the magnitude of the component vector of A r along ĵ î + = 2 5 ) 3 2 ( 2 1 2 ĵ î ). j ˆ 3 î 2 ( â . A = + = + + = r
- 75. 71 Motion in aPlane Solution : F2 F1 2 ˆ F 250j = r , 1 ˆ F 250i = r 2 1 ˆ ˆ F F 250j 250i - = - r r , 2 2 2 1 | F F | 250 250 250 2 N - = + = r r (N-W direction) 135° W N E S 250 tan 1 250 q = = - - ° = q Þ 135 Example9. If a,b r r and c r are unit vectors such that a +b +c = 0 r r r , then find the angle between a r and b r . Solution : Given : a b c 0 + + = r r r c (a b) Þ = - + r r r Also, | a | | b | | c | 1 = = = r r r Let angle between a r and b r = q 2 2 1 1 1 2 1 1 cos = + + ´ ´ ´ q cos 1/ 2 q = - Þ 120 2 / 3 q = ° = p PRODUCT OF TWO VECTORS Scalar or Dot Product The scalar or dot product of two vectors A and B is a scalar, which is equal to the product of the magnitudes of A r and B r and cosine of the smaller angle between them. i.e., A . B r r = A B cosq q X B cosq B A e.g. W F s;P F v = × = × r r r r Properties of Scalar or Dot Product: 1. ˆ ˆ ˆ ˆ ˆ ˆ . . . 1 i i j j k k = = = ˆ ˆ ˆ ˆ ˆ ˆ . . . 0 i j j k k i = = = 2. . A B r r = A (B cosq) = B (A cosq) The dot product of two vectors can be interpreted as the product of the magnitude of one vector and the magnitude of the component of the other vector along the direction of the first vector. Therefore, component vector of A along ˆ ˆ i j + ˆ ˆ 5 i j 5 ˆ ˆ ˆ ˆ (A.a)a (i j) 2 2 2 æ ö + = = = + ç ÷ è ø r Similarly, if b̂ is the unit vector along the direction of ĵ î - , then magnitude of the component vector of A along ĵ î - = 2 1 2 ) 3 2 ( 2 ) ĵ î ( ). ĵ 3 î 2 ( | ĵ î | ) ĵ î ( ). ĵ 3 î 2 ( ) b̂ . A ( - = - = - + = - - + = r Component vector of A along ˆ ˆ i j - ˆ ˆ 1 i j 1 ˆ ˆ ˆ ˆ (A.b)b (i j) 2 2 2 æ ö - = = - = - - ç ÷ è ø r Example5. If k̂ c ĵ 4 . 0 î 3 . 0 + + is a unit vector, then find the value of c. Solution : Unit vector is a vector of unit magnitude. 1 c 4 . 0 3 . 0 2 2 2 = + + 1 c 16 . 0 09 . 0 2 = + + Þ 75 . 0 c 75 . 0 25 . 0 1 c2 = Þ = - = Þ Example6. What is the vector joining the points (3, 1, 14) and (–2, –1, –6) ? Solution : If P and Q be the points represented by the coordinates (3, 1, 14) and (–2, –1, –6) respectivelythen, PQ = p.v. of Q – p.v. of P = k̂ 20 ĵ 2 î 5 ) k̂ 14 ĵ î 3 ( ) k̂ 6 ĵ î 2 ( - - - = + + - - - - and ˆ ˆ ˆ QP PQ 5i 2j 20k = - = + + uuu r uuu r Example7. Find the angle between two vectors of magnitude 12 and 18 units, if their resultant is 24 units. Solution : Magnitude of first vector (A) r = 12; Magnitude of second vector (B) r = 18 and resultant of the given vectors (R) r = 24 24 = 2 2 A B 2ABcos + + q 24 = 2 2 (12) (18) 2 12 18 cos + + ´ ´ q or (24)2 = 144 + 324 + 432 cos q or 432 cos q = 108 or cos q = 108 432 = 0.25 or q = 75°52¢ Example8. Two forces 1 F 250N = r due east and N 250 F2 = r due north th have their common initial point. Find the magnitude and direction of 1 2 F F r r -
- 76. 72 PHYSICS 3. .. A B B A = r r r r Dot product of two vectors is commutative. 4. 2 . A A A = r r 5. .( ) . . A B C A B A C + = + r r r r r r r Dot product is distributive. 6. x y z x y z ˆ ˆ ˆ ˆ ˆ ˆ A.B (A i A j A k).(B i B j B k) = + + + + r r = (Ax Bx + Ay By + Az Bz) Vector or Cross Product The vector product of two vectors is defined as a vector having magnitude equal to the product of two vectors and sine of the angle between them. Its direction is perpendicular to the plane containing the two vectors (direction of the vector is given by right hand screw rule or right hand thumbrule. C A B = ´ u r ur u r = (AB sin q) n̂ The direction of (A B) ´ ur u r perpendicular to the plane containing vectors A ur and B u r in the sense of advance of a right handed screw rotated from A ur to B u r is through the smaller angle between them. ( ) A B C ´ = ur u r u r q B A e.g., ; ; v r r F L r p = w´ t = ´ = ´ r r r r r r r r r Properties of Vector or Cross Product 1. ˆ ˆ ˆ ˆ ˆ ˆ 0 i i j j k k ´ = ´ = ´ = 2. ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ; ; i j k j k i k i j ´ = ´ = ´ = i j k 3. 0 A A ´ = r r 4. A B B A ´ ¹ ´ r r r r (not commutative) [ ] ´ = - ´ r r r r Q A B B A 5. ( ) ( ) ( ) A B C A B A C ´ + = ´ + ´ r r r r r r r (followsdistributivelaw) 6. ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) x y z x y z A B A i A j A k B i B j B k ´ = + + ´ + + r r x y z x y z ˆ ˆ ˆ i j k A A A B B B = = (Ay Bz – Az By) î + (Az Bx – A Ax Bz) ĵ + (Ax By – Ay Bx) k̂ 7. The cross product of two vectors represents the area of the parallelogram formed bythem. k q A B P Q R S | | ( sin ) A B A B ´ = q r r = area ofparallelogram PQRS = 2 (area of DPQR) 8. A unit vector which is perpendicular to A as well as B is sin | | ´ ´ = q ´ uu r uu r r r uu r uu r A B A B AB A B Keep in Memory 1. x x y y z z 2 2 2 2 2 2 x y z x y z A B A B A B A.B cos | A || B | A A A B B B + + q = = + + + + ur u r ur u r 2. tan q = q q = ´ cos B A sin B A B . A | B A | r r r r 3. 2 2 2 2 . | A B | | A B| A B ´ + = r r r r 4. | ) A ( ) B ( | 2 | B – A | | B A | r r r r r r ´ = ´ + 5. If 0 C B A = + + r r r , then A C C B B A r r r r r r ´ = ´ = ´ 6. q = ´ - 2 cos B A | B A | | B . A | 2 2 2 2 r r r r 7. If B . A | B A | = ´ then angle between A and B is 4 p . 8. If B || A then 0 B A = ´ 9. Division by a vector is not defined. Because, it is not possible to divide by a direction. 10. The sum and product of vectors is independent of co-ordinate axes system. CONDITION OF ZERORESULTANT VECTOR AND LAMI'S THEOREM If the three vectors acting on a point object at the same time are represented in magnitude and direction by the three sides of a triangle taken in order, then their resultant is zero and the three vectors are said tobe in equilibrium. i.e. 1 2 3 F F F 0 + + = r r r
- 77. 73 Motion in aPlane P Q R F3 F2 F1 a b g Lami's Theorem It states that if three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. 3 1 2 F F F sin sin sin = = a b g r r r F3 F1 F2 a b g or, 3 1 2 F F F PQ QR PR = = r r r Example10. Calculate the area of a parallelogram formed from the vectors k̂ 3 ĵ 2 î A + + = r and k̂ j 3 î 2 B + - = r , as adjacent sides. Solution : The area of a parallelogram is given by | B A | r r ´ Here, ˆ ˆ ˆ i j k A B 1 2 3 2 3 1 ´ = + - r r = î [(2 × 1) – (–3 × 3)] + ĵ [(3 × 2) – (1 × 1)] + k̂ [(1 × –3)– (2 × 2)] = 11 î + 5 ĵ – 7 k̂ 2 2 2 | A B | (11) (5) ( 7) ´ = + + - ur u r @ 14 Example11. A particle suffers three displacements 4m in the northward, 2 m in the south-east and 1 m in the south-west directions. What is the displacement of the particle and the distance covered by it? Solution : Taking a frame of reference with the x-axis in the eastward and the y-axisin the northward direction Y North (N) X East (E) West (W) South (S) 1 2 ˆ ˆ ˆ s = 4j, s = 2cos 45ºi 2 sin 45º j - ur uu r 3 ˆ ˆ s = 2 cos 45º i 2 sin 45º j - - uu r 1 2 3 ˆ ˆ ˆ ˆ ˆ s = s +s +s = 4j+ 2 i 2 j 2 i 2 j - - - r ur uu r uu r Displacement ˆ s (4 2 2)j = - r ˆ = (1.17)j= (1.17) (northward) And total distance covered = 4 + 2 + 1 = 7m Example12. Prove that vectors ˆ ˆ ˆ A= i +2j+3k r and ˆ ˆ B = 2i j - r are perpendicular to each other. Solution : Here, k̂ 3 ĵ 2 î A + + = r and ĵ î 2 B - = r Two vectors are perpendicular to each other if, 0 B . A = r r Now B . A r r = ) k̂ 3 ĵ 2 î ( + + . ) ĵ î 2 ( - = 0 = 1 × 2+ 2 × (–1) + 3 ×(0) = 2 – 2+ 0 = 0 Since 0 B . A = r r , therefore vectors A r and B r are perpendicular to each other. Example13. Find the angle between the vectors ˆ ˆ ˆ A i j 2k = + - r and ˆ ˆ ˆ B i 2 j k = - + - r . Solution : Here k̂ 2 ĵ î A - + = r , k̂ ĵ 2 î B - + - = r We know that q = cos AB B . A r r or AB B . A cos r r = q Now 2 2 2 2 2 2 A 1 1 ( 2) 6 , B ( 1) 2 ( 1) 6 ˆ ˆ ˆ ˆ ˆ ˆ A.B (i j 2k).( i 2j k) 1 2 3 3 3 3 1 cos or 60 6 2 6 6 = + + - = = - + + - = = + - - + - = - + + = q = = = q = ° ´ r r Example14. A particleis displaced from a point (3, – 4, 5) toanother point (–2, 6, – 4) under a force k̂ ĵ 3 î 2 - + . Find the work doneby the force. Solution : k̂ ĵ 3 î 2 F - + = r The displacement of the particle is s r = position vector of point (–2, 6, – 4) – position vector of point (3, – 4, 5) s r = k̂ 9 ĵ 10 î 5 ) k̂ 5 ĵ 4 î 3 ( ) k̂ 4 ĵ 6 î 2 ( - + - = + - - - + -
- 78. 74 PHYSICS workdone by the force is ) k̂ 9 ĵ 10 î 5 ).( k̂ ĵ 3 î 2 ( s . F W - + - - + = = r r W = (2) (–5) + (3) (10) + (–1) (–9) = 29 units. Example15. Aforce ˆ ˆ F 6i xj = + r acting on a particle displaces it from the point A(3, 4) to the point B (1, 1). If the work done is 3 units, then find value of x. Solution : ˆ ˆ d 2i 3j = - - r , ˆ ˆ F 6i xj = + r W F.d = r r ; 3 = – 12 – 3x Þ x = –5 MOTION IN A PLANE OR MOTION IN TWO DIMENSIONS The motion in which the movement of a body is restricted to a plane is called motion in a plane. Example : A ball is thrown with some initial velocity (u) and makingangle q with harizontal. The general approach to solve problem on this topic is to resolve the motion into two mutually perpendicular co-ordinates. One along X-axis and other along Y-axis. These two motions are independent of each other and can be treated as two separate rectilinear motions. The velocity v and acceleration a can be resolved into its x and y components. u sin q u cos q vy vx P(x,y) v b u O q X Y x y x y ˆ ˆ v v i v j a a i a j = + = + x-componentofmotion y-component ofmotion vx = ux + axt vy = uy + ayt x = ux t+ 1 2 axt2 y = uyt + 1 2 ayt2 vx 2 – ux 2 = 2 axx vy 2 – uy 2 = 2 ayy x= x x u + v t 2 æ ö ç ÷ è ø y = y y u v t 2 + æ ö ç ÷ ç ÷ è ø Velocity The ratio of the displacement and the corresponding time interval is called the average velocity. x y ˆ ˆ v v i v j = + r Average velocity ˆ ˆ ˆ ˆ x y r x y v i j v i v j t t t D D D = = + = + D D D Instantaneous velocity 0 lim inst t r v t D ® D = D O X Y v vy j ^ vx i ^ q The magnitude of v = 2 2 + x y v v The direction of the velocity, y x v tan v q = y 1 x v tan v - q = Acceleration The average acceleration in a x–yplane in time interval Dt isthe change in velocity divided by the time interval. 2 2 ˆ ˆ x y x y a a i a j The magnitude of a a a = + = + r Average acceleration ˆ ˆ y x v v v a i j t t t D D D = = + D D D Instantaneous acceleration inst 0 0 0 ˆ ˆ lim lim lim D ® D ® D ® D D D = = + D D D y x t t t v v v a i j t t t In two or three dimensions, the velocityand acceleration vectors mayhave anyangle between 0°and 180° between them. RELATIVE VELOCITY IN TWO DIMENSIONS If two objects A and B moving with velocities VA and VB with respect to some common frame of reference, then : (i) Relative velocityofAw.r.t B AB A B v v v = - r r r (ii) Relative velocityof B w.r.t. A BA B A v v v = - r r r Therefore, AB BA v v = r r and AB BA v v = r r PROJECTILE MOTION Projectile is the name given to a body thrown with some initial velocity in any arbitrary direction and then allowed to move under the influence of gravity alone. Examples : A football kicked by the player, a stone thrown from the top of building, a bomb released from a plane. The path followed bythe projectile is called a trajectory. The projectile moves under the action of two velocities: (1) A uniform velocity in the horizontal direction, which does not change (if there is no air resistance) (2) A uniformlychanging velocityin the vertical direction due to gravity. Thehorizontal and vertical motions areindependent ofeach other. Types of Projectile: 1. Oblique projectile : In this, the body is given an initial velocity making an angle q with the horizontal and it moves under the infuence of gravity along a parabolic path.
- 79. 75 Motion in aPlane 2. Horizontal projectile : In this, the body is given an initial velocity directed along the horizontal and then it moves under the influence of gravity along a parabolic path. Motionalong x-axis ux = u, ax = 0 x = uxt + 1 2 axt2 o u = u x P(x,y) v = u = u x x x g v v =gt y y x y b u = 0 y x = ut + 0 t = x u ……(1) Motion along y-axis uy = 0, ay = g y = uyt + 1 2 ayt2 Þ 0 + 1 2 gt2 y = 1 2 gt2 ……(2) From equations (1) and (2) we get y= 2 2 g x 2u æ ö ç ÷ è ø which is the equation of a parabola. Velocity at any instant: x y ˆ ˆ v v i v j = + r v = 2 2 2 u g t + If b is the angle made by v r with the horizontal, then tanb = y x v v = gt u Time of flight andhorizontal range: If h is the distance of the ground from the point of projection, T is the time taken tostrikethe ground and Ris the horizontal range of the projectile then T = 2h g and 2h R u g = Case 1: If the projectile is projected from the top of the tower of height 'h', in horizontal direction, then the height of tower h, range x and time of flight t are related as : 2 1 h gt and x vt 2 = = h u x vy vx v 1b Case 2: Ifa particleis projectedat an angle(q) in upward direction from the top of tower of height h with velocity u, then uy = u sin q ay = – g ux = u cos q ax = 0 2 gt 2 1 t . sin u h - q + = and x = u cosq.t x h A B q u sin q u cos u + – q Case 3: Ifa bodyis projected at an angle (q) from the top of tower in downward direction then uy = – u sin q, ux = u cosq, ax = 0 ay = 2 1 g, h u sin .t gt 2 + - = - q - and x = u cosq.t + – x h q u usin q ucos q EquationofTrajectory Let the point from which the projectile is thrown into space is taken as the origin, horizontal direction in the plane of motion is taken as the X-axis, the vertical direction is taken as the Y-axis, Let the projectile be thrown with a velocity u making an angle q with the X-axis. u sin q u cos q vy vx P(x,y) v b u O q X Y The components of the initial velocity in the X-direction and Y- direction are u cos q and u sin q respectively. Then at anyinstant of time t,
- 80. 76 PHYSICS Motion alongx – axis ux = ucosq, ax = 0 2 x x 1 x u t a t 2 = + x = (u cosq) t ...(1) Motion along y–axis uy = u sin q, ay = –g y = uyt + 1 2 ayt2 y = u sinq t + 1 2 gt2 ...(2) From equations (1) and (2) we get y = x tanq 2 2 2 g x 2u cos - q which is the equation of a parabola. Hence the path followed by the projectile is parabolic. Velocity at anyPoint Let vy be the vertical velocity of projectile at time t. (at P) And vx be the horizontal component of velocity at time t. 2 2 sin (1) cos (2) y x x y v u gt v u v v v q q = - ¼¼ = ¼¼ = + 2 2 2 2 2 2 u cos u sin 2gt usin g t = q + q - q + 2 2 2 v u g t 2gt usin = + - q and the instantaneous angle (b) with horizontal is given by q - q = q - q = = b cos u gt tan cos u gt sin u v v tan x y Time of Flight : The time of flight of the projectile is given by 2u sin T 2t g q = = , where 't' is the time of ascent or descent. MaximumHeight : Maximum height attained bythe projectile is given by 2 2 u H sin 2g = q . In case of vertical motion, q = 90º so maximum height attained 2 2 u H g = Horizontal Range : The horizontal range of the projectile is given by 2 u sin 2 R g q = and 2 max u R g = at q = 45º (Q maximum valueof sin2q = 1) Keep in Memory 1. The horizontal range of the projectile is same at two angles of projection for q and (90° – q). 2. The height attained by the projectile above the ground is the largest when the angle of projection with the horizontal is 90° (vertically upward projection). In such a case time of flight is largest but the range is the smallest (zero). 3. Ifthe velocityof projection is doubled. The maximum height attained and the range become 4 times, but the time of flight is doubled. 4. When the horizontal rangeoftheprojectile is maximum, (q= 45°), then the maximum height attained is ¼th of the range. 5. For a projectile fired from the ground, themaximum height is attained after covering a horizontal distance equal tohalf of the range. The velocity ofthe projectile is minimum but not zero at the highest point, and is equal to u cosq i.e. at the highest point of the trajectory, the projectile has net velocity in the horizontal direction (vertical component is zero). Horizontal component of velocity also remains same as the component of g in horizontal direction is zero i.e., no acceleration in horizontal direction. Example16. A boat takes 2 hours to travel 8 km and back in still water lake. If the velocity of water is 4 km h–1, the time taken for going upstream of 8 km and coming back is (a) 2 hours (b) 2 hours 40 minutes (c) 1 hour and 20 minutes (d) can't be estimated with given information Solution : (b) Total distance travelledbyboat in 2hours = 8 + 8 = 16 km. Thereforespeedofboatin stillwater,vb =16/2=8kmh–1. Effective velocity when boat moves upstream = vb – vw = 8 – 4 = 4 km h–1. Therefore time taken to travel 8 km distance = 8/4 = 2h. Effective velocity when boat moves along the stream = vb + vw = 8 + 4 = 12 km h–1. Thetimetakentotravel8 kmdistance=8/12=2/3h= 40min. Total time taken =2h + 40 min = 2hours 40 min. Example17. A boat man can row a boat with a speed of 10 km/h in still water. If the river flows at 5 km/h, the direction in which the boat man should row to reach a point on the other bank directly opposite to the point from where he started (width of the river is 2 km). (a) is in a direction inclined at 120º to the direction of river flow (b) is in a direction inclined at 90º to the direction of river flow. (c) is 60º in the north-west direction (d) is should row directly along the river flow
- 81. 77 Motion in aPlane Solution : (a) Refer to fig., the boat man should go along OC in order to cross the river straight (i.e. along OB). B C O A 2km q b n r n º 30 sin 2 1 10 5 OC CB sin b r = = = n n = = q ; q = 30º; Boat man should go along in a direction inclined at 90º + 30º = 120º to the direction of river flow. Example18. A man swims at an angle q = 120º to the direction of water flow with a speed vmw = 5km/hr relative to water. If the speed of water vw = 3km/hr, find the speed of the man. Solution : r v mw = r v m – r v w r v m = r v mw + r v w Þ vm = | r v mw + r v w| = v v v v mw w mw w 2 2 2 + + . cosq Þ vm = 5 3 2 5 3 120 2 2 + + ( )( )cos º Þ vm = 25 9 15 + - = 19 m/sec. Example19. A gun throws a shell with a muzzle speed of 98 m/sec. When the elevation is 45º, the range is found to be 900 m. How much is the range decreased by air resistance? Solution : Without air resistance, the expected range 8 . 9 ) 98 ( 8 . 9 90 sin ) 98 ( g 2 sin u R 2 2 2 = ´ = q = =980m Decrease in range = 980 m – 900 m = 80 m Example20. A particle is projected with velocity u at an angle q with the horizontal so that its horizontal range is twice the greatest height attained. The horizontal range is (a) u2/g (b) 2 u2/3 g (c) 4 u2/5 g (d) None of these Solution : (c) Greatest height attained g 2 sin u H 2 2 q = ……(1) Horizontal range, g cos sin u 2 g 2 sin u R 2 2 q q = q = ……(2) [Qsin 2q = 2 sin q cosq] Given that R = 2 H g 2 sin u 2 g cos sin u 2 2 2 2 q = q q Solving we get tan q = 2 .....(3) Hence ) 5 / 1 ( cos and ) 5 / 2 ( sin = q = q From eqn. (2) g 5 u 4 5 1 5 2 g u 2 R 2 2 = ´ ´ = Example21. A body is projected downwards at an angle of 30º to the horizontal with a velocity of 9.8 m/s from the top of a tower 29.4 m high. How long will it take before striking the ground? Solution : The situation is shown in fig. 30° u B A – + The time taken bythe bodyis equal to the time taken bythe freelyfalling bodyfrom the height 29.4m with initial velocity u sin q. This is given by 9.8 usin 4.9 m / s 2 q = = Applying the formula, s = u t + 2 1 g t2, we have 29.4 = 4.9 t + 2 1 (9.8) t2 or 4.9 t2 + 4.9 t – 29.4 = 0 (because s, u and g are all in downward direction) t2 + t – 6 = 0or t = 2 or –3 Time taken to reach ground = 2 second Example22. Two boys stationed at A and B fire bullets simultaneously at a bird stationed at C. The bullets are fired from A and B at angles of 53° and 37° with the vertical. Both the bullets fire the bird simultaneously. What is the value of vA if vB = 60 units? (Given : tan 37° = 3/4) Solution : The vertical components must be equal.
- 82. 78 PHYSICS vAcos 53° = vB cos 37° or vA = vB cos 37 cos (90 37 ) ° ° - ° or vA = 60 cot 37° = 60 tan 37° [QvB = 60 units] = 60 4 3 ´ = 80 units PROJECTILE ON AN INCLINED PLANE Let a body is thrown from a plane OA inclined at an angle a with the horizontal, with a constant velocity u in a direction making an angle q with the horizontal. The body returns back on the same plane OA. Hence the net displacement of the particle in a direction normal to the plane OA is zero. a q (q-a) g sin( ) a g cos( ) a O A B u a g y uy x ux = u cos (q – a ) along the incline, + x-axis) uy = u sin (q – a ) along the incline, + y-axis) ax = g sin a along – x-axis, as retardation ay = g cos a along – y-axis, as retardation The time of flight of the projectile is given by 2 at 2 1 ut s + = or 2 1 0 u sin( )T g cos T 2 = q - a - a usin( ) T gcos( ) 2 q - a = a If maximumheight above the inclined plane is H, 2 2 u sin ( ) H 2g q - a = The horizontal range R of the projectile is given by OB = u cos q t = 2 u sin( )cos R gcos 2 q - a q = a The range of the projectile at the inclined plane is given by OA= 2 2 OB u sin( )cos R cos gcos 2 q- a q = = a a Condition for horizontal range R on the inclined plane to be maximum: Since 2 2 u R [2sin( )cos ] gcos = q - a q a 2 2 u [sin(2 ) sin ] gcos = q - a - a a {2 sin Acos B = sin(A+B)+sin(A–B)} R is maximumwhen sin (2q –a) is maximum i.e., sin (2q – a) =1 or 4 2 p a q = + 2 max 2 u R [1 sin ] gcos Þ = - a a or Rmax(on inclined plane) max R (on horizontalplane) 1 sin = + a where Rmax (on horizontal plane) g 2 u2 = . Conditionfor time offlightTto bemaximum: 2 u sin( ) T g cos q - a = a so T is max when sin (q – a) is maximum i.e., sin (q –a) = 1 or 2u T 2 g cos p q = + a Þ = a Itmeans that ifq1 istheanglefor projectilefor which Tismaximum and q2 isthe anglefor which Rismaximum, then q1 = 2q2. Example23. The slopes of wind screen of two cars are b1 = 30° and b2 = 15° respectively. At what ratio v1/v2 of the velocities of the cars will their drivers see the hailstorms bounced by windscreen of their cars in the vertical direction? Assume hailstorms falling vertically. Solution : From the fig tan 1 v v a = and 1 90 2 a = °- b where v is velocity of hail 1 1 1 v tan(90 2 ) cot 2 v ° - b = b = b1 –v1 b1 b1 v a Similarly, 2 2 v cot 2 v b = 1 2 2 1 v cot 2 cot30 3. v cot 2 cot 60 b = = = b
- 83. 79 Motion in aPlane Example24. A particle is projected up an inclined plane of inclination a to the horizontal. If the particle strikes the plane horizontally then tan a = ... . Given angle of projection with the horizontal is b. (a) 1/2 tanb (b) tanb (c) 2 tanb (d) 3 tanb Solution : (a) If the projectile hits the plane horizontallythen b a A B C plane horizontalplane 1 T T 2 = or 2usin( ) usin gcos g b - a b = a 2 sinb cos a –2 cosb sina = cosa sinb or sinb cosa = 2 cosb sina or tan tan 2 b a = Keep in Memory 1. Equation of trajectory of an oblique projectile in terms of range (R) is y = x tan q x 1 R æ ö - - ç ÷ è ø 2. There are two unique times at which the projectile is at the same height h(< H) and the sum of these two times. Since, h = (u sin q)t 2 1 gt 2 - is a quadratic in time, so it has two unique roots t1 and t2 (say) such that sum of roots (t1 + t2) is 2u sin g q and product (t1t2) is 2h g . The time lapse (t1– t2) is 2 2 2 4u sin 8h . g g q - UNIFORM AND NON-UNIFORM CIRCULAR MOTION Uniform Circular Motion An object moving in a circle with a constant speed is said to be in uniform circular motion.Ex. Motion of the tip of the second hand of a clock. Angular displacement : Change in angular position is called angular displacement (dq). B DS dq O q1 q2 A Angular velocity : Rateofchangeofangular displacement is called angular velocity w i.e., d dt q w = Relationbetweenlinear velocity(v)andangular velocity(w). In magnitude, v r v r = w ´ = w r r r Angular acceleration : Rateof changeofangular velocityis called angular accelaeration. i.e., 2 2 d d dt dt w q a = = Relation between linear acceleration and angular accelaration. In magnitude, a r a r = a ´ = a r u r r Centripetal acceleration : Acceleration acting on abody moving in uniform circular motion is called centripetal acceleration. It arises due to the change in the direction of the velocity vector. Magnitude of certipetal acceleration is 2 2 c v a r r = = w 2 1 2 frequency T T p æ ö w = = pu u = = ç ÷ è ø Q 2 2 4 c a r = p u This acceleration is always directed radiallytowards the centre of the circle. Centripetal force: The force required to keep a body moving in uniform circular motion is called centripetal force. 2 2 c mv F mr r = = w It is always directed radiallyinwards. Centrifugal force : Centrifugal force is a fictitious force which acts on abody in rotating (non-inertial frames)frame of reference. Magnitude of the centrifugal force 2 mv F r = . This force is always directed radially outwards and is also called corolious force.
- 84. 80 PHYSICS Non-uniform CircularMotion : An object moving in a circle with variable speed is said to be in non-uniform circular motion. If the angular velocity varies with time, the object has two accelerations possessed by it, centripetal acceleration (ac) and Tangential accelaration (aT) and both perpendicular to each other. b a aT ac Net acceleration 2 2 2 4 2 2 4 2 ( ) and, tan c T c T a a a a r r a r a a = + = w + a = w + a b = Keep in Memory 1. Angular displacement behaves like vector, when its magnitude is very very small. It follows laws of vector addition. 2. Angular velocity and angular acceleration are axial vectors. 3. Centripetal acceleration always directed towards the centre of the circular path and is always perpendicular to the instantaneous velocity of the particle. 4. Circular motion is uniform if aT = ra = 0, that is angular velocity remains constant and radial acceleration 2 2 c v a r r = = w is constant. 5. When aT or a is present, angular velocity varies with time and net acceleration is 2 2 c T a a a = + 6. If aT = 0 or a = 0, no work is done in circular motion. Example25. A sphere of mass 0.2 kg is attached to an inextensible string of length 0.5 m whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 0.3 m. What will be the speed of the sphere? Solution : Centripetal forceis providedbycomponent T sin q, therefore ; r m sin T 2 n = q and, T cos mg; q = so, 2 2 2 2 2 1/2 2 2 2 1/2 1/2 1/2 v r tan ; rg r r tan r r g v ( r ) 0.09 10 (0.25 0.09) = 1.5 m/s. q = = - é ù q = ê ú ê ú - ë û é ù = ê ú - ë û é ù ´ = ê ú - ë û l Q l l O B A T cos q T Tsin q l mg q r q Example26. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? Solution : Speed, v = 27 km/h = 1 1 5 27 ms 7.5ms 18 - - ´ = centripetal acceleration, 2 c v a r = or 2 2 2 c (7.5) a ms 0.7ms 80 - - = = v ac at a q P P is the point at which cyclist applies brakes. At this point, tangential acceleration at, being negative, will act opposite to v r . Total acceleration, 2 2 c t a a a = + 2 2 2 2 c t or, a (0.7) (0.5) ms 0.86ms a 0.7 tan 1.4 a 0.5 54 28 - - = + = q = = = q = ° ¢
- 85. 81 Motion in aPlane
- 86. 82 PHYSICS 1. Itis found that . A B A + = This necessarilyimplies, (a) B = 0 (b) A, B areantiparallel (c) A, B are perpendicular (d) A.B £ 0 2. Which one of the following statements is true? (a) A scalar quantity is the one that is conserved in a process. (b) A scalar quantityis the one that can never take negative values. (c) A scalar quantity is the one that does not vary from one point to another in space. (d) A scalar quantity has the same value for observers with different orientations ofthe axes. 3. Two balls are projected at an angle q and (90º – q) to the horizontal with the same speed. Theratio oftheir maximum vertical heights is (a) 1: 1 (b) tanq : 1 (c) 1 : tanq (d) tan2q : 1 4. Which of the following is not correct ? (a) A B B A ´ = - ´ r r r (b) A B B A ´ ¹ ´ r r r (c) ( ) A B C A B A C ´ + = ´ + ´ r r r r r r r (d) ( ) ( ) A B C A B C ´ + = ´ + r r r r r r 5. The greatest height to which a man can through a ball is h. What is the greatest horizontal distance to which he can throw the ball? (a) 2h (b) 4 h (c) 2 h (d) None of these 6. If A and B are two vectors, then the correct statement is (a) A+ B= B +A (b) A – B = B – A (c) A× B= B ×A (d) None of these 7. Three vectors A, B and C satisfy the relation A · B = 0 andA · C = 0. The vector Ais parallel to (a) B (b) C (c) B · C (d) B× C 8. If n̂ is a unit vector in the direction of the vector A, then (a) A n̂ = A (b) n̂ = A A (c) A n̂ = A (d) ˆ ˆ n = n×A 9. A projectile thrown with a speed v at an angle q has a range R on the surfaceofearth. For same v and q, its range on the surface of moon will be Earth moon g g 6 é ù = ê ú ë û (a) R/6 (b) R (c) 6R (d) 36R 10. Given that A+ B = R and A2 + B2 = R2. The angle between A and B is (a) 0 (b) p/4 (c) p/2 (d) p 11. Given that A + B = R and A = B = R. What should be the angle between A and B ? (a) 0 (b) p/3 (c) 2p/3 (d) p 12. Let A = iAcos q + jAsin q be anyvector. Another vector B, which is normal toA can be expressed as (a) i B cos q – j B sin q (b) i B cos q + j B sin q (c) i B sin q – j B cos q (d) i B sin q + j B cos q 13. Three particlesA, Band C are projected from the samepoint with same initial speeds making angles 30º, 45º and 60º respectively with the horizontal. Which of the following statements is correct? (a) A, B and C have unequal ranges (b) Ranges of A and C are equal and less than that of B (c) Ranges ofAand C are equal and greater than that of B (d) A, B and C have equal ranges 14. Twoparticles of equal masses are revolving in circular paths of radii r1 and r2 respectively with the same period. The ratioof their centripetal force is (a) r1/r2 (b) 1 2 r / r (c) (r1/r2)2 (d) (r2/r1)2 15. A bomb is released from a horizontal flying aeroplane. The trajectory of bomb is (a) a parabola (b) a straight line (c) a circle (d) a hyperbola 16. A projectile can have the same range for two angles of projection. If h1 and h2 are maximum heights when the range in the two cases is R, then the relation between R, h1 and h2 is (a) 1 2 R 4 h h = (b) 1 2 R 2 h h = (c) 1 2 R h h = (d) None of these 17. A bombis dropped from an aeroplane moving horizontally at constant speed. If air resistance is taken into consideration, then the bomb (a) falls on earth exactlybelow the aeroplane (b) falls on the earth exactly behind the aeroplane (c) falls on the earths ahead of the aeroplane (d) flies with the aeroplane 18. Two vectors A and B lie in a plane, another vector C lies outside this plane, then the resultant of these three vectors i.e.,A+ B +C (a) can be zero (b) cannot be zero (c) lies in the plane containingA + B (d) lies in the plane containingA – B
- 87. 83 Motion in aPlane 19. A cannon on a level plane is aimed at an angle q above the horizontalandashellisfiredwitha muzzlevelocityn0 towards a vertical cliff a distance D away. Then the height from the bottom at which the shell strikes the side walls of the cliff is (a) 2 2 2 0 g D D sin 2v sin q - q (b) 2 2 2 0 g D D cos 2 v cos q - q (c) 2 2 2 0 g D D tan 2 v cos q - q (d) 2 2 2 0 g D D tan 2 v sin q - q 20. A projectile thrown with velocity v making angle q with vertical gains maximum height H in the time for which the projectile remains in air, the time period is (a) g / cos H q (b) g / cos H 2 q (c) g / H 4 (d) g / H 8 21. Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally. Theyhit the ground with speeds vA, vB and vC respectively then, (a) vA = vB = vC (b) vA = vB > vC (c) vB > vC > vA (d) vA > vB = vC 22. An aeroplane flying at a constant speed releases a bomb. As the bomb moves away from the aeroplane, it will (a) always be vertically below the aeroplane only if the aeroplane was flying horizontally (b) always be vertically below the aeroplane only if the aeroplanewas flying at an angle of45° to the horizontal (c) always be vertically below the aeroplane (d) graduallyfall behind the aeroplane ifthe aeroplane was flying horizontally. 23. In uniform circular motion, the velocity vector and acceleration vector are (a) perpendicular to each other (b) same direction (c) opposite direction (d) not related to each other 24. The time offlight ofa projectileon an upward inclinedplane depends upon (a) angle of inclination of the plane (b) angle of projection (c) the value of acceleration due to gravity (d) all of the above. 25. At the highest point on the trajectory of a projectile, its (a) potential energyis minimum (b) kinetic energyis maximum (c) total energyismaximum (d) kinetic energyis minimum. 1. Aprojectileis projected with a kineticenergyE. Its rangeisR. It will have the minimum kinetic energy, after covering a horizontal distance equal to (a) 0.25R (b) 0.5R (c) 0.75R (d) R 2. The range of a projectile when launched at an angle of 15º with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of45º to the horizontal (a) 1.5km (b) 3.0km (c) 6.3km (d) 0.75km 3. A gun fires two bullets at 60º and 30º with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio (a) 2: 1 (b) 3 : 1 (c) 4: 1 (d) 1: 1 4. The angular speed of a fly-wheel making 120 revolutions/ minute is (a) p rad/sec (b) 4p rad/sec (c) 2p rad/sec (d) 4p2 rad/sec 5. Consider two vectors 1 2 ˆ ˆ ˆ ˆ F 2i 5kandF 3j 4k = + = + r r . The magnitude of the scalar product of these vectors is (a) 20 (b) 23 (c) 33 5 (d) 26 6. Ifrangeisdoublethemaximum heightofa projectile, then qis (a) tan–14 (b) tan–1 1/4 (c) tan–1 1 (d) tan–1 2 7. A bodyis projected such that its KE at the top is 3/4th of its initial KE. What istheangle ofprojectile with the horizontal? (a) 30º (b) 60º (c) 45º (d) 120º 8. Consider a vector F = 4 i – 3 j. Another vector that is perpendicular to F is (a) 4 i + 3j (b) 6i (c) 7k (d) 3 i – 4 j 9. From a point on the ground at a distance 2 meters from the foot of a vertical wall, a ball is thrown at an angle of 45º which just clears the top of the wall and afterward strikes the ground at a distance 4 m on the other side. The height of the wall is (a) m 3 2 (b) m 4 3 (c) m 3 1 (d) m 3 4
- 88. 84 PHYSICS 10. Thevelocity of projection of a body is increased by 2%. Other factors remaining unchanged, what will be the percentage change in the maximum height attained? (a) 1% (b) 2 % (c) 4 % (d) 8 % 11. A ball is thrown from the ground with a velocity of 3 20 m/smaking an angleof60º with thehorizontal. The ball will be at a height of40 m from the ground after a time t equal to (g = 10 ms–2) (a) 2 sec (b) 3 sec (c) 2 sec (d) 3 sec 12. Forces of4 N and 5 N are applied at origin along X-axis and Y-axis respectively. The resultant force will be (a) ÷ ø ö ç è æ - 4 5 tan , N 41 1 (b) ÷ ø ö ç è æ - 5 4 tan , N 41 1 (c) ÷ ø ö ç è æ - - 4 5 tan , N 41 1 (d) ÷ ø ö ç è æ - - 5 4 tan , N 41 1 13. Aparticlemoves in a circleof radius25cm at tworevolutions per second. The acceleration of the particle in meter per second2 is (a) p2 (b) 8 p2 (c) 4 p2 (d) 2 p2 14. The vector sum of the forces of 10 N and 6 N can be (a) 2N (b) 8 N (c) 18N (d) 20N 15. A bomb is dropped on an enemypost byan aeroplane flying horizontallywith a velocityof60 km h–1 and at a height of 490 m. At the time of dropping the bomb, how far the aeroplane should be from the enemy post sothat the bomb maydirectly hit the target ? (a) m 3 400 (b) m 3 500 (c) m 3 1700 (d) 498m. 16. A projectile is thrown horizontally with a speed of 20 ms–1. If g is 10 ms–2, then the speed of the projectile after 5 second will be nearly (a) 0.5 ms–1 (b) 5 ms–1 (c) 54 ms–1 (d) 500 ms–1 17. A ball is projected at such an angle that the horizontal range is threetimes themaximumheight. The angleofprojection of the ball is (a) ÷ ø ö ç è æ - 4 3 sin 1 (b) ÷ ø ö ç è æ - 3 4 sin 1 (c) ÷ ø ö ç è æ - 3 4 cos 1 (d) ÷ ø ö ç è æ - 3 4 tan 1 18. A body is projected horizontally from a point above the ground and motion of the body is described by the equation x = 2t, y = 5t2 where x, and y are horizontal and vertical coordinates in metre after time t. The initial velocityof the body will be (a) 29 m/s horizontal (b) 5 m/s horizontal (c) 2 m/s vertical (d) 2 m/s horizontal 19. A vector 1 P uu r is along the positive x-axis. If its vector product with another vector 2 P is zero then 2 P could be (a) ĵ 4 (b) î 4 - (c) ) k̂ ĵ ( + (d) ) j ˆ î ( + - 20. A wheel rotates with constant acceleration of 2.0 rod/s2, if the wheel starts from rest the number of revolutions it makes in the first ten seconds will be approximately (a) 32 (b) 24 (c) 16 (d) 8 21. A projectile ofmass m is fired with velocityu making angle q with thehorizontal. Its angular momentum about thepoint of projection when it hits the ground is given by (a) 2 2musin cos g q q (b) 3 2 2mu sin cos g q q (c) 2 musin cos 2g q q (d) 3 2 mu sin cos 2g q q 22. A bucket, full of water is revolved in a vertical circle of radius 2 m. What should be the maximum time-period of revolution so that the water doesn’t fall out of the bucket? (a) 1 sec (b) 2 sec (c) 3 sec (d) 4 sec 23. If | a b | | a b | + = - r r r r then angle between a & b r r is (a) 45º (b) 30º (c) 90º (d) 180º 24. If the sum of two unit vectors is a unit vector, then the magnitude oftheir difference is (a) 3 (b) 2 (c) 5 (d) 2 1 25. Out of the following sets of forces, the resultant of which cannot be zero ? (a) 10,10, 10 (b) 10,10,20 (c) 10,20, 20 (d) 10,20,40
- 89. 85 Motion in aPlane 26. The magnitude of the vector product of two vectors is 3 times the scalar product. The angle between vectors is (a) 4 p (b) 6 p (c) 3 p (d) 2 p 27. IfA= B + C and the magnitudes ofA, B and C are5, 4 and 3 units, the angle between A and C is (a) cos–1 (3/5) (b) cos–1 (4/5) (c) 2 p (d) sin–1 (3/4) 28. A rectangular sheet of material has a width of 3 m and a length of 4 m. Forces with magnitudes of 3 N and 4N. respectively, are applied parallel to two edges of the sheet, as shown in the figure below. 3m 4m 4N 3N F q Athird force F, is applied to the centre of the sheet, along a line in the plane of the sheet, at an angle q = tan 0.75 with respect to the horizontal direction. The sheet will be in translational equilibrium when F has what value? (a) F = 3 N (b) F= 4N (c) F = 5 N (d) F= 7N 29. The linear velocityof a rotating body is given by : v r = w ´ uu r uu r ur If k̂ 3 ĵ 4 r and k̂ 2 j ˆ 2 î - = + - = w , then the magnitude of v is (a) units 29 (b) units 31 (c) units 37 (d) units 41 30. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are (a) 12 N, 6 N (b) 13 N, 5 N (c) 10 N, 8 N (d) 16N, 2N. 31. For any two vectors A r and B r , if A . B | A B | = ´ r r r r , the magnitude of C A B = + r r r is (a) A+ B (b) AB 2 B A 2 2 + + (c) 2 2 B A + (d) 2 AB B A2 + + 2 32. The angles which the vector ˆ ˆ ˆ A 3i 6j 2k = + + r makes with the co-ordinate axes are : (a) 7 1 cos , 7 4 cos , 7 3 cos 1 1 1 - - - (b) 7 2 cos , 7 6 cos , 7 3 cos 1 1 1 - - - (c) 7 3 cos , 7 5 cos , 7 4 cos 1 1 1 - - - (d) None of these 33. The resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled. The angle between the two forces is (a) 60º (b) 120º (c) 70º (d) 180º 34. Let C A B = + r r r (a) | C | r is always greater than | A | r (b) It is possible to have | C | | A | < r r and | C | | B | < r r (c) C r is always equal to A B + r r (d) C r is never equal to A B + r r 35. At what angle must the two forces (x + y) and (x – y) act so that the resultant may be 2 2 (x y ) + ? (a) 1 2 2 2 2 cos [ (x y )/ 2(x y )] - - + - (b) 1 2 2 2 2 cos [ 2(x y )/(x y )] - - - + (c) 1 2 2 2 2 cos [ (x y ) /(x y )] - - + - (d) 1 2 2 2 2 cos [ (x y )/(x y )] - - - + 36. A force of k̂ F – acts on O, the origin of the coordinate system. The torque about the point (1, –1) is X Y Z O (a) ) j ˆ î ( F - (b) ) ĵ î ( F + - (c) ) ĵ î ( F + (d) ) j ˆ î ( F - - 37. The component of vector ˆ ˆ a 2i 3j = + r along the vector i + j is (a) 2 5 (b) 2 10 (c) 2 5 (d) 5
- 90. 86 PHYSICS 38. Abodyis projected at an angle of 30º to the horizontal with speed 30 m/s. What is the angle with the horizontal after 1.5 seconds? Take g = 10 m/s2. (a) 0º (b) 30º (c) 60º (d) 90º 39. A projectile is moving at 60 m/s at its highest point, whereit breaks into two equal parts due to an internal explosion. One part moves vertically up at 50 m/s with respect to the ground. The other part will move at a speed of (a) 110m/s (b) 120m/s (c) 130m/s (d) s / m 61 10 40. A particle having a mass 0.5 kg is projected under gravity with a speed of 98 m/sec at an angle of 60º. The magnitude ofthe change in momentum (in N-sec)oftheparticle after 10 seconds is (a) 0.5 (b) 49 (c) 98 (d) 490 41. A large number ofbullets are fired in all directions with the same speed v. What is the maximum area on the ground on which these bullets will spread? (a) g v2 p (b) 2 4 g v p (c) 2 2 2 g v p (d) 2 4 2 g v p 42. A force of ˆ ˆ (3i + 4j) N acts on a body and displaces it by ˆ ˆ (3i + 4j) m. The work done by the force is (a) 5 J (b) 30J (c) 25J (d) 10J 43. A person aims a gun at a bird from a point at a horizontal distance of 100 m. If the gun can impact a speed of 500 ms– 1 to the bullet. At what height above the bird must he aim his gun in order to hit it? (g = 10 ms–2) (a) 10.4cm (b) 20.35cm (c) 50cm (d) 100cms 44. If ˆ ˆ ˆ ˆ ˆ ˆ a i j k and b 2i j 3k = - + = + + r r , then theunit vector along a b ® ® + is (a) 5 k 4 i 3 + (b) 5 k 4 i 3 + - (c) 5 k 4 i 3 - - (d) None of these 45. A body is thrown with a velocity of 9.8 ms–1 making an angleof30º with the horizontal. It will hit the ground after a time (a) 3.0 s (b) 2.0 s (c) 1.5 s (d) 1 s 46. Two bullets are fired horizontally, simultaneously and with different velocities from the same place. Which bullet will hit the ground earlier ? (a) It would depend upon the weights of the bullets. (b) The slower one. (c) The faster one. (d) Both will reach simultaneously. 47. A cricket ball is hit with a velocity 25 1 s m - , 60° above the horizontal. How far above the ground, ball passes over a fielder 50 m from the bat (consider the ball is struck very close to the ground)? Take 7 . 1 3 = and g = 10 ms–2 (a) 6.8m (b) 7m (c) 5m (d) 10m 48. The equation of a projectile is 2 gx x 3 y 2 - = The angle of projection is given by (a) 3 1 tan = q (b) 3 tan = q (c) 2 p (d) zero. 49. A body is thrown horizontally with a velocity gh 2 from the top of a tower of height h. It strikes the level ground through thefoot ofthe tower at a distance x from thetower. The value of x is (a) gh (b) gh 2 (c) 2h (d) 2gh 3 50. A plane flying horizontally at a height of 1500 m with a velocityof 200 ms–1 passes directlyoverhead on antiaircraft gun. Then the angle with the horizontal at which the gun should be fired from the shell with a muzzle velocityof 400 ms–1 to hit the plane, is (a) 90° (b) 60° (c) 30° (d) 45° 51. AprojectileA is thrown at an angle of 30° to the horizontal from point P.At the same time, another projectileB is thrown with velocity v2 upwards from the point Q verticallybelow the highest point. For B to collide with A, 1 2 v v should be Highest point P Q 30° A B v1 v2 (a) 1 (b) 2 (c) 2 1 (d) 4
- 91. 87 Motion in aPlane 52. The velocity of projection of oblique projectile is 1 s m ) j ˆ 8 î 6 ( - + . The horizontal rangeof the projectile is (a) 4.9m (b) 9.6m (c) 19.6m (d) 14m 53. A gun is aimed at a horizontal target. It takes 2 1 s for the bullet to reach the target. The bullet hits the target x metre below the aim. Then, x is equal to (a) m 4 8 . 9 (b) m 8 8 . 9 (c) 9.8m (d) 19.6m. 54. The equation of trajectory of projectile is given by 2 x gx y 20 3 = - , where x and yare in metre. The maximum range ofthe projectile is (a) 8 m 3 (b) 4 m 3 (c) 3 m 4 (d) 3 m 8 55. A bullet is fired with a speed of 1500 m/s in order to hit a target 100 m away. If g = 10 m/s2. The gun should be aimed (a) 15 cm above the target (b) 10 cm above the target (c) 2.2 cm above the target (d) directly towards the target 56. A projectile of mass m is thrown with a velocityv making an angle60° with the horizontal. Neglecting air resistance, the changein momentum fromthe departureAtoitsarrival at B, along the vertical direction is 60° A v B (a) 2mv (b) 3mv (c) mv (d) mv 3 57. A person sitting in the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with velocityof 20 m/s. Aperson standing outside on the ground also observes the ball. How will the maximum heights (ym) attained and the ranges (R) seen bythe thrower and the outside observer compare with each other? (a) Same ym different R (b) Same ym and R (c) Different ym same R (d) Different ym and R 58. A particle moves in a circle of radius 4 cm clockwise at constant speed 2 cm/s. If x̂ and ŷ are unit acceleration vectors along X and Y-axis respectively (in cm/s2), the acceleration of the particle at the instant halfway between P and Q is given by (a) ˆ ˆ 4(x y) - + (b) ˆ ˆ 4(x y) + (c) ˆ ˆ (x y)/ 2 - + O y x P Q (d) ˆ ˆ (x y)/ 4 - 59. A projectile is thrown in the upward direction making an angleof 60° with the horizontal direction with a velocityof 147 ms–1. Then the time after which its inclination with the horizontal is 45°, is (a) 15 s (b) 10.98s (c) 5.49s (d) 2.745s 60. A cyclist moving at a speed of 20 m/s takes a turn, if he doubles his speed then chance of overturn (a) is doubled (b) is halved (c) becomes four times (d) becomes 1/4 times 61. A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed ofswimming is 0.5 m/s at an angle of 120º with the direction of flow of water. The speed of water is (a) 1.0 m/s (b) 0.5m/s (c) 0.25m/s (d) 0.43m/s 62. The position vector of a particle is . ĵ ) t sin a ( î ) t cos a ( r w + w = r The velocity of the particle is (a) directed towards the origin (b) directed awayfrom the origin (c) parallel to the position vector (d) perpendicular to the position vector 63. A ball whose kinetic energyis E is thrown at an angle of 45º with horizontal. Its kinetic energyat highest point of flight will be (a) E (b) E/2 (c) 2 E (d) O 64. Twoprojectiles are fired from the same point with the same speed at angles of projection 60º and 30º respectively. Which one of the following is true? (a) Their maximum height will besame (b) Their rangewill be same (c) Their landing velocity will be same (d) Their time of flight will be same 65. A bodyof 3kg. moves in X-Y plane under the action of force given by ĵ t 4 î t 6 + . Assuming that the bodyis at rest at time t = 0, the velocity of body at t = 3 sec is (a) j ˆ 6 î 9 + (b) ĵ 6 î 18 + (c) j ˆ 12 î 18 + (d) ĵ 68 î 12 +
- 92. 88 PHYSICS 66. Froma 10 m high building a stone A is dropped, and simultaneously another identical stone B is thrown horizontally with an initial speed of 5 ms–1. Which one of the following statements is true? (a) It is not possible to calculate which one of the two stones will reach the ground first (b) Both the stones (A and B) will reach the ground simultaneously (c) A stone reaches the ground earlier than B (d) B stone reaches the ground earlier than A 67. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces (a) cannot be predicted (b) are equal to each other (c) are equal to each other in magnitude (d) are not equal to each other in magnitude 68. A particle moves along a circle of radius m 20 ÷ ø ö ç è æ p with constant tangential acceleration. It the velocity of particle is 80 m/sec at end of second revolution after motion has begun, the tangential acceleration is (a) 40 pm/sec2 (b) 40 m/sec2 (c) 640 pm/sec2 (d) 160 pm/sec2 69. If B . A 3 | B A | r r r r = ´ then the value of | A B| + r r is (a) ½ 2 2 ) AB 3 B A ( + (b) ½ 2 2 ) AB B A ( + + (c) ½ 2 2 3 AB B A ÷ ÷ ø ö ç ç è æ + + (d) A+ B 70. If a vector 2 3 8 i j k Ù Ù Ù + + is perpendicular to the vector 4 4 j i k Ù Ù Ù - +a , then the value of a is (a) 1/2 (b) –1/2 (c) 1 (d) –1 Directions for Qs. (71 to 75) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethe correctanswer (ONLYONE option is correct ) from the following- (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 71. Statement -1 : F r r r r ´ = t and r F r r r ´ ¹ t Statement -2 : Cross product of vectors is commutative. 72. Statement -1 : If dot product and cross product of A r and B r are zero, it implies that one of the vector A r and B r must be a null vector Statement -2 : Null vector is a vector with zero magnitude. 73. Statement-1 Two stones are simultaneously projected from level ground from same point with same speeds but different angles with horizontal. Both stones move in same vertical plane.Then the twostones maycollide in mid air. Statement-2 : For twostones projected simultaneouslyfrom same point with same speed at different angles with horizontal, their trajectories mayintersect at some point. 74. Statement-1 : K.E. of a moving bodygiven byas2 where s is the distance travelled in a circular path refers to a variable acceleration. Statement-2 :Acceleration varies with direction onlyin this case of circular motion. 75. Statement-1 : Centripetal andcentrifugal forcescancel each other. Statement-2 : This is because they are always equal and opposite. Exemplar Questions 1. The angle between ˆ ˆ A i j = + and ˆ ˆ B i j = - is (a) 45° (b) 90° (c) –45° (d) 180° 2. Which one of the following statements is true? (a) A scalar quantity is the one that is conserved in a process (b) A scalar quantityis the one that can never take negative values (c) A scalar quantity is the one that does not vary from one point to another in space (d) A scalar quantity has the same value for observers with different orientation of the axes 3. Figure shows the orientation of two vectors u and v in the xy-plane. X Y u v O If ˆ ˆ u ai bj = + and ˆ ˆ v pi qj = + Which of the following is correct? (a) a and p are positive while b and q are negative (b) a, p and b are positive while q is negative (c) a, q and b are positive while p is negative (d) a, b, p and q are all positive
- 93. 89 Motion in aPlane 4. Thecomponent ofa vector ralong X-axiswill havemaximum value if (a) r is along positive Y-axis (b) r is along positive X-axis (c) r makes an angle of 45° with the X-axis (d) r is along negative Y-axis 5. The horizontal range of a projectilefired at an angleof 15° is 50 m. Ifit is fired with the same speed at an angle of 45°, its range will be (a) 60m (b) 71m (c) 100m (d) 141m 6. Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are : (a) impulse, pressure and area (b) impulseand area (c) area and gravitational potential (d) impulse and pressure 7. In a two dimensional motion, instantaneous speed v0 is a positive constant. Then, which of the following are necessarily true? (a) The average velocity is not zero at any time (b) Average acceleration must always vanish (c) Displacements in equal time intervals are equal (d) Equal path lengths are traversed in equal intervals 8. In a two dimensional motion, instantaneous speed v0 is a positive constant. Then, which of the following are necessarily true? (a) The acceleration ofthe particle is zero (b) The acceleration of the particle is bounded (c) The acceleration of the particle is necessarily in the plane of motion (d) The particle must be undergoing a uniform circular motion NEET/AIPMT (2013-2017) Questions 9. The velocity ofa projectile at the initial pointA is $ ( ) 2 3 i j + $ m/s. It’s velocity (in m/s) at point B is [2013] (a) $ 2 3 i j - + $ (b) $ 2 3 i j - $ (c) $ 2 3 i j + $ (d) $ 2 3 i j - - $ 10. Vectors , A B r r and C r are such that 0 A B × = r r and 0. A C × = r r Then the vector parallel to A r is [NEET Kar. 2013] (a) and B C r r (b) A B ´ r r (c) B C + r r (d) B C ´ r r 11. A particle is moving such that its position coordinate (x, y) are (2m,3m) attimet = 0 (6m, 7m) at time t = 2 s and (13m, 14m) attimet = 5s. Average velocity vector av (V ) r from t = 0 to t = 5s is : [2014] (a) 1 ˆ ˆ (13i +14j) 5 (b) 7 ˆ ˆ (i + j) 3 (c) ˆ ˆ 2(i + j) (d) 11 ˆ ˆ (i + j) 5 12. A ship A is moving Westwards with a speed of 10 km h–1 and a ship B 100 km South ofA, is moving Northwards with a speed of 10 km h–1. The time after which the distance between them becomes shortest, is : [2015] (a) 5 h (b) 5 2 h (c) 10 2 h (d) 0 h 13. If vectors ˆ ˆ A cos ti sin tj = w + w r and t t ˆ ˆ B cos i sin j 2 2 w w = + r are functions of time, then the value of t at which they are orthogonal to each other is : [2015 RS] (a) t 2 p = w (b) t p = w (c) t = 0 (d) t 4 p = w 14. The position vector of a particle R r as a function of time is given by: ˆ ˆ R 4sin(2 t)i 4cos(2 t)j = p + p r Where R is in meter, t in seconds and î and ĵ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle? [2015 RS] (a) Magnitude of acceleration vector is 2 v R , where v is the velocity of particle (b) Magnitude of thevelocityofparticle is 8 meter/second (c) path of the particle is a circle ofradius 4 meter. (d) Acceleration vector is along - R r 15. A particle moves so that its position vector is given by ˆ ˆ r cos tx sin ty = w + w r . Where w is a constant. Which ofthe following is true ? [2016] (a) Velocityand acceleration both are perpendicular to r r (b) Velocityand acceleration both are parallel to r r (c) Velocity is perpendicular to r r and acceleration is directed towards the origin (d) Velocity is perpendicular to r r and acceleration is directed awayfrom the origin 16. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is : [2016] (a) 0° (b) 90° (c) 45° (d) 180° 17. The x and ycoordinates of the particle at anytime are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is (a) 5 m/s2 (b) –4 m/s2 [2017] (c) –8 m/s2 (d) 0
- 94. 90 PHYSICS EXERCISE -1 1. (b) 2.(d) 3. (d) q = q - q = 2 2 2 2 2 2 1 tan g 2 / ) º 90 ( sin u g 2 / sin u H H 4. (d) 5. (a) 6. (a) In vector addition, the commutative law is obeyed i.e., A B B A r r r r + = + Vector subtraction does not follow commutative law. 7. (d) Since B . A 0 C . A r r r r = = , itmeans that A r isperpendicular to both B & C r r , hence A r is parallel to ) C B ( r r ´ or ) B C ( r r ´ . 8. (a) The unit vector of any vector A r is defined as | A | A Â r r = 9. (c) On earth, R = u2 sin 2q/g. On moon, g' = g/6 R' = u2 sin 2q/g' = 6u2 sin2q/g = 6R. 10. (c) B A 2 B A R cos 2 2 2 - - = q 0 B A 2 R R 2 2 = - = 2 / p = q 11. (c) ] cos AB 2 B A [ R 2 2 2 q + + = q + + = cos R 2 R R R 2 2 2 2 2 / 1 cos or cos R 2 R 2 2 - = q q = - or 3 / 2p = q 12. (c) The dot product should be zero. 13. (c) ) g / u ( 2 3 g º 60 sin u R 2 2 º 30 = = g / u g 90 sin u R 2 2 º 45 = = ) 2 / 3 ( g u g º 30 cos u g º 120 sin u R 2 2 2 º 60 = = = so 30º 60º 45º R R R = > or A C B R R R = > 14. (a) 2 1 1 mr F w = ; 2 2 2 mr F w = since period T is same, so w is same, because w p = 2 T . Hence ÷ ÷ ø ö ç ç è æ = 2 1 2 1 r r F F 15. (a) A parabola 16. (a) 2 2 1 u sin h 2g q = 2 2 2 u sin (90 ) h 2g -q = , 2 u sin 2 R g q = Range R is same for angle q and (90° – q) 2 2 2 2 1 2 u sin u sin (90 ) h h 2g 2g q - q = ´ 4 2 2 2 u (sin ) sin (90 ) 4g q ´ -q = [ sin(90 ) cos ] - q = q Q 4 2 2 2 u (sin ) cos 4g q ´ q = [ sin 2 2sin cos ] q = q q Q 4 2 2 u (sin cos ) 4g q q = 4 2 2 u (sin 2 ) 16g q = 2 2 2 2 (u sin 2 ) R 16 16g q = = or, R2 = 16 h1h2 or 1 2 R 4 h h = 17. (b) Ifthere isnoresistance, bomb will drop ata placeexactly below the flying aeroplane. But when we take into account air resistance, bomb will face deceleration in its velocity. So, it will fall on the earth exactly behind the aeroplane. 18. (b) 19. (c) From the resultant path of the particle, when it is projected at angle q with its velocity u is q - q = 2 2 2 cos u gx 2 1 tan x y y B D O q A C a n o n Where y denotes the instantaneous height of particle when it travels an instantaneous horizontal distance x. herex = D, u = vo so 2 2 2 0 1 gD y D tan 2 v cos = q - q 20. (d) Max. height = g 2 ) 90 ( sin v H 2 2 q - = .....(i) Time of flight, g ) 90 sin( v 2 T q - = ...(ii) Hints & Solutions
- 95. 91 Motion in aPlane Horizontal Vertical v q From (i), g H 2 g cos v = q ,From (ii), 2H 8H T = 2 g g = 21. (a) For A: It goes up with velocity u will it reaches its maximumheight (i.e. velocitybecomeszero) andcomes back to O and attains velocity u. Using as 2 u v 2 2 + = Þ gh 2 u v 2 A + = O vx u = vX u = A v B v C v u u h 2 2 c x y v v v = + For B, going down with velocity u Þ gh 2 u v 2 B + = For C, horizontal velocityremains same, i.e. u.Vertical velocity= gh 2 0 + = gh 2 The resultant C v = 2 y 2 x v v + = gh 2 u2 + . Hence C B A v v v = = 22. (c) Since horizontal component ofthevelocityof thebomb will be the same as the velocity of the aeroplane, therefore horizontal displacements remain the same at anyinstant of time. 23. (a) In uniform circular motion speed is constant. So, no tangential acceleration. It hasonlyradial acceleration R v a 2 R = [directedtowards center] anditsvelocityisalwaysin tangential direction. Sothese two areperpendicular toeach other. 24. (d) 2usin ( ) T gcos q - a = a 25 (d) Velocityand kinetic energyis minimum at the highest point. q = 2 2 cos v m 2 1 E . K EXERCISE - 2 1. (b) K.E. isminimum at the highestpoint. So,the horizontal distanceishalfoftherangeRi.e., 0.5R. 2. (b) g 30 sin u g 2 sin u R 2 2 1 = q = or 3 g u or g 2 u 5 . 1 2 2 = = km 3 g u g 90 sin u R 2 2 2 = = = 3. (b) The bullets are fired at the same initial speed º 30 sin º 60 sin º 30 sin u g 2 g 2 º 60 sin u H H 2 2 2 2 2 2 = ´ = ¢ 2 2 ( 3 / 2) 3 1 (1/ 2) = = 4. (b) 5. (a) 20 ) k̂ 4 j ˆ 3 ).( k̂ 5 î 2 ( F . F 2 1 = + + = r r 6. (d) g 2 sin u 2 g cos sin 2 u 2 2 2 q ´ = q q or tan q = 2 7. (a) 2 2 1 3 1 m(u cos ) m u 2 4 2 q = ´ or 4 3 cos 2 = q or º 30 cos 2 3 cos = = q . 8. (c) 9. (d) g u 6 or g u R 2 2 = = ; q - q = 2 2 2 cos u 2 x g tan x y or ÷ ø ö ç è æ - = 6 1 º 45 cos 2 4 º 45 tan 2 h 2 m 3 4 3 2 2 = - = 10. (c) We know that, g 2 ) sin u ( H y 2 m q = = g 2 sin u 2 2 q = . u u 2 H H D = D Given % 2 u u = D H 2 2 4% H D = ´ = 11. (c) As, 2 gt 2 1 t sin u s - q = so 2 t 10 2 1 t ) 2 / 3 ( 3 20 40 ´ ´ - ´ = or 5t2 – 30t + 40 = 0 or t2 – 6t + 8 = 0 or t = 2 or 4. Theminimum timet = 2s. 12. (a) 2 2 5 4 R + = N 41 = q R 5N 4N
- 96. 92 PHYSICS The angleq will begiven by tan 4 5 = q or 1 5 tan 4 - æ ö q= ç ÷ è ø 13. (c) Here sec 2 1 T = the required centripetal acceleration for moving in a circle is 2 2 2 2 C v (r ) a r r (2 / T) r r w = = = w = ´ p so 2 2 2 c a 0.25 (2 / 0.5) 16 .25 4.0 = ´ p = p ´ = p 14. (b) max R (10 6) 16N = + = , min R (10 6) 4N = - = Þ Values can be from 4N to 16N 15. (b) Time taken for vertical direction motion 2h 2 490 t 100 10s g 9.8 ´ = = = = The same time is for horizontal direction. 5 500 x vt 60 10 m 18 3 æ ö = = ´ ´ = ç ÷ è ø 16. (c) Even after 5 second, the horizontal velocity x v will be 20 1 s m - . The vertical velocity y v is given by 1 y s m 50 5 10 0 v - = ´ + = Now, 2 2 2 2 1 x y v v v 20 50 54 ms- = + = + » 17. (d) Given 2 2 2 u sin 2 u sin 3 g 2g q q = Þ ÷ ø ö ç è æ = q - 3 4 tan 1 18. (d) The horizontal velocity of the projectile remains constant throughout the journey. Since the body is projected horizontally, the initial velocity will be same as the horizontal velocity at any point. Since, x = 2t , dx 2 dt = Horizontal velocity= 2 m/s Initial velocity= 2 m/s 19. (b) Vector product of parallel vectors is zero. 20. (c) For circular angular motion, the formula for angular displacement q and angular acceleration a is 2 1 t t 2 q =w + a where w= initial velocity or 2 1 0 t 2 q = + a or 2 1 (2)(10) 2 q = ´ or q = 100 radian 2p radian are covered in 1 revolution 1 radian is covered in 1 2p revolution or 100 radian are covered in 100 2p revolution Number of revolution 50 16 3.14 = = 21. (b) L R p = ´ r r Where R = range 2 u sin 2 g q = The angle between R r and p r = q Also, p = m u. Hence, 2 u sin 2 L mu sin( ) g q = ´ q 3 2 2mu sin cos g q q = 22. (c) Let its angular velocity be w at all points (uniform motion). At the highest point weight of the body is balanced by centrifugal force, so 2 g m r mg r w = Þ w = 2 r 2 2 T 2 2 g 10 5 p p = = p = p = w . sec 3 1 . 2 14 . 3 2 = ´ = 23. (c) | a b | | a b | + = - Þ r r r r 2 2 | a b | | a b | + = - r r r r 2 2 2 2 | a | | b | 2a . b | a | | b | 2a . b Þ + + = + - r r r r 24. (a) 25. (d) R2 = P2 + Q2 – PQ cos q (40)2 = (10)2 + (20)2 – 2 × 10 × 20 × cosq 400cos q = 500 – 1600 = – 1100 4 11 400 1100 cos - = - = q which is not possible. In this way, the set of forces given in option (d) can not be represented both in magnitude and direction bythe sides of a triangle taken in the same order. Thus their resultant can not be zero. 26. (c) Here, | A B | 3 | A.B| ´ = r r r r ABsin 3ABcos Þ q = q tan 3 Þ q = 60 3 p Þ q = ° = 27. (a) See fig. ClearlyA is the resultant ofB and C. Further B is perpendicular to C B C=3 4 A 5 4 q cos q = 3/5 or q = cos–1 (3/5)
- 97. 93 Motion in aPlane 28. (c) A body is in translational equilibrium when the components of all external forces cancel. Forthesheet :Fcos q=4 N, Fsinq =3 N.Themagnitude of F is found byadding the squares ofthe components: F2cos2 q+ F2 sin2 q = F2 = 42 + 32 = 25 N2. Therefore F = 5 N. The F vector points in the proper direction, since tan q = 0.75 = 3/4. 29. (a) ˆ ˆ ˆ i j k v 1 2 2 0 4 3 = - - r v r ] 0 4 [ k̂ ] 3 0 [ j ˆ ] 8 6 [ î - + + + - = v r k̂ 4 j ˆ 3 î 2 + + - = | v | 29 units Þ = uu r . 30. (b) Use tan q + q = a cos P Q sin P Þ tan 90° ¥ = q + q = cos P Q sin P Q + P cos q = 0 Þ P cos q = -Q. R= q + + cos PQ 2 Q P 2 2 R = 2 2 2 Q 2 Q P - + or R = 2 2 Q P - = 12 144 = (P + Q)(P - Q) or P - Q = 144/18 = 8. P = 13 N and Q = 5 N. 31. (b) | A.B| | A B | = ´ r r r r ABcos ABsin Þ q = q 1 tanθ =1 cosθ = 2 Þ Þ Now, 2 2 | A B | A B 2ABcos + = + + q r r 2 2 1 A B 2AB. 2 = + + 2 2 A B 2AB = + + 32. (b) ˆ ˆ ˆ 3i bj 2k 3 6 2 ˆ ˆ ˆ Â i j k 7 7 7 9 36 4 + + + æ ö = = + + ç ÷ + + è ø . If a, b and g are angles made by A with coordinate axes, then 7 2 cos and 7 6 cos , 7 3 cos = g = b = a . 33. (b) Solve two equations : R2 = 9P2 + 4P2 + 12P2 cosq and 4R2 = 36P2 + 4P2 + 24P2 cosq. 34. (b) 35. (a) 2 2 2 2 2 x y (x y) (x y) 2(x y)(x y)cos æ ö + = + + - + + - q ç ÷ è ø 2 2 2 2 2 2 2 2 x y x y 2xy x y 2xy 2(x y )cos Þ + = + + + + - + - q or 2 2 2 2 2(x y ).cos (x y ) - q = - + 2 2 1 2 2 (x y ) cos 2(x y ) - é ù - + Þ q = ê ú - ê ú ë û 36. (c) Torque ˆ ˆ ˆ r F (i j) ( Fk) t = ´ = - ´ - u r r ur ˆ ˆ ˆ ˆ F[ i k j k] = - ´ + ´ ˆ ˆ ˆ ˆ F(j i) F(i j) = + = + ˆ ˆ ˆ ˆ ˆ ˆ Since k i j and j k i é ù ´ = ´ = ë û 37. (a) Component of a r along a . b b | b | = r r r r 38. (a) x u 30cos30º 30 3 /2, = = , uy = 30 sin 30°, y v 30sin30 gt = °- 0 5 . 1 10 º 30 sin 30 vy = ´ - = As vertical velocity = 0, Angle with horizontal a= 0º It is a state, when a particle reach to a highest point of its path. 39. (c) From conservation of linear momentum, 1 2 m m mv v v 2 2 = + r r r q 120 m/s y Explode = 50m/sec v1 v2 50 m/s (here we take particle & earth as a system so in this case external force is zero & linear momentum is conserved) Where v r is velocity of particle before explosion & 1 2 v ,v r r are velocity of its equal pieces. here ˆ v 60i = r (in x direction), 1 ˆ v 50j = r (in ydirection) so 2 ˆ ˆ v 120i 50 j = - r or 2 | v | 130m/sec = r & [From conservation of linear momentum] 120 50 tan - = q 40. (b) There is no change in horizontal velocity, hence no change in momentum in horizontal direction. The vertical velocity at t = 10sec is 10 ) 8 . 9 ( º 60 sin 98 v ´ - ´ = = –13.13 m/sec sochange in momentum in vertical direction is = (0.5 98 3/2) [ (0.5 13.13)] ´ ´ - - ´ =42.434+6.56 =48.997»49 41. (b) Maximum possible horizontal range = v2/g Maximum possible area of the circle 2 2 4 2 v v g g æ ö p = p = ç ÷ ç ÷ è ø 2 v Here r g é ù = ê ú ê ú ë û
- 98. 94 PHYSICS 42. (c)Here , ˆ ˆ F (3i 4j)N = + r ˆ ˆ d (3i 4j)m = + r W F.d (9 16)J 25J = = + = r r 43. (b) Let the gun be fired with velocityu from point O on the birdat B,makingan angleqwiththehorizontaldirection. Thereforethe height oftheaims oftheperson is at height BA (=h) above the bird. O B A h u q 100 m Here, horizontal range = 100 g 2 sin u2 = q or 2 2 500 sin2 100 or sin 2 10 100 10 1 sin14 250 (500) q = q ´ = = = ¢ or 2q = 14' or q = 7' = 180 60 7 p ´ radian As, arc angle radius = OB AB = q or AB = q× OB = 7 (100 100) cm 20.35 cm 60 180 p ´ ´ ´ = 44. (a) ˆ ˆ a b 3i 4k + = + r r Required unit vector = 2 2 ˆ ˆ a b 3i 4k | a b | 3 4 + + = + + r r r r ˆ ˆ 3i 4k 5 + = . 45. (d) Time offlight = g sin u 2 q = . sec 1 2 1 2 8 . 9 º 30 sin 8 . 9 2 = ´ = ´ ´ 46. (d) Theinitial velocityin theverticallydownward direction is zero and same height has to be covered. 47. (c) 2 2 2 1 gx y x tan 2 u cos = q - q ° ´ ´ ´ ´ ´ - ° = 60 cos 25 25 2 50 50 10 60 tan 50 y 2 =5m 48. (b) Comparing the given equation with 2 2 2 gx y x tan 2u cos = q - q , we get 3 tan = q 49. (c) x h 2gh uy = 0, sy = –h, ay = –g, ty = ? 2 1 s ut at 2 = + 2 1 h gt 2 - = - 2h t g Þ = velocity = x t 2h x 2gh 2h g = ´ = 50. (b) Horizontal distance covered should be same for the time of collision. 200 cos 400 = q or 2 1 cos = q or ° = q 60 51. (c) This happen when vertical velocity of both are same. 2 1 v v sin30 = ° or 2 1 v 1 v 2 = 52. (b) j ˆ 8 î 6 v + = 8 10 6 q Comparing with x y ˆ ˆ v v i v j = + uu r , we get 1 x s m 6 v - = and 1 y s m 8 v - = Also, 2 y 2 x 2 v v v + = = 36+ 64= 100 or 1 s m 10 v - = 10 6 cos and 10 8 sin = q = q g cos sin v 2 g 2 sin v R 2 2 q q = q = 8 6 1 R 2 10 10 10 10 10 = ´ ´ ´ ´ ´ =9.6m 53. (b) 2 gt 2 1 x = 1 1 1 9.8 9.8 m 2 2 2 8 = ´ ´ ´ =
- 99. 95 Motion in aPlane 54. (b) Comparing the given equation with the equation of trajectoryof a projectile, 2 2 2 gx y x tan 2u cos = q - q we get, 1 tan 30 3 q = Þ q = ° and 2 2 2 2 20 2u cos 20 u 2cos q = Þ = q 2 10 cos 30 = ° 2 10 3 2 = æ ö ç ÷ ç ÷ è ø 40 3 = Now, 2 max u 40 R g 3 10 = = ´ 4 m 3 = 55. (c) Thebullet performs a horizontal journeyof100cmwith constant velocityof 1500 m/s. The bullet also performs a vertical journey of h with zero initial velocity and downward acceleration g. For horizontal journey, time (t) Distance Velocity = 100 1 t sec 1500 15 = = …(1) The bullet performs vertical journeyfor this time. For vertical journey, 2 1 h ut gt 2 = + 2 1 1 h 0 10 2 15 æ ö = + ´ ´ç ÷ è ø or, 10 10 100 h m cm 2 15 15 2 15 15 ´ = = ´ ´ ´ ´ or, 20 h cm 2.2cm 9 = = The gun should be aimed 20 9 æ ö ç ÷ è ø cm above the target. 56. (b) 60° 60° A B v P = mv sin 60° A P = mvsin 60° B As the figure drawn above shows that at points A and B the vertical component of velocity is v sin 60° but their directions are opposite. Hence, change in momentum is given by: p mvsin 60 ( mvsin60 ) 2mvsin60 D = ° - - ° = ° 3 2mv 3mv 2 = = 57. (a) The motion of the train will affect only the horizontal component of the velocity of the ball. Since, vertical component is same for both observers, the ym will be same, but R will be different. 58. (c) 2 v a r = = 1 cm/s. Centripetal acceleration is directed towards the centre. Its magnitude = 1. Unit vector at the mid point on the path between P and Q is ˆ ˆ (x y)/ 2 - + . 59. (c) 60° 45° v cos 45° v v sin 45° u cos 60° u u sin 60° O B A Velocityof projectile u = 147 ms–1 angle of projection a = 60° Let, the time taken by the projectile from O to A be t where direction b = 45°. As horizontal component of velocityremains constant during the projectile motion. Þ v cos 45° = u cos 60° Þ v × 1 1 147 2 2 = ´ Þ 1 147 v ms 2 - = For Vertical motion, vy = uy – gt Þ vsin 45° = 45sin 60° – 9.8 t Þ 147 1 3 147 9.8t 2 2 2 ´ = ´ - Þ 9.8 t = 147 ( 3 1) 2 - Þ t = 5.49 s 60. (d) When a cyclist moves on a circular path, it experiences a centrifugal force which is equal to mv2 / r. It tries to overturn the cyclist in outward direction. If speed increases twice, the value of centrifugal force too increases to 4 times its earlier value. Therefore the chance of overturning is 1/4 times. 61. (c) Here v= 0.5 m/sec. u = ? so Þ = q v u sin 2 1 5 . u = or u = 0.25 ms–1 u v 120º 30º C B A direction of flow river
- 100. 96 PHYSICS 62. (d)j ˆ ) t sin a ( î ) t cos a ( r w + w = r dt d dt ) r ( d v = = r } j ˆ ) t sin a ( î ) t cos a {( w + w ĵ ) t cos a ( î ) t sin a ( w w + w w - = ] j ˆ ) t cos a ( î ) t sin a [( w + w - w = r.v 0 = r r velocity is perpendicular to the displacement. 63. (b) Since m E 2 v mv 2 1 E 2 = Þ = Now at highest point of flight, the vertical component of velocity is zero & onlyhorizontal component is non zero. SoK.E. at highest point is 2 1 E ' m(v cos 45º ) 2 = =E /2 64. (b) Given, u1 = u2 = u, q1 = 60º, q2 = 30º In Ist case, we know that range g ) 30 90 sin( u g 120 sin u g ) 60 ( 2 sin u R 2 2 2 1 ° + ° = ° = ° = 2 2 u (cos 30 ) 3u g 2g ° = = In IInd case when ° = q 30 2 , then g 2 3 u g 60 sin u R 2 2 2 = ° = Þ R1 = R2 (we get same value of ranges). 65. (a) ˆ ˆ F = 6ti + 4tj or x y 6t 4t a ,a 3 3 = = so 2 t u = a dt = t o x x ò (u ) = 9m/sec x t=3 Þ and 2 2t t u = a dt = y y o 3 ò sec / m 6 ) u ( 3 t y = Þ = (because ux & uy = 0 at t = 0 sec) 66. (b) 2 gt 2 1 s = , s and g are same for both the balls, so time of fall ‘t’ will also be the same for both of them (s is vertical height) 67. (c) B A sum vector P r r r + = = Q vector differences A B = = - r r r Since P r and Q r are perpendicular 0 Q . P = r r Þ (A B).(A B) 0 + - = r r r r Þ A A2 = B2 Þ B A = 68. (b) Circumferenceofcircleis 2pr = 40m Total distance travelled in tworevolution is 80m. Initial velocityu =0, final veloctiyv = 80 m/sec so from v2 =u2+2as Þ (80)2 =02+2×80×a Þ a = 40m/sec2 69. (b) q = ´ sin B A | B A | r r q = cos B A B . A r r B . A 3 | B A | r r r r = ´ Þ AB sin q = Ö3 AB cos q or tan q = Ö3 q = 60º 2 2 | A B| A B 2ABcos60º + = + + r r AB B A 2 2 + + = 70. (b) For two vectors to be perpendicular to each other A B ® ® × = 0 (2 3 8 i j k Ù Ù Ù + + )· (4 4 j i k Ù Ù Ù - +a )= 0 –8 + 12 + 8a = 0 or a = - = - 4 8 1 2 71. (d) 72. (c) 73. (d) 74. (d) 75. (a) EXERCISE - 3 Exemplar Questions 1. (b) Given, ˆ ˆ A i j = + ˆ ˆ B i j = - As we know that | || | cos A B A B × = q r r 2 2 2 2 ˆ ˆ ˆ ˆ ( ) ( ) ( 1 1 )( 1 1 )cos i j i j + × - = + + q ( )( ) 2 2 cos i j i j + - = ´ q where q is the angle between A and B 1 0 0 1 cos 0 2 2 - + - q = = q = 90° 2. (d) A scalar quantity does not depend on direction so it has the same value for observers with different orientations of the axes. 3. (b) From the diagram, ˆ ˆ u ai bj = + As u is in the first quadrant, so both components a and b will be positive. For ˆ ˆ, v pi qj = + as it is in positive x-direction and located downward so x-component p will be positive and y-component q will be negative. Hence, a, b and p are positive but q is negative. 4. (b) Let r makes an angle q with positive x-axis component of r along x-axis. | | cos x r r = q maximum maximum ( ) | | (cos ) x r r = q | | cos0 | | r r = ° = ( cosq Q is maximumofq=0°) q = 0°. Hence the vector r has maximum value along positive x-axis.
- 101. 97 Motion in aPlane 5. (c) Consider, projectile is fired at an angle q. According to question, q = 15° and R = 50 m R X Y u q Range, 2 sin 2 u R g q = 2 sin(2 15 ) 50m u R g ´ ° = = 2 2 1 50 sin30 2 g u u ´ = ° = ´ 50 × g × 2 = u2 2 50 9.8 2 100 9.8 980 u = ´ ´ = ´ = 980 u = = 31.304 m/s 14 5 = 2 ( 9.8 m/s ) = g Q Now, 45 ; q = ° 2 2 sin 2 45 u u R g g ´ ° = = 2 (14 5) 14 14 5 100 m 9.8 R g ´ ´ = = = 6. (b) As we know that, Impulse, D æ ö = D = D = D ç ÷ è ø D p I F t t p t where F is force, Dt is time duration and Dp is change in momentum.AsDp is a vector quantity, henceimpulse is also a vector quantity. Sometimes area can also be treated as vector. 7. (d) As speed is a scalar quantity, hence it will be related with path length (scalar quantity) only. Hence, Speed 0 total distance travelled time taken v = So, total distance travelled = Path length = (speed) × time taken Hence, path length which is scalar and traversed in equal intervals. 8. (c) As given that in two dimensional motion the instanteous speed v0 is positive constant and we know that acceleration is rate of change of velocity or instantaneous speed and hence it will also be in the plane of motion. NEET/AIPMT (2013-2017) Questions 9. (b) At point B the direction of velocity component of the projectile along Y - axis reverses. Hence, B V ® = 2i 3j - $ $ 10. (d) Vector triple product ( ) ( ) ( ) 0 A B C B A C C A B ´ ´ = × - × = r r r r r r r r r Þ || ( ) A B C ´ r r r [ 0 and 0] × = × = r r r r Q A B A C 1. (a) 2 2 ( ) ( ) A B C + = r r r 2 2 2 . A B A B C 2 Þ + + = r r 2 2 3 4 2 . 5 A B 2 Þ + + = r r 2 . 0 A B Þ = r r or . 0 A B Þ = r r A B ^ r r Here A2 + B2 = C2. Hence, A B ^ uu r uu r 11. (d) av v r = r (displacement) t (time taken) D D r = ˆ ˆ (13 2)i (14 3)j 5 0 - + - - = 11 ˆ ˆ (i j) 5 + 12. (a) ( ) A V 10 –i = ur $ ( ) B V 10 j = ur $ BA ˆ V 10 j 10 i 10 2 km / h = + = ur $ DistanceOB = 100 cos 45° = 50 2 km O S 100 km w 10 km/h A 45° 100 km N(j) $ BA V 10 2 km / h = B Time taken to reach the shortest distance between A and BA OB B V = uuuur 50 2 5h 10 2 = = 13. (b) Two vectors are A r = ˆ ˆ cos ti sin tj w + w B r = t t ˆ ˆ cos i sin j 2 2 w w +
- 102. 98 PHYSICS For twovectors A r and B r to be orthogonal A.B r r = 0 A.B r r = 0 = cos wt.cos t 2 w + sin wt.sin t 2 w = cos t t t cos 2 2 w w æ ö æ ö w - = ç ÷ ç ÷ è ø è ø So, t 2 2 w p = t = p w 14. (b) Here,x= 4sin(2pt) ...(i) y= 4cos(2pt) ...(ii) Squaring and adding equation (i) and (ii) x2 + y2 = 42 Þ R= 4 Motion of the particle is circular motion, acceleration vector is along – R ur and its magnitude = 2 V R Velocityof particle, V= wR= (2p)(4)= 8p 15. (c) Given: Position vector r r = cos wt x̂ + sin wt ŷ Velocity, v r = – wsin wt x̂ + wcos wt ŷ and acceleration, a r = –w2 cos wt x̂ – w2sin wt ŷ = – w2 r r r r . v r = 0 hence r v ^ r r and a r is directed towards the origin. 16. (b) A B A B + = - r r r r Squaring on both sides 2 2 A B A B + = - r r r r Þ A·A 2A ·B B·B + + r r r r r r = A·A – 2A ·B B·B + r r r r r r Þ 4A ·B 0 4AB cos 0 = Þ q = r r Þ cos q = 0 Þ q = 90° 17. (b) Given: x = 5t – 2t2 y= 10t vx = dx dt = 5 – 4t vy = dy 10 dt = ax = x dv dt = – 4 ay = y dv 0 dt = x y a a i a j = + r 2 a 4i m / s = - r Hence, acceleration ofparticleat (t = 2s) = –4m/s2
- 103. ARISTOTLE’S FALLACY According toAristotelianlawan external force is required to keep a body in motion. However an external force is required to overcome the frictional forces in case ofsolids and viscous forces in fluids which are always present in nature. LINEAR MOMENTUM(p) Linear momentum of a body is the quantity of motion contained in the body. Momentum p mv = r r It is a vector quantity having the same direction as the direction of the velocity. Its SI unit is kg ms–1. NEWTON’S LAWS OF MOTION First law : A body continues to be in a state of rest or of uniform motion, unless it is acted upon by some external force to change its state. Newton’sfirst lawgives thequalitativedefinition offorceaccording to which force is that external cause which tends to change or actually changes the state of rest or motion of a body. Newton’s first lawof motion is the same as law of inertia given by Galileo. Inertia is the inherent property of all bodies because of which they cannot change their state of rest or of uniform motion unless acted upon by an external force. Second law :The rate of change of momentum of a body isdirectly proportional to the external force applied on it and the change takes place in the direction of force applied. i.e., = = = dp mdv F ma dt dt r r r r This is the equation of motion of constant mass system. For variable mass system such as rocket propulsion ( ) = d mv F dt r r And, ( ) = m dv dm F v dt dt + r r r The SI unit of force is newton. (One newton force is that much force which produces an acceleration of 1ms–2 in a bodyof mass 1kg. The CGS unit of force is dyne. (1N = 105 dyne) The gravitational unit of force is kg-wt (kg-f) or g-wt (g-f) 1 kg-wt (kg-f) = 9.8 N, 1 g-wt (g-f) = 980dyne Third law : To every action there is an equal and opposite reaction. For example – walking , swimming , a horse pulling a cart etc. = – AB BA F F r r Action and reaction act on different bodies and hence cannot balance each other. Action and reaction occur simultaneously. Forces always occur in pairs. EQUILIBRIUM OF A PARTICLE A body is said to be in equilibrium when no net force acts on the body. i.e., = 0 F S r Then x y F 0, F 0 S = S = and z F 0 S = Stable equilibrium: Ifa bodyisslightlydisplaced from equilbrium position, it has the tendency to regain its original position, it is said tobe in stable equilibrium. In this case, P.E. is minimum. 2 2 d u ve dr æ ö = + ç ÷ ç ÷ è ø So, the centre of gravity is lowest. Unstable equilibrium: If a body, after being displaced from the equilibrium position, moves in the direction of displacement, it is said to be in unstable equilibrium. In this case, P.E. is maximum. 2 2 d u ve dr æ ö = - ç ÷ ç ÷ è ø So, the centre of gravity is highest. Neutral equilibrium : If a body, after being slightly displaced from the equilibrium position has no tendencytocome back or to movein the direction ofdisplacement the equilibrium is known to be neutral. In this case, P.E. is constant 2 2 d u constant dr æ ö = ç ÷ ç ÷ è ø The centre of gravity remains at constant height. COMMON FORCES IN MECHANICS 1. Weight : It is the force with which the earth attracts a body and is called force of gravity, For a body of mass m, where acceleration due to gravity is g, the weight W= mg 5 Laws of Motion
- 104. 100 PHYSICS 2. Tension:Theforce exerted bythe ends ofaloaded/stretched string (or chain) is called tension. The tension has a sense of pull at its ends. Case 1 T T Case 2 2T 2T T T m1 m2 m1g m2g T T Massless pulley Case 3 T T T T' T' T m a T T1 T – T = ma 1 If m = 0, T = T 1 i.e tension is same The tension in a string remains the same throughout the string if (a) string is massless, (b) pulley is massless or pulley is frictionless Case 4 : String having mass Let the total mass of the string be M and length be L. Then mass per unit length is L M Let x be the distanceof thestring from themass m. Then the mass of the shaded portion of string is ÷ ø ö ç è æ ´ x L M If the string is at rest then the tension T has to balance the wt of shaded portion of string and weight of mass m. g x L M m T ÷ ø ö ç è æ + = Þ as x increases, the tension increases. Thus tension is non- uniform in a string having mass. 3. Normal force : It measures how strongly one bodypresses the other body in contact. It acts normal to the surface of contact. Case 1 N N = mg mg Case 2 N a m N – mg = ma Þ N = m(g + a) mg Case 3 q q N = mg cos q N mg cos q mg mg sinq 4. Spring force : Ifan object is connected byspring and spring is stretched or compressed by a distance x, then restoring force on the object F = – kx where k is a spring contact on force constant. 5. Frictional force : It is a force which opposes relative motion between the surfaces in contact. f = mN This will be discussed in detail in later section. 6. Pseudoforce : If a bodyof mass m is placed in a non-inertial frame having aceleration a r , then it experiences a Pseudo force acting in a direction opposite to the direction of a r . pseudo F – ma = r r Negativesign shows that the pseudoforce is always directed in a direction opposite tothe direction of the acceleration of theframe. x y z a Fpseudo m CONSTRAINT MOTION : When the motion of one body is dependent on the other body, the relationship of displacements, velocities and accelerations of the two bodies are called constraint relationships. Case 1 Pulley string system : X x F Block Step 1 : Findthe distance ofthetwobodies from fixedpoints. Step 2 : The length of the string remain constant. (We use of this condition) Therefore X + (X – x) = constant Þ 2X – x = constant dt dx dt dX 2 0 dt dx – dt dX 2 = Þ = Þ p B p dX 2V v V velocity of pulley dt é Þ = = = ê ë Q
- 105. 101 Lawsof Motion B dx v velocityof block dt ù = = ú û Again differentiating we get, 2ap = aB B p B dv dVp a and a dt dt é ù = = ê ú ë û ap = acceleration of pulley, aB = acceleration of block Case 2 Here = + + y x h 2 2 constt. On differentiating w.r.t ‘t’ F x y h q 1 2 [Negative sign with dy/dt shows that with increase in time, y decreases] 2 2 1 2x dx dy 0 dt dt 2 h x ´ - = + Þ cos q (v1 – v2) = 0 2 2 x cos h x é ù q = ê ú ê ú + ë û Q Case 3 Wedge block system : Thin lines represents thecondition of wedge block at t = 0 and dotted lines at t = t ax c Ax Ax A q q B ay ax ay Ax = acceleration of wedge towards left ax, ay = acceleration of block as shown From DABC, y x x a tan a A q = + Frame of Reference : Reference frames are co-ordinate systems in which an event is described. There are two typesof reference frames (a) Inertial frame of reference: These are frames of reference in which Newton’s laws hold good. These frames are at rest with each other or which are moving with uniform speed with respect to each other. All reference frames present on surface of Earth are supposed to be inertial frame of reference. (b) Non – inertial frame of reference: Newton’s law do not hold good in non-inertial reference frame. All accelerated and rotatory reference frames are non – inertial frameof reference. Earth isa non-intertial frame. When the observer is in non-inertial reference framea pseudo force is applied on the body under observation. Free Body Diagram (FBD) : Free body diagram of a mass is a separate diagram of that mass. All forces acting on the mass are sketched. A FBD is drawn to visualise the direct forces acting on a body. Case 1 : Masses M1 and M2 are tied to a string, which goes over a frictionless pulley (a) If M2 > M1 and they move with accelerationa 1 M M2 M g 2 M g 1 T T a a FBD of M1, FBD of M2 T M1 1 M g a T M2 M g 2 a 1 1 T M g M a - = 2 2 M g T M a - = where T is the tension in the string. It gives 2 1 1 2 M M a g M M - = + and 1 2 1 2 2M M T g M M = + (b) If the pulley begins to move with acceleration f, downwards 2 1 1 2 ( ) M M a g f M M - = - + uu r uu r uu r and 1 2 1 2 2 ( ) M M T g f M M = - + ur uu r uu r Case 2 : Three massesM1, M2 and M3 are connected with strings as shown in the figure and lie on a frictionless surface. They are pulled with a force F attached to M1. M3 M2 M1 F T 1 T1 T2 T2 The forces on M2 and M3 are as follows 2 3 1 1 2 3 M M T F M M M + = + + and 3 2 1 2 3 M T F M M M = + + ; Acceleration of the system is 1 2 3 F a M M M = + + Case 3 : Two blocks of masses M1 and M2 are suspended vertically from a rigid support with the help of strings as shown in the figure. The mass M2 is pulled down with a force F.
- 106. 102 PHYSICS M1 T1 T1 T2 T2 M2 M2g M1g F The tensionbetween the masses M1 and M2 will be T2 = F + M2g Tension between the support and the mass M1 will be T1 = F + (M1 + M2)g Case 4 : Two masses M1 and M2 are attached to a string which passes over a pulley attached to the edge of a horizontal table. The mass M1 lies on the frictionless surface of the table. 1 M 2 M T a T M g 2 Let the tension in the string be T and the acceleration of the system be a. Then T = M1a ...(1) M2g – T = M2a ...(2) Adding eqns. (1) and (2), we get 2 1 2 M a g M M é ù = ê ú + ë û and 1 2 1 2 M M T g M M é ù = ê ú + ë û Case 5 : Two masses M1 and M2 are attached to the ends of a string, which passes over a frictionless pulley at the top of the inclined plane of inclination q. Let the tension in the string be T. q M1 M g 1 M g cos 1 q M g sin 1 q q N M2 M2g (i) WhenthemassM1 movesupwardswithaccelerationa. From the FBD of M1 and M2, T – M1g sin q = M1a ...(1) M2g – T = M2a ...(2) Solving eqns. (1) and (2) we get, 2 1 1 2 sin M M a g M M é ù - q = ê ú + ë û FBD of mass M1 R=N T M g cos 1 q M g sin 1 q M g 1 x y 2 1 1 2 (1+sin ) M M g T M M é ù = ê ú + q ë û T M g 2 a FBD of M2 (ii) When the mass M1 moves downwards with acceleration a. Equation of motion for M1 and M2, M1g sin q – T = M1a ...(1) T – M2g = M2a ...(2) Solving eqns. (1) and (2) we get, 1 2 1 2 sin ; M M a g M M é ù q- = ê ú + ë û 2 1 1 2 (1 sin ) M M g T M M é ù = ê ú + + q ë û (a) If (M2/M1 = sinq) then the system does not accelerate. (b) Changing position of masses, does not affect the tension. Also, the acceleration of the system remains unchanged. (c) If M1 = M2 = M (say), then 2 cos sin ; 2 2 2 g a q q æ ö æ ö = - ç ÷ ç ÷ è ø è ø 2 cos sin 2 2 2 Mg T q q æ ö æ ö = + ç ÷ ç ÷ è ø è ø Case 6 : Two masses M1 and M2 are attached to the ends of a string over a pulley attached to the top of a double inclined plane of angle of inclination a and b. Let M2 move downwards with acceleration a and the tension in the string be T then a b 2 M 1 M FBD of M1 a T M g 1 M gcos 1 a a M gsin 1 a M1 Equation of motion for M1 T – M1g sin a = M1a or T = M1g sin a + M1a ...(1)
- 107. 103 Lawsof Motion FBD ofM2 a T M g 2 M gcos 2 b M gsin 2 b M 2 b Equation of motion for M2 M2g sinb – T = M2a or T = M2g sin b – M2a ...(2) Using eqn. (1) and (2) we get, M1g sin a + M1a = M2g sin b – M2a Solving we get, ( ) 2 1 1 2 sin sin M M g a M M b - a = + and 1 2 1 2 [sin sin ] M M g T M M = b + a + Case 7 :Aperson/monkey climbing a rope T Mg a a (a) A person of mass M climbs up a rope with acceleration a. The tension in the rope will be M(g+a). T – Mg = Ma Þ T = M(g + a) (b) If the person climbs down along the rope with acceleration a, thetension in the rope will be M(g–a). a a Mg T Mg – T = Ma Þ T = M(g – a) (c) When the person climbs up or down with uniform speed, tension in the string will be Mg. Case 8 : Abodystarting from rest moves along a smooth inclined plane of length l, height h and having angle of inclination q. h l q FBD of body N=R mg mg sinq mg cosq q (where N=R is normal reaction applied by plane on the body of mass m) For downward motion, along the inclined plane, q = Þ = q sin g a ma sin mg Bywork-energytheorem loss in P.E. = gain in K.E. gh 2 v mv 2 1 mgh 2 = Þ = Þ Also, from the figure, h = l sin q. q = = sin g 2 gh 2 v l (a) Acceleration down the plane is g sin q. (b) Its velocity at the bottom of the inclined plane will be 2 2 sin gh g = q l (c) Time taken to reach the bottom will be 1/ 2 1/ 2 2 1/ 2 2 2 1 1 2 sin sin sin sin 2 h h t g g g g h æ ö æ ö = = = = ç ÷ ç ÷ ç ÷ q q q è ø æ ö è ø qç ÷ è ø l (d) Ifangles ofinclination are q1 and q2 for twoinclined planes Keeping the length constant then ½ 1 2 2 1 sin sin t t æ ö q = ç ÷ q è ø Case 9 : Weight of a man in a lift : (i) When lift is acceleratedupward :In thiscase theman also movesin upwarddirection with an acceleration a r . N a a mg Then from Newton’ second law N – mg = ma or N = m(g + a) or Wapp = m(g + a) (1 / ) o W a g = + (as W = mg) Where Wapp is apparent weight ofthe man in the lift, Wo is thereal weight, N isthe reaction oflift on theman. It is clear that N = Wapp When the lift moves upward and if we measure the weight of the man byany means (such as spring balance) then we observe more weight (i.e., Wapp) than the real weight (Wo) Wapp >Wo (ii) When lift is accelerated downward : In this case from Newton’s second law N a mg mg – N = ma or N = m(g – a) = Wo(1– a/g) or W'app= Wo(1– a/g) { } mg Wo = Q If we measure the weight of man by spring balance, we observe deficiency because Wapp< Wo.
- 108. 104 PHYSICS (iii) Whenliftis at rest or movingwithconstantvelocity :From Newton’s second law N –mg = 0 or N = mg In this case spring balance gives the true weight of the man. Case 10 : Three masses M1, M2 and M3 are placed on a smooth surface in contact with each other as shown in the figure. A force F pushes them as shown in the figure and the three masses move with acceleration a, M3 M2 F2 F1 F2 F1 M1 F a M1 F F1 Þ F – F1 = m1a ...(i) M2 F2 F1 Þ F1 – F2 = m2a ...(ii) F2 M3 Þ F2 = M3 a ...(iii) Adding eqns. (i), (ii) and (iii) we get, 1 2 3 F a M M M = + + Þ 3 2 1 2 3 M F F M M M = + + and 2 3 1 1 2 3 ( ) M M F F M M M + = + + Keep in Memory 1. When a man jumps with load on his head, the apparent weight of the load and the man is zero. 2. (i) If a person sitting in a train moving with uniform velocitythrows a coin verticallyup, then coin will fall back in his hand. (ii) If the train is uniformlyaccelerated, the coin will fall behind him. (iii) Ifthe train is retarded uniformly, then the coin will fall in front ofhim. Example1. A chain of length l is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere. What will be the acceleration a of each element of the chain when its upper end is released? It is assumed that the length of chain R π 2 æ ö <ç ÷ è ø l . Solution : Let m be the mass of the chain of length l. Consider an element of length dlof thechain at an angle q with vertical, R q dl dq From figure, dl = R d q ; Mass of the element, dm = l l d m ; or dm = q d R . m l Force responsible for acceleration, dF = (dm)g sinq ; dF = q q = q ÷ ø ö ç è æ q d sin mgR ) sin g ( d R m l l Net force on the chain can be obtained by integrating the above relation between 0 to a, we have ò a a a - = q - = q q = 0 0 ] cos 1 [ R mg ) cos ( R mg d sin R mg F l l l ú û ù ê ë é - = R cos 1 R mg l l ; Acceleration, F gR a 1 cos m R æ ö = = - ç ÷ è ø l l . Example2. A block slides down a smooth inclined plane to the ground when released at the top, in time t second. Another block is dropped vertically from the same point, in the absence of the inclined plane and reaches the ground in t/2 second. Then find the angle of inclination of the plane with the vertical. Solution : If q is the angle which the inclined plane makes with the vertical direction, then the acceleration of the block sliding down the plane of length l will be g cosq. A C B q l h Using the formula, 2 at 2 1 ut s + = , we have s = l, u = 0, t = t and a = g cos q. so 2 2 t ) cos g ( 2 1 t cos g 2 1 t 0 q = q + ´ = l ...(i) Taking vertical downward motion of the block, we get 4 / gt 2 1 ) 2 / t ( g 2 1 0 h 2 2 = + = ...(ii) Dividing eqn. (ii) by (i), we get q = cos 4 1 h l ] / h cos [ l Q = q or q = q cos 4 1 cos ; or 4 1 cos2 = q ; or 2 1 cos = q or q = 60º
- 109. 105 Lawsof Motion Example3. A largemass M and a small mass m hang at the two ends of a string that passes through a smooth tube as shown in fig. The mass m moves around a circular path in a horizontal plane. The length of the string from mass m to the top of the tube is l, and q is the angle the string makes with the vertical. What should be the frequency (n) of rotation of mass m so that mass M remains stationary? M m l r q T Solution : Tension in the string T = Mg. Centripetal force on the body= mrw2 =mr ( 2pn )2. This is provided by the component of tension acting horizontally i.e. T sinq ( = Mg sinq). mr ( 2pn)2 = Mg sinq = Mgr/l. or 1 Mg 2 m n = p l Example4. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in fig. The force on the pulley by the clamp is given by (a) 2 Mg (b) 2 mg M m (c) 2 2 [ (M m) m ] + + g (d) 2 2 [ (M m) M ] + + g Solution : (c) Force on the pulley by the clamp = resultant of T = (M + m)g and mg acting along horizontal and vertical respectively F 2 2 [(M m)g] (mg) = + + 2 2 [ (M m) m ]g = + + Example5. The masses of 10 kg and 20 kg respectively are connected by a massless spring infig. A force of 200 newton actson the 20 kg mass. At the instant shown, the 10 kg mass has acceleration 12 m/sec2. What is the acceleration of 20 kg mass? 10 kg 20 kg 200 newton Solution : Force on 10 kg mass = 10 × 12 = 120 N Themassof10kgwill pullthemassof20kgin thebackward direction with a force of 120 N. Net force on mass 20 kg = 200 – 120 = 80 N Its acceleration force a mass = 2 s / m 4 kg 20 N 80 = = Example6. Two masses each equal to m are lying on X-axis at (–a, 0) and (+ a, 0) respectively as shownin fig. They are connected by a light string. A force F is applied at the origin and along the Y-axis. As a result, the masses move towards each other. What is the acceleration of each mass? Assume the instantaneous position of the masses as (– x, 0) and (x, 0) respectively m O m (–a, 0) (a, 0) –X X F Solution : O (–x, 0) (x, 0) F T T q B C A From figure F = 2 T cos q or T = F/(2 cos q) The force responsible for motion of masses on X-axis is T sin q q ´ q = q = sin cos 2 F sin T a m ) x a ( x 2 F OA OB 2 F tan 2 F 2 2 - ´ = ´ = q = so, ) x a ( x m 2 F a 2 2 - ´ = Example7. A block of mass M is pulled along horizontal frictionless surface by a rope of mass m. Force P is applied at one end of rope. Find the force which the rope exerts on the block. Solution : The situation is shown in fig M T O m P Let a be the common acceleration of the system. Here T = M a for block P – T = m a for rope P – M a = m a or P = a (M + m) or ) m M ( P a + = ) m M ( P M T + =
- 110. 106 PHYSICS Example8. In thesystem shown below, friction and mass of the pulley are negligible. Find the acceleration of m2 if m1 = 300 g, m2 = 500 g and F = 1.50 N Solution : When thepulleymovesa distance d,m1 will move a distance 2d. Hence m1 will have twice as large an acceleration as m2 has. For mass m1, T1 = m1 (2a) ...(1) For mass m2, F – T2 = m2(a) ...(2) Putting 2 1 T T 2 = in eqn. (1) gives T2 = 4m1a Substituting value of T2 in equation (2), F= 4m1a + m2a = (4m1 + m2)a Hence 2 1 2 F 1.50 a 0.88 m / s 4m m 4(0.3) 0.5 = = = + + LAW OF CONSERVATION OF LINEAR MOMENTUM A system is said to be isolated, when no external force acts on it. For such isolated system, the linear momentum ( ) P mv = r r is constant i.e., conserved. The linear momentum is defined as v m P r r = .....(1) where v r isthe velocityofthe body, whosemass is m.Thedirection of P r is same as the direction of the velocity of the body. It is a vector quantity. From Newton’s second law, P dt d ) v m ( dt d F . ext r r r = = .....(2) i.e., time rate ofchange in momentum ofthe bodyis equal tototal external force applied on the body. If 0 ) P ( dt d 0 F . ext = Þ = r r or P r = constant .....(3) This is called lawofconservation of momentum. Now let us consider a rigid body consisting of a large number of particles moving with different velocities, then total linear momentumof therigidbodyis equal tothesummation ofindividual linear momentum of allparticles i.e., n i 1 2 3 n i 1 p p p p ..........p = å = + + + r r r r r or 1 2 3 1 .......... n total i n i P p p p p p = = å = + + + + r r r r r r where n 2 1 p ..... .......... p , p r r r are individual linear momentum offirst, second and nth particle respectively. If this rigid bodyis isolated i.e., no external force is applied on it, then = total P r constant (from Newton’s second law). Further weknow that internal forces(such as intermolecular forces etc.) alsoact inside the body, but these can onlychange individual linear momentum oftheparticles (i.e., p1, p2.........), but their total momentum total P r remains constant. Gun Firing a Bullet If a gun of mass M fires a bullet of mass m with velocity v. Then from law of conservation of momentum, as initiallybullet & gun areat rest position i.e., initial momentum iszero, so final momentum (gun + bullet) must also be zero. Since on firing, the bullet moves with velocity b v r in forward direction, then from Newton’sthird law, thegun moves in backward direction g v r . So, Initial momentum = finalmomentum b g Momentum Momentum of bullet of gun 0 mv MV = + r r b g mv V M - = uuu r uuu r (–ve sign shows that the vel. of gun will have the opposite direction to that of bullet) IMPULSE According to Newton’s second law the rate of change of momentum of a particle is equal tothe total external force applied on it (particle) i.e., ext F dt P d r r = ...(i) or dt . F P d ext r r = or f i t f i ext t P P P F .dt D = - = ò r r r r ...(ii) Where i P r is momentum of the particle at initial time ti and when we apply some external force ext F r its final momentum is f P r at time tf . The quantity ext F dt × r on R.H.S in equation (ii) is called the impulse. We can write equation (ii) as f i t ext t I F .dt P = = D ò r r ...(iii) So, the impulse of the force ext F r is equal to the change in momentum of the particle. It is known as impulse momentum theorem. Area= impulse t tf ti Fext. (a)
- 111. 107 Lawsof Motion Example10. A hammerof mass M strikes a nail of mass m with velocity of u m/s and drives it ‘s’ meters in to fixed block of wood. Find the average resistance of wood to the penetration of nail. Solution : Applying the law of conservation of momentum, m u = (M + m) v0 Þ u M m M v0 ÷ ø ö ç è æ + = There acceleration a can be obtained using the formula (v2 = u2 + 2as). Here we have 0 – v0 2 = 2as or a = v0 2 /2s s 2 u M m M a 2 2 ÷ ø ö ç è æ + = Resistance = (M + m) a 2 2 M u m M 2 s æ ö = ç ÷ ç ÷ + è ø Example11. A ball of mass 0.5 kg is thrown towards a wall so that it strikes the wall normallywith a speed of 10 ms–1. If the ball bounces at right angles away from the wall with a speed of 8ms–1, what impulse does the wall exert on the ball ? Solution : f 10 8 Approaching wall u = –10 ms–1 Leaving wall v = +8 ms–1 Taking the direction of the impulse J as positive and using J = mv – mu we have 1 1 J 8 ( 10) 9 2 2 = ´ - - = N-s Therefore the wall exerts an impulse of9 N-s on the ball. Example12. Two particles, each of mass m, collide head on when their speeds are 2u and u. If they stick together on impact, find their combined speed in terms of u. Solution : 2u Before impact After impact m u v 2m m Using conservation of linear momentum (in the direction of the velocity 2u) we have (m) (2u) – mu = 2m × V Þ 1 V u 2 = The combined mass will travel at speed u/2. (Note that the momentumofthe second particlebefore impact is negative because its sense is opposite to that specified as positive.) Force vary with time and impulse is area under force versus time curve Area=Fext.Dt t tf ti Fext. Fav Fext. (b) . Force constant with time i.e., . ext F r constant with time (shown by horizontal line) and it would give same impulse to particle in time Dt = tf – ti as time varying force described. It is a vector quantityhaving a magnitude equal to the area under theforce-timecurve asshown in fig. (a). In thisfigure, it isassumed that force varieswith time and is non-zeroin time interval Dt = tf– ti. Fig.(b) shows the time averaged force . ext F r i.e., it is constant in time interval Dt, then equation (iii) can be written as f i t ext. t I F dt = ò r ) t t ( F i f . ext - = r t F I . ext D = r ...(iv) The direction of impulsive vector I is same as the direction of change in momentum. Impulse I has same dimensions as that of momentum i.e, [MLT–1] Rocket propulsion (A case of system of variable mass ) : It is based on principleofconservation of linearmomentum. In rocket, the fuel burns and produces gases at high temperature. These gases are ejected out of the rocket from nozzle at the backside of rocket and the ejecting gas exerts a forward force on the rocket which accelerates it. Let the gas ejects at a rate dt dM r - = and at constant velocity u w.r.t. rocket then from the conservation of linear momentum rt M ru M ru dt dv 0 - = = whereM = M0 - rt and M0 is mass of rocket with fuel and solving this equation, we get ÷ ÷ ø ö ç ç è æ - = rt M M log u v 0 0 e where v = velocity of rocket w.r.t. ground. Example9. Two skaters A and B approach each other at right angles. Skater A has a mass 30 kg and velocity 1 m/s and skater B has a mass 20 kg and velocity 2 m/s. They meet and cling together. Find the final velocity of the couple. Solution : Applying principle of conservation of linear momentum, 2 2 2 2 1 1 2 1 2 2 2 1 ) v m ( ) v m ( v ) m m ( ; p p p + = + + = ( ) ( ) ( ) 2 2 30 20 v 30 1 20 2 50 + = ´ + ´ = 50 v 1 m / s 50 = =
- 112. 108 PHYSICS FRICTION When abody is in motion on a rough surface, or when an object moves through water (i.e., viscous medium), then velocityof the body decreases constantly even if no external force is applied on the body. This is due to friction. So “an opposing force which comes into existence, when two surfaces are in contact with each other and try to move relative to one another, is called friction”. Frictional force acts along the common surface between the two bodies in such a direction so as to oppose the relative movement of the two bodies. (a) The force of static friction fs between book and rough surface is opposite to the applied external force Fext. The force of static friction fs = ext F r . Book R=N W (a) fs Fext. (b) When ext F r . exceeds the certain maximum value of static friction, the book starts accelerating and during motion Kinetic frictional force is present. Book R=N W (b) fk Fext. Body just starts moving (c) A graph ext F r . versus | f | shown in figure. It is clear that fs, ,max > fk =m N s fk=m N k Body starts with acceleration Body is at rest static region kinetic region O |f| (c) (f ) s max Fig.(a) shows a book on a horizontal rough surface. Now if we apply external force . ext F r , on the book, then the book will remain stationaryif ext. F r is not too large. Ifwe increase . ext F r then frictional force f also increase up to s max (f ) (called maximum force of static friction or limiting friction) and s max (f ) = msN. At any instant when . ext F r is slightly greater than s max (f ) then the book moves and accelerates to theright. Fig.(b) when the book is in motion, the retarding frictional force become less than, s max (f ) Fig.(c) s max (f ) is equal tomkN. When the book is in motion, we call the retarding frictional force as the force of kinetic friction fk. Since fk< s max (f ) , soit isclear that,werequire moreforceto start motion than to maintain it against friction. By experiment one can find that s max (f ) and fk are proportional to normal force N acting on the book (byrough surface) and depends on the roughness of the two surfaces in contact. Note : (i) The force of static friction between any two surfaces in contact is opposite to . ext F r and given by N f s s m £ and s max s (f ) N = m (when the body just moves in the right direction). where N = W = weight of book and ms is called coefficient of static friction, fs is called force of static friction and s max (f ) is called limiting friction or maximum value of static friction. (ii) The force of kinetic friction is opposite to the direction of motion and is given by fk = mkN where mk is coefficient of kinetic friction. (iii) The value of mk and ms depends on the nature of surfaces and mk is always less then ms. Friction on an inclined plane : Now we consider a book on an inclined plane & it just moves or slips, then by definition R=N mg cos q mg=W q mgsinq q Book (fs max ) max ( ) = m s s f R Now from figure, q = sin mg f max , s and R = mg cosq Þ ms= tanq or q = tan–1(ms) where angle q is called the angle of friction or angle of repose Some facts about friction : (1) The force of kinetic friction is less than the force of static friction and the force of rolling friction is less than force of kinetic friction i.e., fr < fk < fs or mrolling < mkinetic < mstatic hence it is easyto roll the drum in comparison to sliding it. (2) Frictional force does not oppose the motion in all cases, infact in some cases the body moves due to it. A B Fext In the figure, book B moves to the right due to friction betweenAand B. IfbookAis totallysmooth (i.e., frictionless) then book B does not move to the right. This is because of no force applies on the book B in the right direction.
- 113. 109 Lawsof Motion Laws oflimiting friction : (i) The force of friction is independent of area of surfaces in contact and relative velocity between them (if it is not too high). (ii) The force of friction depends on the nature of material of surfaces in contact (i.e., force of adhesion). m depends upon nature of the surface. It is independent of the normal reaction. (iii) The force of friction is directlyproportional tonormal reaction i.e., F µ N or F = mn. While solving a problem having friction involved, follow the given methodology Check (a) Fapp (b) Limiting friction (f) l If F < f app app l Body does not move and F = frictional force If F = f Body is on the verge of movement if the body is initially at rest Body moves with constant velocity app l Rolling Friction : Thename rollingfriction isa misnomer.Rollingfriction hasnothing todo with rolling. Rolling friction occurs during rolling as well as sliding operation. Cause of rolling friction : When a body is kept on a surface of another bodyit causes a depression (an exaggerated view shown in the figure). When the body moves, it has to overcome the depression. This is the cause of rolling friction. Rolling friction will be zero onlywhen both the bodies incontact are rigid. Rolling friction is verysmall as compared to sliding friction. Work done byrolling friction is zero CONSERVATIVE AND NON-CONSERVATIVE FORCES If work done on a particle is zero in complete round trip, the force is said to be conservative. The gravitational force, electrostatics force, elastic force etc., are conservative forces. On the other hand if the work done on a body is not zero during a complete round trip, the force is said to be non-conservative. The frictional force, viscous force etc. are non-conservative forces. A B C f i Initial position Final position Figure shows three processes A, B and C by which we can reach from an initial position to final position. If force is conservative, then work done is same in all the three processes i.e., independent of the path followed between initial and final position. If force is non conservative then work done from i to f is different in all three pathsA,B and C. Hence it is clear that work done in conservative force depends only on initial & final position irrespective of the path followed between initial & final position. In case of non-conservative forces the work done depends on the path followed between initial and final position. We can say also that there is no change in kinetic energy of the body in complete round trip in case of conservative force. While in case of non conservative forces, when a body return to its initial position after completing the round trip, the kinetic energy of the body may be more or less than the kinetic energy with which it starts. Example13. Pushing force making an angle q to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is f, then determine the magnitude of force required to move the body. Solution : The various forces acting on the block are shown in fig. N F f mg f cosq q F sinq Here, f tan N m = f = ; or f= N tanf...(i) The condition for the block just to move is Fcosq = f = N tanf ...(ii) and F sinq + W = N ...(iii) From(ii) and(iii), F cosq = (W + F sinq ) tan f = W tanf+ F sin q tanf; or F cos q – F sinq sinf/cosf = W sinf/cosf or F (cosq cosf – sinq sinf) = W sinf; or F cos (q + f) = W sinf or F = W sinf / cos (q + f) Example14. An object of weight W is resting on an inclined plane at an angle q to the horizontal. The coefficient of static friction is m. Find the horizontal force needed to just push the object up the plane.
- 114. 110 PHYSICS Solution : Thesituation is shown in fig. F sinq F cosq W cosq W sinq F R q W q q f= R m Let F be the horizontal force needed to just push the object up the plane. From figure R = W cos q + F sin q Now f = mR = m [W cos q + F sin q] ...(1) Further, F cos q = W sin q + f ...(2) F cos q = W sin q + m [W cos q + F sin q] F cos q – m F sin q = W sin q + m W cos q ) sin (cos ) cos (sin W F q m - q q m + q = CASES OF CIRCULAR MOTIONS Motion in a Vertical Circle : Let us consider a particle of mass m attached toa string oflength R let the particle be rotated about its centre O. At t = 0 the particle start with velocity u from the point A(lowest point of vertical circle) and at time t its position is P. Then the tension at point P is given by T P vP q q O B R A u mg mg mg cos q sin q R mv cos mg T 2 P P = q - or R mv cos mg T 2 P P + q = ...(1) So tension at point A(lowest point of vertical circle) is R mv mg T 2 A A = - (Q q = 0º) ...(2) and tension at point B (highest point of vertical circle) is R mv mg T 2 B B = + (Q q=180º) ...(3) Where 2 mv r is centripetal force required for the particletomove in a vertical circle. Now from law of conservation of energy mgR 2 mv 2 1 mv 2 1 2 B 2 A = - or, 2 2 A B v v 4gR - = ...(4) (change in kinetic energyof particle) = (change in potential energy of particle) or (loss in kinetic energyofthe particle) = (gain in potential energy) In conservative force system (such as gravity force) the mechanical energy(i.e., kinetic energy+ potential energy) must be constant. Total energy will be constant Nowfrom eqns.(2) and (3), we get mg R TB vB B A D mg R TA v = v A C B A D ) V V ( R m mg 2 T T 2 B 2 A B A - + = - ) gR 4 ( R m mg 2 + = Þ mg 6 T T B A = - ...(5) or TA =TB + 6mg ...(6) So it is clear from eqn. (6) that tension in string at lowest point of vertical circle is greater then the tension at highest point of vertical circle by 6mg. Condition to complete a vertical circle : Ifwe reducethevelocityvA in equation (2), then TA will be reduce and at some critical velocity vc, TB will be zero, then put TB = 0 and vB = vC in equation (3) and we obtain C B v v gR = = ...(7) In this condition the necessary centripetal force at point B is provided by the weight of the particle [see again equation (3)] then from equation (4), we get 2 4 5 A A v gR gR v gR - = Þ = ...(8) then the tension at the point A will be mg 6 R ) gR 5 ( m mg TA = + = ...(9) Hence if we rotate a particle in a vertical circle and tension in string at highest point is zero, then the tension at lowest point of vertical circle is 6 times of the weight of the particle. Some Facts of Vertical Motion : (i) The body will complete the vertical circle if its velocity at lowest point is equal to or greater then gR 5 (ii) The bodywill oscillate about the lowest point if itsvelocity at lowest point is less then gR 2 . This will happen when the velocity at the halfwaymark, i.e. ú û ù ê ë é = = mgR mv 2 1 0 v 2 A D Q (iii) The string become slack and fails to describe the circle when its velocity at lowest point lies between gR 5 to gR 2
- 115. 111 Lawsof Motion Example15. A massm is revolving in a vertical circle at the end of a string of length 20 cm. By how much does the tension of the string at the lowest point exceed the tension at the topmost point? Solution : The tension T1 at the topmost point is given by, g m 20 v m T 2 1 1 - = Centrifugal force acting outward while weight acting downward The tension T2 at the lowest point, g m 20 v m T 2 2 2 + = Centrifugal force and weight (both) acting downward g m 2 20 v m v m T T 2 1 2 2 1 2 + - = - ; h g 2 v v 2 2 2 1 - = or g 80 ) 40 ( g 2 v v 2 1 2 2 = = - g m 6 g m 2 20 g m 80 T T 1 2 = + = - Example16. A stone of mass 1 kg tied to a light inextensible string of length L = (10/3) m is whirling in a circular path of radius Linaverticalplane.Iftheratioofthe maximumtotheminimum tension in the string is 4 and g = 10 m/s2, then find the speed of the stone at the highest point of the circle. Solution : O T P VP VO Q q q mg cos q mg L The tension T in the string is given by ú ú û ù ê ê ë é + = L v g m T 2 Q max and ú ú û ù ê ê ë é + - = L v g m T 2 P min According to the given problem 4 ) L / v ( g ) L / v ( g 2 P 2 Q = + - + or L v 4 g 4 L v g 2 P 2 Q + - = + or L v 4 g 4 L L g 4 v g 2 P 2 P + - = + + L = (10/3) m and g = 10 m/s2 (given) Solving we get vP = 10 m/s. Negotiating a Curve : Case of cyclist To safelynegotiate a curve of radius r, a cyclist should bend at an angle q with the vertical. q Nsinq N Ncosq Which is given by tan q = rg v2 . Angle q is also called as angle of banking. r mv sin N 2 = q and mg cos N = q Case of car on a levelled road A vehicle can safelynegotiate a curve of radius r on a rough level road when coefficient of sliding friction is related to the velocity as 2 s v rg m ³ . Now consider a case when a vehicle is moving in a circle, the centrifugal force is r mv2 whereas m is mass of vehicle, r = radius of circle and v is its velocity. fs mv r 2 The frictional force is static since wheels are in rolling motion because point of contact with the surface is at rest r mv f 2 s = mg f f s max s m = £ mg r mv s 2 m £ or rg v2 s ³ m Case of banking of road (frictionless) A vehicle can safely negotiate a curve of radius r on a smooth (frictionless) road, when the angle q of banking of the road is given by 2 tan v rg q = . N mg Horizontal q q Vertical
- 116. 112 PHYSICS When thebanked surface issmooth, theforceacting will be gravity and normal force only. N q mv r 2 mg Balancing forces cos N mg q = ...(1) 2 sin mv N r q = ...(2) 2 tan v rg = q ...(3) Case of banking of road (with friction) The maximum velocity with which a vehicle can safelynegotiate a curve of radius r on a rough inclined road is given by v2 = ( tan ) 1 tan rg m + q -m q ; where m is the coefficient of friction of the rough surface on which the vehicle is moving, and q is the angle of inclined road with the horizontal. Suppose a vehicle is moving in a circle of radius r on a rough inclined road whose coefficient of friction is μ M and angle of banking is q. fs fs q N mv r 2 mv r 2 mg mg N Let velocity of object (vehicle) be V. If we apply pseudo force on body, centrifugal force is r mv2 when v is max. and friction force will be acting down the slope. Balancing the forcehorizontally, q + q = sin N cos f r mv s 2 ...(1) Balancing the force vertically, mg sin f cos N s + q = q ...(2) when v = maximum, f = fmax = fs = mN ...(3) From eqn.(2), mg sin N cos N + q m = q mg ) sin (cos N = q m - q Þ or q m - q = sin cos mg N From eqns.(1) and (3), q m - q q + q m = sin cos sin mg cos mg r mv2 Þ q m - q + m = tan 1 ) tan ( mg r mv2 2 max ( tan ) 1 tan v rg m + q Þ = -m q Nowin the case ofminimum velocitywith which bodycould move in a circular motion, the direction of friction will be opposite to that one in maximum velocitycase. fs q N mv r 2 mg and 2 min tan 1 tan v rg æ ö m - q = ç ÷ + m q è ø Keep in Memory 1. Whenever a particle is moving on the circular path then there must be some external force which will provide the necessary centripetal acceleration to the particle. For examples : (i) Motion ofsatellite around aplanet : Here the centripetal force is provided by the gravitational force. i.e. r mv r GMm 2 2 = V (M) Planet Satellite (m) (ii) Motion of electron around the nucleus : Here the required centripetal force is provided by the Coulombian force i.e. r mv r ) e )( ze ( 4 1 2 2 o = pe Nucleus (Ze) r Electron (e) (iii) Motion of a bodyin horizontal and vertical circle: Here the centripetal force is provided by the tension. Horizontal circle r mv T 2 = T V (m)
- 117. 113 Lawsof Motion (d) Thevertical depth h of P below A is independent of the length of the string since from eqn. (1) and (4) h mg T mg T h = Þ = l l but 2 T m = w l Therefore 2 2 m g g m h h w = Þ = w l l which is independent of l. Example17. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2, where k is a constant. Determine the power delivered to the particle by the forces acting on it. Solution : Heretangential acceleration alsoexistswhich requires power. Given that centripetal acceleration ac = k2rt2 also, ac = v2/r ; v2/r = k2rt2 or v2 = k2r2t2 or v = k r t ; Tangential acceleration, r k dt dv a = = Now, force F = ma = m k r ; So, power, P = F v = m k r × k r t = m k2 r2 t. Example18. The string of a pendulum is horizontal. The mass of the bob is m. Now the string is released. What is the tension in the string in the lowest position? Solution : mg v T O Let v be the velocity of the bob at the lowest position. In this position, The P.E. of bob is converted into K.E. Hence, 2 v m 2 1 g m = l or l g 2 v2 = ...(1) If T be the tension in the string, then ÷ ÷ ø ö ç ç è æ = - l 2 v m g m T ...(2) From eqns. (1) and (2). T – m g = 2 m g or T = 3 m g Vertical circle At point A, r mv T 2 A A = ; At point B, r mv mg T 2 B B = + T V B T A B mg mg B V C T C V mg C And at point C, r mv mg T 2 C C = - CONICAL PENDULUM Consider an inextensible string of length l which is fixed at one end, A. At the other end is attached a particle Pof mass m describing a circle with constant angular velocity w in a horizontal plane. Horizontal Plane mg A P r O h P Tsin O 2 r Vertical section As P rotates, the string AP traces out the surface of a cone. Consequently the system is known as a conical pendulum. Vertically, Tcos mg q = ...(1) Horizontally, 2 Tsin mr q = w ...(2) In triangleAOP, r sin = q l ...(3) and h cos = q l ...(4) Several interesting facts can be deduced from these equations : (a) It is impossible for the string to be horizontal. This is seen from eqn. (1) in which mg cos T q = cannot be zero. Hence q cannot be 90°. (b) The tension is always greater than mg. This also follows from eqn. (1) as cos q < 1 (q is acute but not zero). Hence, T > mg (c) The tension can be calculated without knowing the inclination of the string since, from eqn. (2) and (3) 2 Tsin m sin q = q w l Þ 2 T m = w l
- 118.
- 119. 115 Lawsof Motion 1. Arectangular block is placed on a rough horizontal surface in two different ways as shown, then F F (a) (b) (a) friction will be more in case (a) (b) friction will be more in case (b) (c) friction will be equal in both the cases (d) friction depends on the relationsamong its dimensions. 2. Centripetal force : (a) can change speed of the body. (b) is always perpendicular to direction of motion (c) is constant for uniform circular motion. (d) all of these 3. When a horse pulls a cart, the horse moves down to (a) horse on the cart. (b) cart on the horse. (c) horse on the earth. (d) earth on the horse. 4. The force of action and reaction (a) must be of same nature (b) must be of different nature (c) may be of different nature (d) may not have equal magnitude 5. A body is moving with uniform velocity, then (a) no force must be acting on the body. (b) exactly two forces must be acting on the body (c) body is not acted upon by a single force. (d) the number of forces acting on the body must be even. 6. The direction of impulse is (a) same as that of the net force (b) opposite to that of the net force (c) same as that of the final velocity (d) same as that of the initial velocity 7. A monkeyis climbing up a rope, then the tension in therope (a) must be equal to the force applied by the monkey on the rope (b) must be less than the force applied by the monkey on the rope. (c) must be greater than the force applied by the monkey on the rope. (d) may be equal to, less than or greater the force applied by the monkey on the rope. 8. Auniform ropeoflength L restingon a frictionlesshorizontal surface is pulled at one end by a force F. What is the tension in the rope at a distance l from the end where the force is applied. (a) F (b) F (1 + l/L) (c) F/2 (d) F (1 – l/L) 9. A particle of mass m is moving with velocity v1, it is given an impulse such that the velocity becomes v2. Then magnitude of impulse is equal to (a) ) v v ( m 1 2 r r - (b) ) v v ( m 2 1 r r - (c) ) v v ( m 1 2 r r - ´ (d) ) v v ( m 5 . 0 1 2 r r - 10. A constant force F = m2g/2 is applied on the block of mass m1 as shown in fig. The string and the pulley are light and the surface ofthe table is smooth. The acceleration of m1 is F m1 m2 (a) right towards ) m m ( 2 g m 2 1 2 + (b) left towards ) m m ( 2 g m 2 1 2 - (c) right towards ) m m ( 2 g m 1 2 2 - (d) left towards ) m m ( 2 g m 1 2 2 - 11. A mass is hanging on a spring balance which is kept in a lift. Thelift ascends. The spring balance will showin its readings (a) an increase (b) a decrease (c) no change (d) a change depending on its velocity 12. A cart of mass M has a block of mass m attached to it as shown in fig. The coefficient of friction between the block and the cart is m. What is the minimum acceleration of the cart so that the block m does not fall? (a) mg (b) g/m M m (c) m/g (d) M mg/m
- 120. 116 PHYSICS 13. Aparticle ofmass mmoving eastwardwith a speed vcollides with another particle of the same mass moving northward with the same speed v. The two particles coalesce on collision. The newparticleofmass 2mwill movein thenorth- external direction with a velocity: (a) v/2 (b) 2v (c) 2 / v (d) None of these 14. A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v1 and v2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then : (a) v1/v2 = m1/m2 (b) v1/v2 = m2/m1 (c) v1/v2 = –m2/m1 (d) v1/v2 = –m1/m2 15. Two mass m and 2m are attached with each other by a rope passing over a frictionless and massless pulley. If the pulley is accelerated upwards with an acceleration ‘a’, what is the value of T? (a) 3 a g + (b) 3 a g - (c) 3 ) a g ( m 4 + (d) 3 ) a g ( m - 16. A rider on a horse back falls forward when the horse suddenly stops. This is due to (a) inertia of horse (b) inertia of rider (c) large weight of the horse (d) losing of the balance 17. A ball of mass m is thrown verticallyupwards. What is the rate at which the momentum of the ball changes? (a) Zero (b) mg (c) Infinity (d) Data is not sufficient. 18. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is thesame in all cases.At the highest point of the track, the normal reaction ismaximumin (a) v (b) v (c) v (d) v 19. A weight W rests on a rough horizontal plane. If the angle of friction be q , the least force that will move the body along the plane will be (a) q cos W (b) q cot W (c) q tan W (d) q sin W 20. A particle starts sliding down a frictionless inclined plane. If n S is the distance traveled byit from time t = n – 1 sec to t = n sec, the ratio Sn/Sn+1 is (a) 1 n 2 1 n 2 + - (b) n 2 1 n 2 + (c) 1 n 2 n 2 + (d) 1 n 2 1 n 2 - + 21. Ablock iskepton a inclinedplaneofinclination qoflength l . The velocity of particle at the bottom of inclined is (the coefficient offriction is m ) (a) 2 / 1 )] sin cos ( g 2 [ q - q m l (b) ) cos (sin g 2 q m - q l (c) ) cos (sin g 2 q m + q l (d) ) sin (cos g 2 q m + q l 22. A bird is in a wire cage which is hanging from a spring balance . In the first case, the bird sits in the cage and in the second case, thebird fliesabout insidethecage. Thereading in the spring balance is (a) more in the first case (b) less in first case (c) unchanged (d) zero in second case. 23. In an explosion, a bodybreaks up into twopieces ofunequal masses. In this (a) both parts will have numerically equal momentum (b) lighter part will have more momentum (c) heavier part will have more momentum (d) both parts will have equal kinetic energy 24. A block of mass m on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block the coefficient offriction between block and surface is m q F1 F2 (a) q + q + cos F mg sin F F 2 2 1 (b) q - + q sin F mg F cos F 2 2 1 (c) q + q + sin F mg cos F F 2 2 1 (d) 1 2 2 F sin F mg F cos q - - q 25. A plate of mass M is placed on a horizontal of frictionless surface (see figure), and a body of mass m is placed on this plate. The coefficient of dynamic friction between this body and the plate is m . If a force 2 m mg is applied to the body ofmass m along the horizontal, theacceleration of the plate will be M m 2 mg m (a) g M m m (b) g ) m M ( m + m (c) g M m 2m (d) g ) m M ( m 2 + m .
- 121. 117 Lawsof Motion 1. Anobject of mass 10 kg moves at a constant speed of 10 ms–1. A constant force, that acts for 4 sec on the object, gives it a speed of 2 ms–1 in opposite direction. The force acting on the object is (a) –3 N (b) –30N (c) 3N (d) 30N 2. A solid sphere of2 kg is suspended from a horizontal beam by two supporting wires as shown in fig. Tension in each wireis approximately(g = 10 ms–2) (a) 30N (b) 20N (c) 10N mg 30º 30º T T (d) 5N 3. A toy gun consists of a spring and a rubber dart of mass 16 g. When compressed by 4 cm and released, it projects the dart to a height of 2 m. If compressed by 6 cm, the height achieved is (a) 3m (b) 4m (c) 4.5m (d) 6m 4. A player stops a football weighting 0.5 kg which comes flying towards him with a velocity of 10m/s. If the impact lasts for 1/50th sec. and the ball bounces backwith a velocity of 15 m/s, then the average force involved is (a) 250N (b) 1250N (c) 500N (d) 625N 5. A car travelling at a speed of30 km/h is brought to a halt in 4 mbyapplying brakes.Ifthesamecaristravellingat60km/h, it can be brought to halt with the same braking power in (a) 8m (b) 16m (c) 24m (d) 32m 6. A bodyof mass 4 kg moving on a horizontal surface with an initial velocity of 6 ms–1 comes to rest after 3 seconds. If one wants to keep the body moving on the same surface with the velocity of 6 ms–1, the force required is (a) Zero (b) 4N (c) 8N (d) 16N 7. A machine gun has a mass 5 kg. It fires50 gram bullets at the rate of 30 bullets per minute at a speed of 400 ms–1. What force is required to keep the gun in position? (a) 10N (b) 5N (c) 15N (d) 30N 8. A force time graph for the motion of a bodyis shown in Fig. Change in linear momentum between 0 and 8s is 2 0 1 2 4 6 7 8 x t (s) F (N) (a) zero (b) 4 N-s (c) 8 Ns (d) None of these 9. Fig. shows a uniform rod of length 30 cm having a mass of 3.0 kg. Thestrings shown in thefigure arepulled byconstant forces of20 N and 32 N. All the surfaces are smooth and the strings and pulleys are light. The forceexerted by20 cm part of the rod on the 10 cm part is 10 cm 20 cm 20 N 32 N (a) 20N (b) 24N (c) 32N (d) 52N 10. A force of10 N acts on a bodyof mass 20 kg for 10 seconds. Changein its momentum is (a) 5 kg m/s (b) 100 kg m/s (c) 200 kg m/s (d) 1000 kg m/s 11. Consider the system shown in fig. The pulleyand the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m/s2) 1 kg 1 kg (a) 0N (b) 1N (c) 2N (d) 5N 12. Theelevator shown in fig. is descending with an acceleration of2 m/s2. Themass oftheblockA= 0.5 kg. The forceexerted by the block A on block B is (a) 2N (b) 4N (c) 6N A 2 m/s 2 B (d) 8N
- 122. 118 PHYSICS 13. Twoblocks of masses 2 kg and 1 kg are placed on a smooth horizontal tablein contact with each other. Ahorizontal force of 3 newton is applied on the first so that the block moves with a constant acceleration. The force between the blocks would be (a) 3 newton (b) 2 newton (c) 1 newton (d) zero 14. A 4000 kg lift is accelerating upwards. The tension in the supporting cable is 48000 N. If 2 s m 10 g - = then the acceleration of the lift is (a) 1 2 s m - (b) 2 2 s m - (c) 4 2 s m - (d) 6 2 s m - 15. Arocket hasa mass of100kg. Ninetypercent ofthisis fuel. It ejects fuel vapors at the rate of 1 kg/secwith a velocityof500 m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial upthrust on the rocket when it just starts moving upwards is (a) zero (b) 500 newton (c) 1000 newton (d) 2000 newton 16. A 0.1 kg block suspended from a massless string is moved first vertically up with an acceleration of 2 ms 5 - and then moved vertically down with an acceleration of 2 ms 5 - . If 1 T and 2 T are the respective tensions in the two cases, then (a) 1 2 T T > (b) 2 1 T T - = 1 N, if 2 ms 10 g - = (c) 1 2 T T 1kgf - = (d) 1 2 T T 9.8N, - = if 2 g 9.8 ms- = 17. The coefficient of friction between two surfaces is 0.2. The angleof friction is (a) ) 2 . 0 ( sin 1 - (b) ) 2 . 0 ( cos 1 - (c) ) 1 . 0 ( tan 1 - (d) ) 5 ( cot 1 - 18. A man weighing 80 kg is standing on a trolleyweighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley along the rails at a speed of one metre per second, then after 4 seconds, his displacement relative to the ground will be : (a) 5 metres (b) 4.8 metres (c) 3.2 metres (d) 3.0 metres 19. Starting from rest, a bodyslides down a 45º inclined plane in twice the time it takes toslide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is: (a) 0.33 (b) 0.25 (c) 0.75 (d) 0.80 20. A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. Ifthe time ofcontact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is : (a) 2000 newton (b) 1000 newton (c) 5000 newton (d) 125 newton 21. The mass of the lift is 100 kg which is hanging on the string. The tension in the string, when the lift is moving with constant velocity, is (g = 9.8 m/sec2) (a) 100 newton (b) 980 newton (c) 1000 newton (d) None of these 22. In the question , the tension in the strings, when the lift is accelerating up with an acceleration 1 m/sec2, is (a) 100 newton (b) 980 newton (c) 1080 newton (d) 880 newton 23. A block of mass 5 kg resting on a horizontal surface is connected bya cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is : (g = 9.8 m/sec2) 5 kg 5 kg A B (a) 49N (b) Zero (c) 36.75N (d) 2.45N 24. A40 kg slab rests on frictionless floor as shown in fig. A10 kg block rests on the top of the slab. The static coefficient offriction between the block and slabis0.60 while the kinetic friction is0.40. The 10 kg block is actedupon bya horizontal force of 100 N. If g = 9.8 m/s2, the resulting acceleration of the slab will be: 40 kg 100 N No friction (a) 0.98 m/s2 (b) 1.47 m/s2 (c) 1.52 m/s2 (d) 6.1m/s2 25. Two blocks are connected over a massless pulleyas shown in fig. The mass of blockAis10 kg and the coefficient ofkinetic frictionis0.2.BlockAslidesdowntheinclineatconstantspeed. The massof block Bin kg is: A B 30º (a) 3.5 (b) 3.3 (c) 3.0 (d) 2.5
- 123. 119 Lawsof Motion 26. Twotrolleys of mass m and 3m are connected by a spring. They were compressed and released at once, they move off in opposite direction and come to rest after covering a distance S1, S2 respectively. Assuming the coefficient of friction to be uniform, ratio of distances S1 : S2 is : (a) 1: 9 (b) 1: 3 (c) 3: 1 (d) 9: 1 27. A particle of mass 10 kg is moving in a straight line. If its displacement, x with time t is given by x = (t3 – 2t – 10) m, then the force acting on it at the end of 4 seconds is (a) 24N (b) 240N (c) 300N (d) 1200N 28. When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed then the acceleration of the particle is (a) F1/m (b) F2F3/mF1 (c) (F2 – F3)/m (d) F2/m 29. One end of massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms–2) can a man of 60 kg moves downwards on the rope? [Take g = 10 ms–2] P C (a) 16 (b) 6 (c) 4 (d) 8 30. A force k̂ 10 ĵ 6 î 8 F - - = r newton produces an acceleration of 1 ms–2 in a body. The mass of the body is (a) 10kg (b) 2 10 kg (c) 3 10 kg (d) 200kg 31. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freelyfrom the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ? (a) 12J (b) 3.6J (c) 7.2J (d) 1200J 32. A body of mass 1 kg moving with a uniform velocity of 1 ms 1 - . If the value of g is 2 ms 5 - , then the force acting on the frictionless horizontal surface on which the body is moving is (a) 5N (b) 1N (c) 0N (d) 10N 33. A trailer of mass 1000 kg is towed by means of a rope attached to a car moving at a steady speed along a level road. The tension in the rope is 400 N. The car starts to accelerate steadily. Ifthe tension in the rope is now 1650 N, with what acceleration is the trailer moving ? (a) 1.75ms–2 (b) 0.75ms–2 (c) 2.5 ms–2 (d) 1.25ms–2 34. Arocket of mass5000 kg is to beprojected verticallyupward. The gases are exhausted verticallydownwards with velocity 1000 ms–2 with respect tothe rocket. What is the minimum rate of burning the fuel so as to just lift the rocket upwards against gravitational attraction ? (a) 49 kg s–1 (b) 147 kg s–1 (c) 98 kg s–1 (d) 196 kg s–1 35. BlocksAand Bof masses 15 kg and 10 kg, respectively, are connected bya light cable passing over a frictionless pulley as shown below. Approximately what is the acceleration experienced by the system? (a) 2.0m/s2 (b) 3.3m/s2 (c) 4.9m/s2 A B (d) 9.8m/s2 36. A50 kg ice skater, initiallyat rest, throwsa 0.15 kg snowball with a speed of35 m/s. What is the approximate recoil speed of the skater? (a) 0.10m/s (b) 0.20m/s (c) 0.70m/s (d) 1.4m/s 37. Block A is moving with acceleration A along a frictionless horizontal surface. When a second block, B is placed on top of Block Athe acceleration of the combined blocks drops to 1/5 the original value. What is the ratio of the mass ofA to the mass of B? (a) 5: 1 (b) 1: 4 (c) 3: 1 (d) 2: 1 38. A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. 60° F M What is the work done bythe force F? (Note : sin 60° = 0.87; cos60° = 0.50. Ignore friction andthe weightsofthe pulleys) (a) 50J (b) 100J (c) 174J (d) 200J 39. A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m/s. Togive an initial upward acceleration of 20 m/s2, the amount of gas ejected per second to supply the needed thrust will be (Takeg = 10 m/s2) (a) 127.5kg/s (b) 137.5kg/s (c) 155.5kg/s (d) 187.5kg/s
- 124. 120 PHYSICS 40. Abullet is fired from a gun. The force on the bullet is given byF = 600 – 2 × 105 t Where, F is in newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet? (a) 1.8 N-s (b) Zero (c) 9 N-s (d) 0.9 N-s 41. A rifle man, whotogether with his rifle has a mass of100 kg, stands on a smooth surface and fires 10 shots horizontally. Each bullet hasa mass10 gand a muzzlevelocityof 800 ms– 1. The velocity which the rifle man attains after firing 10 shots is (a) 8 1 ms- (b) 0.8 1 ms- (c) 0.08 1 ms- (d) – 0.8 1 ms- 42. A block of mass 4 kg rests on an inclined plane. The inclination to the plane is gradually increased. It is found that when the inclination is 3 in 5, the block just begins to slidedown the plane. The coefficient of friction between the block and the plane is (a) 0.4 (b) 0.6 (c) 0.8 (d) 0.75. 43. The minimum velocity(in ms-1)with which a car driver must traverse a flat curveofradius 150m and coefficient offriction 0.6 to avoid skidding is (a) 60 (b) 30 (c) 15 (d) 25 44. A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the bodyis 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2] (a) 8N (b) 10 N (c) 20N (d) 2.5N 45. Abag ofsand of mass m is suspended bya rope. A bullet of mass 20 m is fired at it with a velocity v and gets embedded into it. The velocity of the bag finally is (a) v 21 20 ´ (b) 20v 21 (c) v 20 (d) v 21 46. For thearrangement shown in the Figure thetension in the string is [Given : 1 tan (0.8) 39 - = ° ] 39° m = 1 kg m = 0.8 (a) 6N (b) 6.4N (c) 0.4N (d) zero. 47. A1kg block anda 0.5 kgblock movetogether on a horizontal frictionless surface . Each block exerts a force of6 N on the other. The block move with a uniform acceleration of a 1 kg 0.5 kg F (a) 3 2 ms- (b) 6 2 ms- (c) 9 2 ms- (d) 12 2 ms- 48. A body of mass 32 kg is suspended by a spring balance from the roof of a vertically operating lift and going downward from rest. At the instant the lift has covered 20 m and 50 m, the spring balance showed 30 kg and 36 kg respectively. Then the velocity of the lift is (a) decreasing at 20 m, and increasing at 50 m (b) increasing at 20m and decreasing at 50 m (c) continuously decreasing at a steady rate throughout the journey (d) constantly increasing at constant rate throughout the journey. 49. An object at rest in space suddenly explodes into three parts of same mass. The momentum of the two parts are î p 2 and ĵ p . The momentum of the third part (a) will have a magnitude 3 p (b) will have a magnitude 5 p (c) will have a magnitude p (d) will have a magnitude 2p. 50. A triangular block of mass M with angles 30°, 60°, and 90° rests with its 30°–90° side on a horizontal table. A cubical block of mass m rests on the 60°–30° side. The acceleration which M must have relative to the table to keep m stationary relativeto the triangularblock assuming frictionless contactis (a) g (b) 2 g (c) 3 g (d) 5 g 51. Abodyofmass 1.0 kg is falling with an acceleration of 10 m/ sec2. Its apparent weight will be (g = 10 m/sec2) (a) 1.0 kg wt (b) 2.0 kg wt (c) 0.5 kg wt (d) zero 52. In the figure a smooth pulley of negligible weight is suspended by a spring balance. Weight of 1 kg f and 5 kg f are attached to the opposite ends of a string passing over the pulley and move with acceleration because of gravity, During their motion, the spring balance reads a weight of (a) 6 kg f (b) less then 6 kg f (c) more than 6 kg f 1 kg 5 kg (d) may be more or less then 6 kg f
- 125. 121 Lawsof Motion 53. Aparticle moves so that its acceleration is always twice its velocity. If its initial velocityis 0.1 ms–1, its velocity after it has gone0.1 m is (a) 0.3 ms–1 (b) 0.7 ms–1 (c) 1.2 ms–1 (d) 3.6 ms–1 54. An object is resting at the bottom of two strings which are inclined at an angleof120°with each other. Each string can withstand a tension of 20N. The maximum weight of the object that can be supported without breaking the string is (a) 5N (b) 10N (c) 20N (d) 40N 55. On a smooth plane surface (figure) two block A and B are accelerated up byapplying a force 15 N on A. Ifmass of Bis twice that ofA, the force on B is (a) 30N (b) 15N (c) 10N (d) 5N B A 15 N 56. A 10 kg stone is suspended with a rope of breaking strength 30kg-wt. Theminimumtimein which thestonecan beraised through a height 10 m starting from rest is (Take kg / N 10 g = ) (a) 0.5 s (b) 1.0 s (c) 3 / 2 s (d) 2 s 57. A ball of mass 0.4 kg thrown up in air with velocity30 1 ms- reaches the highest point in 2.5 second . The air resistance encountered by the ball during upward motion is (a) 0.88N (b) 8800N (c) 300 dyne (d) 300N. 58. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to (a) 150N (b) 3N (c) 30N (d) 300N 59. In the system shown in figure, the pulley is smooth and massless, the string has a total mass 5g, and the two suspended blocks have masses 25 g and 15 g. The system is released from state 0 = l and is studied upto stage 0 '= l During the process, the acceleration of block Awill be (a) constant at g 9 (b) constant at g 4 A 25 g B 15 g l l' (c) increasing by factor of 3 (d) increasing by factor of 2 60. A horizontal force F is applied on back of mass m placed on a rough inclined plane of inclination q . The normal reaction N is F (a) q cos mg (b) q sin mg (c) mgcos Fcos q- q (d) q + q sin F cos mg 61. The coefficient of friction between the rubber tyres and the road way is 0.25. The maximum speed with which a car can be driven round a curve of radius 20 m without skidding is (g = 9.8 m/s2) (a) 5 m/s (b) 7 m/s (c) 10m/s (d) 14m/s 62. A bucket tied at the end ofa 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill when the bucket is at the highest position? (a) 4 m/sec (b) 6.25m/sec (c) 16 m/sec (d) None of the above 63. A cane filled with water is revolved in a vertical circle of radius 4 meter and the water just does not fall down. The time period of revolution will be (a) 1 sec (b) 10 sec (c) 8 sec (d) 4 sec 64. Acircular road of radius r in which maximum velocity is v, has angle of banking (a) ÷ ÷ ø ö ç ç è æ - rg v tan 2 1 (b) ÷ ÷ ø ö ç ç è æ - 2 1 v rg tan (c) ÷ ÷ ø ö ç ç è æ - rg v tan 1 (d) ÷ ø ö ç è æ - v rg tan 1 65. A small sphere is attached to a cord and rotates in a vertical circle about a point O. If the average speed of the sphere is increased, the cord is most likely to break at the orientation when the mass is at C O D m l A B (a) bottom point B (b) the point C (c) the point D (d) top point A
- 126. 122 PHYSICS 66. Aperson with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circle of radius 50 m. What is his inclination to vertical : (g = 10 m/sec2) (a) tan–1(½) (b) tan–1 (1/5) (c) tan–1 (3/5) (d) tan–1(1/10) 67. When the road is dry and the coefficient of the friction is m, the maximum speed of a car in a circular path is 10 ms–1. If the road becomes wet and 2 ' m = m , what is the maximum speed permitted? (a) 5 ms–1 (b) 10 ms–1 (c) 1 ms 2 10 - (d) 1 ms 2 5 - 68. Two pulleyarrangements of figure given are identical. The mass ofthe ropeis negligible. In fig (a), the mass m is lifted by attaching a mass 2m to the other end of the rope. In fig (b), m is lifted up by pulling the other end ofthe rope with a constant downward force F = 2mg. The acceleration of m in the two cases are respectively m 2m (a) m (b) F = 2 mg (a) 3g, g (b) g /3, g (c) g /3, 2g (d) g, g /3 69. The linear momentum p ofa bodymoving in one dimension varies with time according tothe equating P = a + bt2 where a and b are positive constants. The net force acting on the body is (a) proportional to t2 (b) a constant (c) proportional to t (d) inversely proportional to t 70. Three blocks of masses m1, m2 and m3 are connected by massless strings, as shown, on a frictionless table. Theyare pulled with a forceT3 =40 N. Ifm1 =10 kg, m2 =6 kgandm3 = 4kg, the tension T2 will be M1 M2 M3 T1 T2 T3 (a) 20N (b) 40N (c) 10N (d) 32N 71. Aball of mass 400gm is dropped froma height of5 m.A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m.Thetime for which theball remainsin contact with the bat is (g = 10 m/s2) (a) 0.12s (b) 0.08s (c) 0.04s (d) 12 s 72. Block Aof weight 100 kg rests on a block Band is tied with horizontal string to the wall at C. Block B is of 200 kg. The coefficient of friction between A and B is 0.25 and that between Band surface is 1 3 . The horizontal force F necessary to move the block B should be (g = 10 m/s2) A C B F (a) 1050N (b) 1450N (c) 1050N (d) 1250N 73. An open topped rail roadcar of massM has an initial velocity v0 along a straight horizontal frictionless track. It suddenly starts raising at time t = 0. The rain drops fall verticallywith velocityu and add a mass m kg/sec ofwater. The velocityof car after t second will be (assuming that it is not completely filled with water) (a) 0 u v m M + (b) 0 mv M mt + (c) 0 Mv ut M ut + + (d) 0 mut v M ut + + 74. A ball mass m falls verticallyto the ground from a height h1 and rebounds to a height h2. The change in momentum of the ball of striking the ground is (a) 1 2 m 2g(h h ) + (b) 1 2 n 2g(m m ) + (c) 1 2 mg(h h ) - (d) 1 2 m( 2gh 2gh ) - 75. In the given figure, the pulley is assumed massless and frictionless. Ifthe friction force on the object of mass m is f, then its acceleration in terms of the force F will be equal to m F (a) (F f )/ m - (b) F f / m 2 æ ö - ç ÷ è ø (c) F/m (d) None of these
- 127. 123 Lawsof Motion 76. Asmooth block is released at rest on a 45° incline and then slides a distance ‘d’. The time taken to slide is ‘n’ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is (a) k m = 2 n 1 1- (b) k m = 2 n 1 1- (c) s m = 2 n 1 1- (d) s m = 2 n 1 1- 77. The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by (a) 2 cos f (b) 2 sin f (c) tan f (d) 2 tan f 78. A particle of mass 0.3 kg subject to a force F = – kx with k = 15 N/m . What will be its initial acceleration ifit is released from a point 20 cm awayfrom the origin ? (a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2 79. A block is kept on a frictionless inclined surface with angle ofinclination ‘a ’ . The incline isgiven an acceleration ‘a’ to keep the block stationary. Then a is equal to a (a) g cosec a (b) g / tan a (c) g tan a (d) g 80. Consider a car moving on a straight road with a speed of100 m/s . The distance at which car can be stopped is [µk = 0.5] (a) 1000 m (b) 800 m (c) 400 m (d) 100 m 81. A round uniform bodyof radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle qwith the horizontal. Then its acceleration is (a) 2 gsin 1 MR / I q , (b) 2 gsin 1 I / MR q ∗ (c) 2 gsin 1 MR / I q ∗ (d) 2 gsin 1 I/ MR q , 82. A block of mass m is placed on a smooth wedge of inclination q. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be (a) mg/cos q (b) mg cos q (c) mg sin q (d) mg 83. The coefficient of static friction, ms, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. (g = 10 m/s2) A B 2 kg (a) 0.4kg (b) 2.0kg (c) 4.0kg (d) 0.2kg 84. A body under the action of a force ˆ ˆ ˆ F = 6 i –8 j+10k, r acquires an acceleration of 1 m/s2. The mass of this body must be (a) 10kg (b) 20kg (c) 10 2 kg (d) 2 10 kg 85. Aconveyor belt is moving at a constant speed of 2m/s. Abox is gently dropped on it. The coefficient of friction between them is µ= 0.5. Thedistance that theboxwill moverelativeto belt before coming to rest on it taking g = 10 ms–2, is (a) 1.2m (b) 0.6m (c) zero (d) 0.4m 86. A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s2. If g = 10 ms–2, the tension in the supporting cable is (a) 8600N (b) 9680N (c) 11000N (d) 1200N 87. The upper half of an inclined plane of inclination q is perfectlysmooth while lower half is rough. Ablock starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by (a) m = 2 tan q (b) m = 2 tan q (c) m = tan q (d) m = 1 tan q 88. A bridge is in the from of a semi-circle of radius 40m. The greatest speed with which a motor cycle can cross the bridge without leaving the ground at the highest point is (g = 10 m s–2) (frictional force isnegligiblysmall) (a) 40 m s–1 (b) 20 m s–1 (c) 30 m s–1 (d) 15 m s–1
- 128. 124 PHYSICS 89. Anexplosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms–1 andthe second part of mass 2 kg moves with speed 8 ms–1. If the third part flies off with speed 4 ms–1 then its mass is (a) 5 kg (b) 7 kg (c) 17kg (d) 3 kg 90. Two particles of masses m and M (M > m ) are connected by a cord that passes over a massless, frictionless pulley. The tension T in thestring and the acceleration a of the particles is (a) 2mM Mm T g;a g (M m) (M m) = = - + (b) 2mM M m T g;a g (M m) (M m) æ ö - = = ç ÷ + è + ø (c) m M Mm T g;a g (M m) (M m) æ ö æ ö - = = ç ÷ ç ÷ è + ø è + ø (d) mM 2Mm T g;a g (M m) (M m) æ ö æ ö = = ç ÷ ç ÷ è + ø è + ø 91. A bullet of mass m is fired from a gun of mass M. The recoiling gun compresses a spring of force constant k by a distance d. Then the velocity of the bullet is (a) kd M/ m (b) d km M (c) d kM m (d) kM d m 92. A spring of force constant k is cut into two pieces whose lengths are in the ratio 1 : 2. What is the force constant of the longer piece ? (a) 2 k (b) 3 2 k (c) 2 k (d) 3k 93. A motor cycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motor cycleis ascending on theoverbridge, thenormal force on it (a) increases (b) decreases (c) remains the same (d) fluctuates erratically 94. A bodyof mass M hits normally a rigid wall with velocityV and bounces back with the same velocity. The impulse experienced by the body is (a) MV (b) 1.5MV (c) 2MV (d) zero 95. A heavy uniform chain lies on horizontal table top. If the coefficient offriction between the chain and the tablesurface is 0.25, then themaximum fraction ofthelength of the chain that can hang over one edge of the table is (a) 20% (b) 25% (c) 35% (d) 15% 96. A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutuallyperpendicular directions with speeds of 21 m/s. The velocityof heaviest fragment in m/s will be (a) 2 7 (b) 2 5 (c) 2 3 (d) 2 97. Two bodies of masses m and 4m are moving with equal kinetic energies. The ratioof their linear momenta will be (a) 1: 4 (b) 4: 1 (c) 1: 2 (d) 2: 1 Directions for Qs. (98 to 100) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethe correctanswer (ONLYONE option is correct ) from the following- (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 98. Statement -1 :The work done in bringing a bodydown from the top to the base along a frictionless incline plane is the same as the work done in bringing it down the vartical side. Statement -2 : The gravitational force on the body along the inclined plane is the same as that along the vertical side. 99. Statement -1 : On a rainyday, it is difficult to drive a car or bus at high speed. Statement -2 : The value ofcoefficient offriction is lowered due to wetting of the surface. 100. Statement -1 : The two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. Statement -2 : For same air resistance, acceleration of both the bodies will be same.
- 129. 125 Lawsof Motion Exemplar Questions 1.A ball is travelling with uniform translatory motion. This means that (a) it is at rest (b) the path can be a straight line or circular and the ball travels with uniform speed (c) all parts of the ball have the same velocity(magnitude and direction) and the velocity is constant (d) the centre of the ball moves with constant velocityand the ball spins about its centre uniformly 2. Ametre scale is moving with uniform velocity. This implies (a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale (b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero (c) the total force acting on it need not be zero but the torque on it is zero (d) neither the force nor the torque need to be zero 3. A cricket ball of mass 150 g has an initial velocity 1 ˆ ˆ (3 4 )ms u i j - = + r and a final velocity 1 ˆ ˆ (3 4 )ms v i j - = - + r , after beinghit. The changein momentum (final momentum- initial momentum) is (in kgms–1) (a) zero (b) ˆ ˆ (0.45 0.6 ) i j - + (c) ˆ ˆ (0.9 1.2 ) j j - + (d) ˆ ˆ ˆ 5( ) i j i - + 4. In the previous problem(3), the magnitude ofthemomentum transferred during the hit is (a) zero (b) 0.75 kg-m s–1 (c) 1.5 kg-m s–1 (d) 1.4 kg-m s–1 5. Conservation of momentum in a collision between particles can be understood from (a) Conservation of energy (b) Newton's first law only (c) Newton's second law only (d) both Newton's second and third law 6. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is (a) frictional force along westward (b) muscle force along southward (c) frictional force along sotuh-west (d) muscle force along south-west 7. A body of mass 2 kg travels according to the law 2 3 ( ) x t pt qt rt = + + where, q = 4 ms–2, p = 3 ms–1 and r= 5 ms–3. The force acting on the body at t = 2s is (a) 136N (b) 134N (c) 158N (d) 68N 8. A body with mass 5 kg is acted upon by a force ˆ ˆ ( 3 4 ) N F i j = - + r . If its initial velocity at t = 0 is 1 ˆ ˆ (6 12 ) ms v i j - = - , the time at which it will just have a velocity along the y-axis is (a) never (b) 10 s (c) 2 s (d) 15 s 9. A car of mass m starts from rest and acquires a velocity along east, ˆ ( 0) v vi v = > r in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is (a) 2 mv eastward and is exerted by the car engine (b) 2 mv eastward and is due to the friction on the tyres exerted by the road (c) more than 2 mv eastward exerted due to the engine and overcomes the friction of the road (d) 2 mv exerted by the engine NEET/AIPMT (2013-2017) Questions 10. Three blocks with masses m, 2 m and 3 m are connected by strings as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity) [2013] (a) 2mg (b) 3mg (c) 6mg (d) zero 11. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made bythe wire with the vertical is [NEET Kar. 2013] (a) 0° (b) 3 p (c) 6 p (d) 4 p 12. A balloon with mass ‘m’ is descending down with an acceleration ‘a’ (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration ‘a’? [2014] (a) 2ma g a + (b) 2ma g a - (c) ma g a + (d) ma g a -
- 130. 126 PHYSICS 13. Theforce‘F’acting on a particle ofmass ‘m’ is indicated by the force-timegraph shown below.Thechange in momentum of the particle over the time interval from zero to 8 s is : [2014] 6 3 0 –3 4 6 8 F(N) t(s) 2 (a) 24 Ns (b) 20 Ns (c) 12 Ns (d) 6 Ns 14. A system consists ofthree masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction = m). The pulleyis frictionless and of negligible mass. The downward acceleration of mass m1 is : (Assume m1 = m2 = m3 = m) [2014] (a) g(1 – g ) g m (b) 2g 3 m P m1 m2 m3 (c) g(1 – 2 ) 3 m (d) g(1 – 2 ) 2 m 15. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block then the contact force between A and B is [2015] A B C (a) 6N (b) 8N (c) 18N (d) 2N 16. A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is µk. When the block A is sliding on the table, the tension in the string is [2015] (a) 2 1 1 2 (m – k m )g (m m ) m + (b) 1 2 k 1 2 m m (1 )g (m m ) + m + (c) 1 2 k 1 2 m m (1 – )g (m m ) m + (d) 2 k 1 1 2 (m m )g (m m ) + m + 17. Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in radius r 2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is : [2015 RS] (a) 3 (b) 4 (c) 1 (d) 2 18. A plank with a box on it at one end is graduallyraised about the other end.As the angle of inclination with the horizontal reaches 30º the box starts to slip and slides 4.0 m down the plank in 4.0s. The coefficients of static and kinetic friction between the box and the plank will be, respectively : [2015 RS] mg q (a) 0.6 and 0.5 (b) 0.5 and 0.6 (c) 0.4 and 0.3 (d) 0.6 and 0.6 19. What is the minimum velocitywith which a bodyof mass m must enter a vertical loop ofradius R sothat it can complete the loop ? [2016] (a) gR (b) 2gR (c) 3gR (d) 5gR 20. One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v' the net force on the particle (directed towards centre) will be (T represents the tension in thestring) : [2017] (a) 2 mv T l + (b) T – 2 mv l (c) Zero (d) T 21. Two blocks A and B of masses 3 m and m respectively are connected bya massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediatelyafter the string is cut, are respectively: [2017] (a) g ,g 3 A B 3m m (b) g, g (c) g g , 3 3 (d) g g, 3
- 131. 127 Lawsof Motion EXERCISE -1 1. (c) 2. (b) 3. (d) 4. (a) 5. (c) 6. (a) 7. (a) 8. (d) Let n be the mass per unit length of rope. Therefore, mass of rope = nL. Acceleration in the rope due to force F will be a= F/nL. Mass of rope of length (L – l) will be n (L – l). Therefore, tension in the rope oflength (L – l), is equal to pulling force on it = n (L – l) a = n (L – l) × F/nL = F (1 – l/L) 9. (a) Impulse = change in momentum = 1 2 v m v m r r - 10. (a) Let a be the acceleration of mass m2 in the downward direction. Then T – m2 (g/2) = m1 a ....(i) and m2 g – T = m2 a ....(ii) Adding eqs. (1) and (2), we get (m1 + m2) a = m2g – m2 (g/2) = m2 g/2 ) m m ( 2 g m a 2 1 2 + = 11. (a) Let acceleration of a lift = and let reaction at spring balance = R R mg Applying Newton’s law R– mg = ma ( ) a g m R + = Þ thus net weight increases, So reading of spring balance increases. 12. (b) See fig. ma mN N mg If a = acceleration ofthe cart, then N = ma mN = mg or m ma = mg or a = g/m 13. (c) p1 = mv northwards, p2 = mv eastwards N E S W m m v v Let p = momentum after collision. Then, 2 1 p p p r r r + = or 2 2 ) mv ( ) mv ( p + = r 2 mv v m 2 = ¢ or 2 v v = ¢ m/sec 14. (c) Applying law of conservation of linear momentum m1v1 + m2v2 = 0, 1 2 2 1 v v m m - = or 1 2 2 1 m m v v - = 15. (c) The equations of motion are 2 mg – T = 2ma T– mg = ma Þ T= 4ma &a = g/3 soT =4mg/3 If pulley is accelerated upwards with an accleration a, then tension in string is ) a g ( 3 m 4 T + = 16. (b) Inertia is resistance to change. 17. (b) The time rate of change ofmomentum is force. 18. (a) At the highest point of the track, 2 mv ' N mg r + = mg N where r is the radius of curvature at that point and v¢ is the speed of the block at that point. Now 2 mv ' N mg r = - N will be maximum when r is minimum (v¢ is thesame for all cases). Of the given tracks, (a) has the smallest radius of curvature at the highest point. 19. (c) f W =m f W tan = q [ q = m tan ] 20. (a) ) 1 n 2 ( 2 a Sn - = ) 1 n 2 ( 2 a S 1 n + = + 1 n 2 1 n 2 S S 1 n n + - = + 21. (b) From the F.B.D. N = mg cos q F = ma = mg sin q – N m ) cos (sin g a q m - q = Þ x q sin g q g N m q N m m mg cos Hints & Solutions
- 132. 128 PHYSICS Now using,as 2 u v 2 2 = - or, l ) cos (sin g 2 v2 q m - q ´ = ( l = length ofincline) or, v = ) cos (sin g 2 q m - q l 22. (a) Based on Newton’s third law of motion. 23. (a) If m1, m2 are masses and u1, u2 are velocity then by conservation of momentum m1u1 + m2u2 = 0 or 1 1 2 2 | m u | | m u | = 24. (a) Here, on resolving force F2 and applying the concept ofequilibrium F1 F2sin q F2 cos q mg N f m q + = cos F mg N 2 , and f = µN ] cos F mg [ f 2 q + m = …(i) Also q + = sin F F f 2 1 …(ii) From (i)and(ii) ] cos F mg [ 2 q + m = F1 + F2 sin q Þ q + q + = m cos F mg sin F F 2 2 1 25. (a) The frictional force acting on M is µmg Acceleration = M mg m EXERCISE - 2 1. (b) Here u = 10 ms–1, v = –2 ms–1, t = 4 s, a = ? Using 2 s / m 3 4 10 2 t u v a - = - - = - = Force, F =ma = 10×(–3) = –30 N 2. (b) 2 T cos 60º = mg or T = mg = 2×10 = 20 N. 3. (c) If k is the spring factor, then P.E. of the spring compressedbydistance x ÷ ø ö ç è æ = 2 kx 2 1 will equal to gain in P.E. of the dart ( = mgh) i.e. mgh kx 2 1 2 = 200 g 16 ) 4 ( k 2 1 2 ´ ´ = ....(i) and h g 16 ) 6 ( k 2 1 2 ´ ´ = ...(ii) On solving, (i) and (ii), weget h = 450 cm = 4.5m. 4. (d) Herem = 0.5 kg ; u = – 10 m/s ; t = 1/50 s ; v = + 15 ms–1 Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N 5. (b) As, (1/2)m v2 = Fs So 4 F ) 30 ( m 2 1 2 ´ = and s F ) 60 ( m 2 1 2 ´ = s/4 = (60)2 / (30)2 = 4 or s = 4 × 4 = 16 m. 6. (c) Acceleration, 2 v u 0 6 a 2 ms t 3 - - - = = = - Force= m×a =4×2= 8N 7. (a) taken time momentum in change required Force = = 3 (50 10 30) 400 (5 0) 10 N 60 - ´ ´ ´ - ´ = 8. (a) Changein momentum = Force × time =Area which the force-time curve encloses with time axis. 9. (b) 20cm 10cm 32N(F ) 2 20N(F ) 1 l1 l2 L F F It is clear F2 > F1, so rod moves in right direction with an acceleration a, whereas a is given by (F2–F1)=mL×a................(i) where m is mass of rod per unit length. Now consider the motion of length l1 from first end, then F– F1=ml1a..................(ii) Dividing eq (ii) by(i), we get L l F F F F 1 1 2 1 = - - or 1 1 1 2 F L l ) F F ( F + ´ - = here l1 = 10 cm., L= 30 cm., F1 = 20 N, F2 = 32N so F = 24 N 10. (b) Change in momentum = F × t = 10 × 10 = 100 Ns or 100 kg. m/s 11. (d) See fig. 1 kg 1 kg T T T From figure, 1 g – T = 1 a ...(i) and T = 1 a ....(ii) From eqs. (i) and (ii), we get 1g – 1a = 1a or 2a = g a = (g/2) = (10/2) = 5 m/s2 So, T =ma = 1 ×5 = 5 N 12. (b) R= mg –ma = 0.5 × 10 – 0.5 × 2= 5 – 1= 4 N 13. (c) Seefig. Let Fbetheforce between theblocksand a their common acceleration. Then for 2 kgblock,
- 133. 129 Lawsof Motion 3 N2 kg 1 kg F F 3 – F = 2 a ...(1) for 1 kg block, F = 1 × a = a ....(2) 3 – F = 2 F or 3 F = 3 or F = 1 newton 14. (b) ) a g ( m T + = ) a 10 ( 4000 48000 + = Þ a = 2 2 s m - 15. (b) Initial thrust on the rocket = rel v t m D D = 500 × 1= 500 N where m t D D = rate of ejection of fuel. 16. (b) ) a g ( m T1 + = = 0.1(10 5) 1.5N + = N 5 . 0 ) 5 10 ( 1 . 0 ) a g ( m T2 = - = - = Þ N 1 N ) 5 . 0 5 . 1 ( T T 2 1 = - = - 17. (d) Angle of friction = 1 tan µ - 18. (c) Displacement ofthe man on the trolley = 1 × 4 = 4m Nowapplying conservation oflinear momentum 80 × 1 + 400 v= 0or 5 1 v - = m/sec. The distance travelled by the trolley =– 0.2×4=–0.8m. (In opposite direction to the man.) Thus, the relative displacement of the man with the ground= (4–0.8)= 3.2m. 19. (c) In presence of friction a = (g sinq – mg cos q) Time taken to slide down the plane ) cos (sin g s 2 a s 2 t1 q m - q = = In absence of friction q = sin g s 2 t2 Given : 2 1 t 2 t = 2 2 1 2 t = 4t or q ´ = q m - q sin g 4 s 2 ) cos (sin g s 2 sin q = 4 sinq – 4m cos q 75 . 0 4 3 tan 4 3 = = q = m (since q = 45°) 20. (a) ( ) 3 10 2 5 . 0 2 t mv 2 t mv mv F - ´ ´ = = - - = =2×103 N 21. (b) T = m (g + a) = 100 (9.8 + 0) = 980 N 22. (c) T=m(g+a)=100(9.8+1)= 1080N 23. (c) For block A, T – mN = 5a and N = 5g 5 kg 5 kg B A mN T a T N a 5 kg for block B, 5g – T = 5a Þ T =36.75N, a =2.45 m/sec2 24. (a) Force on the slab (m = 40 kg) = reaction of frictional force on the upper block 40 kg 10 kg 100 N mk × 10 × g 40a = mk × 10 × g or a = 0.98 m/sec2 25. (b) Considering the equilibrium ofB –mBg+ T = mBa Since the block A slides down with constant speed. a = 0. Therefore T = mBg Considering the equilibrium ofA, we get 10a = 10g sin 30º – T – mN where N= 10g cos 30° B A T T 10g cos30º m g B mN 10g 10g sin 30º N a a º 30 cos g 10 T g 2 10 a 10 ´ m - - = but a = 0, T = mBg 2 3 2 . 0 g m g 5 0 B - - = × 10 ×g Þ mB = 3.268 »3.3kg 26. (d) mv1 + 3mv2 = 0 or 3 v v 2 1 - = Now 1 1 2 1 S . mg . S . F mv 2 1 m = = 2 2 2 2 S . mg 3 . S . F v ) m 3 ( 2 1 m = = or 1 9 v v S S 2 2 2 1 2 1 = =
- 134. 130 PHYSICS 27. (b)m = 10 kg, x = (t3 – 2t – 10) m 2 t 3 dt dx 2 - = n = t 6 a dt x d 2 2 = = At the end of 4 seconds, a = 6 × 4 = 24 m/s2 F= ma = 10 × 24 = 240 N 28. (a) m F1 F2 F3 The formula for force is given byF1 = ma Acceleration of the particle m F a 1 = , because F1 is equal to the vector sum of F2 & F3. 29. (c) P C mg T a mg T ma - = 60 10 360 a 60 ´ - = 2 a 4 ms- = 30. (b) 2 2 2 8 ( 6) ( 10) m 1 + - + - = = 10 2kg 31. (b) Mass of over hanging chain m’ kg ) 6 . 0 ( 2 4 ´ = Let at the surfacePE = 0 C.M. of hanging part = 0.3 m belowthe table 30 . 0 10 6 . 0 2 4 gx m Ui ´ ´ ´ - = ¢ - = U m'gx 3.6J D = = = Work done in putting the entire chain on the table 32. (a) Weight of body = m g = 5 N 33. (d) Here, the force of friction is 400N. Fnet = (1650 400) 1250N - = a = 2 ms 25 . 1 1000 1250 - = 34. (a) r dm mg dt v = = 1 s kg 49 1000 8 . 9 5000 - = ´ 35. (a) Two external forces, FA and FB, act on the system and move in oppositedirection. Let’sarbitrarilyassumethat the downward direction ispositive and that FA provides downward motion while FB provides upward motion. FA = (+15 kg) (9.8 m/s2)=147N and FB = (–10 kg)(9.8 m/s2) = – 98 N Ftotal = FA + FB = 147 N + (–98 N) = 49 N The total mass that must be set in motion is 15 kg + 10 kg = 25kg Since , a m F total total = a = Ftotal / mtotal = 49 N / 25 kg @ 2 m/s2 36. (a) Momentum is always conserved. Since the skater and snowball are initially at rest, the initial momentum is zero. Therefore, the final momentum after the toss must also be zero. 0 P P snowball skater = + or 0 v m v m snowball snowball skater skater = + skater snowball snowball skater m / v m v - = (0.15kg)(35m/s) 0.10m/s (50kg) = - = - The negative sign indicates that the momenta of the skater and the snowball are in opposite directions. 37. (b) Apply Newton’s second law FA = FAB, therefore : mA aA = (mA + mB)aAB and aAB = aA / 5 Therefore: mA aA = (mA + mB)aA/5 which reduces to 4 mA =mB or 1 :4 38. (d) Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiplythe net force or the weight mg bythe distance it is raised: 4 kgx 10 m/s2 x 5 m = 200 J. 39. (d) Given : Mass ofrocket (m) = 5000 Kg Exhaust speed (v) = 800 m/s Acceleration of rocket (a) = 20 m/s2 Gravitational acceleration (g) = 10 m/s2 We know that upward force F=m(g+a)= 5000(10+20) =5000×30=150000N. We also know that amount of gas ejected s / kg 5 . 187 800 150000 v F dt dm = = = ÷ ø ö ç è æ 40. (d) Given F = t 10 2 – 600 5 ´ The force is zero at time t, given by t 10 2 – 600 0 5 ´ = 3 – 5 10 3 10 2 600 t ´ = ´ = Þ seconds Impulse –3 t 3 10 5 0 0 Fdt (600 – 2 10 t) dt ´ = = ´ ò ò –3 0 3 10 5 2 2 10 t 600t – 2 ´ é ù ´ = ê ú ê ú ë û 2 3 – 5 3 – ) 10 3 ( 10 – 10 3 600 ´ ´ ´ = 1.8 – 0.9 0.9Ns = = 41. (b) According to law of conservation of momentum, 10 100v 10 800 1000 = - ´ ´ ie, v= 0.8 ms–1.
- 135. 131 Lawsof Motion 42. (d) 5 3 sin= q q 5 3 4 tan 4 3 = q Þ 3 tan 0.75 4 m = q = = 43. (b) The condition to avoid skidding, v = rg m = 10 150 6 . 0 ´ ´ = 30 m/s. 44. (d) Limitingfriction = 0.5 2 10 10N ´ ´ = The applied force is lessthan force offriction, therefore the force of friction is equal to the applied force. 45. (d) Applying law of conservation of momentum Momentum of bullet = Momentum of sand-bullet system m m 21 v m V mV 20 20 20 æ ö = + = ç ÷ è ø 46. (d) Here 8 . 0 tan = q where q is angle of repose ° = = q - 39 ) 8 . 0 ( tan 1 The given angle of inclination is equal to the angle of repose. So the 1 kg block has no tendency to move. = q sin mg force of friction Þ T = 0 47. (d) For 0.5 kg block, 6 = 0.5 a 48. (b) While moving down, when the lift is accelerating the weight will beless and when the lift is decelerating the weight will be more. 49. (b) Total momentum = j ˆ p î p 2 + Magnitude of total momentum = p 5 p 5 p ) p 2 ( 2 2 2 = = + This must be equal to the momentum ofthe third part. 50. (c) 90° 30° M mg sin 30° ma cos 30° 30° 60° ma (pseudo force) a ° = ° 30 sin mg 30 cos ma 3 g a = 51. (d) Apparent weight when mass is falling down is given by W ' m(g a) = - W ' 1 (10 10) 0 = ´ - = 52. (b) Reading of spring balance 2T = 1 2 1 2 4m m 4 5 1 10 m m 6 3 ´ ´ = = + kgf 53. (a) a = 2v (given) Þ dv v 2v ds = or dv 2ds = v 0.1 0 0.1 dv 2[s] 0.2 = = ò v 0.1 0.2 - = Þ 1 v 0.3ms- = 54. (c) IfW is themaximum weight, then W = 2T cos 60° or W= T = 20N 55. (c) The acceleration of both the blocks = 15 5 3x x = Force on 5 B 2x 10 N x = ´ = 56. (b) The maximum acceleration that can be given is a a 10 g 10 g 30 + = Þ 2 ms 20 g 2 a - = = We know that 2 at 2 1 ut s + = 20 10 2 a s 2 t ´ = = = 1 s 57. (a) Let the air resistance be F. Then mg F ma + = Þ ] g a [ m F - = Here 2 ms 12 5 . 2 30 a - = = 58. (c) N 30 1 . 0 ) 20 0 ( 15 . 0 t ) u v ( m F = - = - = 59. (c) Considering the two masses and the rope a system, then Initial net force = [ ] 25 (15 5) g 5g - + = Final net force = ( ) 25 5 15 g 15 g + - = é ù ë û Þ (acceleration)final = 3 (acceleration)initial 60. (d) mg sin q mg cos q F cos q q F sin F mg q N From figure q + q = sin F cos mg N
- 136. 132 PHYSICS 61. (b)r / v m mg 2 = m or r g v m = or s / m 7 ) 20 8 . 9 25 . 0 ( v = ´ ´ = 62. (a) Since water does not fall down, therefore the velocity of revolution should be just sufficient to provide centripetal acceleration at the top of vertical circle. So, ) 16 ( )} 6 . 1 ( 10 { ) r g ( v = ´ = = = 4 m/sec. 63. (d) The speed at the highest point must be rg v ³ Now ( ) T / 2 r r v p = w = rg ) T / 2 ( r > p or ÷ ÷ ø ö ç ç è æ p < p < g r 2 g r r 2 T sec 4 8 . 9 4 2 T = ÷ ø ö ç è æ p = 64. (a) Fromfigure, q sin N = r mv2 .......(i) q cos N = mg ......(ii) Dividing, we get q tan = rg v2 or q = rg v tan 2 1 - 65. (a) In the case of a body describing a vertical circle, O D B T A C q Mg Mg sin q Mg cos q q l 2 m cos mg T n = q - l 2 m cos mg T n + q = Tension is maximum when cos q = +1 and velocity is maximum Both conditions are satisfied at q = 0º (i.e. at lowest point B) 66. (b) Since surface (ice) is frictionless, so the centripetal forcerequiredfor skatingwill beprovided byinclination of boy with the vertical and that angle is given as 2 v tan θ rg = where v is speed of skating & r is radius of circle in which he moves. 67. (d) gr vmax m = 68. (b) Let a and a' be the accelerations in both cases respectively. Then for fig (a), (a) a a mg 2mg T T T –mg = ma …(1) and 2mg – T 2ma = …(2) Adding (1) and (2), we get mg 3ma = g a 3 = For fig (b), (b) a¢ a¢ mg F = 2mg T¢ T¢ T' mg ma' - = …(3) and 2mg T' 0 - = …(4) Solving (3) and (4) a ' g = g a 3 = and a ' g = 69. (c) Linear momentum, 2 P a bt = + dP 2bt dt = (on differentiation) Rate of change of momentum, dP t dt µ By2nd law of motion, dP F dt µ F t µ 70. (d) For equilibrium of all 3 masses, T3 = (m1 + m2 + m3)a or 3 1 2 3 T a m m m = + +
- 137. 133 Lawsof Motion For equilibriumof m1 &m2 2 1 2 T (m m ).a = + or, 1 2 3 2 1 2 3 (m m )T T m m m + = + + Given m1 = 10 kg, m2 = 6 kg, m3 = 4 kg, T3 = 40 N 2 (10 6).40 T 32N 10 6 4 + = = + + 71. (a) Velocity of ball after dropping it from a height of 5m 10 m/sec 20 m/sec (using v2 = u2 + 2gh) v2 = 0 + 2 × 10 × 5 Þ v= 10 m/s Velocity gained by ball by force exerted by bat 0 = u2 – 2gh u2 = 2 × 10 × 20 or u = 20 m/s Changein momentum = m(u + v) =0.4 (20 +10)=12kgm/s P F t D = D or P t F D D = 12 t 0.12sec 100 D = = 72. (d) F1 = Force of friction between B and A 1 1 m g = m = 0.25 × 100 × g = 25 g newton F2 = Force of friction between (A + B) and surface 2 2 2 m g = m = m (mass of Aand B) g 1 300 (100 200)g g 100g 3 3 = + = = newton 1 2 F F F = + = 25g + 100 g = 25g = 125 × 10 N F= 1250N 73. (b) The rain drops falling verticallywith velocity u do not affect the momentum along the horizontal track. A vector has no component in a perpendicular direction Rain drops add to the mass of the car Mass added in t sec = (mt) kg Momentum is conserved along horizontal track. Initial mass of car = M Initial velocity of car = v0 Final velocity of (car + water) = v Mass of (car + water) after time t = (M + mt) final momentum = initialmomentum 0 (M mt)v Mv + = 0 Mv v (M mt) = + 74. (d) Let v1 = velocity when height of free fall is h1 v2 = velocity when height of free rise is h2 2 2 1 1 v u 2gh = + for free fall or For free rise after impact on ground 2 2 2 0 v 2gh = - or 2 2 2 v 2gh = Initial momentum = mv1 Final momentum = mv2 Change in momentum = m(v1 – v2) 2 2 m( 2gh gh ) = - 75. (b) T = tension is the string Applied force F = 2T T = F/2 …(i) m F T T T f For block ofmass m, force of friction due to surface f. For sliding the block T – f = force on the block = mass × acceleration or acceleration of block T f m - = . Put T from (i) Acceleration F f 2 m - = 76. (b) smooth d rough q m - q cos g sin g q sin g ° 45 ° 45 d When surface is When surface is smooth rough d = q 2 1 1 (gsin )t 2 , d = 2 2 1 (gsin gcos )t 2 q -m q q = sin g d 2 t1 , q m - q = cos g sin g d 2 t2 According to question, 1 2 nt t = q sin g d 2 n = q m - q cos g sin g d 2 m , applicable here, is coefficient of kinetic friction as the block moves over the inclined plane. n = k 1 1 m - ÷ ÷ ø ö ç ç è æ = ° = ° 2 1 45 sin 45 cos Q k 2 1 1 n m - = or k 1 m - = 2 n 1 or 2 k n 1 1- = m
- 138. 134 PHYSICS 77. (d)Acceleration ofblock while sliding down upper half = f sin g ; retardation of block while sliding down lower half = – ) cos g sin g ( f m - f For the block tocome to rest at the bottom, acceleration in I half = retardation in II half. ) cos g sin g ( sin g f m - f - = f Þ f = m tan 2 Alternative method : According to work-energy theorem, W = DK = 0 (Since initial and final speeds are zero) Work done by friction + Work done by gravity =0 i.e., (µ mg cos ) mg sin 0 2 - f + f = l l µ or cos sin 2 f = f or µ 2tan = f 78. (c) Mass (m) = 0.3 kg Þ F = m.a = – 15 x a = – x 50 x 3 150 x 3 . 0 15 - = - = a = – 50 × 0.2 = 2 s / m 10 79. (c) From free bodydiagram, a sin mg cos g a a ma mg mg cos + ma sin a a N a a For block to remain stationary, mgsin ma cos a = a Þ a = a tan g 80. (a) as 2 u v 2 2 = - or s ) g ( 2 u 0 k 2 2 m - = - s 10 2 1 2 1002 ´ ´ - ´ = - s= 1000 m 81. (b) This is a standard formula and should be memorized. 2 gsin a I 1 MR q = + 82. (a) N = m a sin q + mg cos q ......(1) also m g sin q = m a cos q ......(2) from (2) a = g tan q q + q q = cos mg cos sin mg N 2 , or q = cos mg N m acosq m g sin q mgcosq ma cosq q N ma mg 83. (a) mBg = ms mAg {Q mAg = ms mAg} Þ mB = ms mA or mB = 0.2 × 2 = 0.4 kg 84. (c) ˆ ˆ ˆ F = 6 i –8 j+10k, r | F| 36 64 100 10 2 N = + + = ( ) 2 2 2 x y z F F F F = + + Q a = 1 ms–2 Q F = ma 0 2 m 10 2 1 1 = = kg 85. (d) Frictional force on the box f = mmg Acceleration in the box a = mg = 5 ms–2 v2 = u2 + 2as Þ 0 = 22 + 2 × (5) s Þ s = – 2 5 w.r.t. belt Þ distance = 0.4 m 86. (c) a = 1 m = 1000 kg Total mass = (60 + 940)kg = 1000 kg Let T be the tension in the supporting cable, then T – 1000g = 1000 × 1 ÞT= 1000×11= 11000N 87. (b) S/2 sin q S/2 sin q Smooth Rough S/2 S/2 q q For upper half of inclined plane v2 = u2 + 2a S/2 = 2 (g sin q) S/2 = gS sin q For lower half ofinclined plane 0 = u2 + 2 g (sin q – m cos q) S/2 Þ – gS sin q = gS ( sinq – m cos q) Þ 2 sin q = m cos q Þ m= 2sin cos q q = 2 tan q 88. (b) 1 v gr 10 40 20 m s- = = ´ = 89. (a) 4m /sec m 3 2 kg m2 Presultant 1 kg x y 8 m/sec 12 m/sec m1
- 139. 135 Lawsof Motion Presultant =2 2 12 16 + = 144 256 + = 20 m3v3 = 20 (momentum ofthird part) or, m3 = 20 4 = 5 kg 90. (b) Mg – T = Ma T – mg = ma m mg mg T T M On solving, we get a = (M m)g M m - + and T = 2Mmg M m + 91. (c) Let velocity of bullet be v. If velocity of gun is V, then mv + MV = 0 Þ V = mv M - As spring compresses by d, so 2 2 1 1 kd MV 2 2 = or 2 2 1 1 mv kd M 2 2 M æ ö = ç ÷ è ø Þ v = d kM m 92. (b) Here, l2 = 2l1 As for a spring, force constant 1 k l µ 1 2 1 2 1 2 1 1 1 , , k k k l l l l µ µ µ + 1 1 2 1 k l l k l + = and 2 1 2 2 k l l k l + = or 2 1 1 1 l k k l é ù + = ê ú ë û and 1 2 2 1 l k k l é ù + = ê ú ë û 1 [1 2] 3 k k k = + = [using (i)] 2 3 1 1 2 2 k k k é ù = = + ê ú ë û [using (i)] 93. (a) R = mg cos q – 2 mV r V car R O mg q mg cos q Over bridge q When q decreases, cosq increases i.e. R increases 94. (c) Impulse experienced by the body = change in momentum = MV– (–MV) = 2MV. 95. (a) The force of friction on the chain lying on the table should be equal to the weight of the hanging chain. Let r = mass per unit length of the chain µ = coefficient of friction l = length of the total chain x = length of hanging chain Now, µ(l – x) rg = xrg or µ(l – x) = x or µl = (µ + 1)x or x = µl/(µ+ 1) 0.25 0.25 0.2 (0.25 1) 1.25 l l x l = = = + 0.2 20% x l = = 96. (a) Masses of the pieces are 1, 1, 3 kg. Hence 2 2 2 (1 21) (1 21) (3 ) V ´ + ´ = ´ That is, 7 2 V = m/s 97. (c) 2 1 1 2 2 2 1 2 K p m K m p = ´ [Q p = mv Þ 2 = 2 p K m ] Hence, 1 1 2 2 1 1 4 2 p M p M = = = 98. (d) Work done in moving an object against gravitational force depends only on the initial and final position of the object, not upon the path taken. But gravitational force on the bodyalong the inclined plane is not same as that along the vertical and it varies with angle of inclination. 99. (b) On a rainy day, the roads are wet. Wetting of roads lowers the coefficient of friction between the types and the road. Therefore, grip on a road of car reduces and thus chances of skidding increases. 100. (a) The force acting on the body of mass M are its weight Mg acting vertically downward and air resistance F acting vertically upward. Acceration of the body , M F g a - = Now M > m, therefore, the bodywith larger mass will have great acceleration and it will reach the ground first. EXERCISE - 3 Exemplar Questions 1. (c) In a uniform translatory motion if all the parts of the body moves with (same velocity in same straight line, so the velocity is constant. A v v v The situation is shown in (figure) where a body A is in unfirom translatory motion.
- 140. 136 PHYSICS 2. (b)According to question we have to apply Newton's second law of motion, in terms of force and change in momentum. We know that dp F dt = As given that the meter scale is moving with uniform velocity, hence Force (F) = m × 0 = 0 No change in its velocityi.e., acceleration of it zero by Newton's second law. Hence, net or resultant force must act on body is zero. , r F t = ´ r r r so, As all part ofthe scale is moving with uniform velocity and total force is zero, hence, torque will also be zero. 3. (c) As given that, Mass of the ball = 150 g = 0.15 kg ˆ ˆ (3 4 ) m/s u i j = + r ˆ ˆ (3 4 ) m/s v i j = - + r (Dp) Change in momentum = Final momentum – Initial momentum mv mu = - r r ˆ ˆ ˆ ˆ ( ) (0.15)[ (3 4 ) (3 4 )] m v u i j i j = - = - + - + r r ˆ ˆ (0.15)[ 6 8 ] i j = - - ˆ ˆ [0.15 6 0.15 8 ] i j = - ´ + ´ ˆ ˆ [0.9 1.20 ] i j = - + ˆ ˆ [0.9 1.2 ] p i j D = - + Hence verifies option (c). 4. (c) From previous solution ˆ ˆ ˆ ˆ (0.9 1.2 ) 0.9 1.2 p i j i j D = - + = - - Magnitude of 2 2 | | ( 0.9) ( 1.2) p D = - + - 0.81 1.44 = + 2.25 1.5 = = kg-m s–1 Verifies the option (c). 5. (d) By Newton's second law : ext dp F dt = r As ext F r in law of conservation of momentum is zero. If 0, 0 ext F dp p = = Þ = constant Hence, momentum of a system will remain conserve if external force on the system is zero. In case of collision between particle equal and opposite forces will act on individual particle by Newton's third law. So, 12 21 F F = - r r ( 0) ext F = r Q Total force on the system will be zero. 12 21 dp dp dt dt = - r r or 12 21 dp dp = - r r 12 21 ( 0) dp dp Þ + = r r So prove the law of conservation of momentum and verifies the option (d). 6. (c) Consider the adjacent diagram. A B B O E S N P2 P2 P1 – ( ) W P1 The force on player is due to rate of change of momentum. The direction offorce acting on player will be the same as the direction of change in momentum. Let OA = P1 i.e., Initial momentumof player northward AB = P2 i.e., Final momentum of player towards west. Clearly, OB =OA +OB Changein momentum = P2 – P1 = AB – OA = AB + (–OA) = Clearlyresultant AR will be along south-west. So, it will be also the direction of force on player. 7. (a) As given that, mass = 2 kg p = 3 m/s, q = 4 m/s2, r = 5 m/s3 As given equation is 2 3 ( ) x t pt qt rt = + + 2 ( ) 2 3 ds t v p qt rt dt = = + + 2 2 ( ) 0 2 6 dv d x t a q rt dt dt = = = + + 2 2 ( 2) ( ) 2 12 t d x t q r dt = é ù = + ê ú ë û 2 12 q r = + =2 ×4 +12 ×5 = 8 + 60 = 68 m/s2 Force acting on body ( ) = F ma r = 2 × 68= 136 N. 8. (b) As given that mass = m = 5 kg Acting force = ˆ ˆ ( 3 4 ) N F i j = - + r
- 141. 137 Lawsof Motion Initial velocityatt = 0, ˆ ˆ (6 12 ) m/s u i j = - r Retardation, 2 ˆ ˆ 3 4 ˆ m/s 5 5 F i j a m æ ö = = - + ç ÷ è ø r As the final velocity along Y-component only. So its x- component must be zero. From , x x x v u a t = + for X-component only, , ˆ 3 ˆ 0 6 5 i i t - = + ˆ 3 ˆ ( ) 6 5 = i t i 5 6 10 s 3 t ´ = = t = 10 sec. Hence verifies the option (b). 9. (b) As given that mass of the car = m As car starts from rest, u = 0 Velocity acquired along east ˆ ( ) v vi = r Time (t) =2 s. We know that, v u at = + ˆ 0 2 vi a Þ = + ´ ˆ 2 Þ = v a i r (Force, byengine is internal force) ˆ 2 mv F ma i = = r r Hence, force acting on the car is 2 mv towards east due to force of friction is ˆ 2 mv i which moves the car in eastward direction. Hence, force by engine is internal force. NEET/AIPMT (2013-2017) Questions 10. (d) 2m 3m 2mg mg mg m v m 2m 3m T T' T" F T T' mg 2mg 3mg 6 mg From figure F= 6 mg, As speed is constant, acceleration a = 0 6 mg= 6ma= 0, F = 6mg T = 5mg , T¢ = 3 mg T² = 0 Fnet on block of mass 2 m = T – T' – 2 mg = 0 ALTERNATE: Q v = constant so, a = 0, Hence, Fnet = ma = 0 11. (d) Given; speed = 10 m/s; radius r = 10 m Angle made bythe wire with the vertical 2 2 10 tan 1 10 10 v rg q = = = ´ Þ 45 4 p q = ° = 12. (a) Let upthrust of air be Fa then For downward motion of balloon Fa = mg – ma mg – Fa = ma For upward motion Fa – (m – Dm)g= (m – Dm)a Therefore Dm = 2ma g a + 13. (c) Changein momentum, Dp = Fdt ò = Area of F-t graph = ar of D – ar of + ar of = 1 2 6 3 2 4 3 2 ´ ´ - ´ + ´ = 12 N-s 14. (c) Acceleration = Net force in the direction of motion Total mass of system = 1 2 3 1 2 3 m g (m m )g m m m - m + + + = g (1 2 ) 3 - m (Q m1 = m2 = m3 = m given) 15. (a) Acceleration of system net total F a M = = 2 14 14 2 m / s 4 2 1 7 = = + + A 4kg B 2kg C 1kg 14N The contact force between A and B = (mB + mC) × a = (2+ 1) ×2 = 6N 16. (b) For the motion of both the blocks m1a = T – mkm1g m2g – T = m2a
- 142. 138 PHYSICS m1 m2 mk m g 2 a mg 1 mk T a a = 2 k 1 1 2 m g – m g m m m + m2g – T = (m2) 2 k 1 1 2 m g – m g m m æ ö m ç ÷ + è ø solving we get tension in the string T = 1 2 k 1 2 m m g (1 )g m m + m + 17. (d) According to question, two stones experience same centripetal force i.e. 1 2 C C F F = or, 2 2 1 2 mv 2mv r (r / 2) = or, 2 2 1 2 V 4V = So, V1 = 2V2 i.e., n = 2 18. (a) Coefficient of static friction, ms = tan 30°= 1 3 = 0.577@ 0.6 S = ut + 2 1 at 2 4 = 1 2 a(4)2 Þ a = 1 2 = 0.5 [Q s = 4m and t = 4s given] a = gsinq – mk(g) cosq Þ mk = 0.9 3 = 0.5 19. (d) To complete the loop a bodymust enter a vertical loop ofradius R with the minimum velocity v= 5gR . 20. (d) Net force on particle in uniform circular motion is cen- tripetal force 2 mv l æ ö ç ÷ ç ÷ è ø which is provided by tension in string so the net force will be equal to tension i.e., T. 21. (a) m mg Before cutting the string kx=T +3mg ...(i) T = mg ...(ii) Þ kx=4mg After cutting the string T = 0 A 4mg 3mg a 3m - = 4mg A g a 3 = and B mg a g m = =
- 143. WORK Work Done bya Constant Force Work done (W) by a force F ur in displacing a body through a displacement x is given by W = F.x r r = Fx cos q F x Body F cos q q Fsinq Where q istheanglebetween theappliedforce F r and displacement x r . The S.I. unit ofwork isjoule, CGS unit is erg and itsdimensions are[ML2T–2]. 1 joule = 107 erg (a) When q = 0° then W = Fx (b) When q is between 0 and p/2 then W = Fx cos q = positive (c) When q = p/2 then W = Fx cos 90° = 0 (zero) Work done bycentripetal force is zero as in this case angle q =90° (d) When q is between / 2 and p p then W = Fx cos q = negative Work Done by a Variable Force When the force is an arbitrary function of position, we need the techniques of calculus toevaluate the work done byit. The figure shows Fx as function of the position x. We begin by replacing the actual variation of the force by a series of small steps. The area under each segment of the curveis approximatelyequal tothe area ofa rectangle. The height ofthe rectangle is a constant valueof force, and its width is a small displacement Dx. Thus, the step involves an amount of work DWn = Fn Dxn. The total work done is approximately given by the sum of the areas of the rectangles. i.e., W » n F å Dxn. As the size of the steps is reduced, the tops of the rectangle more closelytrace the actual curve shown in figure. Ifthe limit Dx ®0, which is equivalent to letting the number ofsteps tend to infinity, the discrete sum is replaced by a continuous integral. 0 lim n n n x x W F x F dx D ® = D = å ò Thus, the work done bya force Fx from an initial point A tofinal point B is n A x A B x x W F dx ® = ò The work done bya variable force in displacing a particle from x1 to x2 ò = 2 1 x x Fdx W = area under force displacement graph CAUTION : When we find work, we should be cautious about the question, work done bywhich force? Let us take an example to understand this point. Suppose you are moving a body up without acceleration. Work done by applied force app app app W F x F x · = = uu r uu r Work done by gravitational force x mg Fapplied g grav W F x mgx = = - uuu r uu r uu r 6 Work, Energy and Power
- 144. 140 PHYSICS ENERGY It isthe capacity of doing work. Its units and dimensions are same as that of work. Potential Energy The energy possessed by a body by virtue of its position or configuration is called potential energy. Potential energy is defined only for conservative forces. It does not exist for non conservative forces. (a) Elastic potential energy (Potential energy of a spring) : Let us consider a spring, its one end is attached to a rigid wall and other is fixed to a mass m. We apply an external force . ext F r on mass m in the left direction, so that the spring is compressed by a distance x. l m Fext x Ifspring constant is k, then energystored in spring is given by P.E. of compressed spring = ½kx2 Now if the external force is removed, the mass m is free to move then due to the stored energy in the spring, it starts oscillating (b) Gravitational potential energy : When a bodyis raised to some height, above the ground, it acquires some potential energy, due to its position. The potential energy due to height is called gravitational potential energy. Let us consider a ball B, which is raised by a height h from the ground. h Ground mg Fapp B In doing so, we do work against gravity and this work is stored in the ball B in the form of gravitational potential energy and is given by W = Fapp. h = mgh = gravitational potential energy ...(i) Further ifball B has gravitational P.E. (potential energy) Uo at ground and at height h, Uh, then Uh–Uo =mgh ...(ii) If we choose Uo= 0 at ground (called reference point) then absolute gravitational P.E of ball at height h is Uh = mgh ...(iii) In general, if two bodies of masses m1 and m2 are separated by a distance r, then the gravitational potential energy is 1 2 m m U G r = - Kinetic Energy The energy possessed by a body by virtue of its motion is called kinetic energy. The kinetic energy Ek is given by Ek = ½ mv2 ...(i) Where m is mass of body, which is moving with velocity v. We knowthat linear momentum (p) ofa bodywhich is moving with a velocity v is given by p = mv ...(ii) Sofrom eqs. (i) and (ii), we have 2 2 k p E m = ...(iii) This is the relation between momentum and kinetic energy. The graph between k E and p is a straight line k p 2m E = Q p Ek The graph between k E and p 1 is a rectangular hyperbola Ek 1 p The graph between Ek and 1 m is a rectangular hyperbola Ek 1 m p is constant Keep in Memory 1. Work done by the conservative force in moving a body in a closed loop is zero. Work done by the non-conservative force in moving a body in a closed loop is non-zero. 2. If the momenta of two bodies are equal then the kinetic energyof lighter bodywill be more. Q 2 2 1 1 2 1 E m 2 E m 2 or p p = =
- 145. 141 Work, Energy andPower 1 2 2 1 m m E E = 3. If the kinetic energies of two bodies are same then the momentum of heavier bodywill be more. 2 1 E E = Q 2 1 2 1 m m p p = WORK-ENERGY THEOREM Leta number offorcesacting on a bodyofmass mhave a resultant force . ext F r And by acting over a displacement x (in the direction of . ext F r ), . ext F r does work on the body, and there bychanging itsvelocityfrom u (initial velocity) tov (final velocity). Kinetic energy of the body changes. So, work done by force on the body is equal to the change in kinetic energy of the body. 2 2 W ½ mv ½mu = - This expression is called Work energy (W.E.) theorem. LAW OF CONSERVATION OF MECHANICAL ENERGY The sum of the potential energy and the kinetic energy is called the total mechanical energy. The total mechanical energy of a system remains constant if only conservative forces are acting on a system of particles and the work done by all other forces is zero. i.e., DK+ DU= 0 or Kf – Ki + Uf – Ui = 0 or Kf + Uf = Ki + Ui = constant VARIOUS FORMS OF ENERGY : THE LAW OF CONSERVATION OF ENERGY Energy is of many types – mechanical energy, sound energy, heat energy, light energy, chemical energy, atomic energy, nuclear energy etc. In manyprocesses that occur in natureenergymaybe transformed from one form to other. Mass can also be transformed intoenergy and vice-versa. This is according to Einstein’s mass-energy equivalence relation, E = mc2. In dynamics, we are mainly concerned with purely mechanical energy. Law of Conservation of Energy : The study of the various forms of energy and of transformation of one kind of energy into another has led to the statement of a very important principle, known as the law of conservation of energy. "Energy cannot be created or destroyed, it may only be transformed from one form into another. As such the total amount of energy never changes". Keep in Memory 1. Work done against friction on horizontal surface = mmgx and work done againstforce of friction on inclined plane = (mmg cosq) x where m = coefficient of friction. 2. If a body moving with velocity v comes to rest after covering a distance ‘x’ on a rough surface having coefficient of friction m, then (from work energytheorem), 2m gx = v2. Here retardation is a g =-m u r ur 3. Work done by a centripetal force is always zero. 4. Potential energy of a system decreases when a conservative force does work on it. 5. If the speed of a vehicle is increased by n times, then its stopping distance becomes n2 times and if momentum is increased by n times then its kinetic energy increases by n2 times. 6. Stopping distance of the vehicle force Stopping energy Kinetic = 7. Two vehicles of masses M1 and M2 are moving with velocities u1 and u2 respectively. When they are stopped by the same force, their stopping distance are in the ratio as follows : Since the retarding force F is same in stopping both the vehicles. Let x1 and x2 are thestopping distances ofvehicles of masses M1 & M2 respectively, then ) M mass the stopping in done work ( F.x ) M mass the stopping in done work ( x . F 2 2 1 1 2 k 1 1 1 2 k 2 2 2 E ½M u E ½M u = = ....(i) where u1 and u2 are initial velocity of mass M1 & M2 respectively & final velocity of both mass is zero. 2 1 k k 2 1 E E x x = Þ ....(ii) Let us apply a retarding force F on M1 & M2, a1 & a2 are the decelerations ofM1 &M2 respectively. Then from third equation of motion ( ) 2 2 v u 2ax = + : 1 2 1 1 1 1 2 1 x 2 u a x a 2 u 0 = Þ - = ....(iiia) and 2 2 2 2 2 2 2 2 x 2 u a x a 2 u 0 = Þ - = ....(iiib) If t1 & t2 are the stopping time of vehicles of masses M1 & M2 respectively, then from first equation of motion (v= u+at) 1 1 1 1 1 1 a u t t a u 0 = Þ - = ....(iva) and 2 2 2 2 2 2 a u t t a u 0 = Þ - = ....(iv b)
- 146. 142 PHYSICS Then byrearrangingequation (i), (iii) & (iv), we get ÷ ÷ ø ö ç ç è æ ´ ÷ ÷ ø ö ç ç è æ = = 1 2 2 1 2 2 1 1 2 1 u u x x a / u a / u t t ÷ ÷ ø ö ç ç è æ = ´ = Þ 2 2 1 1 1 2 2 2 2 2 1 1 2 1 u M u M u u ) u M (½ ) u M (½ t t (a) If 2 1 2 1 2 1 M M t t u u = Þ = (b) If 2 1 2 1 2 1 u u t t M M = Þ = (c) If M1u1 = M2u2 Þ t1 = t2 and 2 1 1 1 2 1 2 2 2 1 2 1 2 2 x (M u ) M x M x M x M (M u ) = ´ Þ = (d) Consider two vehicles of masses M1 & M2 respectively. Iftheyare moving with same velocities, then the ratio of their stopping distances bythe application of same retarding force is given by 2 1 2 1 M M x x = and let M2 > M1 then x1 < x2 Þ lighter mass will cover less distance then the heavier mass And the ratio of their retarding times are as follows : 2 1 2 1 M M t t = i.e 2 1 2 1 t t x x = 8. If kinetic energy of a body is doubled, then its momentum becomes 2 times, 2 k k p E p E 2m = Þ µ 9. If two bodies of masses m1 and m2 have equal kinetic energies, then their velocities are inversely proportional to the square root of the respective masses. i.e. 2 2 1 1 2 2 1 1 m v m v then 2 2 = 1 2 2 1 m m v v = 10. (a) The spring constant of a spring is inversely proportional to the no. of turns i.e. 1 K n µ or kn = const. (b) Greater the no. of turns in a spring, greater will be the work done i.e. W µ n (c) The greater is the elasticity of the spring, the greater is the spring constant. 11. Spring constant : The spring constant of a spring is inversely proportional to length i.e., l 1 K µ or Kl = constant. (a) If a spring is divided into n equal parts, the spring constant of each part = nK. (b) If spring of spring constant K1, K2, K3 .......... are connected in series, then effective force constant ...... K 1 K 1 K 1 K 1 3 2 1 eff + + + = (c) If spring of spring constant K1, K2, K3........... are connected in parallel, then effective spring constant Keff = K1 + K2 + K3 +............. POWER Power of the body is defined as the time rate of doing work by the body. The average power Pav over the time interval Dt is defined by av W P t D = D ...(i) And the instantaneous power P is defined by 0 t W dW P Lim t dt d ® D = = D ...(ii) Power is a scalar quantity The S.I. unit of power is joule per second 1 joule/sec = 1watt The dimensions of power are [ML2T–3] ( . ) . . dW d dS P F S F F v dt dt dt = = = = r r r r r r (force is constant over a small time interval) So instantaneous power (or instantaneous rate of working) of a man depends not only on the force applied to body, but also on the instantaneous velocity of the body. Example1. A metre stick of mass 600 mg, is pivoted at one end and displaced through an angle of 60º. The increase in its potential energy is (g = 10 ms–2) (a) 1.5 J (b) 15 J (c) 150 J (d) 0.15 J Solution : (a) G M O h 60º x l/2 l / 2 G´ A´ A TheC.G. of stick rises from G toG´. Increase in P.E. = mgh = mg (l/2 –x) = (1 cos60º) 2 mg - l J 5 . 1 2 1 1 2 1 x 10 x 6 . 0 = ú û ù ê ë é - =
- 147. 143 Work, Energy andPower Example2. A block of mass M slides along the sides of a flat bottomed bowl. The sides of the bowl are frictionless and the base has a coefficient of friction 0.2. If the block is released from the top of the side which is 1.5 m high, where will the block come to rest if, the length of the base is 15 m ? (a) 1 mfromP (b) Mid point of flat part PQ (c) 2 m fromP (d)At Q Solution : (b) P.E. of the block at top of side = 1.5 mg. This is wasted away in doing work on the rough flat part, 15 m 1.5 m P Q 1.5mg = m mg.x or . m 5 . 7 5 . 1 x = m = i.e, the block comes to rest at mid-point of PQ. Example3. Fig. given below shows a smooth curved track terminating in a smooth horizontal part. A spring of spring constant 400 N/m is attached at one end to the wedge fixed rigidly with the horizontal part. A 40 g mass is released from rest at a height of 4.9 m on the curved track. Find the maximum compression of the spring. 4.9 m Solution : According to the law of conservation of energy, 2 x k 2 1 h g m = wherexis maximumcompression. 2m g h x k æ ö = ç ÷ è ø or . cm 8 . 9 ) m / N 400 ( ) m 9 . 4 ( 0 ) s / m 8 . 9 ( ) kg 4 . 0 ( 2 x 2 = ï þ ï ý ü ï î ï í ì ´ ´ ´ = Example4. The K.E. of a body decreases by 19%. What is the percentage decrease in momentum? Solution : As p 2m K.E = f i f i i i K.E K.E p p 100 100 p K.E 81 100 100 100 10% - - ´ = ´ æ ö - = ´ ç ÷ è ø = - Example5. A particle of mass m is moving along a circular path of constant radius R. The centripetal acceleration varies as a = K2 Rt2, where K is a constant and t is the time elapsed. What is the power delivered to the particle by the force acting on it? Sol. For circular motion, ac = v2 / R here K2 Rt2 = R v2 or v2 = K2 R2 t2 Now, 2 2 2 2 t R K . m 2 1 mv 2 1 KE = = Work done in time t = W = DK (from work energytheorem) 2 2 2 2 2 2 t R K 2 m 0 ) t R K ( 2 m K = - = D 2 2 2 2 dW m Power K .R .2t mK R t dt 2 = = = Example6. An electron of mass 9.0 × 10–28 g is moving at a speed of 1000 m/sec. Calculate its kinetic energy if the electron takes up this speed after moving a distance of 10 cm from rest. Also calculate the force in kg weight acting on it. Solution : K.E. = 2 31 3 2 1 1 m 9 10 (10 ) 2 2 - n = ´ ´ = 4.5× 10–25 J ; From 2 n – u2 =2as, 2 n = 2as Q u = 0 2 3 2 1 (10 ) a 2s 2 10- n = = ´ F = ma = 9 × 10–31 (0.5 × 107) N 24 4.5 10 9.8 - ´ = kgwt. = 0.46 × 10–24 kg wt. Example7. The bob of a simple pendulum of length 1 m is drawn aside until the string becomes horizontal. Find the velocity of the bob, after it is released, at the equilibrium position. Solution : When the bob is raised from A to B the height through which it is raised is the length of the pendulum. h= 1m Taking A as the standard level. P.E. at B= mgh = m × 9.8 × 1 = (9.8) m joule, where m isthe mass of the bob. Since it is at rest at B it has no K.E.
- 148. 144 PHYSICS When thebob reaches A after it is released from B, its energyat Ais kinetic one. P.E. at Ais zero. If v be the velocity at A, from the law of conservation of energy K.E. atA= P.E. at B 1 2 mv2 = mgh or v2 = 2gh Þ v = 2gh 2 9.8 1 196 4.427 m/s = ´ ´ = = The bob has a velocity 4.427 m/s at A. COLLISION Collision between two bodies is said to take place if either of two bodies come in physical contact with each other or even when path of one body is affected by the force exerted due to the other. Collision Elastic collision (e = 1) • Total energy conserved • Total momentum conserved • K.E is conserved Head on collision (Impact parameter b = 0) Inelastic collision Oblique collision (Impact parameter b 0) ¹ Perfectly inelastic collision (e = 0) Inelastic collision (0 < e < 1) • Total energy conserved • Total momentum conserved • K.E is not conserved Head on collision Oblique collision (1) Elastic collision : The collision in which both the momentum and kinetic energy of the system remains conserved is called elastic collision. Forces involved in the interaction of elastic collision are conservative in nature. (2) Inelastic collision : The collision in which only the momentum of the system is conserved but kinetic energy is not conserved is called inelastic collision. Perfectly inelastic collision is one in which the twobodies stick together after the collision. Forces involved in the interaction of inelastic collision are non-conservative in nature. Coefficient of Restitution (or coefficient of resilience) : It is the ratio of velocity of separation after collision to the velocity of approach before collision. i.e., e = | v1 – v2 |/ | u1 – u2 | Here u1 and u2 are the velocities of two bodies before collision and v1 and v2 are the velocities of two bodies after collision. 1. 0 < e < 1 (Inelastic collision) Collision between two ivoryballs, steel balls or quartz ball is nearly elastic collision. 2. For perfectly elastic collision, e = 1 3. For a perfectly inelastic collision, e = 0 Oblique Elastic Collision : When a body of mass m collides obliquely against a stationary body of same mass then after the collision the angle between these two bodies is always 90°. Elastic Collision in One Dimension (Head on) Let two bodies of masses M1 and M2 moving with velocities u1 and u2 along the same straight line, collide with each other. Let u1>u2. Suppose v1 and v2 respectively are the velocities after the elastic collision, then: According to law of conservation of momentum 2 2 1 1 2 2 1 1 v M v M u M u M + = + ...(1) M1 M1 u1 v1 u2 v2 M2 M2 Before collision After collision From law of conservation of energy 2 2 2 2 1 1 2 2 1 1 2 2 1 1 1 1 M u M u M v M v 2 2 2 2 + = + ...(2) 1 2 1 2 u u – (v v ) - = - ...(3) Relative velocity of a Relative velocity of a body before collision body after collision Solving eqs. (1) and (2) we get, ) M M ( u M 2 ) M M ( u ) M – M ( v 2 1 2 2 2 1 1 2 1 1 + + + = ...(4) ) M M ( u M 2 ) M M ( u ) M – M ( v 2 1 1 1 2 1 2 1 2 2 + + + = ...(5) From eqns. (4) and (5), it is clear that : (i) If M1 = M2 and u2 = 0 then v1 = 0 and v2 = u1. Under this condition the first particle comes to rest and the second particle moves with the velocity of first particle after collision. In this state there occurs maximum transfer of energy. (ii) If M1>> M2 and (u2=0) then, v1 = u1, v2 = 2u1 under this condition the velocity of first particle remains unchanged and velocity of second particle becomes double that of first. (iii) If M1 << M2 and (u2 = 0) then v1 = –u1 and v2 = » 1 2 1 u M M 2 0 under this condition the second particle remains at rest while the first particle moves with the same velocity in the opposite direction. (iv) If M1 = M2 = M but u2 ¹ 0 then v1 = u2 i.e., the particles mutuallyexchange their velocities. (v) Ifsecond bodyis at rest i.e., u2 = 0, then fractional decrease in kinetic energy of mass M1, is given by 2 2 1 2 1 2 1 2 1 k k k ) M M ( M M 4 u v 1 E ' E E 1 1 1 + = - = -
- 149. 145 Work, Energy andPower Inelastic Collision : Let two bodies A and B collide inelastically. Then from law of conservation of linear momentum M1u1 + M2u2 = M1v1+M2v2 ...(i) velocity of separation Coefficient of restitution velocityof approach æ ö = -ç ÷ è ø e ) u u ( ) v v ( 2 1 2 1 - - - = ...(ii) From eqns.(i) and (ii), we have, 1 2 2 1 1 2 1 2 1 2 M eM M (1 e) v u u M M M M æ ö æ ö - + = + ç ÷ ç ÷ + + è ø è ø ...(iii) 2 2 1 1 2 1 2 1 1 2 u M M eM M u M M M ) e 1 ( v ÷ ÷ ø ö ç ç è æ + - + ÷ ÷ ø ö ç ç è æ + + = ...(iv) Loss in kineticenergy(–DEk) = initial K.E. – final K.E Þ ú û ù ê ë é + - ú û ù ê ë é + = D - 2 2 2 2 1 1 2 2 2 2 1 1 k v M 2 1 v M 2 1 u M 2 1 u M 2 1 E Þ ( )( )2 2 1 2 2 1 2 1 k u u 1 e M M M M 2 1 E - - ÷ ÷ ø ö ç ç è æ + = D - ...(v) Negative sign indicates that the final kinetic energy is less than initial kinetic energy. Perfectly Inelastic Collision In this collision, theindividual bodiesAand Bmove with velocities u1 and u2 but after collision move as a one single body with velocity v. Sofrom law ofconservation of linear momentum, wehave M1u1+M2u2=(M1+M2)V ...(i) or ÷ ÷ ø ö ç ç è æ + + = 2 1 2 2 1 1 M M u M u M V ...(ii) B A u1 M1 u2 M2 (u > u 1 2) A B V M + M 1 2 Before collision After collision And loss in kinetic energy, –DEk = total initial K.E – total final K.E 2 2 1 2 2 1 1 2 1 2 2 2 2 1 1 M M u M u M ) M M ( 2 1 u M 2 1 u M 2 1 ÷ ÷ ø ö ç ç è æ + + + - + = or, 2 2 1 2 1 2 1 k ) u u ( ) M M ( M M 2 1 E - + = D - ...(iii) Oblique Collision : This is the case ofcollision in twodimensions.After the collision, the particles move at different angle. A B Before collision 1 u 0 u2 = A B 1 v 2 v q f After collision m1 m1 m2 m2 x axis y axis We will applythe principle of conservation of momentum in the mutuallyperpendicular direction. Along x-axis, m1u1 = m1v1 cosq + m2 v2 cosf Along y-axis, 0 = m1v1 sinq - m2 v2 sinf Keep in Memory 1. Suppose, a body is dropped from a height h0 and it strikes the ground with velocity v0. After the (inelastic) collision let it rise to a height h1. If v1 be the velocitywith which the body rebounds, then the coefficient of restitution. 1 2 1 1 0 0 v 2gh e v 2gh æ ö = = ç ÷ è ø = 2 1 0 h h ÷ ÷ ø ö ç ç è æ h0 v0 v1 h1 2. If after n collisions with the ground, the velocity is vn and the height to which it rises be hn, then 2 1 o n o n n h h v v e ÷ ÷ ø ö ç ç è æ = = 3. When a ball is dropped from a height h on the ground, then after striking the ground n times , it rises to a height hn = e2n ho where e = coefficient of restitution. 4. If a body of mass m moving with velocity v, collides elastically with a rigid ball, then the change in the momentum of the bodyis 2 m v. (i) If the collision is elastic then we can conserve the energy as 2 2 2 2 1 1 2 1 1 v m 2 1 v m 2 1 u m 2 1 + = (ii) Iftwoparticles having same mass and moving at right angles to each other collide elastically then after the collision theyalso move at right angles to each other. (iii) If a bodyA collides elastically with another body of same mass at rest obliquely, then after the collision the two bodies move at right angles to each other, i. e. (q+ f)= 2 p 5. In an elastic collision of two equal masses, their kinetic energies are exchanged.
- 150. 146 PHYSICS 6. Whentwo bodies collide obliquely, their relative velocity resolved along their common normal after impact is in constant ratio to their relative velocity before impact (resolved along common normal), and is in the opposite direction. 1 m 2 m 1 q 2 q b 2 m 1 m 1 u 2 u v1 v2 After collision Before collision 1 1 2 2 1 2 v sin v sin e u sin u sin q - q = - a - b Example8. A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. What will be the ratio of K.E. of colliding body before and after collision? Solution : mv+ 2m × 0 = mv1 +2mv2 ; v = v1 + 2v2 or 2 v v v 1 2 - = ; 2 2 2 1 2 mv 2 2 1 mv 2 1 mv 2 1 + = 2 1 2 1 2 2 2 1 2 2 v v 2 v v 2 v v ÷ ø ö ç è æ - + = + = 2 vv 2 v v v v 1 2 2 1 2 1 2 - + + = or 1 2 2 1 2 1 2 vv 2 v v v 2 v 2 - + + = or 0 v vv 2 v 3 2 1 2 1 = - - ; 3 v v1 - = collision after body colliding of . E . K collision before body colliding of . E . K 2 2 3 v m 2 1 mv 2 1 ÷ ø ö ç è æ = = 9 : 1 Example9. A bullet of mass m moving horizontally with velocity v hits a block of wood of mass M, resting on a smooth horizontal plane. Find the fraction of energy of the bullet dissipated in the collision itself (assume collision to be inelastic). Solution : Applying the law of conservation of momentum, we have, m v = (m + M) v1 ÷ ø ö ç è æ + = M m mv v1 Loss of K.E. 2 2 1 1 1 mv (m M)v 2 2 = - + 2 2 m M v m ) M m ( 2 1 v m 2 1 ÷ ø ö ç è æ + + - = ú û ù ê ë é + = ú û ù ê ë é + - = M m M v m 2 1 M m m 1 v m 2 1 2 2 Fraction of K.E. dissipated . E . K Initial . E . K of Loss = ÷ ø ö ç è æ + = M m M Example10. A smooth sphere of mass 0.5 kg moving with horizontal speed 3 m/sstrikesat right anglesa vertical wall andbounces offthe wall with horizontal speed 2 m/s. Find the coefficient of restitution between the sphere and the wall and the impulses exerted on the wall at impact. Solution : At impact Just after impact Just before impact ////////////////// 2m/s 3m/s J J e = separation speed : approach speed = 2 : 3 Therefore the coefficient of restitution is 2/3. Using impulse = change in momentum for the sphere we have : = 0.5 × 2 – 0.5 (–3) = 2.5 The equal and opposite impulse acting on the wall is therefore 2.5 N s.
- 151. 147 Work, Energy andPower
- 152. 148 PHYSICS 1. Themagnitude of work done by a force : (a) depends on frame of reference (b) does not depend on frame of reference (c) cannot be calculated in non-inertial frames. (d) both (a) and (b) 2. Work done by a conservative force is positive if (a) P.E. of the body increases (b) P.E. of the body decreases (c) K.E. of the body increases (d) K.E. of the body decreases 3. A vehicle is moving with a uniform velocity on a smooth horizontal road, then power delivered byits enginemust be (a) uniform (b) increasing (c) decreasing (d) zero 4. Which of the following force(s) is/are non-conservative? (a) Frictional force (b) Spring force (c) Elasticforce (d) All of these 5. A ball ofmass m and a ball B ofmass 2m are projected with equal kinetic energies. Then at the highest point of their respective trajectories. (a) P.E. of A will be more than that ofB (b) P.E of B will be more than that of B (c) P.E of A will be equal to that of B (d) can’t be predicted. 6. In case of elastic collision, at the time of impact. (a) total K.E. of colliding bodies is conserved. (b) total K.E. of colliding bodies increases (c) total K.E. of colliding bodies decreases (d) total momentum of colliding bodies decreases. 7. The engine of a vehicle delivers constant power. If the vehicle is moving up the inclined plane then, its velocity, (a) must remain constant (b) must increase (c) must decrease (d) mayincrease, decrease or remain same. 8. A ball projected from ground at a certain angle collides a smooth inclined plane at the highest point of its trajectory. Ifthe collision is perfectlyinelastic then after the collision, ballwill (a) come to rest (b) move along the incline (c) retrace its path. 9. The vessels A and B of equal volume and weight are immersed in water todepth h. The vesselAhas an opening at the bottom through which water can enter. If the work done in immersing A and B are WA and WB respectively, then (a) WA = WB (b) WA < WB (c) WA > WB (d) WA > < W WB 10. A metallic wire of length L metre extends by l metre when stretchedbysuspending aweightMgfrom it. Themechanical energystored in the wire is (a) 2 Mg l (b) Mg l (c) Mg 2 l (b) Mg 4 l 11. A body of mass m accelerates uniformly from rest to v1 in time t1.As a function of t, theinstantaneous power delivered to the body is (a) 1 2 m t t n (b) 2 1 1 m t t n (c) 2 1 1 m t t n (d) 2 1 2 1 m t t n 12. A block is acted upon by a force, which is inversely proportional to the distance covered (x). The work done will be proportional to (a) x (b) x1/2 (c) x2 (d) None of these 13. A small bodyisprojectedin a direction inclined at45º tothe horizontal with kinetic energyK.At the top of its flight, its kinetic energy will be (a) Zero (b) K/2 (c) K/4 (d) K/ 2 14. A motor cycle is moving along a straight horizontal road with a speed v0. If the coefficient of friction between the tyres and the road is m, the shortest distance in which the car can be stopped is (a) 2 0 v 2 g m (b) 2 v m (c) 2 0 v g æ ö ç ÷ m è ø (d) 0 v g m 15. Consider the following twostatement: I. Linear momentum of a system ofparticles is zero. II. Kinetic energy of a system of particles is zero. Then (a) I implies II but II does not implyI. (b) I does not implyII but II implies I. (c) I implies II and II implies I. (d) I does not imply II and II does not imply I. 16. Which ofthefollowingmust be known in order todetermine the power output of an automobile? (a) Final velocity and height (b) Mass and amount of work performed (c) Force exerted and distance of motion (d) Work performed and elapsed time ofwork
- 153. 149 Work, Energy andPower 17. The work done in stretching a spring of force constant k from length 1 l and 2 l is (a) ) ( k 2 1 2 2 l l - (b) ) ( k 2 1 2 1 2 2 l l - (c) ) ( k 1 2 l l - (d) ) ( 2 k 1 2 l l + 18. If the force acting on a bodyis inverselyproportional to its velocity, then the kinetic energy acquired by the body in time t is proportional to (a) t0 (b) t1 (c) t2 (d) t4 19. Theengine ofa truckmoving along astraight road delivers constant power. The distance travelled bythe truck in time t is proportional to (a) t (b) t2 (c) t (d) t3/2 20. A bullet of mass ‘a’ and velocity ‘b’ is fired into a large block of wood of mass ‘c’. The bullet gets embedded into the block of wood. The final velocity of the system is (a) b c a b ´ + (b) a b a c + ´ (c) a b a c ´ + (d) a c b a + ´ 21. A ball is dropped from a height h. If the coefficient of restitution be e, then towhat height will it rise after jumping twice from the ground? (a) e h/2 (b) 2 e h (c) e h (d) e4 h 22. A ball of mass m moving with a constant velocity strikes against a ball of same mass at rest. If e = coefficient of restitution, then what will be the ratio of velocity of two balls after collision? (a) 1 e 1 e - + (b) e 1 e 1 - + (c) 1 e 1 e + - (d) 2 e e 1 + - 23. Which one of the following physical quantities is represented by the shaded area in the given graph? Fo rc e Distance (a) Torque (b) Impulse (c) Power (d) Work done 24. A particle of mass m1 moving with velocityv collides with a mass m2 at rest, then they get embedded. Just after collision, velocity of the system (a) increases (b) decreases (c) remains constant (d) becomes zero 25. A mass m1 moves with a great velocity. It strikes another mass m2 at rest in a head on collision. It comes back along its path with low speed, after collision. Then (a) m1 >m2 (b) m1 <m2 (c) m1 =m2 (d) cannot say 1. A particle describe a horizontal circle ofradius 0.5 m with uniform speed. The centripetal force acting is 10 N. The work done in describing a semicircle is (a) zero (b) 5 J (c) 5 pJ (d) 10pJ 2. A cord is used to lower vertically a block of mass M, a distance d at a constant downward acceleration of g/4. The work done by the cord on the block is (a) d Mg 4 (b) d 3Mg 4 (c) d 3Mg 4 - (d) Mg d 3. A boy pushes a toybox 2.0 m along the floor bymeans of a force of 10 N directed downward at an angle of 60º to the horizontal. The work done by the boy is (a) 6 J (b) 8 J (c) 10J (d) 12J 4. A particlemoving in the xyplane undergoes a displacement of ) ĵ 3 î 2 ( s + = r while a constant force ) ĵ 2 î 5 ( F + = r N acts on the particle. The work done by the force F is (a) 17 joule (b) 18 joule (c) 16 joule (d) 15 joule 5. A simple pendulum 1 metre long has a bob of 10 kg. If the pendulum swings from a horizontal position, the K.E. ofthe bob, at the instant it passes through the lowest position of its path is (a) 89 joule (b) 95 joule (c) 98 joule (d) 85 joule 6. A particle moves under the effect of a force F = cx from x = 0 to x = x1, the work done in the process is (a) 2 1 cx (b) 2 1 1 cx 2 (c) 2 1 2 cx (d) zero
- 154. 150 PHYSICS 7. Amotorof 100 H.P. moves a load with a uniform speed of72 km/hr. The forward thrust applied by the engine on the car is (a) 1111N (b) 3550N (c) 2222N (d) 3730N 8. Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of linear momentum of B toAis (a) 1: 3 (b) 3: 1 (c) 1: 3 (d) 3 :1 9. When a U238 nucleus, originallyat rest, decays byemitting an a-particle, saywith speed of vm/sec, the recoil speed of the residual nucleus is (in m/sec.) (a) – 4 v/234 (b) – 4 v/238 (c) 4v/238 (d) – v/4 10. Calculate the K.E and P.E. of the ball half way up, when a ball of mass 0.1 kg is thrown vertically upwards with an initial speed of 20 ms–1. (a) 10 J, 20 J (b) 10 J, 10 J (c) 15 J, 8 J (d) 8 J, 16 J 11. A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is (a) 16J (b) 8 J (c) 32J (d) 24J. 12. Ifthe linear momentumisincreased by5%, the kineticenergy will increase by (a) 50% (b) 100% (c) 125% (d) 10% 13. A cord is usedtolower verticallya block ofmass M, through a distance d at a constant downward acceleration of g/8. Then the work done by the cord on the block is (a) Mg d/8 (b) 3 Mg d/8 (c) Mg d (d) – 7 mg d/8 14. Aforce F (5i 3j 2k)N = + + r r r r is appliedover a particlewhich displaces it from its origin to the point r (2i j)m. = - r r r The work done on the particle in joule is (a) +10 (b) +7 (c) –7 (d) +13 15. A spring ofspring constant 5 × 103 N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is (a) 18.75J (b) 25.00J (c) 6.25J (d) 12.50J 16. Two solid rubber balls A and B having masses 200 & 400 gm respectively are moving in opposite direction with velocity of A equal to 0.3 m/sec. After collision the two balls come to rest when the velocity of B is (a) 0.15m/sec (b) 1.5m/sec (c) –0.15m/sec (d) None of these 17. A bomb of mass 9 kg explodes into the pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 m/s. The kinetic energy of mass 6 kg in joule is (a) 96 (b) 384 (c) 192 (d) 768 18. A long string is stretched by 2 cm and the potential energy isV. If the spring is stretched by10 cm, its potential energy will be (a) V/25 (b) V/5 (c) 5V (d) 25V 19. When the kinetic energy of a body is increased to three times, then the momentum increases (a) 6 times (b) 1.732times (c) 2 times (d) 2 times 20. Twobodies of masses 2 m and m have their KE in the ratio 8 : 1. What isthe ratio oftheir momenta ? (a) 8: 1 (b) 4: 1 (c) 2: 1 (d) 1: 1 21. A body of mass 5 kg initially at rest explodes into 3 fragments with mass ratio3: 1 :1. Twoof fragments each of mass ‘m’ are found tomove with a speed 60m/s in mutually perpendicular directions. The velocityofthird fragment is (a) 60 2 (b) 20 3 (c) 10 2 (d) 20 2 22. A machine, which is 75% efficient, uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in m/s) (a) 24 (b) 12 (c) 18 (d) 9 23. A body accelerates uniformly from rest to a velocity of 1 ms–1 in 15 seconds. The kinetic energyof the body will be 9 2 J when 't' is equal to [Take mass of body as 1 kg] (a) 4s (b) 8s (c) 10s (d) 12s 24. A machine gun fires a bullet of mass 40 g with a velocity 1200 ms–1. Theman holding it can exert a maximum forceof 144 N on the gun. Howmany bullets can he fire per second at the most? (a) 2 (b) 4 (c) 1 (d) 3 25. A crane is used to lift 1000 kg of coal from a mine 100 m deep. The time taken bythe crane is 1 hour. The efficiency of the crane is 80%. If g = 10 ms–2, then the power of the crane is (a) 104 W (b) 105 W (c) W 8 36 104 ´ (d) W 8 36 105 ´ 26. In figure, a carriagePis pulled up fromAto B. The relevant coefficient of friction is 0.40. The work done will be (a) 10kJ (b) 23kJ (c) 25kJ P 50 kg 50 30 m B A q C m (d) 28kJ
- 155. 151 Work, Energy andPower 27. A neutron with velocity v strikes a stationary deuterium atom, its K.E. changes by a factor of (a) 16 15 (b) 2 1 (c) 1 2 (d) None of these 28. A bodymoves a distance of10 m along a straight line under the action of a force of 5 newtons. If the work done is 25 joules, the angle which the force makes with the direction of motion of body is (a) 0º (b) 30º (c) 60º (d) 90º 29. A sphere of mass 8m collides elastically(in one dimension) with a block ofmass 2m. If the initial energyof sphere is E. What is the final energy of sphere? (a) 0.8E (b) 0.36E (c) 0.08E (d) 0.64E 30. Johnny and his sister Jane race up a hill. Johnny weighs twice as much as jane and takes twice as long as jane to reach the top . Compared to Jane (a) Johnnydid more work and delivered more power. (b) Johnnydid more work and delivered the same amount of power. (c) Johnny did more work and delivered less power (d) Johnnydid less work and johnnydelivered less power. 31. A bodyof mass m moving with velocityv makes a head on elastic collision with another body of mass 2m which in initially at rest. The loss of kinetic energy of the colliding body (mass m ) is (a) 2 1 ofits initial kinetic energy (b) 9 1 ofits initial kinetic energy (c) 9 8 ofits initial kinetic energy (d) 1 4 ofits initial kinetic energy 32. In the non-relativistic regime, ifthe momentum, is increase by 100% , the percentage increase in kinetic energy is (a) 100 (b) 200 (c) 300 (d) 400 33. A body is dropped from a height of 20m and rebounds to a height 10m. The loss of energy is (a) 10% (b) 45% (c) 50% (d) 75% 34. A moving body with a mass m1 and velocity u strikes a stationarybodyofmass m2. The masses m1 and m2 should be in the ratio m1/m2 so as to decrease the velocity of the first body to 2u/3 and giving a velocityofu to m2 assuming a perfectly elastic impact. Then the ratio m1/m2 is (a) 5 (b) 5 / 1 (c) 25 / 1 (d) 25. 35. A body of mass m is suspended from a massless spring of natural length l. It stretches the spring through a vertical distance y. The potential energy of the stretched spring is (a) ) y ( mg + l (b) ) y ( mg 2 1 + l (c) mgy 2 1 (d) mgy 36. Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over s = 0 to 20 m will be 15 10 5 0 0 5 10 20 x(m) f(N) (a) 250J (b) 200J (c) 150J (d) 10J 37. Ten litreof water per second is lifted from a well through 10 m and delivered with a velocity of 10 ms–1. If g = 10 ms–2 , then the power of the motor is (a) 1 kW (b) 1.5kW (c) 2 kW (d) 2.5kW 38. A nucleus ruptures into twonuclear parts which have their velocity ratio equal to 1 : 2 . The ratio of their respective nuclear sizes (nuclear radii )is (a) 1: 2 (b) 2 : 1 (c) 1 : 21/3 (d) 1: 8 39. The rest energyofan electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energywill be (a) 0.026MeV (b) 0.051MeV (c) 0.08MeV (d) 0.105MeV 40. A one-ton car moves with a constant velocity of 15 ms–1 on a rough horizontal road. The total resistance to the motion of the car is 12% of the weight of the car. The power required to keep the car moving with the same constant velocity of 15ms–1 is [Take g = 10 ms–2] (a) 9 kW (b) 18kW (c) 24kW (d) 36kW 41. Hail storms are observed to strike the surface of the frozen lake at 300 with the vertical and rebound at 600 with the vertical. Assume contact to be smooth, the coefficient of restitution is (a) 1 e 3 = (b) 1 e 3 = (c) e 3 = (d) e = 3
- 156. 152 PHYSICS 42. Abodystartsfrom rest and acquires a velocityV in time T. Thework done on the bodyin time t will be proportional to (a) V t T (b) 2 2 V t T (c) 2 2 V t T (d) 2 2 2 V t T 43. Aball isallowedtofall froma height of10 m. Ifthere is 40% loss of energydue to impact, then after one impact ball will go up to (a) 10m (b) 8m (c) 4m (d) 6m 44. The potential energy of a conservative system is given by U = ay2 – by, where yrepresents the position of the particle and a as well as bare constants. What is the force acting on the system ? (a) – ay (b) – by (c) 2ay – b (d) b – 2ay 45. An automobile engine ofmass M accelerates and a constant power p is applied bythe engine. The instantaneous speed of the engine will be (a) 1/ 2 [Pt / M] (b) 1/ 2 [2Pt / M] (c) 1/ 2 [Pt / 2M] (d) 1/ 2 [Pt / 4M] 46. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion ? (a) 2.0cm (b) 3.0cm (c) 1.0cm (d) 1.5cm 47. A bomb of mass 16kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velolcity of the 12 kg mass is 4ms–1. The kinetic energy of the other mass is (a) 144J (b) 288J (c) 192J (d) 96J 48. Given that a force F̂ acts on a bodyfor time t, and displaces the body by d̂ . In which of the following cases, the speed of the body must not increase? (a) F > d (b) F < d (c) ˆ F̂ d = (d) ˆ F̂ d ^ 49. A body is attached to the lower end of a vertical helical spring and it isgraduallyloweredto its equilibrium position. This stretches the spring by a length x. If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case? (a) x (b) 2x (c) 3x (d) x/2 50. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity V and gets caught in the bag. Then for the combined (bag + bullet) system (a) mvM momentum M m = + (b) 2 mV kinetic energy 2 = (c) mV(M m) momentum M + = (d) 2 2 m v kinetic energy 2(M m) = + 51. Aparticle, initiallyat rest on a frictionless horizontal surface, is acted upon by a horizontal force which is constant in magnitude and direction. A graph is plotted of the work done on the particle W, against the speed of the particle v. Ifthere are no other horizontal forces acting on the particle, the graph would look like (a) W v (b) W v (c) W v (d) W v 52. A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northwards with the same speed. If two particles coalesce on collision, the new particle of mass 2 m will move in the north-east direction with a velocity (a) v / 2 (b) v 2 (c) v / 2 (d) None of these 53. A small block ofmass m is kept on a rough inclined surface of inclination q fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t as seen by the observer on the inclined p lane will be (a) zero (b) mgvt cos2q (c) mgvt sin2 q (d) mgvt sin 2q 54. A nucleus moving with a velocity v r emits an a-particle. Let the velocities ofthea-particle andtheremainingnucleus be 1 v r and 2 v r and their masses be m1 and m2. (a) v r , 1 v r and 2 v r must be parallel to each other (b) None of the two of v r , 1 v r and 2 v r should be parallel to each other (c) 1 1 2 2 m v m v + r r must be parallel to 1 2 (m m )v + r . (d) None of these 55. A shell is fired from a cannon with a velocityV at an angle q with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediatelyafter the explosion is (a) 3 V cos q (b) 2 V cos q (c) 3 Vcos 2 q (d) V cos q
- 157. 153 Work, Energy andPower 56. The bobAofa simple pendulum is released when the string makes an angle of 45º with the vertical. It hits another bob B of the same material and same mass kept at rest on the table. If the collision is elastic 45º A B (a) both A and B rise to the same height (b) both A and B come to rest at B (c) both A and B move with the velocity of A (d) A comes to rest and B moves with the velocity of A 57. Twomasses ma and mb moving with velocities va and vb in opposite direction collide elasticallyand after the collision ma and mb move with velocities Vb and Va respectively. Then the ratioma/mb is (a) a b a b V V V V - + (b) a b a m m m + (c) 1 (d) 1 2 58. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (a) x (b) ex (c) x2 (d) loge x 59. A particle ofmass 1 m moving with velocityv strikes with a mass m2 at rest, then the condition for maximumtransfer of kineticenergy is (a) m1 >>m2 (b) m2 >>m2 (c) m1 =m2 (d) m1 =2m2 60. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is (a) 30m/s (b) 20m/s (c) 10m/s (d) 5 m/s 61. A mass m is moving with velocity v collides inelastically with a bobofsimplependulum ofmass mand getsembedded into it. The total height to which the masses will rise after collision is (a) g 8 v2 (b) g 4 v2 (c) g 2 v2 (d) g v 2 2 62. A motor drives a bodyalong a straight line with a constant force. The power P developed by the motor must varywith time t according to (a) P t (b) P t (c) P t (d) P t 63. A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finallycomes torest is (a) t en (b) t e2 (c) e t 1 e 1+ é ù ê ú - ë û (d) ú û ù ê ë é + - 1 e 1 e t 64. A weight suspended from the free end of a vertically hanging spring produces an extension of 3 cm. The spring is cut into two parts so that the length of the longer part is 3 2 of the original length, If the same weight is now suspended from the longer part of the spring, the extension produced will be (a) 0.1cm (b) 0.5cm (c) 1cm (d) 2cm 65. A 10 m long iron chain of linear mass density0.8 kg m–1 is hanging freelyfrom a rigid support. If g= 10 ms–2, then the power required to left the chain uptothe point of support in 10 second (a) 10 W (b) 20W (c) 30 W (d) 40 W 66. A smooth sphere ofmass M moving with velocityu directly collides elastically with another sphere of mass m at rest. After collision, their final velocities areV and vrespectively. The value of v is (a) m uM 2 (b) M um 2 (c) M m 1 u 2 + (d) 2u M 1 m + 67. The kinetic energyofparticle moving alonga circle ofradius R depends upon the distance covered S and is given by K = aS where a is a constant. Then the force acting on the particle is (a) R aS (b) R ) aS ( 2 2 (c) 2 2 R aS (d) R aS 2
- 158. 154 PHYSICS 68. Aramp is constructed in parabolic shape such that the height y of any point on its surface is given in terms of the point's horizontal distance x from the bottom ofthe ramp be y = x2/2L. A block of granite is to be set on the ramp; the coefficient of static friction is 0.80. What is the maximum x coordinate at which the block can be placed on the ramp and remain at rest, ifL = 10 m? y x (a) 8m (b) 8.4m (c) 9m (d) 9.4m 69. The force F acting on a bodymoving in a circle of radius r is always perpendicular to the instantaneous velocity v. The work done by the force on the body in half rotation is (a) Fv (b) F·2pr (c) Fr (d) 0 70. The negative of the distance rate of change of potential energy is equal to (a) force acting on the particle in the direction of displacement (b) acceleration of the particle, perpendicular to displacement (c) power (d) impulse. 71. n small balls each ofmass m impingeelasticallyeach second on a surface with velocity v. The force experienced by the surface will be (a) mnv 2 1 (b) 2mnv (c) mnv (d) 2mnv 72. A horse drinks water from a cubical container of side 1 m. The level of the stomach ofhorse is at 2 m from theground. Assume that all the water drunk by the horse is at a level of 2m from the ground. Then minimum workdone bythehorse in drinking the entire water of the container is (Take rwater = 1000 kg/m3 and g = 10 m/s2 ) – (a) 10 kJ (b) 15kJ (c) 20kJ (d) zero 73. The ball rolls down without slipping (which is at rest at a) along ab having friction. It rolls to a maximum height hc where bc has no friction. Ka, Kb and Kc are kinetic energies at a, b and c. Which of the following is correct ? a c hc ha b (a) Ka = Kc, ha = hc (b) Kb > Kc, ha = hc (c) Kb > Kc, ha < hc (d) Kb > Kc, ha > hc 74. A body falls freely under gravity. Its velocity is v when it has lost potential energy equal to U. What is the mass of the body ? (a) U2/v2 (b) 2U2/v2 (c) 2U/v2 (d) U /v2 75. If v be the instantaneous velocity of the body dropped from the top of a tower, when it is located at height h, then which of the following remains constant ? (a) 2 gh v + (b) 2 v gh 2 + (c) 2 v gh 2 - (d) gh – v2 76. The coefficient offriction between the tyres and the road is m. A car is moving with momentum p. What will be the stopping distance due to friction alone ? The mass of the car is m. (a) p2/2mg (b) p2/2mmg (c) p2/2m2mg (d) p2/2mg 77. A particle moves in the X–Y plane under the influence ofa force F r such that its instantaneous momentum is ˆ ˆ p i2cos t j2sin t = + r . What is the angle between the force and instantaneous momentum ? (a) 0° (b) 45° (c) 90° (d) 180° 78. A particle of mass 10 kg moving eastwards with a speed 5 ms–1 collideswith another particle ofthe same mass moving north-wards with thesame speed 5 ms–1. The two particles coalesce on collision. The new particle of mass 20 kg will move in the north-east direction with velocity (a) 10 ms–1 (b) 5 ms–1 (c) 1 (5/ 2)ms- (d) none of these 79. A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. Ifafter the collision, the twoballs move together, the loss in kinetic energy due to collision is (a) 140J (b) 100J (c) 60J (d) 40J 80. Aforceactsona30gmparticleinsuchawaythattheposition of the particle as a function of time is given by x = 3t – 4t2 + t3,wherexisinmetresand tisinseconds.Theworkdone during the first 4 seconds is (a) 576mJ (b) 450mJ (c) 490mJ (d) 530mJ 81. Arubberballisdroppedfromaheightof5monaplane,where the acceleration due to gravityis not shown. On bouncing it rises to 1.8 m. The ball loses its velocityon bouncing by a factor of (a) 25 16 (b) 5 2 (c) 5 3 (d) 25 9
- 159. 155 Work, Energy andPower 82. A 3 kg ball strikes a heavyrigid wall with a speed of 10 m/ satan angleof60º. Itgetsreflectedwiththesamespeedand angleasshownhere.Iftheball isincontactwiththewallfor 0.20s,whatistheaverageforceexertedontheballbythewall? (a) 150N 60º 60º (b) Zero (c) N 3 150 (d) 300N 83. Abombofmass1kgisthrownverticallyupwardswithaspeed of100m/s.After 5secondsitexplodesintotwofragments. One fragment of mass 400 gm is found to go down with a speed of25 m/s. What will happen tothe second fragment just after the explosion? (g = 10 m/s2) (a) It will go upward with speed 40 m/s (b) It will goupward with speed 100 m/s (c) It will go upward with speed 60 m/s (d) It will alsogodownward with speed 40m/s 84. In a simple pendulum of length l the bob is pulled aside from its equilibrium position through an angle q and then released. The bob passes through the equilibrium position with speed (a) ) cos 1 ( g 2 q + l (b) q sin g 2 l (c) l g 2 (d) ) cos 1 ( g 2 q - l 85. A stationaryparticle explodesintotwoparticlesofmasses m1andm2whichmoveinoppositedirectionswithvelocities v1 and v2. The ratio of their kinetic energies E1/E2 is (a) m1v2/m2v1 (b) m2/m1 (c) m1/m2 (d) 1 86. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearlyweightless spring of force constant k = 50 N/m. The maximum compression of the spring would be (a) 0.5m (b) 0.15m (c) 0.12m (d) 1.5m 87. A ball moving with velocity 2 m/s collides head on with another stationaryball ofdouble the mass. Ifthe coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be (a) 0,1 (b) 1,1 (c) 1,0.5 (d) 0,2 88. A body projected verticallyfrom the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest (a) at the highest position of the body (b) at the instant just before the body hits the earth (c) it remains constant all through (d) at the instant just after the body is projected 89. A mass m moving horizontally (along the x-axis) with velocityv collidesand sticks tomass of3m moving vertically upward(alongthe y-axis) with velocity2v. Thefinal velocity of the combination is (a) 1 3 ˆ ˆ vi v j 4 2 + (b) 1 2 ˆ ˆ vi v j 3 3 + (c) 2 1 ˆ ˆ vi v j 3 3 + (d) 3 1 ˆ ˆ vi v j 2 4 + 90. The potential energy of particle in a force field is 2 A B U r r = - , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is (a) B / 2A (b) 2A / B (c) A / B (d) B / A 91. A uniform force of ˆ ˆ (3 ) i j + newton acts on a particle of mass 2 kg. The particle is displaced from position $ (2 ) i k $ + meter to position $ $ (4 3 ) i j k $ + - meter. The work done by the force on the particle is (a) 6 J (b) 13J (c) 15J (d) 9 J 92. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms–1 and thesecond part ofmass 2 kg moves with speed 8ms–1. If the third part flies off with speed 4 ms–1 then its mass is (a) 5 kg (b) 7 kg (c) 17kg (d) 3 kg 93. If the kinetic energy of a body is increased by 300%, the momentum of the body is increased by (a) 300% (b) 200% (c) 100% (d) 50% 94. If the mass of the body is halved and velocitygets doubled then final kinetic energywould be .........ofinitial. (a) same (b) four times (c) double (d) eight times 95. A train of weight 107 N is running on a level track with uniform speed of36 km h–1. Thefrictionalforce is0.5kg per quintal. If g = 10 m/s2, then power of engine is (a) 500kW (b) 50kW (c) 5 kW (d) 0.5kW DIRECTIONS(Qs.96to100):Eachquestioncontains Statement- 1andStatement-2.Choosethecorrectanswer(ONLYONEoption is correct ) from the following. (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 isnot a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 96. Statement-1 : A quick collision between two bodies is more violent than slow collision, even when initial and final velocities are identical. Statement -2 : The rate of change of momentum determines that the force is small or large.
- 160. 156 PHYSICS Exemplar Questions 1.An electron and a proton are moving under the influence of mutual forces. In calculating thechangein the kineticenergy ofthe system during motion, one ignores themagnetic force of one on another. This is, because (a) the two magnetic forces are equal and opposite, so they produce no net effect (b) the magnetic forces do not work on each particle (c) the magnetic forces do equal and opposite (but non- zero) work on each particle (d) themagnetic forces are necessarilynegligible 2. A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is (a) same as the same force law is involved in the two experiments (b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens (c) more for the case of a positron, as the positron moves away a larger distance (d) same as the work done by charged particle on the stationary proton. 3. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is (a) constant and equal to mg in magnitude (b) constant and greater than mg in magnitude (c) variable but always greater than mg (d) at first greater than mg and later becomes equal to mg 4. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is (a) +2000J (b) – 200 J (c) zero (d) – 20,000 J 5. A bodyis falling freelyunder the action of gravity alone in vaccum. Which ofthe following quantities remain constant during the fall? (a) Kinetic energy (b) Potential energy (c) Total mechanical energy (d) Total linear momentum 6. During inelastic collision between twobodies, which of the following quantities always remain conserved? (a) Total kinetic energy (b) Total mechanical energy (c) Total linear momentum (d) Speed of each body 7. Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. Which of the following statement is correct? h I A B q1 q2 II (a) Both the stones reach the bottom at the same time but not with the same speed (b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II (c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I (d) Both the stones reach the bottom at different times and with different speeds 8. The potential energyfunction for a particle executing linear SHM is given by 2 1 ( ) 2 V x kx = where k is the force constant of the oscillator (Fig.). For k = 0.5 N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ±xm. If V and K indicate the PE and KE, respectively of the particle at x = +xm, then which of the following is correct? x V x ( ) –xm xm (a) V = O, K = E (b) V = E, K = O (c) V < E, K = O (d) V = O, K < E 97. Statement -1 : If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Statement -2 : During collision intermolecular space decreases and hence elastic potential energy increases. 98. Statement -1 :Awork done byfriction is always negative. Statement -2 : If frictional force acts on a body its K.E. may decrease. 99. Statement -1 : An object of mass m is initially at rest. A constant force F acts on it. Then the velocity gained by the object during a fixed displacement is proportional to 1/ m . Statement -2 : For a given forceand displacement velocity is always inversely proportional to root of mass. 100. Statement -1 : Mechanical energy is the sum of macroscopic kinetic & potential energies. Statement - 2 : Mechanical energy is that part of total energy which always remain conserved.
- 161. 157 Work, Energy andPower 9. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on byanother ball bearing of the same mass moving initiallywith a speed v as shown in figure. 1 2 3 v Ifthe collision is elastic, which of the following (figure) is a possible result after collision? (a) 1 v = 0 v/2 (b) 1 2 3 v v = 0 (c) 1 2 3 v/3 (d) 1 2 3 v/1 v/2 v/3 10. A bodyof mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m–1/2s–1. The work done by the net force during its displacement from x = 0 to x = 2 m is (a) 15J (b) 50J (c) 10J (d) 100J 11. A body is moving unidirectionallyunder the influence of a source of constant power supplying energy. Which of the diagrams shown in figure correctlyshown the displacement- time curve for its motion? (a) t d (b) t d (c) t d (d) t d 12. Which of the diagrams shown in figure most closelyshows the variation in kinetic energyof the earth as it moves once around the sun in its elliptical orbit? (a) KE t (b) KE (c) KE (d) KE 13. Which ofthe diagrams shown in figure represents variation of total mechanical energyof a pendulum oscillating in air as function of time? (a) E t (b) t E (c) t E (d) t E 14. A mass of 5kg is moving along a circular path ofradius 1 m. Ifthe massmoveswith 300 rev/min, itskinetic energywould be (a) 250p2 (b) 100p2 (c) 5 p2 (d) 0 15. A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4)h. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground? (a) PE KE t h (b) PE KE t h/4 h (c) PE KE t h x O (d) PE KE t h O 16. In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of1 m s–1 at45°from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s–2, the kinetic energy of the shotput when it just reaches the ground will be (a) 2.5J (b) 5.0J (c) 52.5J (d) 155.0J
- 162. 158 PHYSICS 17. Whichofthe diagrams in figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity? (a) KE Depth t O (b) KE Depth t O (c) KE Depth t O (d) KE Depth t O 18. A cricket ball of mass 150 g moving with a speed of 126 km/ h hits at the middle of the bat, held firmlyat its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completelyelastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be (a) 10.5N (b) 21N (c) 1.05× 104 N (d) 2.1 × 104 N NEET/AIPMT (2013-2017) Questions 19. A uniform force of ˆ ˆ (3 ) i j + newton acts on a particle of mass 2 kg. The particle is displaced from position $ (2 ) i k $ + meter to position $ $ (4 3 ) i j k $ + - meter. The work done by the force on the particle is [2013] (a) 6 J (b) 13J (c) 15J (d) 9 J 20. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms–1 and the second part of mass 2 kg moves with speed 8 ms–1. If the third part flies off with speed 4 ms–1 then its mass is (a) 5 kg (b) 7 kg [2013] (c) 17kg (d) 3 kg 21. A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocityof 800 ms–1. The final velocityacquired by the person and the average force exerted on the person are [NEET Kar. 2013] (a) –1.6 ms–1; 8 N (b) –0.08 ms–1;16N (c) – 0.8 ms–1; 8 N (d) –1.6 ms–1; 16 N 22. Aparticlewith total energyE ismoving in apotentialenergy region U(x). Motion ofthe particle is restricted tothe region when [NEET Kar. 2013] (a) U(x)> E (b) U(x)< E (c) U(x)= O (d) U(x) £E 23. One coolie takes 1 minute to raise a suitcase through a height of 2 m but the second coolie takes 30 s to raise the same suitcase tothe same height. The powers of twocoolies are in the ratio of [NEET Kar. 2013] (a) 1: 2 (b) 1: 3 (c) 2: 1 (d) 3: 1 24. A bodyof mass (4m) is lying in x-yplane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is : (a) mv2 (b) 2 3 mv 2 [2014] (c) 2 mv2 (d) 4 mv2 25. A particle of mass m is driven bya machine that delivers a constant power of k watts. If theparticle starts from rest the force on the particle at time t is [2015] (a) –1/2 mk t (b) –1/2 2mk t (c) –1/2 1 mk t 2 (d) –1/2 mk t 2 26. Two similar springs Pand Q have spring constants KP and KQ, such that KP > KQ. They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b), respectively [2015] (a) WP = WQ ; WP = WQ (b) WP > WQ ; WQ > WP (c) WP < WQ ; WQ < WP (d) WP = WQ ; WP > WQ 27. A blockof mass 10 kg, moving in x direction with a constant speed of 10ms–1, is subject toa retarding forceF =0.1 × J/m during its travel from x = 20 m to 30 m. Its final KE will be: (a) 450J (b) 275J [2015] (c) 250J (d) 475J 28. The heart ofman pumps 5 litresofbloodthrough the arteries per minute at a pressure of150 mm ofmercury. If the density of mercury be 13.6 ×103 kg/m3 and g = 10m/s2 then the power of heart in watt is : [2015 RS] (a) 2.35 (b) 3.0 (c) 1.50 (d) 1.70
- 163. 159 Work, Energy andPower 29. A ball isthrown verticallydownwards from a height of 20 m with an initial velocity v0. It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocityv0 is : [2015 RS] (Take g = 10 ms–2) (a) 20 ms–1 (b) 28 ms–1 (c) 10 ms–1 (d) 14 ms–1 30. On a frictionless surfacea block of mass M moving at speed v collides elastically with another block of same mass M which is initiallyat rest.After collision thefirst block moves at an angle q to its initial direction and has a speed v 3 . The second block's speed after the collision is : [2015 RS] (a) 3 v 4 (b) 3 v 2 (c) 3 v 2 (d) 2 2 v 3 31. Two particles A and B, move with constant velocities 1 v r and 2 v r . At the initial moment their position vectors are 1 r r and 2 r r respectively. The condition for particlesAand Bfor their collision is: [2015 RS] (a) 1 1 2 2 r .v r .v = r r r r (b) 1 1 2 2 r v r v ´ = ´ r r r r (c) 1 2 1 2 r r v v - = - r r r r (d) 1 2 2 1 1 2 2 1 r r v v | r r | | v v | - - = - - r r r r r r r r 32. A body of mass 1 kg begins to move under the action of a time dependent force 2 ˆ ˆ F (2ti 3t j) = + r N, where î and ĵ are unit vectors alogn x and y axis. What power will be developed by the force at the time t? [2016] (a) (2t2 + 3t3)W (b) (2t2 + 4t4)W (c) (2t3 + 3t4) W (d) (2t3 + 3t5)W 33. Aparticle of mass10 gmovesalong acircleofradius 6.4cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10–4 J by the end of the second revolution after thebeginning ofthe motion ? [2016] (a) 0.1m/s2 (b) 0.15 m/s2 (c) 0.18 m/s2 (d) 0.2m/s2 34. Consider a drop ofrain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s2. The work done by the (i) gravitational force and the (ii) resistive force of air is [2017] (a) (i)1.25J (ii) –8.25J (b) (i)100J (ii) 8.75J (c) (i) 10J (ii) – 8.75 J (d) (i) – 10 J (ii) –8.25J
- 164. 160 PHYSICS EXERCISE -1 1. (a) 2. (b) 3. (d) 4. (a) 5. (c) 6. (c) 7. (a) 8. (b) 9. (b) Because water enters into the vessel A, it becomes heavier. Gravity helps it sink. External work required for immersing A is obviously less than that for immersingB. 10. (c) Weight Mg moves the centre of gravity of the spring through a distance 2 / 2 ) 0 ( l l = + Mechanical energy stored = Work done = Mg l/2. 11. (d) From 1 1 v u at, v 0 at = + = + 1 1 v a t = 1 1 F ma m v / t = = Velocity acquired in t sec = at = 1 1 v t t Power 2 1 1 1 2 1 1 1 m v v t m v t F v t t t = ´ = ´ = 12. (d) W = F × s 0 x W ) x ( x 1 W µ µ 13. (b) At the top of flight, horizontal component of velocity u cos 45º u / 2 = = 2 1 u K.E. m 2 2 æ ö = ç ÷ è ø . K 2 1 2 u m 2 1 2 = ÷ ÷ ø ö ç ç è æ = 14. (a) From 2 2 v u 2a s - = s ) g ( 2 v 0 2 0 m = - g 2 v s 2 0 m = 15. (b) If K 0 L Þ = r .E mayor maynot be zero. IfK.E = 0 Þ 0 L = r . 16. (d) Power is defined as the rate of doing work. For the automobile, the power output is the amount of work done (overcoming friction) divided by the length of timein which the work was done. 17. (b) ) ( k 2 1 k 2 1 k 2 1 W 2 1 2 2 2 1 2 2 l l l l - = - = 18. (b) 1 F (given) v µ Then s . F E W k = = Become = k s E v µ Þ k s E s / t µ Þ t Ek µ 19. (d) 2 2 3 mv ms t t = since both P and m are constants 2 3 s t = constant 20. (c) 1 1 2 2 1 2 m v m v a b 0 a (b) v . m m a c a c + ´ + = = = + + + 21. (d) As 2 / 1 0 n n h h e ÷ ÷ ø ö ç ç è æ = . h e h e h e h 4 2 2 0 n 2 n = = = ´ 22. (a) As u2 = 0 and m1 = m2, therefore from m1 u1 + m2 u2 = m1 v1 + m2 v2 we get u1 = v1 + v2 Also, 2 1 2 1 1 2 1 2 1 1 2 v v v v 1 v / v e u v v 1 v / v - - - = = = + + , which gives 1 2 v 1 e v 1 e - = + 23. (d) Work done Fdx = ò 24. (b) 1 m 2 m 2 1 m m + v u 0 u = 1 v u = applying conservation ofmomentum ( ) 1 2 1 2 1 v ) m m ( 0 m v m + = + v ) m m ( m v 2 1 1 1 + = Þ as ) m m ( m 2 1 1 + < so velocity decreases. 25. (b) 1 2 1 1 1 2 (m m )u v m m - = + As 1 v is negative and less than u1, therefore, m1 < m2. EXERCISE - 2 1. (a) W = F s cos 90º = zero 2. (c) As the cord is trying to hold the motion of the block, work done by the cord is negative. W = – M (g – a) d 4 d g M 3 d 4 g g M - = ÷ ø ö ç è æ - - = 3. (c) W = F s cos q = 10 × 2 cos 60º = 10 J. 4. (c) ) ĵ 3 î 2 ).( j ˆ 2 î 5 ( s . F W + + = = r r = 10 + 6 = 16 J. Hints & Solutions
- 165. 161 Work, Energy andPower 5. (c) From horizontal position to lowest position, height through which the bob falls = l At lowest position, v 2 g = l K.E. at lowest point 2 1 1 m v m (2 g) m g 2 2 = = = l l = 10 × 1 × 9.8 = 98 J. 6. (b) ò ò ú û ù ê ë é = = = 1 1 1 x 0 x 0 2 x 0 x c 2 1 dx x c dx F W 2 1 2 1 x c 2 1 ) 0 x ( c 2 1 = - = 7. (d) Forward thrust, P 100 746 F 3730 N. v 20 ´ = = = 8. (c) As 2 2 A A B B 1 1 m v m v 2 2 = A B B A m v v m = ; B B B A A A P m v P m v = A A A B B B m m m 1 m m m 3 = = = 9. (a) From 1 1 2 2 m v m v 0, + = 1 1 2 2 m v 4 v m 234 - = = - 10. (b) Total energy at the time of projection 2 2 1 1 m v 0.1(20) 20J 2 2 = = ´ = Half way up, P.E. becomes half the P.E. at the top i.e. J 10 2 20 . E . P = = K.E. = 20 –10 =10J. 11. (b) Workdone, W ( ) 2 2 2 1 1 k x x 2 = - ( )2 2 1 k 0.15 (0.05) 2 é ù = - ë û 1 800 0.02 8J 2 = ´ ´ = 12. (d) As m 2 p E 2 = dE dp 2 2 5% 10% E p æ ö = = ´ = ç ÷ è ø 13. (d) Mg 8 7 ) 8 / g g ( M T = - = Work done by the cord = T × d cos 180º . 8 / d g M 7 ) 1 ( d g M 8 7 - = - = 14. (b) x . F W r r = ) ĵ î 2 ).( k̂ 2 j ˆ 3 î 5 ( - + + = joules 7 3 10 = - = 15. (a) 2 3 1 ) 05 . 0 ( 10 5 2 1 W ´ ´ = 2 3 2 ) 10 . 0 ( 10 5 2 1 W ´ ´ = Þ . J 75 . 18 05 . 0 15 . 0 10 5 2 1 W 3 = ´ ´ ´ ´ = D 16. (a) 1 2 1 2 m 0.2 kg, m 0.4kg, v 0.3m /s, v ? = = = = Applying lawof conservation of momentum m1v1 – m2v2 0.2 0.3 0.15 m/s. 0.4 ´ = = 17. (c) 1 1 2 2 m v 3 v 16 8m / s m 6 - - = = ´ = - 2 2 2 2 2 1 1 E m v 6( 8) 192 J. 2 2 = = ´ - = 18. (d) 2 2 ) 2 ( k 2 1 ) x ( k 2 1 V = = or 2 V 4 V 2 k = = V 25 ) 10 ( 2 V 2 1 ) 10 ( k 2 1 V 2 2 = ÷ ø ö ç è æ ´ = = ¢ 19. (b) IfP= momentum, K = kinetic energy, then P1 2 = 2 mK1, 2 2 2 P 2mK = 2 2 2 1 1 1 1 P K 3K 3 P K K æ ö = = = ç ÷ è ø 2 1 P 3 1.732 P 1 = = 20. (b) 1 1 1 2 2 2 p 2m K p 2m K = 21. (d) Applying the principle of conservation of linear momentum, we get 2 2 3m v (m 60) (m 60) ´ = ´ + ´ 2 60 m´ = v 20 2 m/s = 22. (c) Energystored (E) J 9 ) 12 ( 100 75 = ´ = As 2 mv 2 1 E = sec / m 18 1 9 2 m E 2 v = ´ = = 23. (c) The uniform acceleration is 2 1 0 1 a ms 15 15 - - = = Let v be the velocity at kinetic energy J 9 2 therefore 1 2 ms 3 2 v or 9 2 v 1 2 1 - = = ´ ´ Using v = u + at 2 1 0 t t 10s 3 15 = + ´ Þ =
- 166. 162 PHYSICS 24. (d)Let n be the number of bullets that the man can fire in one second. change in momentum per second mv n ´ = = F [ m= mass of bullet, v = velocity] (Q F is the force) 1200 40 1000 144 mv F n ´ ´ = = = 3 25. (d) Power supplied = t mgh Power used by crane = 80 100 t mgh ´ W 8 36 10 80 100 3600 100 10 1000 5 ´ = ´ ´ ´ = 26. (b) Work done against gravity Wg = 50 × 10 × 30 = 15 kJ Work done against friction Wf mgcos s = m q´ = 4 0.4 50 10 50 8 kJ 5 ´ ´ ´ ´ = Total work done kJ 23 kJ 8 kJ 15 W W f g = + = + = 27. (d) Let mass of neutron = m then mass of deuterium = 2m [Q it has double nuclides thus has neutron]. Let initial velocityofneutron = v and final velocities of neutron and deuterium are v1and v2 respectively. m 2m v=0 m 2m v 1 v ® 2 v ® neutron deuterium Applying conservation of momentum 2 1 mv 2 mv ) 0 ( m 2 mv + = + ) i .( .......... v 2 v v 2 1 + = Þ applying conservation of energy 2 2 2 1 2 1 1 1 mv mv 2m.v 2 2 2 = + 2 2 2 1 2 v v 2v ......(ii) Þ = + from (i)and (ii), 2 2 2 2 2 v 2 ) v 2 v ( v + - = 2 2 2 2 2 2 2 v 2 v v 4 v 4 v v + - + = Þ 0 v v 4 v 6 2 2 2 = - 3 v v & 3 v 2 v 1 2 - = = Þ now fractional change in kinetic energy i f i K K K - = 2 2 1 2 mv 2 1 mv 2 1 mv 2 1 - = 2 2 2 v v 8 9 9 v - = = 28. (c) W = F s cos q, , 2 1 10 5 25 s F W cos = ´ = = q q = 60º. 29. (b) For elastic collision in one dimension 1 v = 2 1 2 2 m m u m 2 + + ) m m ( u ) m m ( 2 1 1 2 1 + - As mass 2m, is at rest, So 2 u = 0 Þ 1 v = m 2 m 8 u ) m 2 m 8 ( + - = u 5 3 Final energy of sphere = f .) E . K ( = 2 5 u 3 ) m 8 ( 2 1 ÷ ø ö ç è æ = 2 2 5 3 u ) m 8 ( 2 1 ÷ ø ö ç è æ ´ = E 25 9 =0.36E 30. (b) The work is done against gravity so it is equal to the change in potential energy. W= Ep = mgh For a fixed height, workisproportional toweight lifted. Since Johnnyweighs twice as much as Jane he works twice as hard to get up the hill. Power is work done per unit time. For Johnny this is W/Dt. Jane did half the work in half the time, (1/2 W)/ (1/2 Dt) = W/Dt which is the same power delivered by Johnny. 31. (c) Fraction of energytransferred = 9 8 ) 2 1 ( 2 4 2 = + ´ 32. (c) m 2 p E 2 = Þ 2 2 2 1 2 1 p p E E = Þ 4 E E 1 2 ´ = 1 1 2 E 3 E E = - 33. (c) Since the new height gained is half, therefore there is 50% loss of energy. 34. (a) u m 3 u 2 m u m 2 1 1 + = (By condition of linear momentum) Þ 1 2 1 m u m v 3 = ......(i) Also | u u | | v v | e 1 2 2 1 - - = Þ 2u v u 3 - = Þ 5 v u 3 = ......(ii) From (i) and (ii), u m 3 5 u m 3 1 2 1 = Þ 5 m m 2 1 = 35. (c) At Equilibrium , ky= mg Þ y mg k = mgy 2 1 y y mg 2 1 U 2 = ÷ ÷ ø ö ç ç è æ = 36. (a) Loss in K.E =Area under the curve
- 167. 163 Work, Energy andPower 37. (b) In this case, P = t mv 2 1 mgh 2 + Þ ú ú û ù ê ê ë é + = 2 v gh t m P 2 Þ W 2 10 10 10 10 1 10 P ú û ù ê ë é ´ + ´ = =1500W 38. (c) Byconservation of linear momentum, 1 2 2 1 m v m v = or 2 1 r 3 4 r 3 4 3 2 x 3 x 1 = p r p r Þ 3 / 1 2 1 2 1 r r = 39. (c) 2 2 2 0 0 2 2 2 2 m m mc c c v (0.5c) 1 1 c c = = - - 2 0 2 0 2 0 c m 15 . 1 c 75 . 0 m c 25 . 0 1 m = = - = Change in energy = 1.15 m0c2 – m0c2 = 0.15 × 9.1 × 10–31 × (3× 108)2 = 12.285 × 10–15 J MeV 10 6 . 1 10 285 . 12 13 15 - - ´ ´ = =0.07678MeV 40. (b) 12 F 1000 10 N 1200 N 100 = ´ ´ = P= Fv= 1200 N × 15 ms–1 = 18 kW. 41. (b) Components of velocity before and after collision parallel to the plane are equal, So vsin 60° = u sin 30°.......(1) Componentsofvelocitynormal tothe plane are related to each other v cos 60° = e u (cos 30°) ........(2) Þ cot 60° = e cot 30° Þ cos60 e cot30 ° = ° Þ 3 3 1 e = Þ . 3 1 e = 42. (d) Work done on the body is gain in the kinetic energy. Acceleration of the body is a = V/T. Velocityacquired in time t is v = at V t T = K.E. acquired µ v2. That is work done 2 2 2 V t T µ 43. (d) Kinetic energyof ball when reaching the ground = mgh = mg× 10 Kinetic energyafter the impact 60 mg 10 6mg 100 = ´ ´ = If the ball rises to a height h, then mgh = 6 mg. Hence, h = 6 m. 44. (d) dU F b 2ay dy = - = - 45. (b) Fv = P . or dv M v P dt = That is P vdv dt M = ò ò Hence 1/ 2 v [2Pt / M] = 46. (c) cm 3 u 2 u 0 v = x Case I : 3 . a . 2 u 2 u 2 2 = - ÷ ø ö ç è æ or – 4 u 3 2 = 2. a. 3 Þ a = – 8 u2 Case II : 2 2 u 0 ÷ ø ö ç è æ - = 2. a. x or – 4 u2 = 2. ÷ ÷ ø ö ç ç è æ - 8 u2 × x Þ x=1cm Alternative method : Let K be the initial energy and F be the resistive force. Then according towork-energy theorem, W = DK i.e., 2 2 1 1 v 3F mv m 2 2 2 æ ö = - ç ÷ è ø 2 1 1 3F mv 1 2 4 æ ö = - ç ÷ è ø 2 3 1 3F mv 4 2 æ ö = ç ÷ è ø ...(1) and 2 2 1 v 1 Fx m m(0) 2 2 2 æ ö = - ç ÷ è ø i.e., 2 1 1 mv Fx 4 2 æ ö = ç ÷ è ø ...(2) Comparing eqns. (1) and (2) F=Fx or x=1cm
- 168. 164 PHYSICS 47. (b)Let the velocity and mass of 4 kg piece be v1 and m1 and that of 12 kg piece be v2 and m2. Applying conservation of linear momentum m2v2 = m1v1 1 1 ms 12 4 4 12 v - = ´ = Þ J 288 144 4 2 1 v m 2 1 . E . K 2 1 1 1 = ´ ´ = = 48. (d) Velocitywill increase when forceis along the direction ofdisplacement i.e. d̂ F̂ = . 49. (b) When body is lowered gradually, its weight acts at C.G. of the spring. When same bodyis allowed to fall freely, the same weight acts at lower end of the spring. In the latter case, original length (L) ofspring is double. As DL µ L, therefore, DL becomes twice in second casei.e. 2x. 50. (d) If V is velocity of combination (bag + bullet), then from principle ofconservation of linear momentum m v (m M)V m v or V (m M) + = = + 2 2 2 1 m v K.E. (m M)V 2 2(m M) = + = + 51. (c) Workdone W = [ML2 T–2] It shows that 1 2 2 W (LT ) i.e. W v . - µ µ graph between W & v is a parabola. Alternatively: According to work energy theorem 2 2 2 mv 2 1 mu 2 1 mv 2 1 W = - = (Q u = 0) Þ W µ v2 52. (c) Applyingprinciple ofconservation of linear momentum 2 2 (2m)V (mv) (mv) mv 2 = + = v 2 v V 2 2 = = 53. (a) Since block does not slide on wedge so displacement is zero & hence work done by force is zero. f m gsinq q u mg 54. (c) If mass of nucleus is m, mass of a particle is m1 & mass of remaining nucleus is m2, then from the lawof conservation of momentum. nucleus remaining of momentum 2 2 particle of momentum 1 1 nucleus of momentum initial v m v m v m r r r + = a 55. (a) Let M be the mass of shell. Applying law of conservation of linear momentum MV cosq = MV M cos v 2 2 æ ö - q + ç ÷ è ø ie, MV cos q + M 2 Vcosq = M 2 v or v = 3 Vcosq . 56. (d) As bob B is of same material and same mass as the bob A, therefore, on elastic collision, their velocities are exchanged. BobAcomes to rest and B moves with the velocity of A. 57. (c) As velocities are exchanged on perfectly elastic collision, therefore masses of two objects must be equal. . m m or 1 m m b a b a = = 58. (c) a kx = - dv kx dt Þ = - Also dv dx dt or v dt dx = = v x 2 v 0 1 vdv kx v dv kxdx dx = - Þ = - ò ò ( ) 2 2 2 2 1 kx v v 2 - = - ( ) 2 2 2 2 1 1 1 kx m v v m 2 2 2 æ ö - Þ - = ç ÷ è ø 2 K x D a 59. (c) 2 i 1 1 1 K m u , 2 = 1 2 1 2 1 1 2 1 1 f u m m m m v , v m 2 1 K + - = = Fractional loss 2 1 1 2 1 1 2 1 1 i f i u m 2 1 v m 2 1 u m 2 1 K K K - = - ( ) ( )2 2 1 2 2 1 2 1 2 1 m m m m 1 u v 1 + - - = - = ( )2 2 1 2 1 m m m m 4 + = m m2 = ; nm m1 = ( )2 i f n 1 n 4 K K 1 + = - Energytransfer is maximum when 0 Kf = ( )2 4n 1 1 n = + 2 4n 1 n 2n Þ = + + Þ 0 n 2 1 n2 = - + ( ) 0 1 n 2 = - 1 n = ie. m m2 = , m m1 = Transfer will be maximumwhen both masses are equal and one is at rest.
- 169. 165 Work, Energy andPower 60. (c) s / m 10 4 10 0 4 14 10 Vc = + ´ + ´ = ; since spring force is internal force. 61. (a) For inelastic collision, linear momentum is conserved Þ 1 mv = 2 mv 2 Þ 2 v v 1 2 = Lossin K.E. = Gain in P.E. = mgh 2 v ) m 2 ( 2 1 mv 2 1 2 2 2 1 = - Þ 4 mgh = 2 1 mv – 2 mv2 1 = 2 mv2 1 = 2 mv2 Þ h = g 8 v2 62. (d) P F v = ´ Þ P = F a t P t µ 63. (c) g h 2 tAB = 1 BC CB 2 2h t t 2 g 2e h 2h 2 2e g g + = = = h h A B h 1 2 D C g h 2 e 2 t t 2 DB BD = + Total time taken by the body in coming to rest ......... g h 2 e 2 g h 2 e 2 g h 2 2 + + + = 2 2h 2h 2e [1 e e .........] g g = + + + + = e 1 1 g h 2 e 2 g h 2 - ´ + 2h 1 e 1 e t g 1 e 1 e + + é ù æ ö = = ç ÷ ê ú - - ë û è ø 64. (d) Initially, mg k 3 = or 3 mg k = New force constant of longer part 2 mg 3 mg 2 3 k 2 3 ' k = ´ = = Finally, mg y ' k = cm 2 2 mg mg ' k mg y = ´ = = 65. (d) kg 8 kg 8 . 0 10 m = ´ = height ofiron chain = 5m mgh 8 10 5 P W 40W t 10 ´ ´ = = = 66. (c) Bylaw of conservation ofmomentum, Mu = MV + mv ....(i) Also e = MV Mv Mu | u u | | v v | 2 1 2 1 - = Þ - - ....(ii) From (i) and (ii), 2Mu = (M + m)v Þ 2uM 2u v v m M m 1 M = Þ = + + 67. (d) Centripetal force = R mv2 R aS 2 R K 2 R 2 mv 2 1 2 = = ÷ ø ö ç è æ = 68. (a) As the block is at rest at P. q m = q cos mg sin mg dx dy slope tan = = q = m L x L 2 x dx d 2 = ÷ ÷ ø ö ç ç è æ = mg mg cosq mg sinq q N P x y 8 . 0 L x = Þ x = 0.8× 10 = 8m 69. (d) Work done by centripetal force is zero. 70. (a) dU F dx = - 71. (c) Thechangein momentum in the ball after the collision with surface is m(0–v) = –mv Since n balls impinge elastically each second on the surface, then rate ofchange of momentum of ball per second is mvn (consider magnitudeonly) Now According to Newton’s second law rate of change ofmomentum per second ofball = force experienced by surface. 72. (d) h=1.5m The mass of water is m = 1 × 103 kg The increase in potential energy of water is = mgh =(1 × 103) (10)(1.5) = 15 kJ
- 170. 166 PHYSICS 73. (d) 74.(c) 2 U (1/ 2)Mv = 75. (b) P. E + K.E = constant, mass being constant gh + v2/2 = constant 76. (c) 2 K p / 2m mgx = = m Hence, 2 2 x p / 2m g. = m 77. (c) dp ˆ ˆ F i2sin t j2cot t dt = = - + r r Hence F.p 0 = r r , hence angle between F r and p r is 90° 78. (c) Here ˆ ˆ imv jmv 2mV + = r That is v ˆ ˆ V (i j) 2 = + r Hence v v V 2 2 2 = ´ = . Hence v = 5 ms–1 79. (c) Apply conservation of momentum, m1v1 = (m1 + m2)v v = ) m m ( v m 2 1 1 1 + Here v1 = 36 km/hr = 10 m/s, m1 = 2 kg, m2 = 3 kg s / m 4 5 2 10 v = ´ = K.E.(initial) = J 100 ) 10 ( 2 2 1 2 = ´ ´ K.E. (Final) = J 40 ) 4 ( ) 2 3 ( 2 1 2 = ´ + ´ Loss in K.E. = 100 – 40 = 60 J Alternativelyuse the formula ( ) ( )2 1 2 k 1 2 1 2 m m 1 E u u 2 m m -D = - + 80. (a) x = 3t –4t2 + t3 2 t 3 t 8 3 dt dx + - = Acceleration = t 6 8 dt x d 2 2 + - = Acceleration after 4 sec = –8 + 6 × 4 = 16 Displacement in 4 sec = 3 ×4 – 4 × 42 + 43 = 12 m Work = Force × displacement = Mass × acc. × disp. = 3 × 10–3 × 16 × 12 = 576 mJ 81. (b) According to principle of conservation of energy Potential energy = kinetic energy Þ gh 2 v mv 2 1 mgh 2 = Þ = If h1 and h2 are initial and final heights, then Þ 2 2 1 1 gh 2 v , gh 2 v = = Loss in velocity, 2 1 2 1 gh 2 – gh 2 v – v v = = D fractional loss in velocity 1 v v D = = 1 2 1 2 1 h h – 1 gh 2 gh 2 – gh 2 = 6 . 0 – 1 36 . 0 – 1 5 8 . 1 – 1 = = = 5 2 4 . 0 = = 82. (c) Changein momentum along the wall = mv cos60º – mv cos 60º = 0 Change in momentum perpendicular to the wall = mvsin60º – (– mv sin60º) = 2mv sin60º Applied force = Time momentum in Change = 20 . 0 º 60 sin mv 2 = 2 3 10 3 2 0.20 ´ ´ ´ ´ = 50 3 3 ´ = 3 150 newton 83. (b) Speed of bomb after 5 second, v = u – gt = 100 –10×5 = 50m/s Momentum of 400 g fragment = ) 25 ( 1000 400 - ´ (downward) Momentum of600g fragment = v 1000 600 Momentum of bomb just before explosion = 1 × 50= 50 From conservation ofmomentum Total momentum just before collision = Total momentum just after collision Þ v 1000 600 25 1000 400 50 + ´ - = Þ v= 100 m/s (upward) 84. (d) If l is length of pendulum and q be angular amplitude then height. A B P C q l h AC – AB h = = l – l cos q = l(1 – cos q) At extreme position, potential energyis maximum and kinetic energy is zero;At mean (equilibrium) position potential energy is zero and kinetic energy is maximum, so from principle ofconservation ofenergy. B at ) PE KE ( P at ) PE KE ( + = + 0 mv 2 1 mgh 0 2 + = + v 2gh 2g (1– cos ) Þ = = q l
- 171. 167 Work, Energy andPower 85. (b) From law of conservation of momentum, before collision and after collision linear momentum (p) will be same. or initial momentum = final momentum. 2 p E 2m = According to question, 2 1 1 2 2 2 1 2 E p 2m E 2m p = ´ 1 2 2 1 m m E E = Þ [p1 = p2] 86. (b) 2 2 kx 2 1 mv 2 1 = Þ mv2 = kx2 or 0.5 × (1.5)2 = 50×x2 x=0.15m 87. (a) Clearly v1 = 2 ms–1, v2 = 0 m1 = m (say), m2 =2m v1' = ?, v'2 = ? e = 1 2 2 1 v ' v ' v v - - ....(i) Byconservation of momentum, 2m= mv1' + 2mv2' ...(ii) From (i), 0.5 = 2 1 v ' v ' 2 - v2' = 1 + v1' From (ii), 2 = v1'+ 2 + 2 v1' Þ v1 = 0 and v2 = 1 ms–1 88. (b) Power exerted by a force is given by P= F.v When the body is just above the earth’s surface, its velocityis greatest.At this instant, gravitational force is also maximum. Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth. 89. (a) As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved. 2v 3m v x m ˆ ˆ mvi 3m(2v)j (4m)v r + = v 6 ˆ ˆ v i vj 4 4 = + r = v 3 ˆ ˆ i vj 4 2 + 90. (b) For equilibrium 0 dU dr = Þ 3 2 2 0 A B r r - + = r = 2A B for stable equilibrium 2 2 d U dr should be positive for the value of r. here 2 2 4 3 6 2 d U A B dr r r = - is +ve value for 2A r B = So 91. (d) Given : F r = 3i j + $ $ 1 r u r = $ ( ) 2i k + $ , 2 r uu r = ( ) 4i 3j k + - r $ $ r r = 2 1 r r - uu r u r = ( ) 4i 3j k + - r $ $ – $ ( ) 2i k + $ or r r = 2i 3j + $ $ – $ 2k So work done by the given force w = f.r r r = ( ) $ ( ) 3i j . 2i 3j 2k + + - $ $ $ $ = 6 + 3 = 9J 92. (a) 4m /sec m 3 2 kg m2 Presultant 1 kg x y 8 m/sec 12 m/sec m1 Presultant = 2 2 12 16 + = 144 256 + = 20 m3v3 = 20 (momentum ofthird part) or, m3 = 20 4 = 5 kg 93. (c) p 2mK = p' 2m[K 3K] = + = 2p p 100 2p p 100 100% p p D ´ - = ´ = 94. (c) Let m and v be the mass and the velocity of the body. Then initial K.E. , Ki = 2 1 2 mv Now, mass = 2 m velocity = 2v
- 172. 168 PHYSICS FinalK.E.= 2 1 (2 ) 2 2 m v æ ö ç ÷ è ø = 2 1 (2 ) 2 mv = 2 1 2 2 mv æ ö ç ÷ è ø = 2 Ki 95. (a) Power of engine = Force × velocity = Fv Here, mass of engine = 7 10 10 kg = 106 kg F = frictional force = 0.5 kgf per quintal = (0.5 × 10) N per quintal = 5 N per quintal = 6 5 10 100 N æ ö ´ ç ÷ è ø = (5 × 104) N and v = 36 km h–1 = 1 36 1000 60 60 kmh- ´ ´ = 10 ms–1 Power of engine= (5 × 104 ×10)W = 5 × 105 W = 500kW 96. (b) 97. (b) 98. (a) When frictional force is opposite to velocity, kinetic energywill decrease. 99. (c) 100. (d) EXERCISE - 3 Exemplar Questions 1. (b) When electron and proton are moving under influence of their mutual forces, then according to the flemings left hand rule, the direction of force acting on a charge particle is perpendicular to the direction of motion. In magnetic field, work-done = F. s. cosq = F . s. cos 90° = 0. So magnetic forces do not work on moving charge particle. 2. (c) Forces between two protons is same as that ofbetween proton and a positron. As positron is much lighter than proton, it moves away through much larger distance compared to proton. Work done = Force × Distance As forces are same in case of proton and positron but distance moved bypositron is larger, hence, work done on positron will be more than proton. 3. (d) When the man squatting on the ground he is tilted somewhat, hence he also has to applyfrictional force besides his weight. R (reactional force) = friction force (f) + mg i.e. R > mg When the man does not squat and gets straight up in that case friction ( f ) » 0 R (Reactional force) » mg Hence, the reaction force (R) is larger when squatting and become equal to mg when no squatting. 4. (c) According to the question, work done bythe frictional force on the cycle is : = 200 × 10= –2000 J As the road is not moving, hence work done by the cycle on the road is zero. 5. (c) As the body is falling freely under gravity and no external force act on body in vaccum so law of conservation, the potential energy decreases and kinetic energy increases because total mechanical energy(PE + KE) of the body and earth system will be remain constant. 6. (c) According to the question, consider the two bodies as system, the total external force on the system will be zero. Hence, in an inelastic collision KE does not conserved but total linear momentum of the system remain conserved. 7. (c) As the (inclined surface) are frictionless, hence, mechanical energy will be conserved. As both the tracks havingcommon height, h (and no external force acts on system). KE & PE of stone I at top = KE + PE at bottom of I. From conservation of mechanical energy, 2 1 1 0 0 2 mv mgh + = + 1 2 v gh Þ = similarly 2 2 v gh = Hence, speed is same for both stones. For stone I, acceleration along inclined plane a1 = g sin q1 Similarly, for stone II a2 = g sin q2 sin q1 < sin q2 Thus, q2 > q1 hence a2 > a1. a2 is greater than a1 and both length for track II is also less hence, stone II reaches earlier than stone I. 8. (b) Total Mechanical energyis E = PE + KE at anyinstant. When particle is at x = xm i.e., at extreme position, partical returns back and its velocity become zero for an instant. Hence, at x = xm; x = 0, K.E. = 0. FromEq.(i), E = PE + 0 = PE = V(xm) = 2 1 2 m kx but at mean position at origin V(xm) = 0. 9. (b) Iftwobodies of equal masses collides elastically, their velocities are interchanged. When ball 1 collides with ball-2, then velocityof ball- 1, v1 becomes zero and velocity of ball-2, v2 becomes v, i.e., similarlythen its own all momentum is mV. So, 1 2 1 2 0 , 0, v v v P P mV = Þ = = = Now ball 2 collides to ball 3 and its transfer it's momentum is mV toball 3 and itselfcomes in rest. So, 2 3 2 3 0 , 0, v v v P P mV = Þ = = = So, ball 1 and ball 2, become in rest and ball 3 move with velocity v in forward direction.
- 173. 169 Work, Energy andPower 10. (b) As we know that, W.D. 2 2 1 1 0 x x x x F dx ma dx = × = × ò ò uu r uu r r r As given that, m = 0.5 kg, a = 5 m–1/2 s–1, work done (W) = ? 3/2 v ax = We also know that, Acceleration, 3/ 2 3/2 0 ( ) dv dv d a v ax ax dt dx dx = = × = 3/ 2 1/2 2 2 3 3 2 2 ax a x a x = ´ ´ ´ = Now, Force = 2 2 0 3 2 ma m a x = From(i), Work done 2 0 x x Fdx = = = ò 2 2 2 0 3 2 ma x dx é ù = ê ú ë û ò 2 3 2 0 3 2 3 x ma æ ö = ´ç ÷ ç ÷ è ø 2 1 8 2 ma = ´ 1 (0.5) (25) 8 50 J 2 = ´ ´ ´ = 11. (b) As given that power = constant As we know that power (P) dW F dx F dx P dt dt dt × = = = uu r r As the body is moving unidirectionally. Hence, cos 0 F dx Fdx Fdx × = ° = constant Fdx P dt = = ( P = Q constant by question) Now, by dimensional formula 0 F v × = [F] [v] = constant [MLT–2] [LT–1] = constant [ML2T–3] = constant 3 2 T L M = (As mass of body constant) 2 3 3/ 2 L T L T µ Þ µ Þ Displacement 3/2 ( ) d t µ Verifies the graph (b). 12. (d) The speed of earth around the sun can never be zero or negative, so the kinetic energy of earth cannot be zero and negative. So, option (b) and (c) represents wrongly the variation in kinetic energy of earth. A B SUN When the earth is closest to the sun, speed of the earth is maximum, hence, KE is maximum. When the earth isfarthest from the sun speed isminimum hence, KEisminimum. So, K.E. of earth increases (B to A) and then decrease, variation is correctly represented by option (d). 13. (c) When a pendulum oscillates in air,due toair resistance the force of friction acts between bob of pendulum and air, so it will lose energy continuously in overcoming. Therefore, total mechanical energy(KE + PE) of the pendulum decreases continuously with time and finallybecomes zero. Sum of KE and PE can never be negative. So, option (a) and (d) are incorrect. Hence option (c) is verifies. 14. (a) As given that, mass (m) = 5 kg, n = 300 revolution Radius (R) =1 m t = 60 sec 2 (300 2 )rad / 60s n t p æ ö w = = ´ ´p ç ÷ è ø 600 rad/s = 10 rad/s 60 ´p = p linear speed (v) = wR = (10p× 1) v = 10pm/s KE 2 1 2 mv = 2 1 5 (10 ) 2 = ´ ´ p 2 1 100 5 2 = p ´ ´ 2 250 J = p So, verifies the option (a). 15. (b) P.E. is maximum when drop start falling at t = 0 as it fall is P.E. decrease gradually to zero. So, it rejects the graph (a), (c) and (d). K.E. at t = 0 is zero as drop falls with zero velocity, its velocity increases (gradually), hence, first KE also increases. After sometime speed (velocity) is constant this is called terminal velocity, so, KE also become constant. It happens when it falls 3 4 æ ö ç ÷ è ø height or remains at 4 4 æ ö ç ÷ è ø from ground, then PE decreases continuously as the drop is falling continuously. Thevariation in PE and KE is best represented by(b).
- 174. 170 PHYSICS 16. (d)Asgiven that,h=1.5m,v= 1m/s, m=10 kg,g=10ms–2 By the law of conservation of mechanical energy as no force acts on shotput after thrown. (PE)i + (KE)i = (PE)f +(KE)f 2 1 0 (KE) 2 i i f mgh mv + = + (KE)f = 2 1 2 i i mgh mv + Total energy when it reaches ground, so 2 1 (KE) 10 10 1.5 10 (1) 2 f = ´ ´ + ´ ´ E = 150+5= 155J. 17. (b) First velocity of the iron sphere 2 V gh = after sometime its velocity becomes constant, called terminal velocity. Hence, according first KE increases and then becomes constant due to resistance of sphere and water which is represented by (b). 18. (c) As given that, 150 150 g kg 0.15 kg 1000 m = = = Dt = timeof contact = 0.001 s 126×1000 126km/h = m/s 60 60 u = ´ 5 126 35 m/s 18 = ´ = v = –126 km/h 5 126 35 m/s 18 = - ´ = - So, final velocity is acc. to initial force applied by batsman. So, change in momentum of the ball 3 ( ) ( 35 35) kg-m/s 20 p m v u D = - = - - 3 21 ( 70) 20 2 = - = - As we know that, force 4 21/ 2 N = –1.05 × 10 N 0.001 p F t D - = = D Hence negative sign shown that direction of force will be opposite to initial velocity which taken positive direction. Hence verify the option (c). NEET/AIPMT (2013-2017) Questions 19. (d) Given : F r = 3i j + $ $ 1 r u r = $ ( ) 2i k + $ , 2 r uu r = ( ) 4i 3j k + - r $ $ r r = 2 1 r r - uu r u r = ( ) 4i 3j k + - r $ $ – $ ( ) 2i k + $ or r r = 2i 3j + $ $ – $ 2k So work done by the given force w = f.r r r = ( ) $ ( ) 3i j . 2i 3j 2k + + - $ $ $ $ = 6 + 3 = 9J 20. (a) 4m /sec m 3 2 kg m2 Presultant 1 kg x y 8 m/sec 12 m/sec m1 Presultant = 2 2 12 16 + = 144 256 + = 20 m3v3 = 20 (momentum ofthird part) or, m3 = 20 4 = 5 kg 21. (c) According to law of conservation of momentum MV + mnv = 0 Þ 0.01 kg 10 800 m/s 100 mNv V M - - ´ ´ = = Þ – 0.8 m/s According to work energy theorem, Average work done =Changein average kineticenergy i.e., 2 av av rms 1 2 F S mV ´ = Þ 2 av max rms 1 2 2 2 F V t V m = Þ Fav = 8 N 22. (d) As the particle is moving in a potential energyregion. Kinetic energy ³ 0 And, total energyE = K.E. + P.E. Þ U(x) £ E 23. (a) Q Power w P t = Þ 1 2 2 1 30 30 1 1 minute 60 2 P t s s P t s = = = = (t1 = 1 minute; t2 = 30 second given) 24. (b) Byconservation of linear momentum 2mv1 = 1 v 2mv v 2 Þ = m v v m 2m v1 As two masses of each ofmass m move perpendicular to each other. Total KE generated = 2 2 2 1 1 1 mv mv (2m)v 2 2 2 1 + +
- 175. 171 Work, Energy andPower = 2 2 2 mv 3 mv mv 2 2 + = 25. (d) As we know power P = dw dt Þ w= Pt = 1 2 mV2 So, v = 2Pt m Hence, acceleration dV 2P 1 a . dt m 2 t = = Therefore, force on the particle at time ‘t’ = ma = 2 –1/2 2Km 1 Km mK . t m 2t 2 2 t = = 26. (b) As we know work done in stretching spring 2 1 w kx 2 = where k = spring constant x = extension Case(a) If extension (x) is same, 2 1 W K x 2 = So, WP > WQ(QKP > KQ) Case (b) If spring force (F) is same 2 F W 2K = So, WQ > WP 27. (d) From, F =ma a = F m = 0.1x dV 0.01x V 10 dx = = So, 2 1 v 30 v 20 x vdV dx 100 = ò ò – 30 2 2 2 20 1 V V x 30 30 20 20 – 2 200 200 200 V ´ ´ = = = 4.5 – 2 = 2.5 ( ) 2 2 2 1 1 m V – V 10 2.5 J – 25J 2 = ´ = FinalK.E. = 2 2 2 1 1 1 1 mv mv – 25 10 10 10 – 25 2 2 2 = = ´ ´ ´ = 500 – 25= 475 J 28. (d) Power F.V PAV ghAV = = r r r r F P and P gh A é ù = = r ê ú ë û Q = 13.6 × 103 ×10× 150×10–3 ×0.5 ×10–3/60 = 102 60 = 1.70 watt 29. (a) When ball collides with the ground it loses its 50% of energy f i KE 1 KE 2 = Þ 2 f 2 i 1 mV 1 2 1 2 mV 2 = or f i V 1 V 2 = or, 2 0 2gh 1 2 V 2gh = + or, 4gh = 2 0 V 2gh + V0 = 20ms–1 M M 30. (d) Here, M1 = M2 and u2 = 0 u1 = V, V1 = V 3 ; V2 = ? u1=V M1 u2=0 M2 M1 M2 q f V1=V/3 V2=? From figure, alongx-axis, M1u1 + M2u2 = M1V1 cosq + M2V2 cosf ...(i) Along y-axis 0 = M1V1 sinq – M2Vs sinf ...(ii) By law of conservation of kinetic energy 2 2 2 2 1 1 2 2 1 1 2 2 1 1 1 1 M u M u M V M V 2 2 2 2 + = + ...(iii) Putting M1 = M2 and u2 = 0 in equation (i), (ii) and (iii) we get q + f= 2 p = 90° and 2 2 2 1 1 2 u V V = + V2 = 2 2 2 V V 3 æ ö + ç ÷ è ø 1 1 V u V and V 3 é ù = = ê ú ë û Q or, V2 – 2 V 3 æ ö ç ÷ è ø = 2 2 V 2 2 V V 9 - = 2 2 V or 2 2 V = 2 8 V 9 Þ V2 = 2 2 3 V 31. (d) For collision B/A V ur should be along ( ) A/B r B A ® r
- 176. 172 PHYSICS So, 2 11 2 2 1 1 2 r r V V V V r r - - = - - ur ur r r A B B A 1 V 2 V 32. (d) Given force 2 ˆ ˆ F 2ti 3t j = + r According to Newton's second law of motion, 2 dv ˆ ˆ m 2ti 3t j dt = + r (m= 1kg) Þ v 0 dv ò r r = ( ) t 2 0 ˆ ˆ 2ti 3t j dt + ò Þ 2 3 ˆ ˆ v t i t j = + r Power P = 2 2 3 ˆ ˆ ˆ ˆ F·v (2t i 3t j)·(t i t j) = + + r r = (2t3 + 3t5)W 33. (a) Given: Mass of particle, M = 10g = 10 kg 1000 radius of circle R= 6.4 cm Kinetic energyE of particle = 8 × 10–4J acceleration at = ? 2 1 mv 2 = E Þ 2 1 10 v 2 1000 æ ö ç ÷ è ø = 8 ×10–4 Þ v2 = 16 × 10–2 Þ v= 4 × 10–1 = 0.4 m/s Now, using v2 = u2 + 2ats(s= 4pR) (0.4)2 = 02 + 2at 22 6.4 4 7 100 æ ö ´ ´ ç ÷ è ø Þ at = (0.4)2 × 7 100 8 22 6.4 ´ ´ ´ = 0.1 m/s2 34. (c) From work-energytheorem, Wg + Wa = DK.E or, mgh +Wa = 2 1 mv 0 2 - 3 3 3 2 a 1 10 10 10 W 10 (50) 2 - - ´ ´ + = ´ ´ Þ Wa = –8.75 J which is the work done due to air resistance Work done due to gravity = mgh =10–3 ×10 ×103 =10J
- 177. CENTRE OF MASS Centreof mass for a system of particles is defined as that point where the entire mass of the system is imagined to be concentrated. If 3 2 1 r , r , r ......... be the position vectors of masses m1, m2, m3 ....... respectivelyfrom the origin O, y O x m1 m2 m4 m3 r4 r1 r2 r3 then the centre of mass of the system is 1 1 2 2 3 3 1 2 3 ( .....) ( ......) + + + = + + + ur uu r uu r uuu r cm m r m r m r r m m m 1 1 n i i i m r M = = å uu r where M is the total mass of the system of particles. The product ofmass ofthe particles and its position vector w.r.t. the reference point is called moment of mass i.e., moment of mass m r = ´ r MOTION OF CENTRE OF MASS The motion of the centre of mass is governed bythe equation cm ext MA F = r r where 2 2 ( ) = r r cm d rcm A dt Momentumconservation of asystemof particles : In the absence of external forces, the velocity of the centre of mass remains constant. We have, MAcm = Fext If Fext = 0 ( ) cm d M v dt = 0 vcm = constant. Hence, momentum (Mvcm = constant) of the centre of mass system is conserved. Rigid Bodies If a body does not undergo any change in shape by the action of a force, it is said to be rigid. If such body undergoes some displacement, every particle in it undergoes the same displacement. No real bodycan be perfectly rigid. Rotatory Motion A body rotating about a fixed axis then every particle of the body moves in a circle and the centres of all these circles lie on axis of rotation. The motion of the body is said to possess rotational motion. Keep in Memory 1. The centre of mass of a system of two identical particles lies in between them on the line joining the particles. 2. If m1 =m2 1 1 2 2 1 2 cm 1 2 (m r m r ) (r r ) r (m m ) 2 + + = = + uu r uu r ur uu r r so, for particles of equal masses the centre of mass is located at the mean position vector of the particles. 3. The position of centre of mass remains unchanged in pure rotatory motion. But it changes with time in translatory motion or rolling motion. 4. The position of centre of mass of a body is independent of the choice of co-ordinate system. 5. If we take the centre of mass at the origin, then the sum of the moments of the masses at of the system about the origin i i m r S r is zero. 6. In pure rotatorymotion, the axis of rotation passes through the centre of mass. 7. If external force is zero then the velocity of the centre of mass of a body remains constant. 8. The centre of mass and centre of gravity of a body coincide, if the value ofg is same throughout the dimension of the body. 9. In kinematics and dynamics, whole of the mass of a body can be assumed to be concentrated at the centre of mass. 10. The location of the centre of mass depends on the shape and nature of distribution of mass of the body. (a) The position of centre of mass of continuous bodies can be found using integration as 7 System of Particles and Rotational Motion
- 178. 174 PHYSICS Relationship betweenangular velocity and linearvelocity: v r w P x y z q Þ v = rw sin q r r r v = ×r w where q is the angle between w & r. MOMENT OF INERTIA AND RADIUS OF GYRATION A rigid body having constituent particles of masses m1, m2, ....mn and r1, r2 ... rn be their respective distances from the axis of rotation then moment of inertia is given by, I = m1 r1 2 + m2 r2 2 + ... + mn rn 2 2 1 n i i i m r = = å The moment of inertia of continuous mass distribution is given by 2 I r dm = ò where r is the perpendicular distance of the small mass dm from the axis of rotation. Its SI unit is kgm2. It is a tensor. Radius of gyration : The radius of gyration of a body about its axis of rotation may be defined as the distance from the axis of rotation at which, if the entire mass of the body were concentrated, its moment of inertia about the given axis would be same as with its actual distribution of mass. Radius of gyration k is given by, I = MK2 or 1 2 1 2 2 I n i i i i m r K M m é ù S ê ú é ù ê ú = = ê ú S ê ú ë û ë û where, M= Smi r1 m1 r2 m2 r3 m3 k Sm r4 m4 X Y Also, 2 2 2 2 1 2 3 ..... n r r r r k n + + + = Therefore, radius of gyration (k) equals the root mean square of the distances of particles from the axis of rotation. ò ò ò = = = dm z M 1 Z , dm y M 1 Y , dm x M 1 X cm cm cm where, x, y and z are the co-ordinates of small mass dm and M is the total mass of the system. (b) The C.M. of a uniform rod of length L is at its middle point. (c) Centre of mass of a uniform semicircular wire is at ÷ ø ö ç è æ p R 2 , 0 , where Ris the radiusof thesemicircular wire. It does not depend on mass. O Y X R (d) For symmetrical bodies of uniform mass distribution, the C. M. lies at the geometrical centre ANGULAR VELOCITY AND ANGULARACCELERATION The angular velocity is defined as the angle covered by the radius vector per unit time. It is denoted by w. O q r P s Q X Y Average angular velocity t Dq w = D The unit of angular velocity is rad/sec. The instantaneous angular velocity w (similar to instantaneous linear velocity) is defined as o lim D ® Dq q w = = D t d t dt The angular acceleration is the rate of change of angular velocity. It is denoted by a. The average angular acceleration aavg. of a rotating bodyis 2 1 . 2 1 w - w Dw a = = - D avg t t t In analogy to linear acceleration a r , the instantaneous angular acceleration is defined as lim d ® Dw w a = = D t o d t dt The unit of angular acceleration is rad/sec2.
- 179. 175 System of ParticlesandRotationalMotion GENERAL THEOREMS ON MOMENT OF INERTIA Theorem of perpendicular axis : According to this theorem “ the moment of inertia of a plane lamina (a plane lamina is a 2-dimensional body. Its third dimension is so small that it can be neglected.) about an axis, perpendicular to the plane of lamina is equal to the sum of the moment of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other at the point, where the perpendicular axes passes through it. If Ix and Iy bethe moment ofinertia of a planelamina (or 2D rigid body) about the perpendicular axis OX and OY respectively, which lie in plane of lamina and intersect each other at O, then moment of inertia (Iz) about an axis passing through (OZ) and perpendicular to its plane is given by X p(x, y) x Y y o Z r Ix + Iy = Iz Let us consider a particle of mass m at point P distance r from origin O, where 2 2 y x r + = so Ix + Iy = Smy2 + Smx2 = Smr2 i.e., Iz = Ix + Iy Theorem of parallel axes : (Derived by Steiner) This theorem is true for both plane laminar bodyand thin 3D body. It states that “the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the distance between two axes. P A Q B M c.m O r Let ABbe the axis in plane of paper about which, the moment of inertia (I) of plane lamina is to be determined and PQ an axis parallel toAB, passing through centre of mass Ooflamina is at a distance ‘r’fromAB. Consider a masselement m oflamina at point Pdistant x fromPQ. Nowthe moment of inertia of the element aboutAB = m (x + r)2 so moment of inertia of whole lamina aboutAB is I= Sm(x+r)2 = Smx2 + Smr2 +2Smxr Where first term on R.H.S is S mx2 = Ic.m. moment of inertia of lamina about PQ through its centre of mass, second term on R.H.S. is Smr2 = r2Sm = Mr2, M is whole mass of lamina, third term on R.H.S is (Smx)r = 0, becauseSmx is equal tomoments of all particles oflamina about an axis PQ,passing through itscentre of mass. Hence I = Ic.m. + M.r2 i.e., the moment of inertia of lamina about AB = its moment of inertia about a parallel axis PQ passing through its centre ofmass + mass of lamina×(distance between two axes)2 Example1. Three rings each of mass P and radius Q are arranged as shown in fig. What will be the moment of inertia of the arrangement about YY’? Y Y’ 3 Q Q Q 1 2 P P P Solution : Moment of inertia of each ring about its diameter = 2 PQ 2 1 . So total moment of inertia of all three rings about Y Y' is Itotal = I1 + I2+ I3 Using theorem of parallel axes (for rings 1 and 2), we get Itotal = 2 2 2 2 2 PQ 2 1 PQ PQ 2 1 PQ PQ 2 1 + ÷ ø ö ç è æ + + ÷ ø ö ç è æ + = 2 7 PQ 2 Example2. Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. Find MI of this system about an axis passing through one of the particles and perpendicular to plane of disc. Solution : According to the theorem of parallel axes, MI of disc about an axis passing through K and perpendicular to plane of disc, is = 2 2 2 MR 2 3 MR MR 2 1 = + R R O K Particle 2 Particle 3 Particle 4 Particle 1 Total MI of the system = 2 2 2 2 ) R 2 ( m ) R 2 ( m ) R 2 ( m MR 2 3 + + + 2 MR 19 2 =
- 180. 176 PHYSICS Example3. Three identicalrods, each of length l, are joined to from a rigid equilateral triangle. Find its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle. Solution : B A 60 o D l C 2 2 2 AB AC BC D ml ml I I I I (I mh ) 3 3 = + + = + + + = ú û ù ê ë é + + = + + 4 3 12 1 3 2 m ) 60 sin ( m 12 m m 3 2 2 2 2 2 l l l l = 2 2 k ) m 3 ( m 2 3 = l or 2 k = l Example4. A uniform rod of mass m and length l makes a constant angle q with the axis of rotation which passes through one end of the rod. Determine its moment of inertia about this axis. Solution : Mass of element of uniform rod = dx m ÷ ø ö ç è æ l x sinq x dx q Axis Moment of inertia of the element about the axis = 2 ) sin x ( dx m q ÷ ø ö ç è æ l . 2 2 2 2 0 m m I sin θ. x dx sin θ 3 = = ò l l l Example5. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown. Find the position of centre of mass of the remaining portion. Solution : O m2 C1 m1 m C2 Area of whole plate = p(56/2)2 = 784 psq. cm. Area of cut portion = p (42/2)2 = 441 p sq. cm. ; Area of remainingportion = 784p– 441p= 343 pcm2; As mass µ area. 1 2 m Mass of cut portion 441 9 Mass of remaining portion m 343 7 p = = = p Let C2 be centre of mass of remaining portion and C1 be centre of mass of cut portion. O is centre of mass of the whole disc.; OC1 = r1 = 28 -21 = 7 cm. OC2 = r2 = ?; Equating moments of masses about O, we get m2 × r2 = m1 × r1 Þ 9 7 7 9 r m m r 1 2 1 2 = ´ = ´ = Centre of mass of remaining portion is at 9 cm to the left of centre of disc. Example6. Showthat the centre of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length. x O x dx dm Solution : Bysymmetry, we see that yCM = zCM = 0 if the rod is placed along the x axis. Furthermore, if we call the mass per unit length l(thelinear mass density),then l=M/L fora uniform rod. If we divide the rod into elements of length dx, then the massofeachelementisdm= λdx.Sincean arbitraryelement of each element is at a distance x from the origin, equation gives xCM = L L 2 0 0 1 1 L x dm x dx M M 2M l = l = ò ò Because l = M/L, this reduces to xCM = 2 L M L 2M L 2 æ ö = ç ÷ è ø One can also argue that by symmetry, xCM = L/2.
- 181. 177 System of ParticlesandRotationalMotion MOMENT OF INERTIA AND RADIUS OF GYRATION OF DIFFERENT OBJECTS Shape of body Rotational axis Figure Momentof Radiusof inertia gyration (1) Ring (a) Perpendicular to plane M = mass passing through centre R = radius of mass MR2 R (b) Diameter in the plane 1 2 MR2 R 2 (c) Tangent perpendicular to plane 2MR2 2 R (d) Tangent in the plane 3 2 MR2 3 R 2 (2) Disc (a) Perpendicular to plane passing through centre of mass 2 1 MR 2 R 2 (b) Diameter in the plane 2 MR 4 R 2 (c) Tangent in the plane 5 4 MR2 5 2 R M R cm C I M R cm B I M R cm CM 'd I I M R cm D I Ic R M cm Id Y Z Ic R M cm Id
- 182. 178 PHYSICS (d)Tangent perpendicular toplane 3 2 MR2 3 2 R (3) Thinwalled (a)Geometrical axis MR2 R cylinder (b) Perpendicular to length 2 2 R L M 2 12 æ ö + ç ÷ è ø 2 2 R L 2 12 + passing through centre of mass (c) Perpendicular to length 2 2 R L M 2 3 æ ö + ç ÷ è ø 2 2 R L 2 3 + passing through one end (4) Solid cylinder (a)Geometrical axis 2 MR 2 R 2 (b) Perpendicular to length 2 2 R L M 4 12 æ ö + ç ÷ è ø 2 2 R L 4 12 + passing through centre of mass (c) Perpendicular to length 2 2 R L M 4 3 é ù + ê ú ê ú ë û 2 2 R L 4 3 + passing through one end Ic Id R M cm Ic M cm L R Ic M cm L R Ic M cm L R I d Ic M cm R L M L Ic cm R I M L Ic cm Id R
- 183. 179 System of ParticlesandRotationalMotion (5)Annular disc (a) Perpendicular to plane passing through centre M 2 [ 2 2 1 2 R R + ] 2 2 1 2 R R 2 + of mass (b) Diameter in the plane 2 2 1 2 M[R R ] 4 + 2 2 1 2 R R 4 + (6) Hollowcylinder (a)Geometrical axis 2 2 1 2 R R M 2 é ù + ê ú ê ú ë û 2 2 1 2 R R 2 + (b) Perpendicular to length passing through centre 2 2 2 1 2 (R R ) L M 12 4 é ù + + ê ú ê ú ë û 2 2 2 1 2 R R L 12 4 + + of mass (7) Solid sphere (a)Along the diameter 2 5 MR2 2 5 R (b) Along the tangent 7 5 MR2 7 5 R Ic R1 R2 C M Ic R1 R2 C M I M cm L R1 R2 M cm L R1 R2 I Ic M cm R Ic M cm R Id
- 184. 180 PHYSICS (8) Thinspherical shell (a)Along the diameter Ic M R 2 3 MR2 2 3 R (b) Along the tangent I I c d M R 5 3 MR2 5 3 R (9) Hollow sphere Along the diameter Cavity Hollow sphere R r I 2 5 M 5 5 3 3 R r R r é ù - ê ú - ê ú ë û 5 5 3 3 2 (R r ) 5 (R r ) - - (10) Thin rod (a) Perpendicular to length 2 ML 12 L 2 3 passing through centre of mass (b) Perpendicular to length 2 ML 3 L 3 passing through one end M cm L Ic M cm L Ic Id
- 185. 181 System of ParticlesandRotationalMotion (11) Rectangularplate (a) Perpendicular to length in 2 Ma 12 a 2 3 the plane passing through centre of mass a b O AB A B C D I (b) Perpendicular to breadth in 2 Mb 12 b 2 3 the plane passing through centre of mass a b O CD A B C D I (c) Perpendicular to plane passing through centre of mass a b O CM A B C D I 2 2 M(a b ) 12 + 2 2 a b 2 3 + (12) Square Plate (a) Perpendicular to plane passing I1 M a a through centre of mass I1 = 2 Ma 6 a 6 (b) Diagonal passing through I2 I3 M a a 2 2 3 Ma I I 12 = a 2 3 centre of mass (13) Cube (a) Perpendicular toplanepassing I1 M a 2 1 Ma 6 I = 6 a through centre of mass (b) Perpendicular to plane passing M a I2 2 2 2Ma I = 3 2 3 a through one end
- 186. 182 PHYSICS TORQUE, ANGULARMOMENTUM AND ANGULAR IMPULSE Torque : The moment of force is called torque. It is defined as the product of force and the perpendicular distance of the force from the axis of rotation. O X Y q q r F i.e., r F t = ´ r r r or, sin rF t = q where q is the angle between r r and F r . Its S.I. unit is (N-m). The dimensions of torque [ML2T–2] are the same as that of energy but it is not energy. Note : If the line of action of a force passes through axis of rotation then no torque will be formed. Angular Momentum: The angular momentum ofa particle about an arbitrrypoint 'O' is themoment of linear momentum taken about that point. r p O X Y q q It is given as L r p = ´ r r r or, sin L rp = q where q is the angle between r r and F r . Angular Impulse : 2 1 2 1 t t J dt L L = t = - ò (= Change in angular momentum) CONSERVATION OF ANGULAR MOMENTUM From equation dt L d ext = t r . If constant L 0 ext = Þ = t r ...(i) This is called law of conservation of angular momentum. According to this “if resultant external torque ext t r acting on the system is zero then total angular momentum of the system is constant. The magnitude of angular momentum for a system is given by i i i n 1 i v ) r m ( | L | = å = w å = = ) r m ( 2 i i n 1 i [Q v = rw] Þ w = w ´ å = = I ) I ( | L | i n 1 i ...(ii) Where Ii is the moment ofinertia of the ith particle ofthat system and I is total moment of inertia of the system n 3 2 1 i n 1 i I .......... .......... I I I I I + + + + = å = = or 2 n n 2 2 2 2 1 1 r m .......... .......... r m r m I + + + = ...(iii) So if a system undergoes a redistribution of its mass, then its moment of inertia changes but since noexternal torque is applied on the system so total angular momentum is constant before and after the distribution of mass, even if moment of inertia of the system is changed. Initial Final L L = uu r uu r ...(iv) or, I1w1 = I2w2 ...(v) where L initial denote the state previous to the redistribution of mass and final denote the state after the redistribution of mass in that system. Acomparisonof useful relationsin rotational andtranslational or linear motion : Rotational motion about Linear motion a fixed axis Angular velocity d dt q w = Linear velocity dx v dt = Angular acceleration d dt w a = Linear acceleration dv a dt = Resultant torque t = Ia Resultant force F = ma Equations of rotational Equations of linear motion motion 0 2 0 0 2 2 0 0 t constant then t ½ t 2 ( ) w =w + a ì ï ï a = q -q = w + a í ï w =w + a q -q ï î 2 2 2 v u at 1 a constant then s ut at 2 v u 2as ì = + ï ï = = + í ï ï - = î where (s = x – x0) Work o W d q q = t q ò r r work o x x W Fdx = ò r r Kinetic energy Ek = ½ Iw2 Kinetic energy Ek=½ mv2 Power P = tw Power P = Fv rr Angular momentum L = Iw rr Linear momentum p= mv Torque dL dt t = Force dp F dt = Work EnergyTheoreminRotational Motion: According to this theorem “the work done by external forces in rotating a rigid body about a fixed axis is equal to the change in rotational kinetic energy of the body.”
- 187. 183 System of ParticlesandRotationalMotion Since, we can express the torque as t = Ia dt d . d d I dt d I q q w = w = q w w = t d d I but tdq = dW Þ tdq = dW = Iwdw By integrating the above expression, we get total work done by all external force on the body, which is written as 2 1 2 2 I 2 1 I 2 1 d I d W 2 1 2 1 w - w = w w = q t = ò ò w w q q where the angular velocityof the body changes from w1 to w2 as the angular displacement changes from q1 toq2 due to external force . ext F r on the body. . Rotational Kinetic Energy : Let us consider a rigid body (collection ofsmall particles) ofhigh symmetry which is purely rotating about z-axis with an angular velocity w. Each particle has some energy, determined bymi and vi. The kinetic energy of mi particle is ( ) 2 i i i k v m ½ E = ...(i) x y vi mi ri O w Fig. The total kinetic energy of the body is ½Iw2. Now we know that in rigid bodyeveryparticle moves with same angular velocity, the individual linear velocities depends on the distance ri from the axis of rotation according to the expression (vi = riw). Hence the total kinetic energy of rotating rigid bodyis equal to the sum of kinetic energies of individual particles. 2 i i i k k v m ½ ) E ( E S = S = 2 ) r m ( 2 2 i i w S = 2 ½ k E I = w ...(ii) where 2 i ir m I S = is moment of inertia of the rigid body. . Now consider a rigid body which is rolling without slipping. In this case it possesses simultaneous translatory motion and rotatory motion and the total kinetic energy of the rigid body K.ETotal = rotation K.E. + translational K.E. ofC.M. Ek = nal translatio k rotational k ) E ( ) E ( + ...(iii) Ek = 2 cm 2 Mv ½ I ½ + w ... (iv) ) K / r 1 ( I ½ ) r / K 1 ( Mv ½ E 2 2 2 2 2 2 cm k + w = + = ... (v) where, vcm = linear velocity of the centre of mass of the rigid body K = radius of gyration ) r M MK I ( 2 i i 2 S = = r = radius of moving rigid body I = moment of inertia of the rigid body about centre of mass ½ Iw2 = rotational kinetic energyabout the centre of mass. Hence it is clear from the expression that total kinetic energy of rolling body is equal to the sum of rotational kinetic energyabout centreofmass(C.M.)andtranslational kineticenergyof thecentre of mass of body. BodyRolling without SlippingonanInclinedPlane : When a body performed translatory as well as rotatory motion then we can say that the body is in rolling motion. Acceleration for body rolling down an inclined plane without slipping. Let M is the mass of the body, Ris its radius and I is the moment ofinertia about the centre ofmass and K is the radius of gyration. q Mgsinq Mg Mgcosq q R f Force equation, Ma f sin Mg = - q ...(1) Torque equation, a = I fR ...(2) Also, 2 MK I = ...(3) Þ R MK f 2 a = ...(4) Adding, eqn. (1) &(4) ÷ ÷ ø ö ç ç è æ + a = q a R K M sin Mg 2 R a = a (Q Motion is pure rolling) 2 2 2 2 sin sin 1 K a g Mg M a a R K R æ ö q q = + Þ = ç ÷ ç ÷ è ø + Now assume in fig. that a body (which has high symmetry such as cylinder, sphere etc.) is rolling down an incline plane without slipping. This is possible onlyif friction is present between object and incline plane, because it provides net torque to the body for rotating about the centre of mass (since the line of action of the other forces such as mg and R pass through the centre of mass of rigid body, hence they do not produce torque in a body about the centre of mass). M R mgsinq x contact point mg h q w f= R ms vc c. m
- 188. 184 PHYSICS A roundobject rolling down an incline, mechanical energy is conserved if no slipping occurs. But mechanical energy of the body remains constant despite of friction because the contact point is at rest relative to the surface at any instant. For pure rolling motion vc = rw so 2 c 2 c k Mv ½ ) r / v ( I ½ E + = 2 c 2 c k v M r I 2 1 E ÷ ÷ ø ö ç ç è æ + = ...(i) As body rolls down an incline, it loses potential energy Mgh (h is the height of incline). Since body starts from rest at top, hence its total kinetic energy at bottom given by eqn.(i) must be equal to Mgh at top i.e., Mgh V M r I 2 1 2 c 2 c = ÷ ÷ ø ö ç ç è æ + 2 / 1 2 c c MR / I 1 gh 2 v ÷ ÷ ø ö ç ç è æ + = Þ ...(ii) since h =x sin q, where x is length of incline and Ic =MK2 (K is radius of gyration), then ½ 2 2 c R / K 1 sin gx 2 v ú û ù ê ë é + q = ....(iii) For rolling down an inclined plane without slipping, the essential condition is : q + ³ m tan I MR I cm 2 cm s ; ms is the coefficient of static friction where Icm is moment of inertia of the body about its C.M. DIFFERENT TYPES OF MOTION Pure translational motion : In this case thevelocities at all three points : (i) top most point P, (ii) C.M. and (iii) contact point O are same. (A) P v C.M (v) v O Pure rotational motion : In this case the velocityof (i) top most point P is Rw (ii) C.M. is zero (iii) contact point O is – Rw (B) Rw –Rw w O C.M Vcm = 0 Rolling Motin Combination of translatory and rotatory motion : (i) In pure rolling motion, the contact point O remains at rest. (ii) In pure rolling, the velocity of top most point is, V = Vcm + wR= 2Vcm V Vcm Vcm V + R cm w O wR Contact point C.M P R (iii) In rolling without slipping Vcm = wR. 2 R = 2V w cm Vcm O Surface at rest Keep in Memory 1. The axis of the rolling body is parallel to the plane on which the body rolls in case of sphere, disc, ring. 2. Let (Ek)r = rotational kinetic energy (Ek)t = translation kinetic energy (a) For solid sphere, (Ek)r = 40% of (Ek)t (b) For shell (Ek)r = 66% of (Ek)t (c) For disc, (Ek)r = 50% of (Ek)t (d) For ring, (Ek)r = (Ek)t 3. Translational kinetic energy is same for rolling bodies having same M, R and w irrespective of their shape. 4. Total energyis minimum for solid sphere and maximum for ring having same mass and radius. 5. Rotationalkineticenergyismaximumfor ring and minimum for solid sphere of same mass and radius. 6. (i) The acceleration down the inclined plane for different shapes of bodies of same mass and radius are as follows : Sphere > disc > shell > ring (ii) The velocity down the plane is related as follows: Sphere > disc > shell > ring (iii) The time taken to reach the bottom of the inclined plane is related as follows : Ring > shell > disc > sphere.
- 189. 185 System of ParticlesandRotationalMotion This is because (i) For ring 1 R K 2 2 = (ii) For disc 2 1 R K 2 2 = (iii) For sphere 5 2 R K 2 2 = (iv) For shell 3 2 R K 2 2 = 7. The angular speed of all particles of a rotating/revolving rigid body is same, although their linear velocities maybe different. 8. (i) Two particles moving with angular speed w1 and w2 on the same circular path and both in anti-clockwise direction then their relative angular speed will be wr = w1 – w2. 1 p r 2 p w1 w2 (ii) In the above case one particle will complete one revolution more or less as compared to the other in time 1 2 2 1 2 1 2 1 r T 1 T 1 T 1 or T T T T 2 2 T - = - = w - w p = w p = 9. Let two particles move on concentric circles having radius r1 and r2 and their linear speeds v1, v2 both along anti- clockwise direction then their relative angular speed will be 1 2 r 1 2 | v v | ω | r r | - = - ur ur r r 1 r 2 r 2 v 1 v 10. As I µ K2 and I= MK2 , hencegraph between Iand Kwillbe a parabola. However graph between log I and log K will be straight line. Y I K X Y log I log K X 11. Rotational KE, 2 r k I 2 1 ) E ( w = Er µ w2 graph between r k ) E ( and w will be as below Y X w Er 12. Angular momentum L= Iw, hence Lµ w. Graph between L and wis a straight line. Y X L w 13. The moment of inertia is not a vector quantity because clockwise or anti-clockwise, direction is not associated with it. It is a tensor. 14. When a spherical/circular/cylindrical bodyis given a push, it only slips when the friction is absent, It may roll with slipping if friction is less than a particular valueand it may roll without slipping if the friction is sufficient. (i.e. cm 2 cm s I MR ( tan I + q ³ m ) 15. When a bodyrollswithout slipping nowork is done against friction. Example7. A thin circular ring of mass M and radius R rotating about its axis with a constant angular speed w. Two blocks, each of mass, m are attached gently to opposite ends of a diameter of the ring. Find the angular speed of the ring. Solution : As L = Iw = constant. Therefore, 1 1 2 2 w I = w I or 2 1 1 2 I w I = w = m 2 M M K ) m 2 M ( MK 2 2 + w = + w Example8 A solid sphere of mass 500 g and radius 10 cm rolls without slipping with a velocity of 20 cm/s. Find the total K.E. of the sphere.
- 190. 186 PHYSICS Solution : TotalK.E. = 2 2 1 1 Iω mv 2 2 + = 2 2 2 1 2 1 m r ω mv 2 5 2 ´ + As, Isphere = 2 2 mr 5 = 2 2 2 1 1 7 mv mv mv 5 2 10 + = = J 014 . 0 5 1 1000 500 10 7 2 = ÷ ø ö ç è æ ´ ´ . Example9. A body of radius R and mass M is rolling horizontally without slipping with speed v, it then rolls up a hill to a maximum height h. If h = 3v2/4 g, (a) what is the moment of inertia of the body? (b) what might be the shape of the body? Solution : (a) 2 2 total trans. rot. 1 1 K K K M v Iω 2 2 = + = + 2 2 2 2 2 1 1 v v I M v I M 2 2 2 R R æ ö é ù = + = + ç ÷ ê ú ç ÷ ë û è ø When it rolls up a hill to height h, the entire kinetic energy is converted into potential energy M g h Thus 2 2 2 v I v M Mg h Mg 3 2 4g R é ù é ù + = = ê ú ê ú ë û ê ú ë û or 2 2 I 3 MR M M I 2 2 R é ù + = = ê ú ë û (b) The body may be a circular disc or a solid cylinder. Example10. The angular velocity of a body changes from w1 to w2 without applying torque but by changing moment of inertia. What will be the ratio of initial radius of gyration to the final radius of gyration? Solution : Since I1w1 = I2w2 or 2 2 1 1 2 2 M K M K w = w Þ ÷ ÷ ø ö ç ç è æ w w = 1 2 2 1 K K Example11. A rod of mass M and length l is suspended freely from its end and it can oscillate in the vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position with angular speed w. What is the kinetic energy while passing through the mean position? Solution : K.E, 2 1 2 = Iw and I = (M l2/12) + M (l/2)2 = 3 M 2 l 2 2 2 2 M 6 1 3 M 2 1 K w = w ú ú û ù ê ê ë é = l l Example12. A cord iswound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, what will be the angular velocity of the wheel? Solution : 2 2 1 1 mgh Ιω mv 2 2 = + 2 2 2 r m 2 1 2 1 w + w I = or 2 2 ) r m ( h g m 2 w + I = 1/ 2 2 2m g h mr é ù w = ê ú I + ê ú ë û Example13. Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic energy of the earth. Suppose the earth’s radius decrease by 2%, keeping all other quantities same, then (a) g decreases by 2% and K decreases by 4% (b) g decreases by 4% and k increases by 2% (c) g increases by 4% and K decreases by 4% (d) g decreases by 4% and K increases by 4% Solution : (c) We know that, 2 R M G g = Differentiating, ÷ ø ö ç è æ - = R dR 2 g dg ....(1) Further, 2 2 2 R M 5 3 2 1 I 2 1 K w ú û ù ê ë é = w = or ÷ ø ö ç è æ ´ w = R dR 2 M 10 3 K dK 2 ....(2) When radius decreases by2%, then g increases by 4% and K decreases by 4%. Example14. A small solid ball rolls without slipping along the track shown in fig. The radius of the circular part of track is R. If the ball starts from rest at a height 8 R above the bottom, what are the horizontal and vertical forces acting on it at P? 8 R R R O P Ball Solution : Suppose m is the mass of the ball of radius r. On reaching P, the net height through which the ball descends is 8 R – R= 7R, (from the fig.) decrease in P.E. of ball = mg × 7 R This appears as total KE of ball at P.
- 191. 187 System of ParticlesandRotationalMotion Thus mg × 7R = KE oftranslation + KE ofrotational 2 2 1 1 mv Iω 2 2 = + 2 2 2 mr 5 2 2 1 mv 2 1 w ÷ ø ö ç è æ ´ + = 2 mv 10 7 = 2 v = 10 g R (wherev = rw and r is radius of solid ball) The horizontal force acting on the ball, Fh = centripetal force towards O mg 10 R ) R g 10 ( m R mv2 = = = Vertical force on the ball, Fv = weight of the ball = mg. Example15. A solid cylinder first rolls and then slides from rest down a smooth inclined plane. Compare the velocities in the two cases when the cylinder reaches the bottom of the incline. Solution : (i) We know that acc. of a bodyrolling down an inclined plane is 2 2 1 R / K 1 sin g a + q = For a solid cylinder 2 R K 2 2 = ; q = + q = sin g 3 2 ) 2 / 1 1 ( sin g a1 ; From 2 v = u2 + 2 a s 2 1 2 v 0 2 gsinθ 3 æ ö = + ´ ç ÷ è ø l 4 gsin 3 = q ´ l ...(i) (ii) Acc. of the sliding cylinder, a2 =g sin q From 2 v = u2 + 2 a s; 2 2 v 0 2gsinθ = + ´ l ...(ii) 2 1 2 2 v 4gsin θ 2 3 2gsin θ 3 v ´ = = ´ ´ l l ; 1 2 v 2 / 3 v = . Example16. A disc of mass M and radius R is rolling with angular speed w on a horizontal plane (fig.). Determine the magnitude of angular momentum of the disc about the origin O. Y O X M R w Solution : The angular momentum L is given by cm L Ι ω M vR = + R ) R ( M R M 2 1 2 w + w ÷ ø ö ç è æ = w = 2 R M 2 3
- 192.
- 193. 189 System of ParticlesandRotationalMotion 1. Centre of mass of the earth and the moon system lies (a) closer to the earth (b) closer to the moon (c) at the mid-point oflinejoining theearth and themoon (d) cannot be predicted 2. Four particles ofmasses m1,m2,m3 and m4 are placed at the vertices A,B,C and D as respectively of a square shown. The COM of the system will lie at diagonal AC if (a) m1 = m3 (b) m2 = m4 A B D C m1 m2 m3 m4 (c) m1 = m2 (d) m3 = m4 3. Two spheres A and B of masses m and 2m and radii 2R and R respectively are placed in contact as shown. The COM of the system lies A B 2R R (a) insideA (b) inside B (c) at the point of contact (d) None of these 4. Moment of inertia does not depend upon (a) distribution of mass (b) axis of rotation (c) point of application of force (d) None of these 5. Adisc isgiven alinear velocityon arough horizontal surface then itsangular momentum is (a) conserved about COM only (b) conserved about the point of contact only (c) conserved about all the points (d) not conserved about any point. 6. A body cannot roll without slipping on a (a) rough horizontal surface (b) smooth horizontal surface (c) rough inclined surface (d) smooth inclined surface 7. A body is projected from ground with some angle to the horizontal. The angular momentum about the initial position will (a) decrease (b) increase (c) remains same (d) first increase then decrease 8. A ball tied toa string is swung in a vertical circle. Which of the following remains constant? (a) tension in the string (b) speed of the ball (c) centripetal force (d) earth's pull on the ball 9. Angular momentum of a system of a particles changes, when (a) force acts on a body (b) torque acts on a body (c) direction of velocity changes (d) None of these 10. Angular momentum is (a) a polar vector (b) an axial vector (c) a scalar (d) None of these 11. If a running boy jumps on a rotating table, which of the following is conserved? (a) Linearmomentum (b) K.E (c) Angular momentum (d) None of these 12. Agymnast takesturnswith her arms&legs stretched. When she pulls her arms & legs in (a) the angular velocity decreases (b) the moment of inertia decreases (c) the angular velocity stays constant (d) the angular momentum increases 13. Moment of inertia ofa circular wire of mass Mand radius R about its diameter is (a) MR2/2 (b) MR2 (c) 2MR2 (d) MR2/4. 14. A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected ? (a) Angular velocity (b) Angular momentum (c) Moment of inertia (d) Rotational kinetic energy 15. One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively A I and B I Such that (a) A B I I < (b) A B I I > (c) A B I I = (d) A A B B I d I d = where A d and B d are their densities.
- 194. 190 PHYSICS 16. Angularmomentum is (a) moment of momentum (b) product of mass and angular velocity (c) product of M.I. and velocity (d) moment of angular motion 17. Theangular momentum ofa system of particleis conserved (a) when no external force acts upon the system (b) when no external torque acts upon the system (c) when no external impulse acts upon the system (d) when axis of rotation remains same 18. Analogue of mass in rotational motion is (a) moment of inertia (b) angular momentum (c) gyration (d) None of these 19. Moment of inertia does not depend upon (a) angular velocity of body (b) shape and size (c) mass (d) position of axis of rotation 20. A hollow sphere is held suspended. Sand is now poured into it in stages. SAND The centre of gravity of the sphere with the sand (a) rises continuously (b) remains unchanged in the process (c) First rises and then falls to the original position (d) First falls and then rises to the original position 21. A block Q of mass M is placed on a horizontal frictionless surface AB and a body P of mass m is released on its frictionless slope. As P slides bya length L on this slope of inclination q, the block Q would slide by a distance A B M Q q m P (a) (m/M) L cos q (b) mL/(M+m) (c) (M + m)/(m L cos q) (d) (m L cos q) / (m + M) 22. A solid sphere and a hollow sphere of the same material and of same size can be distinguished without weighing (a) by determining their moments of inertia about their coaxialaxes (b) byrolling them simultaneouslyon an inclined plane (c) by rotating them about a common axis of rotation (d) by applying equal torque on them 23. A stick of length L and mass M lies on a frictionless horizontal surface on which it is free to move in anyway.A ball of mass m moving with speed vcollides elasticallywith the stick as shown in fig. L M m v Ifafter the collision ball comes to rest, then what should be the mass of the ball? (a) m= 2M (b) m= M (c) m = M/2 (d) m = M/4 24. A flywheel rotates about an axis. Due to friction at the axis, it experiences an angular retardation proportional to its angular velocity. Ifits angular velocity falls to half while it makes n rotations, how many more rotations will it make before coming to rest? (a) 2n (b) n (c) n/2 (d) n/3 25. A raw egg and a hard boiled egg are made tospin on a table with the same angular momentum about the same axis. The ratio of the time taken by the two to stop is (a) =1 (b) <1 (c) >1 (d) None of these 1. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. The K.E. associated with the rotation of the cylinder is (a) 3025J (b) 3225J (c) 3250J (d) 3125J 2. What is the moment of inertia of a solid sphere of density r and radius R about its diameter? (a) r 5 R 176 105 (b) r 2 R 176 105 (c) r 5 R 105 176 (d) r 2 R 105 176 3. Two particlesAand B, initiallyat rest, moves towards each other under a mutual forceofattraction.At the instant when the speed of A is v and the speed of B is 2 v, the speed of centre of mass is (a) zero (b) v (c) 1.5 v (d) 3 v 4. Point masses 1, 2, 3 and 4 kg are lying at the points (0, 0, 0), (2, 0, 0), (0, 3, 0) and (–2, –2, 0) respectively. Themoment of inertia of this system about X-axis will be (a) 43 kg –m2 (b) 34kg–m2 (c) 27 kg – m2 (d) 72 kg – m2
- 195. 191 System of ParticlesandRotationalMotion 5. A body having moment of inertia about its axis of rotation equal to 3 kg-m2 is rotating with angular velocityequal to3 rad/s. Kinetic energy of this rotating body is the same as that of a body of mass 27 kg moving with a speed of (a) 1.0 m/s (b) 0.5m/s (c) 1.5m/s (d) 2.0m/s 6. A particle moves in a circle of radius 0.25 m at two revolutions per second. The acceleration of the particle in metre per second2 is (a) p2 (b) 8p2 (c) 4p2 (d) 2p2 7. The radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Then, its radius of gyration about a parallel axis through its centre of mass will be (a) 80cm (b) 8cm (c) 0.8cm (d) 80m 8. A particle of mass m is moving in a plane along a circular path of radius r. Its angular momentum about the axis of rotation is L. The centripetal force acting on the particle is (a) L2 /mr (b) L2 m/r (c) L2 /m r3 (d) L2 /m r2 9. A small discofradius 2 cm is cut from a discofradius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc? (a) 0.4cm (b) 2.4cm (c) 1.8cm (d) 1.2cm 10. A solid cylinder of mass m & radius R rolls down inclined plane without slipping. The speed of its C.M. when it reaches the bottom is (a) gh 2 h (b) 4gh/3 (c) gh 4 / 3 (d) gh 4 11. Theratioofmoment ofinertia ofcircular ring& circular disc having the same mass &radii about on axis passing the c.m & perpendicular to plane is (a) 1: 1 (b) 2: 1 (c) 1: 2 (d) 4: 1 12. A rod of length L is pivoted at one end and is rotated with a uniform angular velocityin a horizontal plane. Let T1and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted end. Then (a) T1 > T2 (b) T2 > T1 (c) T1 – T2 (d) the relation between T1 and T2 depends on whether the rod rotates clockwise and anticlockwise. 13. A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a uniform speed v. The centrifugal force on it is (a) r mv2 towards the centre (b) r mv2 away from the centre (c) r mv2 along the tangent through the particle (d) zero 14. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Yof radius4Ris made from an iron plate of thickness t 4 . Then the relation between the moment of inertia X I and Y I is (a) Y X Ι 32Ι = (b) Y X Ι 16Ι = (c) Y X Ι Ι = (d) Y X Ι 64 Ι = 15. A particles performing uniform circular motion. Its angular frequencyis doubled and its kinetic energyhalved, then the newangular momentumis (a) L 4 (b) 2L (c) 4L (d) L 2 16. A simple pendulum is vibrating with angular amplitude of 90° as shown in figure. q For what value of q is the acceleration directed (i) verticallyupwards (ii) horizontally (iii) verticallydownwards (a) ° ° - 90 , 3 1 cos , 0 1 (b) ° ° - 90 , 0 , 3 1 cos 1 (c) ° ° - 0 , 3 1 cos , 90 1 (d) ° ° - 0 , 90 , 3 1 cos 1 17. A mass is tied to a string and rotated in a vertical circle, the minimum velocity of the body at the top is (a) gr (b) r / g (c) 2 / 3 r g ÷ ø ö ç è æ (d) gr
- 196. 192 PHYSICS 18. Acouple is acting on a two particle systems. The resultant motion will be (a) purelyrotational motion (b) purelylinear motion (c) both a and b (d) None of these 19. A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing. 20. A smooth sphere A is moving on a frictionless horizontal plane with angular speed wand centre of mass velocityv. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere.After the collision, their angular speeds are wA and wB , respectively. Then (a) wA < wB (b) wA = wB (c) wA = w (d) wB =w 21. A particle moves in a circle of radius 4 cm clockwise at constant speed 2 cm s–1 . If x̂ and ŷ are unit acceleration vectors along X and Y respectively (in cm s–2), the acceleration of the particle at the instant half waybetween P and Q is given by X Y P Q O (a) ) ŷ x̂ ( 4 + - (b) ) ŷ x̂ ( 4 + (c) 2 / ) ŷ x̂ ( + - (d) 4 / ) ŷ x̂ ( - 22. A particle is confined torotate in a circular path decreasing linear speed, then which of the following is correct? (a) L r (angular momentum) is conserved about the centre (b) onlydirection of angular momentum L r is conserved (c) It spirals towards the centre (d) its acceleration is towards the centre. 23. Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through centre as 1 : 8. The value of n is (a) 2 (b) 2 2 (c) 4 (d) 2 1 24. A cylinder rolls down an inclined plane of inclination 30°, the acceleration of cylinder is (a) 3 g (b) g (c) 2 g (d) 3 g 2 25. A uniform bar of mass M and length L is horizontally suspended from the ceiling by two vertical light cables as shown. Cable A is connected 1/4th distance from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A? Cable B Cable A 1 4 L L (a) 1/4 Mg (b) 1/3 Mg (c) 2/3 Mg (d) 3/4 Mg 26. ABC is a triangular plateof uniform thickness.The sides are in the ratio shown in the figure. IAB, IBC and ICA are the moments ofinertia oftheplateaboutAB, BCandCAasaxes respectively.Which one ofthe following relations is correct? (a) BC AB I I > (b) AB BC I I > B A C 3 90° 3 5 (c) CA BC AB I I I = + (d) CA I ismaximum 27. In carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.12 × 10–10 m. The distance ofthe centre of mass, from the carbon atom is (a) 0.64 × 10–10 m (b) 0.56 × 10–10 m (c) 0.51 × 10–10 m (d) 0.48 × 10–10 m 28. A weightless ladder 20 ft long rests against a frictionless wall at an angleof 60ºfrom thehorizontal.A150 pound man is 4 ft from thetop of the ladder.Ahorizontal force is needed tokeep it from slipping. Choosethe correct magnitudefrom the following. (a) 175lb (b) 100lb (c) 120lb (d) 17.3lb 29. The moment of inertia of a disc of mass M and radius R about an axis, which is tangential to thecircumferenceofthe disc and parallel toits diameter, is (a) 2 MR 2 3 (b) 2 MR 3 2 (c) 2 MR 4 5 (d) 2 MR 5 4 30. A tube one metre long is filled with liquid ofmass 1 kg. The tube is closed at both the ends and is revolved about one end in a horizontal plane at 2 rev/s. The force experienced by the lid at the other end is (a) 4p2N (b) 8p2N (c) 16p2N (d) 9.8 N 31. Ifthe linear density(mass per unit length) ofa rod oflength 3m isproportional tox, wherex is thedistancefrom oneend of the rod, the distance of the centre of gravity of the rod from this end is (a) 2.5m (b) 1m (c) 1.5m (d) 2m
- 197. 193 System of ParticlesandRotationalMotion 32. A composite disc is to be made using equal masses of aluminiumandiron sothat it has ashigh a moment ofinertia as possible. This is possible when (a) thesurfaces ofthediscare madeofiron with aluminium inside (b) the whole ofaluminium is kept in the core and the iron at the outer rim of the disc (c) the whole of the iron is kept in the core and the aluminium at the outer rim of the disc (d) the whole disc is made with thin alternate sheets of iron and aluminium 33. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energywill be (a) 2 2 2 R K R + (b) 2 2 2 R R K + (c) 2 2 R K (d) 2 2 2 R K K + 34. When a bucket containing water is rotated fast in a vertical circle of radius R, the water in the bucket doesn't spill provided (a) thebucket is whirled with a maximum speedof 2gR (b) the bucket is whirled around with a minimum speed of ] gR ) 2 / 1 [( (c) the bucket is having a rpm of ) R /( g 900 2 p (d) the bucket is having a rpm of ) R /( g 3600 2 p 35. A coin placed on a gramophone record rotating at 33 rpm flies off the record, if it is placed at a distance of more than 16 cm from the axis of rotation. If the record is revolving at 66 rpm, the coin will fly off if it is placed at a distance not less than (a) 1cm (b) 2cm (c) 3cm (d) 4cm 36. Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are 5.0 kg m2 and 20.0 kg m2 respectively. Wheel Ais madeto rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become (a) 2 5 rps (b) 0.5rps (c) 2 rps (d) None of these 37. A wheel is rolling straight on ground without slipping. If the axis of the wheel has speed v, the instantenous velocity of a point P on the rim, defined by angle q, relative to the ground will be (a) ÷ ø ö ç è æ q 2 1 cos v (b) ÷ ø ö ç è æ q 2 1 cos v 2 q P (c) ) sin 1 ( v q + (d) ) cos 1 ( v q + 38. Acertain bicycle can go up a gentle incline with constant speed when the frictional force of ground pushing the rear wheel is F2 = 4 N. With what forceF1 must the chain pull on the sprocket wheel if R1=5 cm and R2 = 30 cm? Road Chain F1 R2 R1 F = 4N 2 Horizontal (a) 4N (b) 24N (c) 140N (d) 4 35 N 39. Auniform rod of length l is free torotate in a vertical plane about a fixed horizontal axis through O. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle q, its angular velocity w is given as q O (a) q sin g 6 l (b) 2 sin g 6 q l (c) 2 cos g 6 q l (d) q cos g 6 l 40. A sphere of mass 2000 g and radius 5 cm is rotating at the rate of 300 rpm .Then the torque required to stop it in 2p revolutions, is (a) 1.6 × 102 dyne cm (b) 1.6 × 103 dyne cm (c) 2.5 × 104 dyne cm (d) 2.5 × 105 dyne cm 41. Five masses are placed in a plane as shown in figure. The coordinates of the centre of mass are nearest to y 2 3 kg 5 kg 2 kg 1 kg 4 kg 1 0 0 2 x 1 (a) 1.2,1.4 (b) 1.3,1.1 (c) 1.1,1.3 (d) 1.0,1.0
- 198. 194 PHYSICS 42. Asolid sphere of mass 1 kg rolls on a table with linear speed 1 ms–1. Its total kinetic energy is (a) 1 J (b) 0.5J (c) 0.7J (d) 1.4J 43. Aboyand aman carrya uniform rodoflength L, horizontally in such a way that the boy gets 1/4th load. If the boy is at one end of the rod, the distance of the man from the other end is (a) L/3 (b) L/4 (c) 2L/3 (d) 3L/4 44. A solid sphere ofmassM andradius Ris pulledhorizontally on a sufficiently rough surface as shown in the figure. Choose the correct alternative. (a) The acceleration of the centre of mass is F/M (b) The acceleration of the centre of mass is 2 F 3 M (c) The friction force on the sphere acts forward (d) The magnitude of the friction force is F/3 45. Three identical particles each of mass 1 kg are placed touching one another with their centres on a straight line. Their centres are marked A, B and C respectively. The distance of centre of mass of the system from A is (a) AB AC BC 3 + + (b) AB AC 3 + (c) AB BC 3 + (d) AC BC 3 + 46. M.I ofa circular loop ofradius R about the axis in figure is Axis of rotation O R/2 x y (a) MR2 (b) (3/4) MR2 (c) MR2/2 (d) 2MR2 47. If the earth is treated as a sphere of radius R and mass M, its angular momentum about the axis of its diurnal rotation with period T is (a) 2 4 MR 5T p (b) 2 2 MR 5T p (c) 2 MR T T (d) 3 MR T p 48. The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 seconds it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is (a) 30 radian / second2 (b) 1880 degrees/ second2 (c) 40 radian / second2 (d) 1980 degree/second2 49. Four masses are fixed on a massless rod as shown in the adjoining figure. The moment of inertia about the dotted axis is about 0.2 m 0.2 m 2kg 5kg 5kg 2kg 0.4 m 0.4 m (a) 2kg × m2 (b) 1kg × m2 (c) 0.5 kg × m2 (d) 0.3 kg × m2 50. The moment of inertia ofa bodyabout a given axis is 1.2 kg m2. Initially, the body is at rest . In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad s–2 must be applied about that axis for a duration of (a) 4 s (b) 2 s (c) 8 s (d) 10 s 51. Fig. shows a disc rolling on a horizontal plane with linear velocityv. Its linear velocityis v and angular velocity is w. Which of the following gives the velocity of the particle P on the rim of the disc v O q M N B P A r w (a) v (1 + cos q) (b) v (1 – cos q) (c) v (1 + sin q) (d) v (1 – sin q) 52. A toycar rolls down the inclined plane as shown in the fig. It loops at the bottom. What is the relation between H and h? (a) H 2 h = r D B H h (b) H 3 h = (c) H 4 h = (d) H 5 h =
- 199. 195 System of ParticlesandRotationalMotion 53. A sphere rolls down on an inclined plane of inclination q. What is the acceleration as the sphere reaches bottom? (a) 5 gsin 7 q (b) 3 gsin 5 q (c) 2 gsin 7 q (d) 2 gsin 5 q 54. In a bicycle, the radius of rear wheel is twice the radius of front wheel. If rF and rF are theradii, vr and vr are the speed of top most points of wheel. Then (a) r F v 2v = (b) F r v 2v = (c) F r v v = (d) F r v v > 55. An annular ring with inner and outer radii 1 R and 2 R is rolling without slipping with a uniform angular speed. The ratio of the forces experienced bythe twoparticles situated on the inner and outer parts of the ring , 2 1 F F is (a) 2 2 1 R R ÷ ÷ ø ö ç ç è æ (b) 1 2 R R (c) 2 R 1 R (d) 1 56. A wheel havingmoment ofinertia 2 kg-m2 about its vertical axis, rotates at the rateof60 rpm about this axis, The torque which can stop the wheel’s rotation in one minute would be (a) Nm 18 p (b) 2 Nm 15 p (c) Nm 12 p (d) Nm 15 p 57. Consider a system of two particles having masses m1 and m2 . Ifthe particle of mass m1 is pushed towards the centre of mass particles through a distance d, by what distance would the particle of mass m2 move so as to keep the mass centre of particles at the original position? (a) d m m 1 2 (b) d m m m 2 1 1 + (c) d m m 2 1 (d) d 58. The ratio of the radii of gyration of a circular disc about a tangential axis in the plane ofthe disc and ofa circular ring of the same radius about a tangential axis in the plane of the ring is (a) 1: Ö2 (b) 1: 3 (c) 2: 1 (d) Ö5: Ö6 59. A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity w about the same axis. The final angular velocity of the combination of discs is (a) 1 2 1 I ) I I ( w + (b) 2 1 2 I I I + w (c) w (d) 2 1 1 I I I + w 60. Three particles, each of mass m gram, are situated at the vertices of an equilateral triangleABC of side/cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular toAB and in the plane of ABC, in gram-cm2 units will be (a) 2 m 2 3 l (b) 2 m 4 3 l (c) 2 ml2 m l l l m m C B A X (d) 2 m 4 5 l 61. Themoment ofinertia ofa uniform circular discof radius‘R’ and mass ‘M’about an axispassing from the edge of thedisc and normal to the disc is (a) MR2 (b) 1 2 2 MR (c) 3 2 2 MR (d) 7 2 2 MR 62. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio (a) 2: 1 (b) 1: 2 (c) 2 1 : (d) 1 2 : 63. A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle q. The frictional force (a) dissipates energy as heat. (b) decreases the rotational motion. (c) decreases the rotational and translational motion. (d) converts translational energy to rotational energy 64. Atube oflength Lis filled completelywith an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with uniform angular speed w. What is the force exerted by the liquid at the other end? (a) 2 ML 2 w (b) 2 MLw (c) 2 ML 4 w (d) 2 ML 8 w
- 200. 196 PHYSICS 65. Acircular disk of moment of inertia It is rotating in a horizontal plane, its symmetryaxis, with a constant angular speed wi . Another disk ofmoment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed f w . The energy lost by the initiallyrotating disk to friction is (a) 1 2 2 2 ( ) w + b i t b I I I (b) 2 2 ( ) w + t i t b I I I (c) 2 ( ) b t i t b I I I I - w + (d) 1 2 2 ( ) b t i t b I I I I w + 66. The moment of inertia ofa thin uniform rod ofmass M and length L about an axis passing through its midpoint and perpendicular toits length is I0. Its moment ofinertia about an axis passing through one of its ends and perpendicular to its length is (a) I0 + ML2/2 (b) I0 + ML2/4 (c) I0 + 2ML2 (d) I0 + ML2 67. Theinstantaneous angular position ofa point on a rotating wheel isgivenbytheequation q(t) = 2t3 – 6t2. The torque on the wheel becomes zero at (a) t = 1s (b) t = 0.5 s (c) t = 0.25 s (d) t = 2s 68. The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through B C D A (a) B (b) C (c) D (d) A 69. Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is (a) 40cm (b) 45cm (c) 50cm (d) 30cm 70. If the angular velocity of a body rotating about an axis is doubled and its moment of inertia halved, the rotational kinetic energy will change bya factor of : (a) 4 (b) 2 (c) 1 (d) 1 2 71. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is madeto rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is (a) 2 1 mR 2 (b) 2 1 mR 4 (c) 2 1 mR 8 (d) 2 2mR Directions for Qs. (72 to 75) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethecorrectanswer (ONLYONE option is correct ) from the following- (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 isnot a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 72. Statement 1 : When you lean behind over the hind legs of the chair, the chair falls back after a certain angle. Statement 2 : Centre of mass lying outside the system makes the system unstable. 73. Statement 1:Arigid disc rolls without slipping on a fixed rough horizontal surfacewith uniform angular velocity.Then the acceleration of lowest point on the disc is zero. Statement 2 : For a rigid disc rolling without slipping on a fixed rough horizontal surface, the velocity of the lowest point on the disc is always zero. 74. Statement 1 : If no external force acts on a system of particles, then the centre of mass will not move in any direction. Statement 2 : If net external force is zero, then the linear momentum of the system remains constant. 75. Statement 1 :Awheel moving down a frictionless inclined plane will slip and not roll on the plane. Statement 2 : It is the frictional force which provides a torque necessary for a body to roll on a surface.
- 201. 197 System of ParticlesandRotationalMotion Exemplar Questions 1. For which ofthe following does the centre ofmass lie outside the body? (a) A pencil (b) A shotput (c) A dice (d) A bangle 2. Which of the following points is the likely position of the centre of mass of the system shown in figure? A B C D Hollow sphere Air R/2 R/2 Sand (a) A (b) B (c) C (d) D 3. A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is (a) ˆ mva x e v z a y (b) ˆ 2mva x e (c) ˆ ymv x e (d) ˆ 2ymv x e 4. When a disc rotates with uniform angular velocity, which of the following is not true? (a) The sense of rotation remains same (b) The orientation of the axis of rotation remains same (c) The speed of rotation is non-zero and remains same (d) The angular acceleration is non-zero and remains same 5. A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind in figure. The moment of inertia about the z- axis is then, Q P y x Q P y x hole (a) increased (b) decreased (c) the same (d) changed in unpredicted manner 6. In problem-5, the CM of the plate is now in the following quadrant of x-y plane. (a) I (b) II (c) III (d) IV 7. The density of a non-uniform rod of length 1 m is given by r(x) = a (1 + bx2) where, a and b are constants and 0 1. x £ £ The centre of mass of the rod will be at (a) 3(2 ) 4(3 ) b b + + (b) 4(2 ) 3(3 ) b b + + (c) 3(3 ) 4(2 ) b b + + (d) 4(3 ) 3(2 ) b b + + 8. A merry-go-round, made of a ring-like platform ofradius R and mass M, is revolving with angular speed w.Aperson of mass M is standing on it. At one instant, the person jumps off the round, radially awayfrom the centre ofthe round (as seen from the round). The speed ofthe round of afterwards is (a) 2w (b) w (c) 2 w (d) 0 NEET/AIPMT (2013-2017) Questions 9. A small object of uniform densityrolls up a curved surface with an initial velocity ‘n’.Itreachesuptoamaximum height of 2 3 4g n with respect to the initial position. The object is a (a) solid sphere (b) hollow sphere [2013] (c) disc (d) ring 10. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal bya massless string tied topoint Q as shown in figure. When string is cut, the initial angular acceleration of the rod is [2013] (a) g /L (b) 2g/L (c) 2 3 g L (d) 3 2 g L
- 202. 198 PHYSICS 11. Twodiscsare rotating about their axes, normal to the discs and passing through the centres of the discs. Disc D1 has 2 kg mass and 0.2 m radius and initial angular velocityof 50 rads–1.DiscD2 has4kgmass,0.1mradiusandinitialangular velocityof 200 rad s–1. The twodiscs are brought in contact faceto face, with their axes of rotation coincident. The final angular velocity (in rad s–1) of the system is (a) 40 (b) 60 [NEET Kar. 2013] (c) 100 (d) 120 12. The ratioof radii ofgyration of a circular ring and a circular disc, of the same mass and radius, about an axis passing through their centres and perpendicular to their planes are (a) 2 :1 (b) 1: 2 [NEET Kar. 2013] (c) 3: 2 (d) 2: 1 13. The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline ofangle‘q’ without slipping and slipping down the incline without rolling is : (a) 5: 7 (b) 2: 3 [2014] (c) 2: 5 (d) 7: 5 14. A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis.A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s– 2 is : (a) 25N (b) 50N [2014] (c) 78.5N (d) 157N 15. Threeidentical spherical shells, each ofmass m and radius r are placed as shown in figure. Consider an axis XX' which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX' axis is [2015] (a) 3mr2 (b) 2 16 mr 5 X X¢ (c) 4mr2 (d) 2 11 mr 5 16. A rod of weight W is supported bytwo parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass ofthe rod is at distance x fromA. The normal reaction on A is [2015] (a) Wd x (b) W(d – x) x (c) W(d – x) d (d) Wx d 17. A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to string which passes through a smooth hole in the plane as shown. v0 m The tension in the string is increased graduallyand finally m moves in a circle of radius 0 R 2 . The final value of the kineticenergy is [2015] (a) 2 0 1 mv 4 (b) 2 0 2mv (c) 2 0 1 mv 2 (d) 2 0 mv 18. A force ˆ ˆ ˆ F i 3j 6k = a + + r is acting at a point ˆ ˆ ˆ r 2i 6j 12k = - - r . The value of a for which angular momentum about origin is conserved is : [2015 RS] (a) 2 (b) zero (c) 1 (d) –1 19. Point masses m1 and m2 are placed at the opposite ends of a rigid rod oflength L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass sothat the work required to set the rod rotating with angular velocity w0 is minimum, is given by: [2015 RS] m1 m2 w0 P x (L–x) (a) 1 2 m x L m = (b) 2 1 m x L m = (c) 2 1 2 m L x m m = + (d) 1 1 2 m L x m m = + 20. An automobile moves on a road with a speed of 54 km h-1. The radius of its wheels is 0.45 m and themoment ofinertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted byits brakes to the wheel is : [2015 RS] (a) 8.58 kg m2 s-2 (b) 10.86 kg m2 s-2 (c) 2.86 kg m2 s-2 (d) 6.66 kg m2 s-2
- 203. 199 System of ParticlesandRotationalMotion 21. A disk and a sphere of same radius but different massesroll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ? (a) Disk [2016] (b) Sphere (c) Both reach at the same time (d) Depends on their masses 22. A uniform circular disc of radius 50 cm at rest is free toturn about an axis which is perpendicular to its plane and passes through its centre. It is subjected toa torque which produces a constant angular acceleration of 2.0 rad s–2. Its net acceleration in ms–2 at the end of2.0s is approximately: (a) 8.0 (b) 7.0 [2016] (c) 6.0 (d) 3.0 23. FromadiscofradiusRandmassM,acircular holeofdiameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ? [2016] (a) 15 MR2/32 (b) 13 MR2/32 (c) 11 MR2/32 (d) 9 MR2/32 24. Which of the following statements are correct ? [2017] (A) Centre of mass of a body always coincides with the centre of gravity of the body (B) Centre ofmass ofa bodyis the point at which the total gravitational torque on the body is zero (C) A couple on a body produce both translational and rotation motion in a body (D) Mechanical advantage greater than one means that small effort can be used to lift a large load (a) (A) and (B) (b) (B) and(C) (c) (C) and(D) (d) (B) and (D) 25. Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities w1 and w2 . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:- [2017] (a) 2 1 2 1 I( ) 4 w - w (b) 2 1 2 I( ) w - w (c) 2 1 2 1 ( ) 8 w - w (d) 2 1 2 1 I( ) 2 w + w 26. A rope is wound around a hollow cylinder ofmass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? [2017] (a) 0.25 rad/s2 (b) 25 rad/s2 (c) 5 m/s2 (d) 25 m/s2
- 204. 200 PHYSICS EXERCISE -1 1. (a) 2. (b) 3. (c) 4. (c) 5. (b) 6. (d) 7. (b) 8. (d) The pull of earth changes onlywhen the body moves sothat g changes. It is an exceptional case and should not be considered unless otherwise mentioned. 9. (b) Ifwe applya torqueon a body, then angular momentum of the body changes according to the relation dL τ dt = r r Þ if t r = 0 then, L = constant 10. (b) Angular momentum L r is defined as L r m(v) = ´ r r r so L r is, an axial vector. . 11. (c) The boy does not exert a torque to rotating table by jumping, so angular momentum is conserved i.e., dL 0 L dt = Þ r r = constant 12. (b) Since no external torque act on gymnast, so angular momentum (L=Iω)isconserved.After pullingher arms & legs, the angular velocity increases but moment of inertia of gymnast, decreases in, such a way that angular momentum remains constant. 13. (a) 14. (b) Angularmomentum willremainthesamesinceexternal torque is zero. 15. (a) 2 2 B A 1 I MR I MR 2 = = A B I I < 16. (a) Angular momentum ( ) r linear momentum = ´ r 17. (b) We know that ext dL τ dt = ifangular momentum is conserved, it means change in angular momentum = 0 0 dL , or = , ext dL 0 τ 0 dt = Þ = Thus total external torque = 0. 18. (a) Analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in translational motion. 19. (a) Basic equation of moment of inertia is given by 2 i i n 1 i r m I = S = i m i r where i m is the mass of th i particle at a distance of i r from axis ofrotation. Thus it does not depend on angular velocity. 20. (d) Initially centre of gravity is at the centre. When sand is poured it will fall and again after a limit, centre of gravitywill rise. 21. (d) Here, the centre of mass of the system remains unchanged. in the horizontal direction. When the mass m moved forward by a distance L cos q, let the mass (m + M) moves by a distance x in the backward direction. hence (M + m) x – m L cos q = 0 x = (m L cos q)/(m + M) 22. (b) Acceleration of solid sphere is more than that of hollow sphere, it rolls faster, and reaches the bottom oftheinclined planeearlier. Hence, solid sphere and hollow sphere can be distinguished by rolling them simultaneously on an inclined plane. 23. (d) Applying the law of conservation of momentum m v= M V ...(1) Byconservation of angular momentum 2 M L m v(L / 2) ω 12 æ ö = ç ÷ ç ÷ è ø ...(2) As the collision is elastic, we have 2 2 2 1 1 1 m v M V Ι ω 2 2 2 = + ...(3) Substituting the values, we get m = M/4 24. (b) a is propotional to w Let a= kw (Qk is a constant) dω kω dt = [also dθ dθ ω dt dt ω = Þ = ] ωdω kω dθ = Þ dw= kdq Now ω/ 2 ω dω k dθ = ò ò 0 θ 1 ω/2 0 ω ω dω k dθ kθ kθ 2 2 = Þ - = Þ - = ò ò (Qq1 = 2pn) θ = 1 θ or 2pn1 = 2pn n1 = n 25. (b) Sorawegg islikea spherical shell &hard bioled egg is like solid sphere. Let 1 I , 2 I be M. I. of raw egg and boiled egg respectively. Given that angular momentum L, is same I1w1= I2w2 Þ w2>w1 QI1>I2 Nowfrom first equation ofangular motion (wf=wi+at) here a is retarding decceleration for both cases & wf = 0 for both case. So 1 1 1 2 2 2 t (raw egg) ω / α t 1 t (hard egg) ω / α t = Þ < Hints & Solutions
- 205. 201 System of ParticlesandRotationalMotion EXERCISE - 2 1. (d) K.E. of rotation = 2 1 2 Iw = 2 2 1 1 2 2 mr æ ö ´ w ç ÷ è ø 2 1 20 (0.25) 100 100 4 = ´ ´ ´ ´ = 3125 J 2. (c) For solid sphere 2 3 2 R R 3 4 5 2 R M 5 2 ÷ ø ö ç è æ r p = = I r = r ´ = 5 5 R 105 176 R 7 22 15 8 3. (a) Force FA on particle A is given by A A A A m v F m a t = = ...(1) Similarly B B B B m 2 v F m a t ´ = = ...(2) Now A B A B m v m 2 v ( F F ) t t ´ = = Q So mA = 2mB For the centre of mass of the system A A B B A B m v m v v m m + = + or B B B B 2 m v m 2 v v 0 2m m - ´ = = + Negative sign is used because the particles are travelling in opposite directions. Alternatively, if we consider the two masses in a system then no external force is acting on the system. Mutual forces are internal forces. Since the centre of mass is initially at rest, it well remain at rest 4. (a) Moment of inertia of the whole system about the axis of rotation will be equal to the sum ofthe moments of inertia ofall the particles. (–2,–2, 0) 4 kg (0, 0, 9) 1 kg (2, 0, 0) 2 kg (0, 3, 0) 3 kg X Y I = I1 + I2 + I3 + I4 = 0 + 0 + 27 + 16 = 43 kg m2 5. (a) J 5 . 13 ) 3 ( 3 2 1 2 1 E 3 2 r = ´ ´ = w I = 2 2 1 1 K.E. m v 27 v 13.5 2 2 = = ´ ´ = v 1 m / s = 6. (c) Centripetal acceleration 2 2 c a 4 r = p n 2 2 2 ms 4 25 . 0 2 2 4 - p = ´ ´ ´ p = 7. (b) From thetheorem ofparallelaxes, themoment ofinertia I = ICM + Ma2 where ICM is moment of inertia about centre of mass and a is the distance of axis from centre. mk2 = m(k1)2 +m(6)2 (Q I = mk2 where k is radius of gyration) or, k2 = (k1)2 + 36 or, 100 = (k1)2 + 36 Þ (k1)2 = 64cm k1 = 8 cm 8. (c) L= m v r or v = L /mr Centripetal force 2 2 m v m(L/ m r) r r = 3 2 r m L = 9. (a) The situation can be shown as : Let radius ofcomplete disc is a and that of small disc is b. Also let centre of mass now shifts to O2 at a distance x2 from original centre. b a O O1 O2 x-axis x2 x1 The position of new centre of mass is given by XCM = 2 1 2 2 . b .x . a . b -s p s p - s p Here, a =6 cm, b= 2 cm, x1 = 3.2 cm Hence, XCM = 2 2 2 (2) 3.2 (6) (2) -s ´ p ´ s ´ p ´ - s ´ p ´ = 12.8 32 p p =–0.4cm. 10. (b) By energy conservation f f i i ) E . P ( ) E . K ( ) E . P ( ) E . K ( + = + 0 ) E . P ( , mgh ) E . P ( , 0 ) E . K ( f i i = = = 2 2 f cm (K.E) ½Iω ½mv = + Where I (moment ofinertia) = ½mR2 (for solid cylinder) so 2 2 2 cm cm 2 v mgh ½(½mR ) ½mv R æ ö = + ç ÷ è ø cm v 4gh/3 Þ =
- 206. 202 PHYSICS 11. (b)Moment of inertia (M.I) ofcircular ring I1 = mr2 moment of inertia (M.I) of circular disc I2 = ½mr2 1 2 I I 2 1 = Þ 12. (a) Tension provides the necessary centripetal force for the rest of rod. 13. (d) Centrifugal force is observed when the observer is in a frameundergoing circular motion. 14. (d) 2 1 mR 2 I = 2 M t R µ µ For disc X, 2 X 1 Ι (m)(R) 2 = = 2 2 1 (πr t).(R) 2 for discY, 2 2 Y 1 Ι [π(4R) .t / 4][4R] 2 = Þ X 3 Y Ι 1 Ι (4) = Þ Y X I 64 I = 15. (a) 1 Angular momentum Angular frequency µ Kinetic energy µ Þ K.E. L ω = r 1 1 2 2 1 2 L K.E ω 4 L ω K.E æ ö = ´ = ç ÷ è ø Þ 2 L L 4 = 16. (a) When q = 0°, the net force is directed vertically upwards. 17. (a) Let velocity at A v A = and velocity at B v B = A T r B B v A v Applying conservation of energy at A & B 2 B 2 A mv 2 1 gmr 2 mv 2 1 = + [Q (P.E)A = mg (2r)] ) i ( .......... gr 4 v v 2 A 2 B + = Nowas it is moving in circular path it has centripetal force. At point A r mv mg T 2 A = + Þ 0 T velocity minimum for ³ 2 2 A A mv or mg v gr r ³ Þ ³ gr vA ³ Þ 18. (a) A couple consists of two equal and opposite forces whose lines of action are parallel and laterally separated by same distance. Therefore, net force (or resultant) of a couple is null vector, hence no translatory motion will be produced and only rotational motion will be produced. 19. (b) Angular momentum of mass mmoving with a constant velocity about origin is line parallel to x-axis v m O y x L = momentum × perpendicular distance of line of action ofmomentum from origin L= mv × y In the given condition mvy is a constant. Therefore angular momentum is constant. 20. (c) Since the spheres are smooth, there will benotransfer of angular momentum from the sphereA tosphere B. The sphereAonlytransfers its linear velocityv to the sphere B and will continue to rotate with the same angular speed w. 21. (c) 2 ) ŷ x̂ ( n̂ + - = –a cos45 x o –a sin45 y o 45o a 2 ˆ ˆ v 4 (x y) ˆ a n r 4 2 - + = = ´ u r 22. (b) Since v is changing (decreasing), L is not conserved in magnitude. Sinceit isgiven that a particle isconfined to rotate in a circular path, it can not have spiral path. Since the particle has two accelerations ac and at thereforethenet acceleration is not towards thecentre. L ac at v The direction of L remains same even when the speed decreases. 23. (a) Themoment ofinertia (I) ofcircular ring whose axis of rotation is passing through its center, 2 1 1 R m I = Also, 2 2 2 ) nR ( m I = Since both rings have same density, 1 1 2 2 A R 2 m A ) nR ( 2 m ´ p = ´ p Þ
- 207. 203 System of ParticlesandRotationalMotion Where A is cross-section of ring, 2 1 A A = (Given) 1 2 nm m = Given 8 1 I I 2 1 = = 2 2 2 1 ) nR ( m R m = 2 1 2 1 ) nR ( nm R m 3 n 1 8 1 = Þ or n = 2 24. (a) Remember that acceleration of a cylinder down a smooth inclined plane is x q sin g q g N m q N m m mg cos ÷ ø ö ç è æ + q = 2 mR I 1 sin g a where 2 mR I 2 = is the moment of Inertia for cylinder ÷ ÷ ø ö ç ç è æ ´ + ° = 2 2 mR 1 2 mR 1 30 sin g a 3 g 2 1 1 2 1 g = + ´ = 25. (c) This is a torque problem. While the fulcrum can be placed anywhere, placing it at the far right end of the bar eliminated cable B from the calculation. There are now only two forces acting on the bar ; the weight that produces a counterclockwise rotation and the tension in cable Athat produces a clockwise rotation. Since the bar is in equilibrium, these twotorques must sum to zero. 1 2 3 4 L L fulcrum W = mg Cable 0 ) L 2 / 1 ( Mg ) L 4 / 3 ( TA = - = t S Therefore 3 / Mg 2 ) L 3 / 4 )( 2 / MgL ( ) 4 / L 3 /( ) 2 / MgL ( TA = = = 26. (b) The intersection of medians is the centre of mass of the triangle. Since distances of centre of mass from the sides are related as : xBC < xAB < xAC therefore IBC > IAB > IAC or IBC > IAB. 27. (a) C O 1.12×10 –10 (12 amu) (16 amu) d c.m. 1.12×10 – d –10 From definition of centre of mass. 12 16 0 12 10 12 . 1 16 d 10 + ´ + ´ ´ = - 28 10 12 . 1 16 10 - ´ ´ = =0.64×10–10m. 28. (d) ABis the ladder, let F be the horizontal force and Wis the weigth of man. Let N1 and N2 be normal reactions of ground and wall, respectively. Then for vertical equilibrium W = N1 .....(1) For horizontal equilibrium N2 = F .....(2) Taking moments aboutA N2(AB sin60°) – W(B Cos 60°) = 0......(3) Using (2) andAB = 20 ft, BC = 4 ft, weget N2 N1 W A B C F O 60° 4 f t 0 2 1 4 w – 2 3 20 F = ÷ ø ö ç è æ ´ ÷ ÷ ø ö ç ç è æ ´ 3 5 150 3 5 w 3 20 2 w 2 F = = ´ = Þ pound (lb) 10 3 10 1.73 17.3 = = ´ = pound 29. (c) Moment of inertia of disc about its diameter is 2 d MR 4 1 I = I Id R MI of disc about a tangent passing through rim and in the plane of disc is 2 2 2 2 G MR 4 5 MR MR 4 1 MR I I = + = + = 30. (b) F= mrw2 = 2 2 1 1 4 2 2 8 N 2 ´ ´ p ´ ´ = p 31. (d) Consider an element of length dx at a distance x from end A. Here : mass per unit length l ofrod ) kx x ( = l Þ µ l dm= ldx=kxdx dx B A x
- 208. 204 PHYSICS Position ofcentre of gravity of rod from end A ò ò = L 0 L 0 CG dm dm x x 3 3 3 3 0 0 CG 3 3 2 2 0 0 x (3) x(kx dx) 3 3 x 2m (3) kx dx x 2 2 é ù ê ú ê ú ë û = = = = é ù ê ú ê ú ë û ò ò 32. (b) Densityof iron > densityof aluminium moment ofinertia = ò dm r2 . r dm Since riron > raluminium sowhole ofaluminium is kept in the core and the iron at the outer rim of the disc. 33. (d) Rotational energy= 2 ) ( I 2 1 w = 2 2 ) mK ( 2 1 w Linear kinetic energy = 2 2 R m 2 1 w Required fraction = 2 2 2 2 2 2 R m 2 1 ) mK ( 2 1 ) mK ( 2 1 w + w w = 2 2 2 R K K + 34. (c) At the highest point 2 2 2 g mg mR4 4 R = p n Þ n = p Revolution per minute 2 900 g 60 R = n = p 35. (d) 2 2 4 mr mg n p = m 2 2 2 2 1 1 1 1 2 2 1 1 2 2 2 2 1 r r r r r 4 n n Þ n = n Þ = = n n cm 4 4 r r 1 2 = = 36. (c) Byconservation of angular momentum, 50 5 10 5 20 2 rps 25 ´ = w + w Þ w = = 37. (b) q q v v ) cos 1 ( v 2 cos v 2 v v v 2 2 2 2 R q + = q + + = 2 cos v 2 q = 38. (b) For no angular acceleration tnet = 0 Þ F1 × 5 = F2 × 30 (given F2 = 4N) N 24 F1 = Þ 39. (c) 2 I 2 1 w = Loss of gravitational potental energy 2 2 1 m mg (1 cos ) 2 3 2 ´ w = + q l l 2 cos g 6 2 cos 2 g 3 2 2 q = w Þ ÷ ø ö ç è æ q = w Þ l l 40. (d) Use t = Ia and aq = w - w 2 2 0 2 Here 2 MR 5 2 I = Given M= 2000g =2kg R= 5cm= 5 ×10–2m 2 2 radians q = p ´ p n = 300 rpm = 5rps I t = a = 2 2 2 0 2 MR 5 2 æ ö w - w æ ö = ç ÷ ç ÷ è ø q è ø = ( ) ( ) 2 2 2 2 2 2 4 5 0 2 2 5 10 5 2 4 - - p ´ ´ ´ ´ p =0.025N –m. = 2.5 × 105 dyne cm. 41. (c) 5 4 3 2 1 1 5 2 4 0 3 2 2 0 1 X . M . C + + + + ´ + ´ + ´ + ´ + ´ = 1 . 1 15 17 15 5 8 4 = = + + = C.M 1 0 2 0 3 2 4 2 5 1 Y 1 2 3 4 5 ´ + ´ + ´ + ´ + ´ = + + + + = 6 8 5 1.3 15 + + = 42. (c) 2 2 I 2 1 Mv 2 1 E w + = J 7 . 0 Mv 10 7 R v MR 5 2 2 1 Mv 2 1 2 2 2 2 2 = = ´ ´ + =
- 209. 205 System of ParticlesandRotationalMotion 43. (a) So couple about the C.M must be zero for rotational equilibrium, It means that 6 L x x 4 W 3 2 L 4 W = Þ ´ = ´ C.M x L/2 L/2 W 4 3W 4 boy man and the distance of man from other end is 2 L – x 3 L 6 L 2 L = - = 44. (b,c) 45. (b) Position of C.M w.r. toA A B C 1 0 1 AB 1 AC AB AC 1 1 1 3 ´ + ´ + ´ + = = + + 46. (b) Use theorem of parallel axes. 47. (a) Angular momentum 2 2 2 I MR . 5 T p = w = 2 4 MR 5T p = 48. (d) 2 1 2 1 t t w -w a = - 1 2 1200 40 60 p´ w = = p ; 2 2 4500 150 60 p´ w = = p 2 110 rad /sec 10 p a = Now, radian 180 p = ° 180 1 rad degree = p 2 11 180 degree/sec p´ a = p = 1980 degree/sec2 49. (b) 2 2 2 I 2 5 (0.2) 2 2 (0.4) 1kg m = ´ ´ + ´ ´ = ´ 50. (b) E = 1500= 1/2× 1.2w2 2 3000 2500 1.2 w = = 50rad / sec w = 50 t 2sec 25 w = = = a 51. (c) Velocity of P (NP) (NM MP) = w = + w r(r sin ) v(1 sin ) = + q w = + q 52. (d) Velocity at the bottom and top of the circle is 5gr and gr . Therefore (1/2)M(5gr) = MgH and (1/2) M (gr) = Mgh. 53. (a) Acceleration of a bodyrolling down an inclined plane is given by , 2 2 gsin a K 1 R q = + In case of a solid sphere , we have 2 2 2 2 2 2 K [(I / M)] I (2 /5)MR 2 5 R R MR MR = = = = Substituting 2 2 K 2 5 R = in equation (1) we get 5 a gsin 7 = q 54. (c) The velocity of the top point of the wheel is twice that of centre of mass. And the speed of centre of mass is same for both the wheels. 1 a 2 a 1 1 R v w = 1 R 2 2 R v w = 2 R 55. (c) 1 2 1 2 1 2 1 2 1 1 R R R R v a w = w = = 2 2 2 2 2 2 R R v a w = = Taking particle masses equal 2 1 2 1 2 1 2 1 R R a a ma ma F F = = = Alternative method : The force experienced by any particle is onlyalong radial direction or the centripetal force. Force experienced bythe particle, F = mw2R 1 1 2 2 F R F R = 56. (d) t × Dt = L0 {Q since Lf = 0} Þ t × Dt = Iw or t× 60 = 2 × 2 × 60p/60 ( f 60rpm = Q 60 2 f 2 60 ö w = p = p´ ÷ ø m N 15 - p = t 57. (c) m1d = m2 d2 2 1 2 m d m d = Þ 58. (d) y1 y¢ 1 Circular disc (1) 4 MR I 2 1 y =
- 210. 206 PHYSICS 2 2 2 1 y MR 4 5 MR 4 MR I= + = ¢ y2 y¢ 2 Circular ring(2) 2 MR I 2 2 y = 2 2 2 2 y MR 2 3 MR 2 MR I = + = ¢ 2 1 1 y MK I = ¢ , 2 2 2 y MK I = ¢ 2 y 1 y 2 2 2 1 I I K K ¢ ¢ = 6 : 5 K : K 2 1 = Þ 59. (d) Angular momentum will be conserved I1w= I1w' + I2w' 2 1 1 I I I + w = w¢ Þ 60. (d) IAX = m(AB)2 + m(OC)2 = ml2 + m (l cos 60º)2 = ml2 + ml2/4 = 5/4 ml2 m l l l m m C B A O X 60º 60º 61. (c) M.I. of a uniform circular disc of radius ‘R’ and mass ‘M’ about an axis passing through C.M. and normal to the disc is I MR C M . . = 1 2 2 From parallel axis theorem I I MR MR MR MR T C M = + = + = . . 2 2 2 2 1 2 3 2 62. (d) K L I = 2 2 Þ L2 = 2KIÞ L KI = 2 L L K K I I 1 2 1 2 1 2 = × = × = K K I I 2 1 2 L1 : L2= 1 : 2 63. (d) Net work done by frictional force when drum rolls down without slipping is zero. R M f q Wnet = 0, Wtrans. + Wrot. = 0 DKtrans. + DKrot. = 0 DKtrans = –DKrot. i.e., converts translation energy to rotational energy. 64. (a) Tube maybe treated as a particle ofmass M at distance L/2 from one end. Centripetal force 2 2 ML Mr 2 = w = w 65. (d) Byconservationofangularmomentum,It i w =(It +Ib ) f w where f w is the final angular velocityof disks f w = t i t b I I I æ ö w ç ÷ + è ø Lossin K.E., K D = Initial K.E. – Final K.E. = 2 1 2 t i I w – 2 1 ( ) 2 t b f I I + w Þ K D = 1 2 2 1 2 t i I w - 2 2 2 ( ) ( ) t t b i t b I I I I I + w + = 1 2 2 i w ( ) t t b t t b I I I I I I + - + = 1 2 2 i w t b t b I I I I + 66. (b) Bytheorem of parallel axes, I = Icm + Md2 I = I0 + M (L/2)2 = I0 + ML2/4 67. (a) When angular acceleration (a) is zero then torque on the wheel becomes zero. q(t) = 2t3 – 6t2 d 6t 12t dt 2 q Þ = - 2 d 12t 12 0 dt2 q Þ a = = - = t = 1 sec.
- 211. 207 System of ParticlesandRotationalMotion 68. (a) According to parallel axis theorem of the moment of Inertia I = Icm + md2 d is maximum for point B so Imax about B. 69. (a) 1 1 2 2 3 3 cm 1 2 3 m x m x m x X m m m + + = + + cm 300 (0) 500(40) 400 70 300 500 400 X ´ + + ´ = + + cm 500 40 400 70 1200 X ´ + ´ = cm 50 70 120 3 3 X + = = =40cm 70. (b) 2 1 K I 2 = w 2 1 1 K' (2 ) 2K 2 2 æ ö = w = ç ÷ è ø 71. (a) For complete disc with mass '4M', M.I. about given axis = (4M)(R2/2) = 2 MR2 Hence, bysymmetry, for the given quarter of thedisc M.I. = 2 MR2 /4 2 1 MR 2 = 72. (d) 73. (a) For a disc rolling without slipping on a horizontal rough surface with uniform angular velocity, the acceleration of lowest point of disc is directed vertically upwards and is not zero (Due to translation part of rolling, acceleration of lowest point is zero. Due to rotational part of rolling, the tangential acceleration of lowest point is zero and centripetal acceleration is non-zero and upwards). Hence statement 1 is false. 74. (a) 75. (b) EXERCISE - 3 Exemplar Questions 1. (d) A bangle is in the form of a ring as shown in figure. The centre of mass ofthe ring lies at its centre, which is outside the ring or bangle. C Centre 2. (c) Centre of mass of a system lies towards the part ofthe system, having bigger mass. In the above diagram, lower part is heavier, hence CM of the system lies belowthe horizontal diameter at C point. 3. (b) The initial velocityis ˆ i y v ve = andafter reflection from thewall, the final velocityis ˆ f y v ve = - . The trajectory is at constant distance a on z axisand as particlemoves along y axis, its y component changes . So position vector (moving along y-axis), ˆ ˆ y z r ye ae = + r Hence, the change in angular momentum is ˆ ( ) 2 f j x r m v v mvae ´ - = r . 4. (d) Angular acceleration d dt w a = where w is angular velocityof the disc and isuniform or constant. 0 d dt w a = = Hence, angular acceleration is zero. 5. (b) According to the question, when the small piece Q removed it is stick at P through axis of rotation passes, but axis of rotation does not passes through Q and It is glued to the centre of the plate, the mass comes closer tothez-axis, hence, moment ofinertiadecreases. 6. (c) Let us consider the diagram of problem 5, there is a line shown in the figure drawn along the diagonal. First, centre of mass of the system was on the dotted line and was shifted towards Q from the centre (1st quadrant). Q y x Q x y x hole When the mass Q is decrease, it will be on the same line but shifted towards other side of Q on the line joining QPor away from the centre so, new CM will lies in IIIrd quadrant in x-y plane. 7. (a) As given that density : 2 ( ) (1 ) x a bx r = + where a and b are constants and 0 1 x £ £ At b = 0, ( ) x a r = = constant In that case, centre of mass will be at x = 0.5 m (i.e., mid-point of the rod) Putting, b = 0 in options. (a) 3 2 1 0.5 m 4 3 2 ´ = =
- 212. 208 PHYSICS (b) 4 2 0.5m 3 3 ´ ¹ (c) 3 3 0.5 m 4 2 ´ ¹ (d) 4 3 0.5 m 3 2 ´ ¹ So, only (a) option gives 0.5. 8. (a) As there is external torque acting on the system, angular momentum should be conserved. HenceIw = constant ...(i) where, I is moment of inertia of the system and w is angular velocity of the system. FromEq.(i), 1 1 2 2 I I w = w where w1 and w2 are angular velocities before and after jumping. 2 ( ) I mr = Q So, given that, m1 = 2M, m2 = M, w1 = w, w2 = ? r1 = r2 = R 2 2 1 1 1 2 2 2 m r m r w = w 2 2 2 2MR MR w = w as mass reduced tohalf, hence, moment of inertia also reduced to half. 2 2 w = w NEET/AIPMT (2013-2017) Questions 9. (c) 2 3V 4g V From law of conservation of mechanical energy 1 2 Iw2 + 0 + 1 2 mv2 = mg × 2 3v 4g Þ 1 2 Iw2 = 3 4 mv2 – 1 2 mv2 = 2 mv 2 3 1 2 æ ö - ç ÷ è ø or, 1 2 I 2 2 V R = 2 mv 4 or, I = 1 2 mR2 Hence, object is a disc. 10. (d) Weight of the rod will produce the torque t = mg L 2 = I a = 2 mL 3 a 2 rod ML I 3 é ù = ê ú ê ú ë û Q Hence, angular acceleration a = 3g 2L 11. (c) Given: m1 = 2 kg m2 = 4 kg r1 = 0.2 m r2 = 0.1 m w1 = 50 rad s–1 w2 = 200 rad s–1 As, I1W1 = I2W2 = Constant 1 1 2 2 1 2 f I W I W W I I + = + = 2 2 1 1 1 2 2 2 2 2 1 1 2 2 1 1 2 2 1 1 2 2 m r w m r w m r m r + + By putting the value of m1, m2, r1, r2 and solving we get = 100 rad s–1 12. (a) Q I = MK2 K = I M Iring = MR2 and Idisc = 2 1 2 MR 2 1 1 2 2 2 2 :1 2 K I MR K I MR = = = æ ö ç ÷ è ø 13. (a) For solid sphere rolling without slipping on inclined plane, acceleration a1 = 2 2 g sin K 1 R q + For solid sphere slipping on inclined plane without rolling, acceleration a2 = g sin q Therefore required ratio = 1 2 a a = 2 2 1 1 5 2 7 K 1 1 5 R = = + +
- 213. 209 System of ParticlesandRotationalMotion 14. (d) Here a = 2 revolutions/s2 = 4p rad/s2 (given) Icylinder = 2 2 1 1 MR (50)(0.5) 2 2 = = 25 4 Kg-m2 As t = Ia so TR= Ia Þ T = 25 (4 ) I 4 N R (0.5) æ ö p ç ÷ è ø a = = 50pN= 157N 15. (c) Moment of inertia of shell 1 along diameter Idiameter = 2 2 MR 3 Moment of inertia of shell 2 = m. i of shell 3 = Itangential = 2 2 2 2 5 MR MR MR 3 3 + = X X¢ 3 2 1 So, I of the system along x x1 = Idiameter + (Itangential) × 2 or, Itotal = 2 2 2 5 MR MR 2 3 3 æ ö + ´ ç ÷ è ø = 2 2 12 MR 4MR 3 = 16. (c) By torque balancing about B NA (d) = W (d – x) A W(d – x) N d = d x d – x B NA NB W A 17. (b) Applying angular momentum conservation V0 m mV0R0 =(m)(V1) 0 R 2 æ ö ç ÷ è ø v1 = 2V0 Therefore, new KE = 1 2 m (2V0)2 = 2 0 2mv 18. (d) From Newton's second law for rotational motion, t r = dL dt r , if L r = constant then t r = 0 So, t r = r F ´ r r = 0 ( ) ˆ ˆ ˆ ˆ ˆ ˆ 2i 6j 12k ( i 3j 6k) 0 - - ´ a + + = Solving we get a = –1 19. (c) Work required to set the rod rotating with angular velocity w0 K.E. = 2 1 I 2 w Workis minimum when I is minimum. I is minimum about the centre ofmass So, (m1) (x) = (m2)(L–x) or,m1x=m2L–m2x x = 2 1 2 m L m m + 20. (d) Given : Speed V = 54 kmh–1 = 15 ms–1 Moment of inertia, I = 3 kgm2 Timet =15s wi = V r = 15 100 0.45 3 = wf = 0 wf = wi + at 0 = 100 3 + (– a) (15) Þ a = 100 45 Average torque transmitted by brakes to the wheel t = (I) (a) = 3 × 100 45 = 6.66 kgm2s–2 21. (b) Time of descent µ 2 2 K R Order of value of 2 2 K R for disc; 2 2 K R = 1 2 = 0.5 for sphere; 2 2 K R = 2 5 = 0.4 (sphere) < (disc) Sphere reaches first 22. (a) Given: Radius ofdisc, R= 50 cm angular acceleration a = 2.0 rads–2; time t = 2s Particle at periphery (assume) will have both radial (one) and tangential acceleration at = Ra = 0.5 × 2 = 1 m/s2
- 214. 210 PHYSICS From equation, w=w0 + at w = 0 + 2 × 2 = 4 rad/sec ac = w2R= (4)2 × 0.5 =16× 0.5 =8m/s2 Net acceleration, atotal = 2 2 t c a a + = 2 2 1 8 + » 8 m/s2 23. (b) Moment of inertia of complete disc about point 'O'. ITotal disc = 2 MR 2 M R R O Mass of removed disc MRemoved = M 4 (Mass µ area) Moment of inertia of removed disc about point 'O'. IRemoved (about same perpendicular axis) = Icm + mx2 = ( )2 M R / 2 4 2 + 2 2 M R 3MR 4 2 32 æ ö = ç ÷ è ø Therefore the moment of inertia ofthe remaining part ofthe disc about a perpendicular axis passing through the centre, IRemaing disc = ITotal – IRemoved = 2 2 2 MR 3 13 MR MR 2 32 32 - = 24. (d) Centre of mass may or may not coincide with centre of gravity. Net torque ofgravitational pull is zero about centre of mass. 0 g i i ig r m t = St = S ´ = Mechanical advantage , M.A.= Load Effort If M.A. > 1 Þ Load > Effort 25. (a) Here, 1 2 I I 2I w + w = w Þ w = 1 2 2 w + w (K.E.)i = 2 2 1 2 1 1 I I 2 2 w + w (K.E.)f = 2 1 2I 2 ´ w = 2 1 2 I 2 w + w æ ö ç ÷ è ø Loss in K.E. f i (K.E) (K.E) = - = 2 1 2 1 I( ) 4 w - w 26. (b) Given, mass ofcylinder m = 3kg R = 40 cm = 0.4 m 40 cm F = 30 N F = 30 N ; a = ? As we know, torque t = Ia F × R = MR2a 2 F R MR ´ a = 2 30 (0.4) 3 (0.4) ´ a = ´ or, 2 25rad / s a =
- 215. NEWTON'S UNIVERSAL LAWOF GRAVITATION Gravitational force is an attractive force between any two point masses M1 and M2 separated by any distance r. It is given by F = G 1 2 2 æ ö ç ÷ è ø M M r where G is the universal gravitational constant. Its value, G = 6.67 × 10–11 Nm2 kg–2 Dimensions of G are [M–1 L3 T–2] (a) The gravitational force is the central force and follows inversesquare law. It acts along the linejoining theparticles. (b) Since the work done by the gravitational force is independent of the path followed and hence it is a conservative force. (c) It is the weakest force in nature. It is 1038 times smaller than nuclear force and 1036 timessmaller than electricforce. Strongest force being nuclear force (for small range) followed by electric force. (d) Gravitation is independent of the presence of other bodies around it. ACCELERATION DUE TO GRAVITY (g) The force of attraction exerted by earth on a body of mass m is the force of gravity. So the force of gravityfrom Newton’s gravitational law is 2 = e GM m F r ……(i) where Me =mass of earth r=distance of the body from the centre of earth. The force of gravity can be written as F = mg ……(ii) where g is called the acceleration due togravity. From the expression (i) and (ii), we get 2 = e GM g r ……(iii) If body is located at the surface of earth i.e., r = Re (radius of earth) then 3 2 2 (4 / 3 ) 4 3 p r = = = pr e e e e e GM G R g GR R R ……(iv) where r is the density of earth. This is the relation between universal gravitational constant 'G' and acceleration due to gravity 'g'. Change in the Value of Acceleration due to Gravity (g) (i) Due to rotation or latitude of earth : Let us consider a particle P at rest on the surface of earth at a latitude f. ( ) 2 2 cos f = - w f e g g R f r R g¢ w At poles f= 90º, so g(f) = g ...(v) At equator f = 0º so g(f) = g –w2Re ...(vi) (ii) Due to shape of earth : The shape of the earth is not perfectlyspherical. It is flattened at poles. The polar radius is 21 km less than the equatorial radius. Hence acceleration due to gravity at poles is greater than at the equator. gp = 9.83 ms–2 = value of g at poles ge = 9.78 ms–2 = values of g at equator i.e. gp > ge (iii) At a depth 'd' below the earth surface : The acceleration due to gravity at the surface of earth. G R 3 4 R GM g e 2 e e ú û ù ê ë é r p ÷ ø ö ç è æ = = (from eqn. iv) ...(vii) d R-d O R P Earth Surface of earth 8 Gravitation
- 216. 212 PHYSICS If abody is taken at a depth d belowthe earth surface, then the body is attracted by inner core of mass M' of earth radius (Re– d). Then acceleration due to gravityat point P is 3 e 2 2 e e G (4 /3) (R d) GM g (R d) (R d) é ù p - r ¢ ë û ¢ = = - - ( ) ( ) r - p = ) d R ( 3 / 4 G e ...(viii) From eqns. (vii) and (viii) we have 1 æ ö æ ö - ¢ = = - ç ÷ ç ÷ è ø è ø e e e R d d g g g R R ...(ix) The value of acceleration due to gravity at the centre of the earth is zero. (iv) At a height 'h' above the earth surface : The acceleration due to gravity at the surface of earth from expression (iv) is defined as 2 e e R GM g = At a height h above the earth surface, the acceleration due to gravity is 2 e 2 e e r GM ) h R ( GM g = + = ¢ ...(x) Þ 2 e e R h R g g - ÷ ÷ ø ö ç ç è æ + = ¢ Þ [from eqs.(ix) &(x)] If h << Re e 2 1 R æ ö » - ¢ ç ÷ è ø h g g ...(xi) Keep in Memory 1. The value of the acceleration due to gravity on the moon is about one sixth of that on the earth and on the sun is about 27 times that on the earth. E s E M g g 27g and g 6 æ ö = = ç ÷ è ø 2. The value of g is minimum on the mercury, among all planets. 3. For h<<R, the rate of decrease of the acceleration due to gravitywith height istwice as compared tothat with depth. 4. The value of g increases with the increase in latitude. Its value at latitude q is given by : gq= g – Rw2 cos2q. 5. Rotation of the earth about its own axis is responsible for decrease in the value of g with latitude. 6. Theweight ofthebodyvaries along with the valueof g (i.e. W = mg) 7. Inertial mass and gravitational mass- (a) Inertial mass (Mi) defined byNewton's law of motion Mi = F/a = External force applied on the body Acceleration produced in it i 1 2 i 2 1 (M ) a (M ) a = (ff applied force on (Mi)1 and (Mi)2 is same.) (b) Gravitational mass Mg defined by Newton's law of gravitation Mg = gravity to due on Accelerati body of Weight g W g Fg = = 8. If two spheres ofsame material, mass and radiusare put in contact, the gravitational attraction between them isdirectly proportional to the fourth power of the radius. 4 2 R 3 4 G 4 1 F ÷ ø ö ç è æ ps = A R R B m 2R ¬ ® m This system maybe considered to be made up of two mass each separated by 2R distance. 9. If the earth stops rotating about its axis, the value of g at the equator will increase by about 0.35%, but that at the poles will remain unchanged. If the earth starts rotating at the angular speed of about 17 times the present value, there will be weightlessness on equator, but g at the poles will remain unchanged. In such a case, the duration of the daywill be about 84 min. 10. If the radius of planet decreases by n%, keeping the mass unchanged, the acceleration due to gravity on its surface increases by 2n%. i.e., R R 2 g g D - = D 11. If the mass of the planet increases by m% keeping the radius constant, the acceleration due to gravity on its surfaceincreases bym% i.e., g M g M D D = where R= constant. 12. Ifthe densityof planet decreases by r % keeping the radius constant, the acceleration due to gravitydecreases by r%. 13. If the radius of the planet decreases by r% keeping the densityconstant, the acceleration due to gravitydecreases by r%. Example1. What will be the acceleration due to gravity on the surface of the moon with its radius as 1/4th the radius of the earth and its mass as 1/80th the mass of earth? Solution : 2 e e R GM g = and 2 m m R GM ' g = ; 5 1 1 4 80 1 R R M M g ' g 2 2 m e e m = ÷ ø ö ç è æ ´ = ÷ ÷ ø ö ç ç è æ = ; 5 / g ' g =
- 217. 213 Gravitation Example2. A mass Msplits into two parts m and (M–m), which are then separated by a certain distance. What will be the ratio m/M which maximizes the gravitational force between the two parts? Solution : 2 x ) m M ( Gm F - = ;For maxima, 2 1 M m or 0 ) m 2 M ( x G dm dF 2 = = - = Example3. Identical point masses each equal to m are placed at x = 0, x = 1, x = 2, x = 4, ... The total gravitational force on mass m at x = 0 due to all other masses is (a) infinite (b) (4/3)Gm2 (c) (4/3GM2) (d) zero Solution : (c) o m m m x=4 x=2 x=1 x=0 .... 4 Gm 2 Gm 1 Gm F 2 2 2 2 2 2 0 + + + = ú û ù ê ë é + + + = ... 4 1 2 1 1 1 Gm F 2 2 2 2 0 It is a G.P. with common ratio1/4. ú û ù ê ë é - = 4 / 1 1 1 Gm F 2 0 3 Gm 4 F 2 0 = Example4. Four identical point masses each equal to m are placed at the corners of a square of side a. The force on a point mass m’ placed at the point of intersection of the two diagonals is (a) (4Gmm¢)/a2 (b) (2Gmm¢)/a2 (c) (Gmm¢)/a2 (d) zero Solution : (d) The forces acting on the mass m¢ placed at the point of intersection of diagonals are balanced by each other as shown in figure. Therefore net force is zero. m m m m m¢ F F F F Example5. A tunnel is dug along the diameter of the earth. Determine the force on a particle of mass m placed in the tunnel at a distance x from the centre. Solution : Force on a mass m placed at a distance x from the centre is equal to the force of gravity due to a mass of spherical volume of earth of radius x. m R x Center of earth 2 3 x m d 3 x 4 G F ÷ ÷ ø ö ç ç è æ p = where d = density of earth = 3 R 4 M 3 p ; x . R GMm F 3 = [M is total mass of earth] Example6. If the radius of the earth were to shrink by two per cent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would (a) decrease by 2% (b) increase by 2% (c) increase by 4% (d) decrease by 4% Solution : (c) As 2 R GM g = ; So, if R decreases then g increases. Taking logarithm of both the sides; log g = log G + log M – 2 log R Differentiating it, we get g dg = 0 + 0 – R dR 2 ; g dg 100 4 100 2 2 = ÷ ø ö ç è æ - - = % increase in 100 g dg g ´ = = % 4 100 100 4 = ´ Example7. An apple of mass 0.25 kg falls from a tree. What is the acceleration of the apple towards the earth? Also calculate the acceleration of the earth towards the apple. Given : Mass of earth = 5.983 × 1024 kg, Radius of earth = 6.378 × 106 metre, G = 6.67 × 10–11 Nm2kg–2. Solution : Mass of apple, m = 0.25 kg, Mass of earth, M = 5.983 × 1024 kg, Radius of earth, R= 6.378 × 106 metre Gravitational constant, G = 6.67 × 10–11 Nm2kg–2 Let F be the gravitational force of attraction between the apple and earth. Then, 2 GmM F R = Let a be the acceleration of apple towards the earth. 2 2 F GmM GM a m mR R = = = Þ 11 24 2 2 6 2 6.67 10 5.983 10 a ms 9.810ms (6.378 10 ) - - - ´ ´ ´ = = ´
- 218. 214 PHYSICS Let a¢be the acceleration of the earth towards the apple. 2 11 2 6 2 25 2 F Gm a M R 6.67 10 0.25 ms (6.378 10 ) 4.099 10 ms . - - - - = = ¢ ´ ´ = ´ = ´ Example8. The acceleration due to gravity at the moon’s surface is 1.67 ms–2. Iftheradius ofthemoon is 1.74× 106 m, calculate the mass of the moon. Use the known value of G. Solution : 2 GM g R = or 2 gR M G = This relation is true not only for earth but for any heavenly body which is assumed to be spherical. Here, 2 6 g 1.67ms , R 1.74 10 m - = = ´ and 11 2 2 G 6.67 10 Nm kg - - = ´ Mass of moon, 6 2 22 11 1.67 (1.74 10 ) M kg 7.58 10 kg 6.67 10- ´ ´ = = ´ ´ GRAVITATIONAL FIELD (or Gravitational Field Intensity) The region or space around a body in which its gravitational influence is experienced by other bodies is the gravitational field of that body. The gravitational field strength Eg, produced by a mass M at any point P is defined as the force exerted on the unit mass placed at that point P. Then 2 2 æ ö = = = ç ÷ è ø g F G Mm GM E m m r r ...(i) where m= test mass r=distance between M and m (i) The direction of Eg always points towards the mass producing it. (ii) The gravitational field can be represented bygravitational lines of force. (iii) The S.I unit of Eg is newton/kg. Its dimensions are [M0LT–2]. Gravitational intensity (Eg) for spherical shell r |E |=0 g Radius of spherical shell |E | g O R 2 r GM | Eg | = (i) |Eg|= 0 at points inside the spherical shell (i.e. r < R) (ii) 2 | | - = g GM E R at the surface of shell (i.e. r = R) (iii) 2 | | - = g GM E r for outside the spherical shell (r >R). It is clear that as r increases, Eg decreases. Gravitational intensity for solid sphere : R r |E | g Solid sphere Eg (i) 3 . | |= - g GM r E R for points inside the solid sphere (r<R) (ii) At the surface of solid sphere (r = R) 2 | |= - g GM E R (iii) Outside the solid sphere 2 | |= - g GM E r (r > R) GRAVITATIONAL POTENTIAL Gravitational potential at any point is the work done by gravitational force in carrying a body of unit mass from infinity to that point in gravitational influence of source. i.e., A o W GM V m r - = = where r is the distance of point from source mass M. The S.I. unit of V is joule/kg. Its dimensions are [M0L2T–2]. It is a scalar quantity. Gravitational potential at the earth surface is - = = - e A e e GM V gR R Gravitational potential due to a uniform ring at a point on its axis 2 2 = - + g GM V a r a O r P q Gravitational potential due to a uniform thin spherical shell O Vg r R GM -
- 219. 215 Gravitation (i) at apoint outside the shell, g GM V r - = (r > R) (ii) at a point inside the shell, g GM V R = - (r < R) where R = radius of spherical shell. Potential due to a uniform spheral shell is constant throughout the cavity of the shell. Gravitational potential due to a uniform solid sphere (i) at an external point =- g GM V r (r³ R) V R solid state (ii) at an internal point 2 2 3 (3 ) 2 =- - g GM V R r R (r < R) O r R (iii) Potential at the centre of solid sphere is 3 2 - = GM V R GRAVITATIONAL POTENTIAL ENERGY Let Vg be thegravitational potential at a point. Ifwe place a mass m at that point, then we saythat the gravitational potential energy possessed by the mass is = ´ g g U V m Þ = - ´ g GM U m r g GM V r é ù = - ê ú ë û We can also therefore saythat the gravitational potential energy of a system containing two masses m1 and m2 placed such that the centre to centre distance between them is r then 1 2 = - g Gm m U r This formula is valid taking into consideration the fact that we have taken the gravitational potential of few at infinity. Please remember that unless otherwise stated it is understood that the gravitational potential is taken to be zero at infinity. Relation between gravitational potential energy and gravitationalpotential. Gravitational potential energy = gravitational potential × mass Keep in Memory 1. The gravitational potential energy of a mass m at a point above the surface of the earth at a height h is given by g GMm U R h - = + . The –ve sign shows that if h increases, the gravitational PE decreases and becomes zero at infinity. 2. (a) If we take reference level to be at the surface of earth (not at infinity) i.e., we assume that the gravitational P.E ofa mass m is zero at thesurface of earth, then the gravitational potential energy at a height h above the surface of earth is (mgh), where h << Re (radius of earth) (b) The gravitational P.E ofmass m on the earth’s surface is e g e GM m U mgR R - = = - If we assume that gravitational P.E of mass m is zero at infinityi.e., we take reference level at infinitythen P.E.of anymass m at a height h above the earth surface ) h R ( m GM U e e h + - = So work done against the gravitational force, when the particle is taken from surface of earth to a height h above the earth surface Work done = change in potential energy Þ Uh–Ug = e e e e R m GM h R m GM + + - ) h R ( R h m GM e e e + ´ @ ÷ ÷ ø ö ç ç è æ » e e e R h R m GM (ifh<<R) e 2 e GM m h R æ ö = ç ÷ ç ÷ è ø So, work done = mgh is stored in the particle as particle potenital energy, when kept at height ‘h’. Example9. Four particles each of mass 1 kg are at the four corners of a square of side 1m. Find the work done to remove one of the particles to infinity. Solution : The gravitational potential energyassociated with anypair of particle of mass m1 and m2 separated by a distance r is r m Gm U 2 1 - = Initial P.E. of system, – U1 = 2 1 1 G 2 1 1 1 G . 4 ´ ´ + ´ ´ or ú ú û ù ê ê ë é + - = 2 2 2 4 G U1
- 220. 216 PHYSICS 1 Kg1 Kg 1 Kg 1 Kg 1m 1m 1m 1m 2 m 2 m When one mass is removed, then P.E. = U2 = ú û ù ê ë é ´ ´ + ´ ´ - 2 1 1 G 1 1 1 G . 2 G 2 1 2 2 U2 ú ú û ù ê ê ë é + - = ; Work done = ú ú û ù ê ê ë é - - + 2 1 2 2 2 2 4 G = ú ú û ù ê ê ë é + 2 1 2 2 G [Q Work done = change in P.E. = –U2–[–U1] = U1–U2] Example10. A solid sphere of uniform density and mass M has a radius of 4m. Its centre is at the origin of coordinate system. Two spheres of radii 1m are taken out so that their centres are at P(0, –2, 0) and Q(0, 2, 0) respectively. This leaves two spherical cavities. What is the gravitational field at the centre of each cavity? O P Q R z y Solution : If the cavities are not made, then the intensityat the point P (or Q), due to whole sphere 32 GM 2 64 GM x . R GM I 3 = ´ = = where P Q R I I I I = + + r r r r Q P R I I I I r r r r - - = ……(i) Where IR is the intensity of gravitational field at a point, distance x from O due toremainder sphere (exclude cavity) Mass of big sphere 3 d ) 4 ( 4 M 3 ´ p = Mass of small sphere P or Q, 3 4π (1) 3 ´ = d m M m 64 = At P, Ip = 0, ÷ ø ö ç è æ = = 64 M . 4 G r Gm I 2 2 Q IR = GM GM 32 1024 - [From eq (i)] R 31GM I 1024 = Example11. Find the gravitational potential energy of a system of four particles, each having mass m, placed at the vertices of a square of side l. Also obtain the gravitational potential at the centre of the square. Solution : The system has four pairs with distance l and two diagonal pairs with distance 2 l . m l 2l m m m 2 2 2 2 Gm Gm 2Gm 1 Gm U 4 2 2 5.41 2 2 æ ö = - - = - + = - ç ÷ è ø l l l l The gravitational potential at the centre of the square is ( ) r 2 / 2 = l V = Algebraic sum of potential due to each particle Þ 4 2Gm V = - l Example12. The radius of earth is R. Find the work done in raising a body of mass m from the surface of earth to a height R/2. Solution : Ifa bodyofmass m is placed at a distance x from the centre of earth, the gravitational force of attraction F between the body and the earth is 2 x m GM F = Small amount of work done in raising a body through a small distance dx is given by dx x m GM dx F dW 2 = = Total work done in raising the body from the surface of earth to a height R/2 is given by W = R R/2 2 R GM m dx x + ò = 3 R/2 3 R/2 2 R R 1 GMm x dx GMm x - æ ö = - ç ÷ è ø ò = 2 1 GMm 3R R é ù - - ê ú ë û = 2 M m gR m 3R 3R G = = 1 mg R 3 [ GM = gR2]
- 221. 217 Gravitation SATELLITES A satellite isany body revolving around a large body under the gravitational influence of the latter. The period of motion T of an artificial satellite of earth at a distance h above the surface of the earth is given by, p 3 2 ( + ) = 2 e e R h T gR where, Re = radius of the earth where, g = acceleration dueto gravityon thesurface ofthe earth. If Re > > h, then e R T 2 g = p ; Re = 6.4 × 106 m ; g = 9.8 ms–2 i.e. T 84.58 minutes 5075sec » ; Assuming Re + h = r, the distance of the satellite from the centre of the earth, 3 / 2 T (r) µ Orbital velocity (v0 ) Let a satellite ofmass m revolve around the earth in circular orbit of radius r with speed v0. The gravitational pull between satellite and earth provides the necessary centripetal force. Centripetal force required for the motion = 2 0 mv r Gravitational force = GMm r² 2 0 2 mv GMm r r = or 2 0 GM v r = or 0 = GM v r ..........(1) or 0 = + g v R R h [Q 2 GM g R = and r = (R + h)] Orbital velocity, e R = = e e GM Vo gR (i) Valueof orbital velocitydoes not depend on the mass of the satellite. (ii) Around the earth the value of orbital velocity is 7.92 km/sec. (iii) Greater is the height of the satellite, smaller is the orbital velocity. (iv) The direction of orbital velocity is along the tangent to the path. (v) The work done by the satellite in a complete orbit is zero. Angularmomentum (L):For satellite motion, angular momentum will be given by GM L mvr mr r = = i.e., L = [m2GMr]1/2 Angular momentum of a satellite depends on both, the mass of orbiting and central body. It also depends on the radius of the orbit. Energy of a satellite Total energy of satellite revolving in an orbit of radius r around the earth can be calculated as follows : (i) The gravitational potential energy of a satellite of mass m is – = g GMm U r , where r is the radius of the orbit. (ii) Kinetic energy of the satellite is 2 0 1 2 2 = = k GMm E mv r (iii) So, total energy of the satellite 2 = + = - g k GMm E U E r The negative sign shows that satellite cannot leave the orbit itself. It requires an energyequal to 2 GMm R , which is called BindingEnergy (B.E.) ofsatellite. (iv) Total energy of a satellite at a height equal to the radius of the earth 1 – – 2( ) 4 4 = = = + GMm GMm mgR R R R GEO-STATIONARY SATELLITE A satelite which appears to be stationary for a person on the surface of the earth is called geostationary satellite. It is also known as parking satellite or synchronous satellite. (i) The orbit ofthesatellitemust becircular and in theequatorial plane of the earth. (ii) The angular velocity of the satellite must be in the same direction as the angular velocityof rotation ofthe earth i.e., from west to east. (iii) The period ofrevolution ofthe satellite must be equal tothe period of rotation of earth about its axis. i.e. 24 hours = 24 × 60 × 60 = 86400 sec. 2 4 ² T r³ GM p = or 1 1 2 2 2 3 3 2 2 2 GMT GM R T r 4 R 4 æ ö æ ö = = ´ ç ÷ ç ÷ p p è ø è ø = 1 3 6 2 (86400)² 9.8 (6.38 10 ) 4 ² é ù ´ ´ ´ ê ú p ë û =42237 ×103 m = 42,237 km.»42000km. h= r –R= 42000–6400= 35600km. (a) Height of geostationary satellite from the surface of the earth is nearly35600 km. (b) The orbital velocity of this satellite is nearly 3.08km/sec. (c) The relative velocity of geostationary satellite with respect to earth is zero. This type of satellite is used for communication purposes. The orbit of a geostationary satellite is called ‘ParkingOrbit’. Polar Satellite : Polar satellites travel around the earth in an orbit that travels around the earth over the poles. The earth rotates on its axis as the satellite goes around the earth. Thus over a period of many orbits it looks down on every part of the earth.
- 222. 218 PHYSICS Different orbitalshapes correspondingtodifferent velocitiesof a satellite : [1] When v< v0 (i) Thepath is spiral. The satellite finallyfalls on the earth (ii) Kinetic energy is less then potential energy (iii) Total energy is negative [2] When v= v0 (i) The path is circular (ii) Eccentricityis zero (iii) Kinetic energy is less than potential energy (iv) Total energy is negative [3] When v0 < v< ve (i) The path is elliptical (ii) Eccentricity< 1 (iii) Kinetic energy is less than potential energy (iv) Total energy is negative [4] When v = ve (i) The path is a parabola (ii) Eccentricity= 1 (iii) Kinetic energy is equal to potential energy (iv) Total energy is zero [5] When v > ve (i) The path is a hyperbola (ii) Eccentricity> 1 (iii) Kinetic energy is greater than potential energy (iv) Total energy is positive If the orbit of a satellite is elliptical (1) The energy GMm E 2a = - = const. with ‘a’as semi-major axis; (2) KE will bemaximum when the satelliteisclosestto thecentral body(at perigee) and minimum when it is farthest from the central body (at apogee) [as for a given orbit L = const., i.e., mvr = const., i.e., vµ 1/r] (3) PE = (E – KE) will be minimum when KE = max, i.e., the satellite is closest to the central body (at perigee) and maximum when KE = min, i.e., the satellite is farthest from the central body (at apogee). ESCAPE SPEED (Ve) It is the minimum speed with which a body should be projected from the surface of a planet so as to reach at infinity i.e., beyond the gravitational field of the planet. If a bodyof mass m is projected with speed v from the surface of a planet of mass M and radius R, then as 2 1 2 = K mv and GMm U R = - , ES = 2 1 2 mv – GMm R Now if v' is the speed of body at ¥, then 2 2 1 1 ( ) 0 ( ) 2 2 ¥ = + = ¢ ¢ E m v m v [as U 0 ¥ = ] So by conservation of energy 2 2 1 1 ( ) 2 2 = - = ¢ GMm mv m v R i.e. 2 2 1 ( ) 2 2 1 = + ¢ GMm mv m v R sov will be minimum when v¢ ® 0, i.e. min 2 2 = = = e GM v v gR R 2 GM as g R é ù = ê ú ë û • The value of escape velocitydoes not depend upon the mass of the projected body, instead it depends on the mass and radius of the planet from which it is being projected. • The value of escape velocity does not depend on the angle and direction of projection. • The value of escape velocityfrom the surface of the earth is 11.2 km/sec. • The minimum energyneeded for escape is = GMm/R. • If the velocity of a satellite orbiting near the surface of the earth is increased by 41.4%, then it will escape away from the gravitational field of the earth. • Ifa bodyfalls freelyfrom infinite distance, then it will reach the surface ofearth with a velocityof 11.2 km/sec. Relationbetweenorbital velocity(V0) andescape speed(Ve) 0 2 2 = = e V gR V Keep in Memory 1. The escape velocity on moon is low 6 æ ö = ç ÷ è ø E m g as g hence there is no atmosphere on moon. 2. If the orbital radius of the earth around the sun be one fourth of the present value, then the duration of the year will be one eighth of the present value. 3. The satellites revolve around the earth in a plane that coincides with the great circle around the earth. KEPLER'S LAWS OF PLANETARY MOTION 1. The law of orbits : Each planet revolves about the sun in an elliptical orbit with the sun at one of the foci of the ellipse. ea ea P x Perihelion Aphelion Sun S¢ y a a S a = Semi-major axis A planet ofmass m moving in an elliptical orbit around the sun. The sun of mass M, is at one focus S¢ of the ellipse. The other focus is S, which is located in empty space. Each focus is at distance ‘ea’ from the ellipse’s centre, with ‘e’ being the eccentricity of the ellipse and ‘a’ semimajor axis of the ellipse, the perihelion (nearest to the sun) distance rmin., and the aphelion (farthest from the sun) distance rmax. are also shown. rmax = a + e a = (1 + e) a rmin = a – e a = (1 – e) a
- 223. 219 Gravitation The distance ofeach focus from the centre of ellipse is ea, where e is the dimensionless number between 0 to 1 called the eccentricity. If e = 0, the ellipse is a circle. For earth e = 0.017. 2. Law of areas : An imaginary line that connects a planet to the sun sweeps out equal areas in the plane of the planet's orbit in equal times; i.e., the rate dA/dt at which it sweeps out area Ais constant. S P4 P3 P2 P1 A1 A2 If 1 2 3 4 = P P P P t t then A A1 = A2 dA 1 (r)(v dt) 1 rv dt 2 dt 2 = = and as L = mvr dA r vdt so 2 = dA L dt m ....(1) But as L = constt., (force is central, so torque = 0 and hence angular momentum of the planet is conserved) areal velocity(dA/dt) = constant which is Kepler's IInd law, i.e., Kepler's IInd law or constancy of areal velocity is a consequence of conservation of angular momentum. 3. Law of periods : The square of the period of revolution of any planet is proportional to the cube of the semi-major axis of the orbit, i.e., T2 µ r3. or, 2 3 1 1 2 3 2 2 = T r T r Focus Semi major axis a Satellite r Perigee KE = Max PE = Min Apogee KE = Min PE = Max rmin rmax b b = Semi minor axis Sun WEIGHTLESSNESS The “weightlessness” you mayfeel in an aircraft occurs anytime the aircraft is accelerating downward with acceleration g. It is possible to experience weightlessness for a considerable length oftime by turning the nose of the craft upward and cutting power so that it travels in a ballistic trajectory. A ballistic trajectory is the common type of trajectory you get by throwing a rock or a baseball, neglecting air friction.At everypoint on the trajectory, the acceleration is equal to g downward since there is nosupport. A considerable amount of experimentation has been done with such ballistic trajectories to practice for orbital missions where you experience weightlessness all the time. Thesatelliteis moving in a circular orbit, it has a radialacceleration 2 0 0 2 v GM GM a as v r r r é ù æ ö = = = ê ú ç ÷ è ø ê ú ë û i.e., it is falling towards earth's centre with acceleration a, so apparent weight of the body in it Wap = m(g´ – a) where g´ is the acceleration due togravityof earth at the position (height) of satellite, i.e. g´= (GM/r2), so that ap 2 2 GM GM W m 0 r r é ù = - = ê ú ë û i.e., the apparent weight of a body in a satellite is zero and is independent of the radius of the orbit. Keep in Memory 1. The moon takes 27.3 days to revolve around the earth. The radius of its orbit is 3.85 × 105 km. 2. Kepler's second law is based on conservation of angular momentum. 3. Perihelion distance is the shortest distance between the sun and the planet. 4. Aphelion distance is the largest distance between the Sun and the planet. perihelion aphelion r r = Vaphelion Vperihelion 5. If e is the eccentricity of the orbit then aphelion perihelion r 1 e 1 e r + = - raphelion + rperihelion = 2r 6. If e > 1 and total energy (KE + PE) > 0, the path of the satellite is hyperbolic and it escapes from its orbit. 7. If e < 1 and total energy is negative it moves in elliptical path. 8. If e = 0 and total energy is negative it moves in circular path. 9. If e = 0 and total energy is zero it will take parabolic path. 10. The path of the projectiles thrown to lower heights is parabolic and thrown to greater heights is elliptical. 11. Kepler’s laws may be applied to natural and artificial satellites as well. 12. Gravitational force does not depend upon medium so no medium can shield it or block it. 13. The escape velocityand the orbital velocity are independent of the mass of the body being escaped or put into the orbit.
- 224. 220 PHYSICS Example13. If theearth is at 1/4 of its present distance from the sun, what would be the duration of the year? Solution : We know that, 3 2 3 1 2 2 2 1 R R T T = Substituting the given values, we get 64 ) 4 ( T 1 or 4 R R T ) 1 ( 3 2 2 3 3 2 2 2 = = ÷ ø ö ç è æ = . year 8 1 T or 64 1 T 2 2 2 = = Example14. Two satellites A and B go round a planet P in a circular orbits having radii 4R and R respectively. If the speed of satellite A is 3v, what will be the speed of satellite B? Solution : We know that v (G M/ r) = Here G M 3v 4R æ ö = ç ÷ è ø and GM v' R æ ö = ç ÷ è ø ; v' G M 4R 2 3v R GM æ ö æ ö = ´ = ç ÷ ç ÷ è ø è ø ; v' 6v = Example15. A spaceship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? Radius of earth = 6400 km and g = 9.8 m/s2. Solution : The orbital velocity of spaceship in circular orbit 0 G M G M v r R = = (Q spaceship is very close to earth, r = R) As 2 R M G g = hence G M = g R2 6 0 v (R g) (6.4 10 ) (9.8) = = ´ ´ = 7.9195 km/s Further, ) 9195 . 7 2 ( Rg 2 ve ´ = = = 11.2 km/s (where ve is escape velocity) Additional velocity required = 11.2000 – 7.0195 = 3.2805 km/s Sothe velocity3.2805 km/s must be added to orbital velocity of spaceship.
- 225.
- 226. 222 PHYSICS 1. Asatellite is orbiting around the earth near its surface. Ifits kinetic energy is doubled, then (a) it will remain in the same orbit. (b) it will fall on theearth. (c) it will revolve with greater speed. (d) it will escapeout of the gravitational field of the earth. 2. There is no atmosphere on the moon because (a) it is closer to the earth and also it has the inactiveinert gases in it. (b) it is too for from the sun and has very low pressure in its outer surface. (c) escape velocity of gas molecules is greater than their root mean square velocity. (d) escape velocity of gas molecules is less than their root mean square velocity. 3. Themaximum kinetic energyofa planet moving around the sun is at a position B A Sun D C (a) A (b) B (c) C (d) D 4. A man waves his arms whilewalking. This is to (a) keep constant velocity (b) ease the tension (c) increase the velocity (d) balance the effect of earth’s gravity 5. A missile is launched with a velocity less than escape velocity. The sum of its kinetic and potential energies is (a) zero (b) negative (c) positive (d) may be positive, negative or zero. 6. Which of the following graphs represents the motion of a planet moving about the sun ? (a) T2 R3 (b) T2 R3 (c) T2 R3 (d) T2 R3 7. Due to rotation of the earth the acceleration due to gravity g is (a) maximum at the equator and minimum at the poles (b) minimum at the equator and maximum at the poles (c) same at both places (d) None of these 8. A planet moves around the sun. At a point P it is closest from the sun at a distance d1 and has a speed v1.At another point Q, when it is farthest from the sun at a distnace d2 its speed will be (a) 2 2 1 2 1 d / v d (b) 1 1 2 d / v d (c) 2 1 1 d / v d (d) 2 1 1 2 2 d / v d 9. The weight of an object in the coal mine, sea level and at the top of the mountain, are respectively W1, W2 and W3 then (a) W1< W2 > W3 (b) W1= W2 = W3 (c) W1< W2 < W3 (d) W1> W2 > W3 10. Two planets of radii r1 and r2 are made from the same material. The ratio of the acceleration due to gravityg1/g2 at the surfaces of the two planets is (a) r1/r2 (b) r2/r1 (c) (r1/r2)2 (d) (r2/r1)2 11. What would be the length of a sec. pendulum at a planet (where acc. due to gravityis g/4) if it’s length on earth is l (a) l/2 (b) 2 l (c) l/4 (d) 4 l 12. Time periodof a simple pendulum inside a satellite orbiting earth is (a) zero (b) ¥ (c) T (d) 2T 13. Theratio ofthe radii ofthe planets R1 and R2 is k. The ratio ofthe acceleration duetogravityis r. The ratioof the escape velocities from them will be (a) kr (b) kr (c) (k /r) (d) (r/k) 14. If e v and 0 v represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then (a) e o v v = (b) e o v 2 v = (c) e o v (1/ 2)v = (d) ve and vo are not related 15. The kinetic energyneeded to project a body of mass m from the earth surface (radius R) toinfinityis (a) mgR/2 (b) 2mgR (c) mgR (d) mgR/4.
- 227. 223 Gravitation 16. The escapevelocity of a body depends upon mass as (a) m0 (b) m1 (c) m2 (d) m3. 17. The radius of a planet is 1/4th of Re and its acc. due to gravity is 2g. What would be the value of escape velocity on the planet, if escape velocity on earth is ve. (a) e v 2 (b) e v 2 (c) 2 ve (d) e v 2 18. If the gravitational force had varied as r–5/2 instead of r–2; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to (a) 1 r- (b) 2 r- (c) 3 / 2 r- (d) 5 / 2 r- 19. Twosatellites revolve round the earth with orbital radii 4R and 16R, if the time period of first satellite is T then that of the other is (a) 4T (b) 42/3 T (c) 8T (d) None of these 20. A planet revolves in an elliptical orbit around the sun. The semi-major and semi-minor axesarea andb. Then thesquare of time period, T is directly proportional to (a) 3 a (b) 3 b (c) 3 2 b a ÷ ø ö ç è æ + (d) 3 2 b a ÷ ø ö ç è æ - 21. Which of the following quantities do not depend upon the orbital radius of the satellite ? (a) R T (b) R T2 (b) 2 2 R T (d) 3 2 R T 22. Theorbital velocityofan artificial satellitein a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitudeof 6Rfrom Earth's surface will be (a) 7 v (b) 6 v (c) v (d) v 6 23. Escapevelocitywhen a bodyof mass m is thrown vertically from the surface of the earth is v, what will be the escape velocity of another body of mass 4 m is thrown vertically (a) v (b) 2v (c) 4v (d) None of these 24. The potential energyofa satellite of mass m and revolving at a height Re above the surface of earth where Re = radius of earth, is (a) – m g Re (b) 2 R g m e - (c) 3 R g m e - (d) 4 R g m e - 25. Energyrequired to movea bodyof mass m from an orbit of radius 2R to 3R is (a) GMm/12R2 (b) GMm/3R2 (c) GMm/8R (d) GMm/6R 1. The escape velocityfrom the earth's surfaceis 11 km/s. The escape velocity from a planet having twice the radius and same mean density as that of earth is (a) 5.5km/s (b) 11km/s (c) 22km/s (d) None of these 2. Two point masses each equal to 1 kg attract one another with a force of 10–10 N. The distance between the two point masses is (G = 6.6 × 10–11 MKS units) (a) 8cm (b) 0.8cm (c) 80cm (d) 0.08cm 3. There are twobodies ofmasses 103 kg and 105 kg separated by a distance of 1 km. At what distance from the smaller body, the intensity of gravitational field will be zero (a) 1/9km (b) 1/10km (c) 1/11km (d) 10/11km 4. Taking the gravitational potential at a point infinte distance away as zero, the gravitational potential at a point A is –5 unit. If the gravitational potential at point infinite distance away is taken as + 10 units, the potential at point A is (a) – 5 unit (b) + 5 unit (c) + 10 unit (d) + 15 unit 5. A planet of mass 3 × 1029 gm moves around a star with a constant speed of 2 × 106 ms–1 in a circle of radii 1.5 × 1014 cm. The gravitational forceexerted on theplanetbythe star is (a) 6.67 × 1022 dyne (d) 8 × 1027 Newton (c) 8 × 1026 N (d) 6.67 × 1019 dyne 6. The mass of the moon is 1/81 of earth’s mass and its radius 1/4 that ofthe earth. If the escape velocityfrom the earth’s surface is 11.2 km/sec, its value from the surface of the moon will be (a) 0.14 kms–1 (b) 0.5kms–1 (c) 2.5kms–1 (d) 5.0kms–1 7. Ifthe mass ofearth is eighty times the mass of a planet and diameter of the planet is one fourth that of earth, then acceleration due to gravity on the planet would be (a) 7.8m/s2 (b) 9.8m/s2 (c) 6.8m/s2 (d) 2.0m/s2
- 228. 224 PHYSICS 8. Twobodies of masses 10 kg and 100 kg are separated by a distance of 2m ( G = 6.67 × 10–11 Nm2 kg–2). The gravitational potential at the mid point on the line joining the two is (a) 7.3 × 10–7 J/kg (b) 7.3 × 10–9 J/kg (c) –7.3 × 10–9 J/kg (d) 7.3 × 10–6 J/kg 9. The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to4timesthepreviousvalue,thenewtimeperiodwillbecome (a) 10 hours (b) 80 hours (c) 40 hours (d) 20 hours 10. At sea level, a bodywill have minimum weight at (a) pole (b) equator (c) 42° south latitude (d) 37° north latitude 11. The distance of neptune and saturn from the sun is nearly 1013 and 1012 meter respectively.Assuming that theymove in circular orbits, their periodic times will be in the ratio (a) 10 (b) 100 (c) 10 10 (d) 1000 12. A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is (a) 5 (b) 10 (c) 6 2 (d) 6 2 13. The escape velocityfor a bodyprojected vertically upwards from the surface of earth is 11 km/s. Ifthe bodyis projected at an angle of45º with the vertical, theescapevelocitywill be (a) 22km/s (b) 11km/s (c) 2 11 km/s (d) 2 11 km/s 14. Ifthe length ofa simple pendulum is increased by 2%, then the time period (a) increases by 2% (b) decreases by 2% (c) increases by 1% (d) decreases by 1% 15. The kinetic energyof a satellite in its orbit around the earth is E. What should be the kinetic energyofthe satellite soas toenable it to escapefrom thegravitational pull ofthe earth? (a) 4E (b) 2E (c) E 2 (d) E 16. The time period ofa satellite in a circular orbit of radius Ris T, the period of another satellite in a circular orbit of radius 4 R is (a) 4T (b) T/4 (c) 8T (d) T/8 17. If the change in the value of g at the height h above the surface of the earth is the same as at a depth ‘x’ below it, then (both x and h being much smaller than the radiusofthe earth) (a) x=h (b) x = 2h (c) x= h/2 (d) x = h2 18. A satellite of mass m revolves around the earth of radius R at a height ‘x’ from its surface. Ifg is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is (a) x R gR2 + (b) x R gR - (c) gx (d) 2 / 1 x R gR2 ÷ ÷ ø ö ç ç è æ + 19. The time period of an earth satellite in circular orbit is independent of (a) both the mass and radius of the orbit (b) radius of its orbit (c) the mass of the satellite (d) neither themassofthesatellite nor theradiusofitsorbit. 20. Suppose the gravitational force varies inverselyas the nth power of distance. Then thetime period ofa planet in circular orbit of radius ‘R’ around the sun will be proportional to (a) Rn (b) ÷ ø ö ç è æ - 2 1 n R (c) ÷ ø ö ç è æ + 2 1 n R (d) ÷ ø ö ç è æ - 2 2 n R 21. An earth satellite ofmass m revolves in a circular orbit at a height h from the surface of the earth. Ris the radius ofthe earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by (a) g R2/(R + h) (b) g R (c) g R/(R – h) (d) 2 gR / (R h) é ù + ë û 22. Mass of the Earth has been determined through (a) use of Kepler's 3 2 R T constancy law and Moon's period (b) sampling the densityofEarth'scrust and using Earth's radius (c) Cavendish'sdetermination ofG andusing Earth radius and g at its surface (d) use of periods of satellites at different heights above Earth's surface and known radius ofEarth 23. Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist Bgoes high up in a balloon. The gravitational field measured by (a) A goes on decreasing and that byBgoes on increasing (b) Bgoes on decreasing and that byAgoes on increasing (c) each decreases at the same rate (d) each decreases at different rates
- 229. 225 Gravitation 24. There are_______ gravitational lines of force inside a sphericallysymmetric shell. (a) infinitelymany (b) zero (c) varying number depending upon surface area (d) varying number depending upon volume 25. A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and ofequal radius. The centre of the ring is directlyabove the centre of the sphere at a distance 3 r as shown in the figure. The gravitational force exerted by the sphere on the ring will be (a) 2 r 8 GMm (b) 2 r 4 GMm r 3r 2r (c) 2 r 8 GMm 3 (d) 3 r 8 GMm 3 26. The gravitational potential difference between the surface ofa planet and a point 20 m above the surface is 2 joule/kg. If the gravitational field is uniform, then the work done in carrying a 5 kg bodyto a height of 4 m above the surface is (a) 2 J (b) 20J (c) 40J (d) 10J 27. The ratio of the kinetic energy required to be given to a satellite so that it escapes the gravitational field of Earth to the kinetic energyrequired to put the satellite in a circular orbit just above the free surface of Earth is (a) 1 (b) 2 (c) 3 (d) 9 28. The radii of two planets are respectively R1 and R2 and their densities are respectively r1 and r2. The ratio of the accelerations due to gravity at their surfaces is (a) 2 2 2 2 1 1 2 1 R : R g : g r r = (b) 2 1 2 1 2 1 : R R g : g r r = (c) 1 2 2 1 2 1 R : R g : g r r = (d) 2 2 1 1 2 1 R : R g : g r r = 29. An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) is E0. Its potential energyis – (a) –E0 (b) 1.5E0 (c) 2E0 (d) E0 30. If value of acceleration due to gravity changes from one place toanother, which ofthe following forces will undergo a change ? (a) Viscous force (b) Buoyant force (c) Magnetic force (d) All of the above 31. The amount of work done in lifting a mass ‘m’ from the surface of the earth to a height 2R is (a) 2mgR (b) 3mgR (c) 2 3 mgR (d) 3 2 mgR 32. The radius of the earth is 4 times that of the moon and its mass is 80 times that ofthe moon. If the acceleration due to gravity on the surface of the earth is 10 m/s2, then on the surface of the moon its value will be (a) 1 ms–2 (b) 2 ms–2 (c) 3 ms–2 (d) 4 ms–2 33. Asatellite ofmass‘m’, moving around theearth in a circular orbitof radiusR,hasangular momentum L.Theareal velocity of satellite is (Me = mass of earth) (a) L/2m (b) L/m (c) 2L/m (d) 2L/Me 34. A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at A, distance 2R from the centre of the sphere. R R A A spherical cavityof radiusR/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle placed at A. The ratio F2/F1 will be (a) 1/2 (b) 3 (c) 7 (d) 1/9 35. A body starts from rest from a point distance R0 from the centre of the earth. The velocity acquired bythe bodywhen it reaches the surface of the earth will be (R represents radius of the earth). (a) ÷ ÷ ø ö ç ç è æ - 0 R 1 R 1 M G 2 (b) ÷ ÷ ø ö ç ç è æ - R 1 R 1 M G 2 0 (c) ÷ ÷ ø ö ç ç è æ - 0 R 1 R 1 M G (d) ÷ ÷ ø ö ç ç è æ - 0 R 1 R 1 M G 2 36. The largest and the shortest distance of the earth from the sun are r1 and r2. Its distance from the sun when it is at perpendicular tothemajor-axisoftheorbit drawnfromthesun (a) 1 2 (r r ) / 4 + (b) 1 2 1 2 (r r ) /(r r ) + - (c) 1 2 1 2 2 r r /(r r ) + (d) 1 2 (r r ) / 3 +
- 230. 226 PHYSICS 37. Theorbitalvelocityofan artificial satellitein a circular orbit just above the centre’s surface is u. For a satellite orbiting at an altitude ofhalfoftheearth’sradius, theorbital velocityis (a) 0 v 3 2 ÷ ÷ ø ö ç ç è æ ÷ ø ö ç è æ (b) 0 v 3 2 (c) 0 v 2 3 (d) 0 v 2 3 ÷ ø ö ç è æ 38. A planet is revolving around the sun in an elliptical orbit. Its closests distance from the sun is rmin. The farthest distance from the sun is rmax. If the orbital angular velocity of the planet when it is nearest to the sun is w, then the orbital angular velocityat the point when it is at the farthest distance from the sun is (a) w ) r / r ( max min (b) w ) r / r ( min max (c) w ) r / r ( 2 min 2 max (d) w ) r / r ( 2 max 2 min 39. Twospherical bodies ofmass M and 5Mand radii R and 2R respectivelyare released in free spacewith initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is (a) 2.5R (b) 4.5R (c) 7.5R (d) 1.5R 40. If the radius of the earth were to shrink byone per cent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would (a) decrease (b) remain unchanged (c) increase (d) None of these 41. If earth is supposed to be a sphere of radius R, if 30 g is valueof acceleration duetogravityat lattitude of30° and g at the equator, the value of g – g30 is (a) R 4 1 2 w (b) R 4 3 2 w (c) R 2 w (d) R 2 1 2 w 42. A ball is dropped from a satelliterevolving around the earth at height of120 km. The ball will (a) continue to move with same speed along a straight line tangentially to the satellite at that time (b) continue to move with same speed along the original orbit of satellite. (c) fall down to earth gradually (d) go far away in space 43. Twoidentical geostationarysatellites aremoving with equal speeds in the same orbit but their sense of rotation brings them on a collision course. The debris will (a) fall down (b) move up (c) begin to move from east to west in the same orbit (d) begin to move from west to east in the same orbit 44. Ifthere were a small gravitational effect, then which ofthe following forces will undergo a change? (a) Viscous force (b) Electrostaticforce (c) Magnetic force (d) Archimedes' uplift 45. The gravitational force of attraction between a uniform sphere of mass M and a uniform rod of length l and mass m oriented as shown is r l m M (a) ) r ( r GMm l + (b) 2 r GM (c) l + 2 Mmr (d) mM ) r ( 2 l + 46. Explorer 38, a radio-astronomy satellite of mass 200 kg, circles the Earth in an orbit of average radius 2 R 3 whereRistheradiusoftheEarth.Assuming thegravitational pull on a mass of 1 kg at the earth's surface to be 10 N, calculate the pull on the satellite (a) 889N (b) 89N (c) 8889N (d) 8.9N 47. Suppose, the acceleration due to gravity at the Earth's surface is 10 m s–2 and at the surface of Mars it is 4.0 m s–2.A60 kg pasenger goes from the Earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part offigure best represents the weight (net gravitational force) of the passenger as a function of time? t0 time D Weight 600 N 200 N A B C (a) A (b) B (c) C (d) D 48. A projectile is fired verticallyfrom the Earth with a velocity kve where ve is the escape velocity and k is a constant less than unity. The maximum height to which projectile rises, as measured from the centre of Earth, is (a) k R (b) 1 k R - (c) 2 k 1 R - (d) 2 k 1 R +
- 231. 227 Gravitation 49. A diametricaltunnel is dug across the Earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the Earth is .... (Given : gravitational potential at the centre of Earth = R GM 2 3 – ) (a) R (b) gR (c) gR 5 . 2 (d) gR 1 . 7 50. A man of mass m starts falling towards a planet of mass M and radius R.As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he seesthat the planet is reallymade of two pieces a spherical shell ofnegligible thickness ofmass 2M/ 3 and a point mass M/3 at the centre. Change in the force of gravity experienced by the man is (a) 2 2 GMm 3 R (b) 0 (c) 2 1 GMm 3 R (d) 2 4 GMm 3 R 51. In a region ofonlygravitational field ofmass 'M' a particle is shifted from AtoB via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then M B A 3 2 1 C (a) W1 = W2 = W3 (b) W1 > W2 > W3 (c) W1 = W2 > W3 (d) W1 < W2 < W3 52. Four similar particles of mass m are orbiting in a circle of radiusr in thesameangular direction becauseoftheir mutual gravitational attractiveforce. Velocityofa particle isgiven by m m m m r (a) 1/ 2 GM 1 2 2 r 4 é ù æ ö + ê ú ç ÷ è ø ê ú ë û (b) 3 GM r (c) ( ) GM 1 2 2 r + (d) 1/ 2 1 GM 1 2 2 r 2 é ù æ ö + ê ú ç ÷ è ø ê ú ë û 53. The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is (a) s 2 m 1 M r 100 M r ´ (b) 2 s 2 m 1 M r 100 M r æ ö ´ ç ÷ è ø (c) 2 1 m 2 s r M 2 100 r M æ ö ´ ç ÷ è ø (d) 2 s 1 2 m M r 100 r M æ ö ´ ç ÷ è ø 54. A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit ofradius r + Dr (Dr << r) with time periodT + D T then, (a) T 3 r T 2 r D D = (b) T 2 r T 3 r D D = (c) T r T r D D = (d) T r T r D D = - 55. A cavity of radius R/2 is made inside a solid sphere of radius R. The centre ofthe cavity is located at a distance R/ 2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is – (opposite to the centre of gravity) [Here g = GM/R², where M is the mass of the sphere] (a) mg 2 (b) 3mg 8 (c) mg 16 (d) None of these 56. A spherical uniform planet is rotating about its axis. The velocityof a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 ofg at poles. The escape velocityof a particle on the pole of planet in terms of V is (a) Ve=2V (b) Ve=V (c) Ve=V/2 (d) e V 3V = 57. The escape velocity from a planet is ve. A tunnel is dug along a diameter of the planet and a small bodyis dropped into it at the surface. When the body reaches the centre of the planet, its speed will be (a) ve (b) e v / 2 (c) ve/2 (d) zero
- 232. 228 PHYSICS 58. A(nonrotating) star collapses onto itself from an initial radius Ri with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration ag on the surface of the star as a function of the radius of the star during the collapse Ri ag a b d R c (a) a (b) b (c) c (d) d 59. The earth is assumed tobe sphere ofradius R.A platform is arranged at a height Rfrom thesurface ofEarth. The escape velocity of a body from this platform is kv, where v is its escape velocity from the surface of the earth. The value of k is (a) 1 2 (b) 1 3 (c) 1 2 (d) 2 60. Four equal masses (each ofmass M)areplacedat thecorners of a square of side a. The escape velocity of a body from the centre O of the square is (a) a GM 2 4 (b) a GM 2 8 (c) a GM 4 (d) a GM 2 4 61. A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum and minimum distance ofthe planet from the sun are r1 and r2 respectively. The time period of planet is proportional to (a) 5 / 2 1 r (b) 2 / 3 2 1 r r ÷ ø ö ç è æ + (c) 2 / 3 2 1 r – r ÷ ø ö ç è æ (d) 2 / 3 r 62. The change in potential energy, when a body of mass m is raised toa height nRfrom the earth’s surface is (R = radius of earth) (a) ÷ ÷ ø ö ç ç è æ 1 – n n mgR (b) nmgR (c) ÷ ÷ ø ö ç ç è æ +1 n n mgR 2 2 (d) ÷ ø ö ç è æ +1 n n mgR 63. Asatellite islaunched intoa circular orbit ofradius Raround the earth. A second satellite is launched into an orbit of radius 1.01 R. The period of second satellite is larger than the first one byapproximately (a) 0.5% (b) 1.0% (c) 1.5% (d) 3.0% 64. If gE and gM are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio electronic charge on the moon to be electronic charge on the earth (a) M E g / g (b) 1 (c) 0 (d) E M g / g 65. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be (a) ½ R (b) 2R (c) 4R (d) 1/4R 66. Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravityon the surface of earth is g and that on the surface of the new planet is g’, then (a) g’ = g/9 (b) g’= 27g (c) g’=9g (d) g’=3g 67. For a satellite moving in an orbit around the earth, theratio of kinetic energy to potential energy is (a) 1 2 (b) 1 2 (c) 2 (d) 2 68. The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then : A B m v S C D (a) t1 = 4t2 (b) t1 = 2t2 (c) t1 = t2 (d) t1 > t2 69. The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satelliteA is 3 V, then the speed of satellite B will be (a) 3V/4 (b) 6V (c) 12V (d) 3V/2
- 233. 229 Gravitation 70. A particleof mass M is situated at the centre of a spherical shell of samemass and radius a. The gravitational potential at a point situated at 2 a distance from the centre, will be (a) 3GM a - (b) 2GM a - (c) GM a - (d) 4GM a - 71. A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance of r2. If v1 and v2 are the linear velocities at these points respectively, then the ratio 1 2 v v is (a) (r1/r2)2 (b) r2/r1 (c) (r2/r1)2 (d) r1/r2 72. A particle of mass m is thrown upwards from the surface of the earth, with a velocityu. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u sothat the particle does not return back to earth, is (a) 2GM R (b) 2 2GM R (c) 2 2gR (d) 2 2GM R 73. Which one of the following graphs represents correctly the variation of the gravitational field intensity (I) with the distance (r) from the centre of a spherical shell of mass M and radius a ? (a) I r = a r (b) I r = a r (c) I r = a r (d) I r = a r Directions for Qs. (74 to 75) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethecorrectanswer (ONLYONE option iscorrect)fromthe following (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true,Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 74. Statement -1 : For the planets orbiting around the sun, angular speed, linear speed and K.E. changes with time, but angular momentum remains constant. Statement -2 : No torque is acting on the rotating planet. So its angular momentum is constant. 75. Statement -1 : Gravitational potentialismaximum at infinity. Statement -2 : Gravitational potential isthe amount ofwork done to shift a unit mass from infinity to a given point in gravitational attraction force field. Exemplar Questions 1. Theearth is an approximatesphere. Ifthe interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity (a) will be directed towards the centre but not the same everywhere (b) will have the same value everywhere but not directed towards the centre (c) will be same everywherein magnitudedirectedtowards the centre (d) cannot be zero at any point 2. As observed from the earth, the sun appears to move in an approximatecircular orbit. For themotion ofanother planet like mercury as observed from the earth, this would (a) be similarlytrue (b) not be true because the force between the earth and mercury is not inverse square law (c) not be true because the major gravitational force on mercury is due to the sun (d) not be true because mercury is influenced by forces other than gravitational force 3. Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the CM (centre of mass) causing translation and a net torque at the CM causing rotation around an axis through the CM. For the earth-sun system (approximating the earth as a uniform densitysphere). (a) the torque is zero (b) the torque causes the earth to spin (c) the rigid body result is not applicable since the earth is not even approximately a rigid body (d) the torque causes the earth to move around the sun
- 234. 230 PHYSICS 4. Satellitesorbitting the earth have finite life and sometimes debris of satellites fall to the earth. This is because (a) the solar cells and batteries in satellites run out (b) the laws of gravitation predict a trajectory spiralling inwards (c) of viscous forces causing the speed of satellite and hence height to gradually decrease (d) of collisions with other satellites 5. Both the earth and the moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon (a) will be elliptical (b) will not be strictly elliptical because the total gravitational force on it is not central (c) is not elliptical but will necessarily be a closed curve (d) deviates considerably from being elliptical due to influence of planets other than the earth 6. In our solar system, the inter-planetary region has chunks ofmatter (much smaller in size compared toplanets) called asteroids. They (a) will not move around the sun, since they have very small masses compared to the sun (b) will move in an irregular way because of their small masses and will drift away into outer space (c) will move around the sun in closed orbits but not obey Kepler's laws (d) will move in orbits likeplanets and obeyKepler's laws 7. Choose the wrong option. (a) Inertial mass is a measure of difficulty of accelerating a bodyby an external force whereas the gravitational mass is relevant in determining the gravitational force on it byan external mass (b) That thegravitational mass andinertial mass areequal is an experimental result (c) That the acceleration due togravityon the earth is the samefor all bodies isduetotheequalityofgravitational mass and inertial mass (d) Gravitational mass of a particlelikeproton can depend on the presence of neighbouring heavy objects but the inertial mass cannot 8. Particles ofmasses 2M, mandM arerespectivelyat pointsA, B and C with 1 ( ) 2 AB BC m = × is much-much smaller than M and at time t = 0, theyare all at rest as given in figure. At subsequent times before any collision takes place. r A B C M m 2M (a) m will remain at rest (b) m will move towards M (c) m will move towards2M (d) m will have oscillatorymotion NEET/AIPMT (2013-2017) Questions 9. Abody of mass ‘m’is taken from the earth’s surface to the height equal to twice the radius (R) ofthe earth. The change in potential energy of body will be [2013] (a) 2 3 mgR (b) 3mgR (c) 1 3 mgR (d) mg2R 10. Infinite number ofbodies, each of mass 2 kg are situated on x-axisat distances1m,2m, 4m, 8m, .....respectively, from the origin. The resulting gravitational potential due to this system at the origin will be [2013] (a) 8 3 - G (b) 4 3 - G (c) – 4 G (d) –G 11. The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. If VP and VE are escape velocities of the planet and the earth, respectively, then [NEET Kar. 2013] (a) VE = 1.5VP (b) VP = 1.5VE (c) VP = 2VE (d) VE = 3VP 12. A particle of mass ‘m’ is kept at rest at a height 3R from the surface ofearth, where ‘R’ is radius of earth and ‘M’ is mass of earth. The minimum speed with which it should be projected, sothat it does not return back, is (g is acceleration due to gravity on the surface of earth) [NEET Kar. 2013] (a) 1 2 GM R æ ö ç ÷ è ø (b) 1 2 2 GM R æ ö ç ÷ è ø (c) 1 2 4 gR æ ö ç ÷ è ø (d) 1 2 2 4 g æ ö ç ÷ è ø 13. Dependence of intensity of gravitational field (E) of earth with distance(r) from centreofearth is correctlyrepresented by: [2014] (a) E O R r (b) E O R r (c) E O R r (d) E O R r 14. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole? [2014] (a) 10– 9 m (b) 10– 6 m (c) 10– 2 m (d) 100m
- 235. 231 Gravitation 15. Twospherical bodiesof mass M and 5 M and radii R and 2 Rreleased in free spacewith initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller bodybefore collision is [2015] (a) 4.5R (b) 7.5R (c) 1.5R (d) 2.5R 16. Kepler's third law states that square of period of revolution (T)of a planet aroundthe sun, is proportional tothird power of average distance r between sun and planet i.e. T2 = Kr3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law ofgravitation force of attraction between them is F = 2 GMm , r here G is gravitational constant. The relation between G and K is described as [2015] (a) GMK= 4p2 (b) K=G (c) K = 1 G (d) GK=4p2 17. A remote - sensing satellite of earth revolves in a circular orbit at a height of0.25 × 106 m above the surface of earth. If earth's radius is 6.38 × 106 m and g = 9.8 ms–2, then the orbital speed of the satellite is: [2015 RS] (a) 8.56 km s–1 (b) 9.13 km s–1 (c) 6.67 km s–1 (d) 7.76 km s–1 18. A satellite S ismovingin an elliptical orbit around theearth. Themass ofthe satellite is verysmall compared to themass of the earth. Then, [2015 RS] (a) the total mechanical energy of S varies periodically withtime. (b) the linear momentum of S remains constant in magnitude. (c) the acceleration of S is always directed towards the centre of the earth. (d) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant. 19. The ratio of escape velocity at earth (ve) to the escape velocityat a planet (vp) whoseradius and mean densityare twice as that of earth is : (a) 1: 2 (b) 1 : 2 2 [2016] (c) 1: 4 (d) 1: 2 20. At what height from the surface of earth the gravitational potential andthe value ofg are –5.4× 107 J kg–1 and 6.0 ms–2 respectively ? Take the radius of earth as 6400 km : [2016] (a) 2600km (b) 1600km (c) 1400km (d) 2000km 21. Two astronauts are floating in gravitation free space after having lost contact with their spaceship. The two will (a) move towards each other. [2017] (b) move away from each other. (c) become stationary (d) keep floating at the same distance between them. 22. The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then [2017] (a) d = 1 km (b) d = 3 km 2 (c) d = 2 km (d) d = 1 km 2
- 236. 232 PHYSICS EXERCISE -1 1. (d) 2. (d) 3. (a) 4. (d) 5. (b) 6. (c) 7. (b) 8. (c) In planetary motion ext. τ r = 0 Þ L r = constant ( ) = ´ = = r r r r L r p mv mrv (Q q=90º) So m1d1v1 = m2d2v2 (here r = d) Þ 1 1 2 2 v d v d = d1 d2 v2 v1 sun 9. (a) At the surface of earth, the value of g = 9.8m/sec2. If we go towards the centre of earth or we go above the surface of earth, then in both the cases the value of g decreases. Hence W1=mgmine, W2=mgsea level, W3=mgmoun So W1< W2 > W3 (g at the sea level = g at the suface of earth) 10. (a) According to Gravitational Law 1 2 2 GM M F r = ……(i) Where M1 is mass of planet & m2 is the mass of any body. Now according to Newton’s second law, bodyof mass m2 feels gravitational acceleration g which is F = m2g ……(ii) Sofrom (i) &(ii), we get 1 2 GM g r = So the ratio of gravitational acceleration due to two planets is 2 3 2 1 1 2 1 2 2 2 3 2 2 1 1 2 g M r (4/3)πr ρ r g M r r (4/3)πr ρ ´ = ´ = ´ ´ 1 1 2 2 g r g r æ ö = ç ÷ è ø (both planet have same material, so density is same) 11. (c) SinceT(time period) = l 2π g for second pendulum T=2sec Nowon theplanet the valueof acceleration dueto gravityis g/4. So for the planet, the length of sec. pendulam l' is 2sec g / 4 / 4 2sec g l l' l l' = ´ Þ = 12. (b) Since T 2π g l = but inside the satellite g = 0 So T = ¥ 13. (b) We know that, e v (2g R) = e P 1 1 1 e P 2 2 2 (v ) (2g R ) (v ) (2g R ) = 1 1 2 2 g R g R æ ö æ ö = ´ ç ÷ ç ÷ è ø è ø k r = 14. (b) e o v 2v = where ve and vo are the escape velocity and orbital velocity respectively. 15. (c) KE= mgR ) gR 2 ( m 2 1 mv 2 1 2 esc 2 = = . 16. (a) vesc = gR 2 , where R is radius of the planet. Hence escape velocity is independent of m. 17. (a) The escape velocity on the earth is defined as e e e v 2g R = Where Re & ge are the radius & acceleration due to gravity of earth. Now for planet gP=2ge, RP=Re/4 So e P P P e e v v 2g R 2 2g R /4 2 = = ´ ´ = 18. (c) Thepotential energyfor a conservativeforceisdefined as dr dU F - = or r U F.dr ¥ = -ò r r ……(i) or r 1 2 1 2 r 2 GM M GM M U dr r r ¥ - = = ò ……(ii) (Q U¥ = 0) If we bring the mass from the infinity to the centre of earth, then we obtain work, ‘so it has negative (gravitational force do work on the object) sign & potential energy decreases. But if we bring the mass from the surface of earth to infinite, then we must do work against gravitational force & potential energyof the mass increases. Now in equation (i) if 1 2 5/ 2 GM M F r = instead of 1 2 2 GM M F r = then r 1 2 1 2 r 5/2 3/2 GM M GM M 2 U dr 3 r r ¥ - = = ò r 3/2 1 U r+ Þ µ Hints & Solutions
- 237. 233 Gravitation 19. (c) 2 / 3 2 2 / 3 2 1 2 1 R 16 R 4 T T R R T T ÷ ø ö ç è æ = Þ ÷ ÷ ø ö ç ç è æ = T 8 T2 = Þ 20.(a) 2 T µ (major axis)3 Þ T2 a a3 21. (d) According to 3rd law of Kepler 3 2 R T µ 3 2 KR T = Þ where K is a constant Thus 3 2 R T does not depends on radius. 22. (a) 1 2 1 1 v , v R 7R µ µ R 7R 2 1 2 1 v v 1 v v v 7 7 7 = Þ = = 23. (a) Escape velocity does not depend upon the mass of the body. 24. (b) At a height h above the surface of earth the gravitational potential energy of the particle of mass m is e h e GM m U R h = - + Where Me & Re are the mass & radius of earth respectively. In this question, since h = Re So 2 2 = - = - = e e e h R e GM m mgR U R 25. (d) Energy required to move a body of mass m from an orbit of radius 2R to 3R is GMm GMm GMm U 3R 2R 2R - D = - - = GMm GMm 3R 6R - = EXERCISE - 2 1. (c) 2. (c) According to Newton’s Gravitation Law 1 2 g 2 GM M F r = here Fg = 10–10 Newton, m1=m2=1kg, G = 6.6 ×10–11 So 11 2 1 2 10 g GM M 6.6 10 1 1 r 0.66 F 10 - - ´ ´ ´ = = = orr =0.8125metre= 81.25cm »80cm 3. (c) 2 5 2 3 ) r 1 ( 10 G ) r ( 10 G - ´ = ´ 2 2 2 1 10 r (1 r) = - r 1 r 10 r 1 10 r 1 - = Þ - = 1 11 = r km 4. (b) The gravitational potential V at a point distant ‘r’ from a bodyof mass m is equal to the amount of work done in moving a unit mass from infinity to that point. ( ) r r V V E.dr GM 1/ r 1/ µ µ - = - = - - µ ò u r r GM r - = dV As E dr - æ ö = ç ÷ è ø r (i) In the first case when Vµ = 0, r GM V 5 r - = = - unit (ii) In the second case Vµ = + 10 unit Vr – 10 = – 5 or Vr = + 5 unit 5. (c) Gravitational force supplies centripetal force 2 29 8 2 14 mv 3 10 (2 10 ) F dynes r 1.5 10 ´ ´ ´ = = ´ = 8 × 1031 dyne = 8 × 1026 N (Q 1N = 105 dyne.) 6. (c) e e e m e e e M 2G 2GM 2 81 v ; v v R R 9 4 = = = = 2/9 × 11.2 kms–1 = 2.5 kms–1 7. (d) Since gravitational acceleration on earth is defined as 2 e e e R GM g = ……(i) mass of planet is e P M M 80 = & radius e P R R 4 = So P P 2 P GM g R = ……(ii) From (i) &(ii), weget 2 e e P P e 2 e P R g M g g M 5 R = ´ = =2m/sec2 (as g=10m/sec2) 8. (c) 1 100 10 67 . 6 1 10 10 67 . 6 – V 11 11 g ´ ´ - ´ ´ = - - = –6.67 × 10–10 – 6.67 × 10–9 = – 6.67 × 10–10 × 11 = –7.3 × 10–9 J/kg
- 238. 234 PHYSICS 9. (c)2 3 T R µ Þ 3 2 1 2 2 1 R R T T ÷ ÷ ø ö ç ç è æ = ÷ ÷ ø ö ç ç è æ Þ 2 / 3 2 / 3 2 1 2 1 4 1 R R T T ÷ ø ö ç è æ = ÷ ÷ ø ö ç ç è æ = ÷ ÷ ø ö ç ç è æ Þ 8 ) 4 ( T T 2 / 3 1 2 = = Þ 2 1 T 8 T 8 5 40 = ´ = ´ = hours 10. (b) At poles, the effect of rotation is zero and also the distance from the centre of earth is least. 11. (c) 2 3 T R µ (According to Kepler’s law) 2 13 3 2 12 3 1 2 T (10 ) and T (10 ) µ µ 2 3 1 1 2 2 2 T T (10) or 10 10 T T = = . 12. (c) According to Kepler’s law of period T2 µ R3 2 3 3 1 1 2 3 3 2 2 (6 ) 8 (3 ) T R R T R R = = = 2 2 24 24 8 T ´ = 2 2 24 24 8 T ´ = = 72=36 ×2 2 6 2 T = 13. (b) Since escape velocity( ) e e v 2gR = is independent of angle of projection, so it will not change. 14. (c) Given that l2 = 1.02 l1 We know that T 2π or T g æ ö = µ ç ÷ è ø l l 2 2 1 1 1 1 T (1.02 ) 101 T æ ö = = = ç ÷ è ø l l l l Thus time period increases by 1%. 15. (b) We know that e 0 v 2 v = , where v0 is orbital velocity. K.E. in the orbit, 2 0 1 E M v 2 = K.E. to escape 2 2 e 0 1 1 E M v M (2 v ) 2 2 = = 2 0 1 M v 2 2E 2 = ´ = 16. (c) 1/ 2 3 2π (R h) T R g é ù + = ê ú ê ú ë û Here (R + h) changes from R to 4 R. Hence period of revolution changes from T to (43)1/2 T = 8 T. 17. (b) h 2h g g 1 R æ ö = - ç ÷ è ø x x g g 1 R æ ö = - ç ÷ è ø Given that, h x g g = x= 2h 18. (d) 2 2 ) x R ( GmM ) x R ( mv + = + also 2 R GM g = 2 2 2 2 mv GM R m (R x) R (R x) æ ö = ç ÷ è ø + + 2 2 2 ) x R ( R mg ) x R ( mv + = + x R gR v 2 2 + = Þ 2 / 1 2 x R gR v ÷ ÷ ø ö ç ç è æ + = 19. (c) 2 2 ) x R ( GmM x R mv + = + x = height ofsatellite from earth surface m = mass of satellite 2 GM GM v or v (R x) R x Þ = = + + 2 (R x) 2 (R x) T v GM R x p + p + = = + which is independent of mass of satellite 20. (c) 2 n MR KR F w = = - ) 1 n ( 2 KR + - = w Þ or 2 ) 1 n ( R ' K + - = w [where K' = K1/2, a constant] 2 ) 1 n ( R T 2 + - a p (n 1) 2 T R + a 21. (d) According to Newton gravitational force mg R m GM 2 e e = & g m ) h R ( m GM 2 e e ¢ = + Þ g' 2 e 2 e 2 e ) h R ( gR ) R / h 1 ( 1 g + = + =
- 239. 235 Gravitation where Me ismass ofearth, G is gravitational constant, Re is radius & earth, h is height of satellite above the surfaceof earth, g is value at the surface of earth, g' is value at height h above the surface of earth. so g m ) h R ( mgR ) h R ( mv 2 e 2 e e 2 ¢ = + = + h R gR v e 2 e + = Þ h Re +Re O Earth earth surface satellite orbit v 22. (c) 23. (d) Both decreases but variation are different. 24. (b) There is no gravitational field in the shell. 25. (c) Thegravitational field duetothe ring at a distance r 3 is given by 2 2 / 3 2 2 r 8 Gm 3 E ] ) r 3 ( r [ ) r 3 ( Gm E = Þ + = Attractive force = 2 r 8 GmM 3 26. (a) For uniform gravitational field Eg = r V - 2 ms 10 1 20 2 - = - - = Now, W = mgh = 1 5 4 2J 10 ´ ´ = 27. (b) 2 0 2 2 e 1 mv 2 1 E , mv 2 1 E = = 0 e 2 0 e 2 1 v 2 v 2 v v E E = = ÷ ÷ ø ö ç ç è æ = Q 28. (d) g R µ r 29. (c) 30. (b) As buoyant force involves ‘g’ in it. 31. (d) Work done = Uf – Ui mgR 3 2 R GmM 3 2 R Gmm R R 2 GMm = = ÷ ø ö ç è æ - - + - = 32. (b) 33. (a) 34. (d) The gravitational force due to the whole sphere at A point is e o 1 2 GM m F (2R) = , where m0 is the assumed rest mass at point A. In the second case, when we made a cavity of radius (R/2), then gravitational force at point Ais 2 o e 2 ) 2 / R R ( m GM F + = F2/F1= 1/9 35. (b) R 2 R0 0 GMm 1 1 P.E. dr GMm R R r é ù = = - - ê ú ë û ò TheK.E. acquired bythe bodyat the surface 2 1 m v 2 = 2 0 1 1 1 mv GMm 2 R R é ù = - - ê ú ë û 0 1 1 v 2G M R R æ ö = - ç ÷ è ø 36. (c) Applying the properties of ellipse, we have 1 2 1 2 1 2 r r 2 1 1 R r r r r + = + = R Instant position of satellite major axis r2 r1 Sun 1 2 1 2 2r r R r r = + 37. (a) GM v r æ ö = ç ÷ è ø where r is radius of the orbit of the satellite. herer =Re + h = e e e R 3 R R 2 2 + = So, 0 e 2GM 2 v v 3R 3 = = , where v0 is the orbital velocity of the satellite, which is moving in circular orbit of radius, r = Re 38. (d) 2 2 1 1 2 2 1 1 2 2 v r v r or r ω r ω = = (QL=mrv= constant) or 2 2 min max r ω r ω¢ = 2 2 min max ω (r / r )ω ¢ = 39. (c) Distance between the surface of the spherical bodies = 12R – R– 2R = 9R Force Mass µ Acceleration Mass µ Distance Acceleration µ Þ 1 2 a M 1 a 5M 5 = = Þ 1 2 2 1 S 5 S 5 1 S S = Þ =
- 240. 236 PHYSICS S1 +S2 = 9 Þ 6S1 = 9Þ 5 . 1 6 9 S1 = = S2 = 1.5×5 = 7.5 Note: Maximum distance will be travelled bysmaller bodies due to the greater acceleration caused by the same gravitational force 40. (c) 2 2 ) R 99 . 0 ( GM ' g ; R GM g = = 2 2 g' R g' g g 0.99R æ ö = Þ > ç ÷ è ø 41. (b) Acceleration due to gravity at lattitude’ l ’ is given by l w - = l 2 2 e e cos R g g At equator, l = 90° Þ l cos = cos 90° = 0 or l g = e g = g (as given in question) At 30°, 2 2 2 30 R 4 3 g 30 cos R g g w - = w - = or, 2 30 R 4 3 g g w = - 42. (b) The orbital speed of satellite is independent of mass of satellite, so ball will behave as a satellite & will continue to move with same speed in original orbit. 43. (a) The total momentum will be zero and hence velocity will be zero just after collisiion. The pull of earth will makeit fall down. 44. (d) Weight of liquid displaced = mg. 45. (a) The force of attraction between sphere and shaded position 2 x l ÷ ø ö ç è æ = dx m GM dF x r l m M r + l 2 2 2 1 2 1 1 2 1 1 ( ) + + + + - + - + + - = = é ù = = ê ú - + ë û é ù é ù = - = - = ë û ê ú + ë û ò ò ò r l r l r r r l r l r r r l r l r r GMm GMm F dx dx lx l x GMm GMm x x dx l l GMm GMm GMm x l l x r r l 46. (a) 2 4 3 1 since 2 æ ö æ ö = - = = + ç ÷ ç ÷ è ø è ø h g R g g h R h R h Force on the satellite = mg 9 4 mgh = N 889 10 200 9 4 » ´ ´ = 47. (c) 2 R 1 g µ so we will not get a straight line. AlsoF = 0 at a point where Force due to Earth = Force due to mars 48. (c) Applying conservation of energy principle, we get r GMm R GMm v mk 2 1 2 e 2 - = - r GMm R GMm R GM 2 mk 2 1 2 - = - Þ R k R 1 r 1 r 1 R 1 R k 2 2 - = Þ - = - Þ 2 2 k 1 R r ) k 1 ( R 1 r 1 - = Þ - = Þ 49. (b) Loss in potential energy= Gain in kineticenergy 2 mv 2 1 R GMm 2 3 R GMm = ÷ ø ö ç è æ - - - gR R GM v mv 2 1 R 2 GMm 2 = = Þ = Þ 50. (a) Change in force of gravity = 2 2 M G m GMm 3 R R - (onlydue tomass M/3 due toshell gravitational field is zero(inside the shell)) = 2 2GMm 3R 51. (a) 52. (a) Centripetal force = net gravitational force 2 0 1 mv 2Fcos45 F r = ° + = 2 2 2 2 2GM 1 Gm 2 ( 2r) 4r + r V0 F F F1 2r 45° 2 2 0 2 mv Gm [2 2 1] r 4r = + 1/2 GM(2 2 1 4r æ ö + Þ ç ÷ è ø
- 241. 237 Gravitation 53. (c) Duringtotal eclipse : Total attraction due to sun and moon, s e m e 1 2 2 1 2 GM M GM M F r r = + When moon goes on the opposite side of earth. Effective force of attraction, s e m e 2 2 2 1 2 GM M GM M F r r = - Change in force, m e 1 2 2 2 2GM M F F F r D = - = Change in acceleration ofearth Da = m 2 e 2 2GM F M r D = Average force on earth, Fav = av s 2 e 1 F GM M r = %age change in acceleration = 2 m 1 2 av s 2 2GM r a 100 100 a GM r D ´ = ´ ´ 2 1 m 2 s r M 2 100 r M æ ö = ´ ç ÷ è ø 54. (a) Since, 3 2 kr T = Differentiating the above equation Þ T r 2 3 T r D D = Þ T 3 r T 2 r D D = 55. (b) Gravitational field at mass m due to full solid sphere 1 0 0 0 r R 1 E ........ 3 6 4 G r r é ù = = e = ê ú e e p ë û r r Gravitational field at mass m due to cavity(–r) 3 3 2 2 2 0 0 ( ) (R / 2) r a E ........ using E 3 R 3 r é ù -r r = = ê ú e e ê ú ë û r r = 3 2 0 0 ( )R R 24 24 R -r -r - = - e e R m R/2 CM Net gravitational field 1 2 0 0 R R E E E 6 24 r r = + = - e e r r r 0 R 8 r = e Net force on m 0 m R F mE 8 r ® = = e r Here, 0 3 M 1 & 4 G (4 /3) R r = e = p p then 3mg F 8 = 56. (a) V= wR 2 0 g g R = - w [g = at equator, g0 = at poles] 2 0 0 g g R 2 = - w ; 2 0 g R 2 w = ; 2 0 g R V 2 = 2 e 0 V 2g R 4V 2V = = = 57. (b) 2 e 1 GMm mv 2 R = ; e GMm v 2gR R = = In tunnel bodywill perform SHM at centre Vmax = Aw (see chapter on SHM) = e v R2 gR 2 R / g 2 p = = p 58. (b) 2 1 g R µ R decreasing g increase hence, curve b represents correct variation. 59. (a) Here, 2GM v R = and 2GM kv R R = + . Solving 1 k 2 = 60. (b) Potential energy of particle at the centre of square GMm 4 a 2 æ ö ç ÷ = - ç ÷ ç ÷ è ø 2 2 GMm 1 8 2 GM 4 mv 0 v a 2 a 2 æ ö ç ÷ - + = Þ = ç ÷ è ø 61. (b) T2 a r3, where r = mean radius = 2 r r 2 1 + 62. (d) ÷ ø ö ç è æ - - + - = - = D R GMm h nR GMm U U U i f mgR 1 n n R GMm . 1 n n + = + = 63. (c) GM ) h R ( 2 T 3 + p = GM ) R 01 . 1 ( 2 T , GM R 2 T 3 2 3 1 p = p = % 5 . 1 100 T T T 1 1 2 = ´ - 64. (b) Electronic charge does not depend on acceleration due to gravity as it is a universal constant. So, electronic charge on earth = electronic charge on moon Required ratio = 1.
- 242. 238 PHYSICS 65. (a) 2 R M G g= also 3 R 3 4 d M p ´ = R d 3 4 g p = at the surface of planet R ) d 2 ( 3 4 gp ¢ p = , R ) d ( 3 4 ge p = ge = gp Þ dR = 2d R' Þ R' = R/2 66. (d) We know that g = GM R2 = 2 3 R R 3 4 G r ÷ ø ö ç è æ p = 4 3 p r GR g g R R R R ' ' = = = 3 3 = g g ' 3 67. (a) K.E. of satellite moving in an orbit around the earth is K = r 2 GMm r GM m 2 1 mv 2 1 2 2 = ÷ ÷ ø ö ç ç è æ = P.E. of satellite and earth system is U GMm r = Þ K U GMm r GMm r = = 2 1 2 68. (b) According to Kepler’s law, the areal velocity of a planet around the sun always remains constant. SCD : A1– t1 (areal velocity constant) SAB : A2 – t2 1 2 1 2 A A , t t = t1 = t2 . 1 2 A , A (given A1 = 2A2) = t2 . 2 2 2A A t1 = 2t2 69. (b) Orbital velocityofa satellite in a circular orbit ofradius a is given by GM v a = Þ v a 1 a Þ 2 1 v v = 1 2 a a v2 = v1 4R R = 2 v1 = 6V 70. (a) Potential at the given point = Potential at the point due to the shell + Potential due to the particle = 2 GM GM a a - - = 3GM a - 71. (b) Angular momentum is conserved L1 = L2 Þ mr1v1 = mr2v2 Þ r1v1 = r2v2 1 2 2 1 v r v r Þ = 72. (a) The velocity u should be equal to the escape velocity. That is, u = 2gR But g = 2 GM R 2 GM u 2· ·R R = 2GM R Þ 73. (d) Intensitywill be zero inside the spherical shell. I = 0 upto r = a and 2 1 I r µ when r > a 74. (b) 75. (b) EXERCISE - 3 Exemplar Questions 1. (d) Let the densityof earth as a sphere is uniform, then it can be treated as point mass placed at its centre then acceleration due to gravity g = 0, at the centre. But if the density of earth is considered as a sphere of non- uniform then value of 'g' will be different at different points 4 3 g GR æ ö = pr ç ÷ è ø Q . So g cannot be zero at any point. 2. (c) Force of attraction between any two objects obeys the inverse square law. As observed from the earth, the sun appears to move in an approximatecircular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law. Due torelative motion between the earth and mercury, the orbit of mercury, as observed from the earth will not be approximately circular, since the major gravitational force on mercury is due to the sun is very large than due to earth and due to the relative motion to sun and earth with mercury. 3. (a) As we know that, the torque on earth due to gravitational attractive force on earth is zero. As the earth is revolving around the sun in a circular motion due to gravitational attraction. The force of attraction will be of radial nature i.e., angle between position vector r and force F is also, zero. So, torque sin0 0 r F rF = t = ´ = ° = 4. (c) As the total (P.E.) ofthe earth satellite orbiting in orbit is negative 2 GM r - æ ö ç ÷ è ø , where r is radius of the satellite and M is mass of the earth. Due to the viscous force acting on satellite, energy decreases continuously and radius of the orbit or height decreases gradually. 5. (b) The major force acting on moon is due togravitational force of attraction by sun and earth and moon is not always in line of joining sun and earth. As observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon. So these two force have different lines of action and it will not be strictly elliptical because total force on the moon is not central.
- 243. 239 Gravitation 6. (d) Asteroidsare also being acted upon by central gravitational forces, hence Asteroid will move in circular orbits like planets and obey Kepler's laws. 7. (d) Gravitational mass of proton is equivalent to its inertial mass and is independent of presence of neighbouring heavy objects so verifies the option (d). 8. (c) Force of Gravitation, 2 g GMm F r = Let AB = r So, force on B due to A 2 (2 ) ( ) BA G Mm F AB = = towards BA. 2 (2 ) 2 g G Mm F r = = and force on B due to C 2 ( ) BC GMm F BC = = towards BC As, (BC) = 2AB 2 2 2 4 (2 ) 4( ) 4 g BC F GMm GMm GMm F AB AB r Þ = = = = As FBA > FBC, hence, m will move towards (BA) i.e., (2M). NEET/AIPMT (2013-2017) Questions 9. (a) Initial P .E.,Ui = GMm R - , Final P.E., Uf = GMm 3R - [QR' = R+ 2R= 3R] Change in potential energy, DU= GMm 3R - + GMm R = GMm R 1 1 3 æ ö - ç ÷ è ø = 2 3 GMm R = 2 3 mgR GMm mgR R æ ö = ç ÷ è ø Q ALTERNATE:DU= mgh h 1 R + By placing the value of h = 2R we get DU= 2 3 mgR. 10. (c) m Gravitational potential V= Gm r - V0 = – G 2 1 ´ – G 2 2 ´ – G 2 4 ´ – G 2 8 ´ – 2G 1 1 1 1 .... 2 4 8 é ù + + + + ¥ ê ú ë û = – 2G × 1 1 1 2 - = – 2G × 1 1 2 = – 4 G. . 11. (c) Escape velocity, 8 3 e V R GP = p Þ Ve µ R Þ 2 P P E E V R V R = = Þ VP = 2VE. 12. (b) As we know, the minimum speed with which a bodyis projected so that it does not return back is called escape speed. 2 2 2 4 = = = + e GM GM GM V r R h R 1 2 ( 3 ) 2 æ ö = = ç ÷ è ø Q GM h R R 13. (b) First when (r < R) E µ r and then when r > R 2 1 E r µ . Hence graph (b) correctly dipicts. 14. (c) From question, Escape velocity = 2GM R = c = speed of light Þ R= 2 2GM c = 11 24 8 2 2 6.6 10 5.98 10 m (3 10 ) - ´ ´ ´ ´ ´ = 10– 2 m 15. (b) 12R–3R=9R 12R M R 2R 5M Before collision M 5M R 2R At the time of collision Let the distance moved by spherical body of mass M is x1 and by spherical body of mass 5m is x2 As their C.M. will remain stationary So, (M) (x1) = (5M)(x2)or, x1 =5x2 and for tauching x1 + x2 = 9R So, x1 =7.5R 16. (a) As we know, orbital speed, orb GM V r =
- 244. 240 PHYSICS Time periodT = orb 2 r 2 r r v GM p p = Squarring both sides, T2 = 2 2 3 2 r r 4 . r GM GM æ ö p p = ç ÷ è ø Þ 2 2 3 T 4 K GM r p = = Þ GMK=4p2. 17. (d) Given: Height of the satellite from the earth's surface h =0.25 ×106m Radius ofthe earth R= 6.38 × 106m Acceleration due to gravity g = 9.8 m/s2 Orbital velocity, V0 = ? V0 = 2 2 GM GM R . (R h) (R h) R = + + = 6 9.8 6.38 6.38 6.63 10 ´ ´ ´ =7.76km/s 2 GM g R é ù = ê ú ë û Q 18. (c) The gravitational force on the satellite will be aiming towards the centre of the earth so acceleration of the satellite will also be aiming towards the centre of the earth. 19. (b) As we know, escape velocity, Ve= 3 2GM 2G 4 · R R R R 3 æ ö = p r µ r ç ÷ è ø e e e p p p V R V R r = r Þ e e e p e e V R V 2R 2 r = r Ratio e p V V =1 : 2 2 20. (a) As we know, gravitational potential (v) and acceleration due to gravity (g) with height V= GM R h - + =–5.4 ×107 …(1) and g = ( )2 GM 6 R h = + …(2) Dividing (1) by(2) ( ) 7 2 GM 5.4 10 R h GM 6 R h - - ´ + = + Þ ( ) 7 5.4 10 6 R h ´ = + Þ R+ h = 9000km so, h = 2600 km 21. (a) Both the astronauts are in the condition of weightlessness. Gravitational force between them pulls towards each other. Hence Astronauts move towards each other under mutual gravitional force. 22. (c) Above earth surface Below earth surface gh = e 2h g 1 R æ ö - ç ÷ è ø gd = e d g 1 R æ ö - ç ÷ è ø According to question, gh = gd e e 2h d g 1 g 1 R R æ ö æ ö - = - ç ÷ ç ÷ è ø è ø Clearly, d = 2h = 2 km
- 245. INTERATOMIC AND INTERMOLECULARFORCE The force between atoms of an element is called interatomic force. The force between molecules of a compound (or element) is called intermolecular force. These forces are electrical in nature. Depending on the distance between the atoms, this force may be attractive or repulsive in nature. These forces are responsible for the definite size or shape of a solid. Detailed calculations as well as deductionsfrom experiment show that the interaction between any isolated pairs of atoms or molecules may be represented by a curve that shows how the potential energy varies with separation between them as shown in the figure. This curvedescribes the interatomic potential. The force between the atoms can be found from the potential energy by using the relation = - dU F dr The resulting interatomic force curve is shown in figure. Force is along the line joining the atoms or molecules, and is shown negative for attraction and positive for repulsion. O F(R) R0 R We see that as the distance decreases, the attractive force first increases and then decreases to zero at a separation where the potential energyis minimum. For smaller distance force is repulsive, because at these distance the negative charge distribution associated with one atom begins to over lap with that associated with the neighboring atom. The force of attraction between molecules may be written as 7 = - a a F r The negative sign shows that Fa is a force of attraction and ‘a’ is a constant which depends on the kind of attractive force between the molecules and the structure of the molecules. The force of attraction results from the creation of induced dipole moment in one molecule bythe neighbouring molecule. When themolecules are brought closer thereisa force ofrepulsion between them. It can be shown that the repulsive force is 9 = + r b F r , where b is a constant like a. The force of repulsion varies very rapidly. Fr is inverselyproportional to the ninth power ofdistance between the molecules. The resultant force acting on the molecules is 7 9 a b F r r = - + The distance between the molecules decides the sign of the resultant force. ELASTICITY The property of the body by virtue of which it tends to regain its original shape and size after removing the deforming force is called elasticity. If the body regains its original shape and size completely, after the removal of deforming forces, then the body is said to be perfectly elastic. · The property of the body by virtue of which it tends to retain its deformed state after removing the deforming force is called plasticity. If the body does not have any tendencyto recover its original shape and size, it is called perfectly plastic. STRESS AND STRAIN Stress : When a deforming force is applied toa body, an internal restoring force comes into play. The restoring force per unit area is called stress. = Restoring force Stress Area 9 Mechanical Properties of Solids
- 246. 242 PHYSICS Its S.I.Unit is Nm-2 Stress is a tensor as its value changes when direction changes. Types of Stress : Longitudinal stress or tensile stress, volumetric stress and tangential stress are the types of stress. Strain : It is defined as the ratio of the change in shape or size to the original shape or size of the body. = Change in dimension Strain Original dimension Strain has no units or dimensions. Types of Strain : Longitudinal strain : It is defined as the ratio of the change in length to the original length. Longitudinal strain = Dl l Volume strain : It is the ratio of the change in volume to the original volume. Volume strain = DV V Shearing strain : It is the angular deformation produced in a body. Shearing strain (q) = Dx h HOOKE’S LAW It is the fundamental law of elasticity given byRobert Hooke in 1679. It states that “the stress is directly proportional to strain provided the strain is small “. i.e. Stress µ strain constant Þ = = stress E strain This proportionality constant is called modulus of elasticity (name given byThomasYoung) or coefficient of elasticity (E). Since stress has same dimensions as that of pressure and strain is dimensionless. So the dimensions of E is same as, that of stress or pressure i.e. [ML–1T–2] The modulus of elasticity depends on the material and on the nature of deformation. There are three type of deformations and therefore three types of modulus of elasticity. (i) Young’s modulus (Y) : It measuresthe resistance ofa solid to elongation. (ii) Shear modulus ) (h or modulus of rigidity : It measures the resistance to motion of the plane of a solid sliding part on each other. (iii) Bulk modulus (B) : It measure the resistance that solid or liquid offer to their volume change. The stress under which the system breaks is called breaking stress. (i) Young's modulus (Y) : Let us consider a long bar (shown in fig.) of cross-sectional area A and length lo, which is clamped at one end. When we apply external force Fl longitudinally along the bar, internal forces in bar resist distortion, but bar attains equilibrium in which its length is greater and in which external force is exactlybalanced by internal forces. The bar is said to be stressed in this condition. l0 Fl Fixed Dl Now if the solid bar obey the Hooke’ law, the Young's modulus, Y is defined as Tensile stress = = D l l l / / F A Y Tensile strain Where Dl is change in the length of bar, when we apply Fl. (ii) Shear modulus (h) : Shear modulus or modulus of rigidityh is h = = D / / t F A Shearing stress Shearing strain x h Ft h Dx q Fixed face (a) As we see, there is no change in volume under this deformation, but shape changes. (b ) q » q = D tan h x (see the figure), where q is shear angle. (iii) Bulk modulus (B) : The Bulk modulus B is defined as -D = = = D D / / / n F A Volume stress P B Volume stain V V V V Fn V V– V D Negative sign comes to make B positive, because with the increase of pressure, the volume of body decreases or vice versa. The reciprocal of the Bulk modulus is called compressibility of material i.e., Compressibility = 1/B.
- 247. 243 Mechanical Properties ofSolids Poisson's Ratio (s) : Lateral strain/longitudinal strain = Poisson's ratio (s). The theoretical value of s lies between –1 and 0.5 and practical value of s lies between 0 and 0.5. STRESS-STRAIN CURVE Plastic region E (Fracture point) B A C D Breaking strength Proportional limit Elastic limit Stress Strain OO¢ (i) Proportional limit : The limitin which Hooke’s lawisvalid i.e, stressisdirectlyproportional tostrain is called proportion limit. Stress µ strain (ii) Elastic limit : It is a maximum stress upto which the body completely recovers its original state after the removal of the deforming forces. (iii) Yield point : The point beyond elastic limit, at which the length of wire starts increasing without increasing stress, is defined as the yield point. (iv) Breaking point : The position when the strain becomes so large that the wire breaks down at last, is called breaking point. At this position the stress acting in that wire is called breaking stress and strain is called breaking strain. • Breaking stress is also known as the tensile strength. • Metalswith small plastic deformation are calledbrittle. • Metalswith largeplasticdeformation are called ductile. Elastic fatigue : This is the phenomenon ofa delayin recovering the original configuration by a body, if it had been subjected to stress for a longer time the body looses the property of elasticity temporarily. Elastic relaxation time : It is the time delay in regaining the original shapeafter removal ofdeformingforces. Elasticrelaxation time for gold, silver and phosphor bronze is negligible. Elastic Hysteresis When the stress applied on a body, is decreased to zero, the strain will not be reduced to zero immediately. For some substances (e.g.-vulcanized rubber), the strain lags behind the stress. This lagging of strain behind stress is called elastic hysteresis. Strain Stress The stress-strain graph for increasing and decreasing load encloses a loop, as shown in figure. The area of the loop gives the energy dissipated during its deformation. Keep in Memory 1. Thermal stress = , T Y A F aD = where a is the coefficient of linear expansion and DT is the change in temperature. 2. (i) The modulus of rigidity(h) for liquids is zero. (ii) For a given tensile force, the increase in length is inverselyproportional to square of its diameter. (iii) The pressure required to stop volume expansion of a piece of metal is P=BgdT, where g = coefficient of volume expansion = 3a (iv) To compare elasticities of different materials, their identical small balls are made and they are dropped from same height on a hard floor. The ball which rises maximum after striking the floor, is most elastic. The order of elasticity of different materials on this basis is as follows : Yivory >Ysteel >Yrubber >Yclay (v) For the construction of rails, bridges, girders and machines, materials with high Young's modulus are used so that theymaynot get permanentlydeformed. For a spring, F = . YA L kx L D = Hence spring constant k = L YA . Here DL = x. 3. When equal force is applied on identical wires of different materials then the wire in which minimum elongation is produced is moreelastic. For the sameload, more elongation is produced in rubber than in steel wire, hence steel is more elastic than rubber. Relation betweenY, B, h and s (i) Y=3B(1 –2s) (ii) Y= 2h(1+ s) (iii) 3 – 2 6 2 B B h s = + h (iv) 9 3 1 Y B = + h Energy storedper unitvolume in a strained body Energy per unit volume = 2 1 stress × strain = 2 1 modulus of elasticity × (strain)2 = 2 1 (stress)2 /modulus of elasticity. . Work done in stretching a wire or work done per unit volume = ´ ´ = ´ ´ 1 1 2 2 stress strain load extension KEEPI
- 248. 244 PHYSICS Torsional rigidityof a cylinder : (i) The torsional rigidityC of a cylinder is given by, 4 2 R C p h = l where h= modulus of rigidity of material of cylinder, , R = radius of cylinder, l = length of cylinder (ii) Restoring couple, 4 2 R C p h f t = f = l (iii) Work done in twisting the cylinder through an angle f , 4 2 4 R W p h f = l joule = 2 1 2 C f Cantilever : A beam fixed at one end and loaded at the other end is called a cantilever. (i) The depression yat a distance x from the fixed end is (when the weight of cantilever is ineffective) ÷ ÷ ø ö ç ç è æ - = 6 x 2 x I Y mg y 3 2 l where mg = load applied, l = length of cantilever, I = geometrical moment of inertia of its cross section. (ii) Maximum depression at free end of cantilever, I Y 3 mg y 3 max l = d = (a) For rectangular beam, 12 bd I 3 = (b) For a circular cross section beam of radius r, 4 r I 4 p = (iv) Depression produced in a beam supported at two ends and loaded at the middle, I Y 48 mg 3 l = d For rectangular beam, 3 3 4 mgl Ybd d = and for circular beam, Y r 12 mgl 4 3 p = d Keep in Memory 1. Wound spring possess elastic potential energy. 2. A material which can be drawn into wires is called ductile and a material which can be hammered into sheet is called malleable. 3. Ductility, brittleness, malleability, etc., are not elastic properties. 4. The substance which breaks just beyond the elastic limit is called brittle. 5. (i) Breaking force of a wire = breaking stress × area of cross section (ii) Breaking stress does not depend on the length of wire (iii) Breaking stress depends on the material of wire. Example1. A uniform rod of mass m, length L, area of cross section A and Young’s modulus Y hangs from a ceiling. Its elongation under its own weight will be (a) 2mg L AY (b) mg L AY (c) mg L 2AY (d) zero Solution (c) Mass of section BC of wire = ) x L ( L m - ; Tension at B, g ) x L ( L m T - = (L–x) A B C x Elongation of element dx at B, dx Y A L g ) x L ( m Y dx A T d - = = l Total elongation ò ò = - = = L 0 A Y 2 L mg dx ) x L ( Y A L mg dl Example2. If the potential energy of the molecule is given by U = 6 12 A B r r - . Then at equilibrium position its potential energy is equal to (a) –A2/4 B (b) A2/4 B (c) 2 A/B (d) A/2 B Solution : (b) ú û ù ê ë é - - = - = 12 6 r B r A dr d dr dU F = ú û ù ê ë é + - 13 7 r B 12 r 6 x A In equilibrium position F = 0. so, 13 7 r B 12 r A 6 = or A B 2 r6 = Potential energy at equilibrium position 2 2 2 2 A B A A A U (2B / A) 2B 4B 4B (2 B / A) = - = - = Example3. The normal density of gold is r and its bulk modulus is K. The increase in density of a lump of gold when a pressure P is applied uniformly on all sides is (a) K/r P (b) P/r K (c) r P/K (d) r K/P Solution : (a) V / V p K D = or K p V V = D ; Also V M = r and V V M ' D - = r ; 1 ' V V 1 ) V / V 1 ( 1 ) V V ( V - ÷ ø ö ç è æ D - = D - = D - = r r
- 249. 245 Mechanical Properties ofSolids Now l l l l D s - = D D D - = s r r or / r / r ; 3 3 10 1 ) 10 2 ( 5 . 0 r r - - ´ - = ´ ´ - = D Further, 0 ) 10 1 ( 2 ) 10 2 ( V V 3 3 = ´ ´ - ´ = D - - % increase in volume is 0. Example6. (a) A heavy machine is to be installed in a factory. To absorb vibrations of the machine, a block of rubber is placed between the machine and floor, which of the two rubbers A and B would you prefer to use for the purpose ? Why ? Strain O Strain O Stress Stress A B (b) Which of the two rubber materials would you choose for a car tyre ? Solution : (a) Rubber B is preferred. The area of the hysteresis loop measures the amount of heat energy dissipated by the material. The area of loop B is more than A. So B can absorb more vibrations. (b) To avoid excessive heating of car tyre, rubber A would be preferred over rubber B. K p 1 V V 1 + = ÷ ø ö ç è æ D + » or K p 1= - r r¢ or K pr = r - r¢ (QDV<<V) Example4. A wire 3 m in length and 1 mm in diameter at 30ºC and kept in a low temperature at –170ºC and is stretched by hanging a weight of 10 kg at one end. Calculate the change in the length of the wire. Given a = 1.2 ×10–5/ºC and Y = 2 × 1011 N/m2. Take g = 10 m/s2. Solution : We know that, 6 11 10 785 . 0 10 2 3 10 10 A Y L F - ´ ´ ´ ´ ´ = = l = 1.91 × 10–3 (whereA= pr2) = 0.785 × 10–6 m2 Contraction in length = –a L D T = – (1.2 × 10–5) (3) (–170 – 30) = 7.2× 10–3 m The resultant change in length = 7.2 × 10–3 – 1.91 × 10–3 = 5.29 mm Example5. A material has Poisson’s ratio 0.5. If a uniform rod of it suffers a longitudinal strain of 2 × 10–3, what is the percentage increase in volume? Solution : Here l l l l l 2 2 2 2 r r r 2 r r ) r ( V V D + D = p p D = D or r r 2 V V D + D = D l l
- 250.
- 251. 247 Mechanical Properties ofSolids 1. Elastomers are the materials which (a) are not elastic at all (b) have verysmall elastic range (c) do not obey Hooke’s law (d) None of these 2. The load versus elongation graph for four wires is shown. The thinnest wire is S R Q P Elongation Load (a) P (b) Q (c) R (d) S 3. Which ofthe following affects the elasticity of a substance? (a) hammering and annealing (b) change in temperature (c) impurity in substance (d) All of these 4. Which of the following has no dimensions ? (a) strain (b) angular velocity (c) momentum (d) angular momentum 5. Minimum and maximum valuesofPossion’sratio for a metal lies between (a) – ¥ to + ¥ (b) 0 to 1 (c) – ¥ to 1 (d) 0 to 0.5 6. In solids interatomic forces are (a) totally repulsive (b) totally attractive (c) both (a) and (b) (d) None of these 7. Which one of the following is not a unit of Young’s modulus ? (a) Nm–1 (b) Nm–2 (d) dyne cm–2 (d) mega pascal 8. Two rods A and B of the same material and length have their radii r1 and r2 respectively. When theyare rigidlyfixed at one end and twisted by the same couple applied at the other end, the ratio Angle of twist at the end of A Angle of twist at the end of B æ ö ç ÷ è ø is (a) 2 2 2 1 r / r (b) 3 2 3 1 r / r (c) 4 1 4 2 r / r (d) 4 2 4 1 r / r 9. The value of tan (90 – q) in the graph gives q Strain Stress (a) Young's modulus of elasticity (b) compressibility (c) shear strain (d) tensile strength 10. The length of a metal is l1 when the tension in it is T1 and is l2 when the tension is T2. The original length ofthe wire is (a) 2 2 1 l l + (b) 2 1 1 2 2 1 T T T T + + l l (c) 1 2 1 2 2 1 T T T T - -l l (d) 2 1 2 1T T l l 11. Uniform rod of mass m, length l , area of cross-section A A hasYoung’s modulusY. Ifit is hanged vertically, elongation under its own weight will be (a) AY 2 mgl (b) AY mg 2 l (c) AY mgl (d) l A mgY 12. Which one of the following affects the elasticity of a substance ? (a) Change in temperature (b) Hammering and annealing (c) Impurity in substance (d) All of the above 13. According to Hooke’s law of elasticity, if stress is increased, then the ratio of stress to strain (a) becomes zero (b) remains constant (c) decreases (d) increases 14. The length of an iron wire is L and area of corss-section is A. The increase in length is l on applying the force F on its two ends. Which of the statement is correct? (a) Increase in length isinverselyproportional toits length (b) Increase in length is proportional to area of cross- section (c) Increase in length is inverselyproportional to area of cross-section (d) Increase in length is proportional toYoung's modulus
- 252. 248 PHYSICS 15. Aand B are two wires. The radius of A is twice that of B. They are stretched by the same load. Then the stress on B is (a) equal to that on A (b) four times that on A (c) two times that on A (d) half that on A 16. Hooke's law defines (a) stress (b) strain (c) modulus of elasticity (d) elasticlimit 17. In case of steel wire (or a metal wire), the limit is reached when (a) the wire just break (b) the load is more than the weight of wire (c) elongation is inversely proportional to the tension (d) None of these 18. A steel ring of radius r and cross sectional area A is fitted onto a wooden disc of radius R (R > r). If the Young’s modulus of steel is Y, then the force with which the steel ring is expanded is (a) AY(R/r) (b) AY(R – r)/r (c) (Y/A)[(R– r)/r] (d) Yr/A R 19. Which of the following relation is true ? (a) 3 (1 ) Y K = - s (b) 9 Y K Y h = + h (c) (6 ) K Y s = + h (d) 05.Y - h s = h 20. For a constant hydraulic stress on an object, the fractional change in the object volume V V D æ ö ç ÷ è ø and its bulk modulus (B) are related as (a) V B V D µ (b) 1 V V B D µ (c) 2 V B V D µ (d) 2 V B V - D µ 21. The diagram shown below represents the applied forces per unit area with the corresponding change X (per unit length) produced in a thin wire of uniform cross section in the curve shown. Theregion inwhich the wirebehaveslikea liquidis (a) ab (b) bc (c) cd F O X a b c d (d) Oa 22. A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends byla and when theweight is immersed completelyin water, the extension is reduced to lw. Then the relative density of material of the weight is (a) w a / l l (b) w a a l l l - (c) ) /( w a w l l l - (d) a w / l l 23. When an elastic material with Young’s modulus Y is subjected to stretching stress S, elastic energy stored per unit volumeof the material is (a) YS/ 2 (b) S2Y/ 2 (c) S2 /2Y (d) S/ 2Y 24. The ratio of shearing stress to the corresponding shearing strain is called (a) bulk modulus (b) Young's modulus (c) modulus of rigidity (d) None of these 25. The Young’s modulus of a perfectly rigid body is (a) unity (b) zero (c) infinity (d) some finite non-zero constant 1. Two wires of same material and length but cross-sections in the ratio 1 : 2 are used to suspend the same loads. The extensions in them will be in the ratio (a) 1: 2 (b) 2: 1 (c) 4: 1 (d) 1: 4 2. Abodyof mass 10 kg is attached to a wire ofradius 3cm. It’s breakingstressis4.8× 107 Nm–2, thearea ofcross-section of thewireis10–6 m2.Whatisthemaximumangularvelocitywith which it can be rotated in the horizontal circle? (a) 1 rad sec–1 (b) 2 rad sec–1 (c) 4 rad sec–1 (d) 8 rad sec–1 3. The Young’s modulus of brass and steel are respectively 1010 N/m2. and2 × 1010 N/m2.Abrass wire and a steel wire of the same length are extended by 1 mm under the same force, the radii of brass and steel wires are RB and RS respectively. Then (a) B S R 2 R = (b) 2 / R R B S = (c) B S R 4 R = (d) 4 / R R B S = 4. A steel wire of length 20 cm and uniform cross-section 1 mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 40ºC to 20ºC. Coefficient of linear expansion for steel a = 1.1 × 10–5/ºC and Y for steel is 2.0 × 1011 N/m2. The change in tension of the wire is (a) 2.2 × 106 newton (b) 16 newton (c) 8 newton (d) 44 newton
- 253. 249 Mechanical Properties ofSolids 5. A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is (a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06 6. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is (a) 0.2J (b) 10J (c) 20J (d) 0.1J 7. Ametal rod ofYoung's modulus 2 × 1010 N m–2 undergoes an elastic strain of0.06%. Theenergyper unit volumestored in J m–3 is (a) 3600 (b) 7200 (c) 10800 (d) 14400 8. A force of 103 newton, stretches the length of a hanging wire by1 millimetre. The force required tostretch a wire of samematerial and length but having four times thediameter by1 millimetreis (a) 4 × 103 N (b) 16 × 103 N (c) 3 1 10 N 4 ´ (d) 3 1 10 N 16 ´ 9. A 2 m long rodofradius 1 cm which is fixed from one end is given a twist of 0.8 radian. The shear strain developed will be (a) 0.002 (b) 0.004 (c) 0.008 (d) 0.016 10. There are twowire ofsame material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be (a) 1: 1 (b) 2: 1 (c) 1: 2 (d) 4: 1 11. For a given material, theYoung's modulus is 2. 4 times that of rigidity modulus. Its Poisson's ratio is (a) 2.4 (b) 1.2 (c) 0.4 (d) 0.2 12. A cube at temperature 0ºC is compressed equally from all sides by an external pressure P. Bywhat amount should its temperature be raised to bring it back to the size it had before the external pressure was applied. The bulk modulus of the material of the cube is B and the coefficient oflinear expansion is a. (a) P/Ba (b) P/3B a (c) 3 pa/B (d) 3B/P 13. The compressibilityofwater is4 × 10–5 per unit atmospheric pressure. The decreasein volume of 100 cm3 ofwater under a pressure of 100 atmosphere will be (a) 0.4cm3 (b) 4 × 10–5 cm3 (c) 0.025cm3 (d) 0.004cm3 14. For the same cross-sectional area and for a given load, the ratio of depressions for the beam of a square cross-section and circular cross-section is (a) 3: p (b) p:3 (c) 1: p (d) p:1 15. A massive stone pillar 20 m high and of uniform cross- section rests on a rigid base and supports a vertical load of 5.0 × 105 N at its upper end. Ifthe compressive stress in the pillar is not to exceed 1.6 × 106 Nm–2, what is theminimum cross-sectional area of the pillar? Density of the stone = 2.5 × 103 kg m–3. (Take g = 10 N kg–1) (a) 0.15m2 (b) 0.25m2 (c) 0.35m2 (d) 0.45m2 16. Acircular tube ofmean radius8 cmandthickness0.04 cm is melted up and recast into a solid rod of the same length. The ratio of the torsional rigidities of the circular tube and the solid rod is (a) 4 4 4 ) 8 . 0 ( ) 98 . 7 ( ) 02 . 8 ( - (b) 2 2 2 ) 8 . 0 ( ) 98 . 7 ( ) 02 . 8 ( - (c) 4 4 2 ) 98 . 7 ( ) 02 . 8 ( ) 8 . 0 ( - (d) 2 3 2 ) 98 . 7 ( ) 02 . 8 ( ) 8 . 0 ( - 17. From a steel wire of densityris suspended a brass block of density rb. The extension of steel wire comes to e. If the brass block is now fullyimmersed in a liquid of densityrl, the extension becomes e'. The ratio ' e e will be (a) l r - r r b b (b) b b r r - r l (c) r - r r - r l b (d) l l r - r r b 18. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and 1 W weight is suspended from looser end. IfA is area of cross-section of the wire, the stress in the wire at a height 4 L from the upper end is (a) a W W1 + (b) a 4 / W 3 W1 + (c) 1 W W / 4 a + (d) a W 3 W 4 1 + 19. A beam of metal supported at the twoedges is loaded at the centre. The depression at the centre is proportional to (a) Y2 (b) Y (c) 1/Y (d) 1/Y 2 20. An iron rod of length 2m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is (a) 2 20 m / N 10 6 . 19 ´ (b) 2 18 m / N 10 6 . 19 ´ (c) 2 10 m / N 10 6 . 19 ´ (d) 2 15 m / N 10 6 . 19 ´ 21. A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is (a) 2Fl (b) Fl (c) 2l F (d) 2 l F
- 254. 250 PHYSICS 22. Ametalicrod of length l and cross-sectional area Ais made ofa material ofYoung modulusY. Ifthe rod is elongated by an amount y, then the work done is proportional to (a) y (b) 1 y (c) y2 (d) 2 1 y 23. On stretching a wire, the elastic energy stored per unit volume is (a) Fl/2AL (b) FA/2L (c) FL/2A (d) FL/2 24. When a pressureof 100 atmosphere is applied on a spherical ball, then its volume reduces to 0.01%. The bulk modulus of the material of the rubber in dyne/cm2 is (a) 10 × 1012 (b) 100 × 1012 (c) 1 × 1012 (d) 10 × 1012 25. What per cent of length of wire increases by applying a stress of 1 kg weight/mm2 on it? (Y = 1 × 1011 N/m2 and 1 kg weight = 9.8 newton) (a) 0.0067% (b) 0.0098% (c) 0.0088% (d) 0.0078% 26. K is the force constant of a spring. The work done in increasing its extension from l1 to l2 will be (a) K(l2 – l1) (b) 2 1 ( ) 2 K l l + (c) 2 2 2 1 ( ) K l l - (d) 2 2 2 1 ( ) 2 K l l - 27. If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is 1 × 103 kg/m3 and g = 10m/s2, then the volume elasticityin N/m2 will be (a) 108 (b) 2× 108 (c) 109 (d) 2× 109 28. The diagram below shows the change in the length X of a thin uniform wire caused by the application of stress F at two different temperatures T1 and T2. The variation shown suggests that (a) T1 > T2 X T2 T1 F (b) T1 < T2 (c) T2 > T1 (d) T1 ³ T2 29. A material has poisson’s ratio 0.50. If a uniform rod of it suffers a longitudinal strain of2 × 10–3, then the percentage change in volume is (a) 0.6 (b) 0.4 (c) 0.2 (d) Zero 30. A5 metrelong wire is fixed to the ceiling.Aweight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by1 mm. The energystored in the wire due to stretching is (a) zero (b) 0.05 joule (c) 100 joule (d) 500 joule 31. Two wires Aand B are of the same material. Their lengths are in the ratioof1 : 2 and the diameter are in the ratio 2 : 1. Ifthey are pulled bythe same force, then increase in length will be in the ratioof (a) 2: 1 (b) 1: 4 (c) 1: 8 (d) 8: 1 32. Twowires aremade ofthe same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount? (a) 4 F (b) 6 F (c) 9 F (d) F 33. A rubber cord catapult has cross-sectional area 25 mm2 and initial length ofrubber cord is 10 cm. It is stretched to 5 cm and then released to project a missile of mass 5 gm. Taking rubber Y =5×108 N/m2.Velocityofprojectedmissile is (a) 20 ms–1 (b) 100 ms–1 (c) 250 ms–1 (d) 200 ms–1 34. The potential energy U between two atoms in a diatomic molecules as a function of the distance x between atoms has been shown in the figure. The atoms are A B C x U O (a) attracted when x lies betweenAand Band are repelled when x lies between B and C (b) attracted when x lies between B and C and are repelled when x lies between A and B (c) are attracted when theyreach B from C (d) are repelled when theyreach B fromA 35. The diagram shows a force - extension graph for a rubber band. Consider the following statements : I. It will be easier to compress this rubber than expand it II. Rubber does not return to its original length after it is stretched III. The rubber band will get heated if it is stretched and released Extension Force Which of these can be deduced from the graph? (a) III only (b) II and III (c) I and III (d) I only
- 255. 251 Mechanical Properties ofSolids 36. A rod of length l and radius r is joined to a rod of length l/2 and radiusr/2 ofsame material. Thefreeend ofsmall rod is fixed toa rigid base and the free end oflarger rod is given a twist of q°, the twist angle at the joint will be (a) q/4 (b) q/2 (c) 5q/6 (d) 8q/48 37. Tobreak a wire, a force of106 N/m2 isrequired. Ifthe density ofthe material is3 × 103 kg/m3, then the length ofthe wire which will break byits own weight will be (a) 34m (b) 30m (c) 300m (d) 3m 38. The upper end ofa wire of diameter 12mm and length 1m is clamped and its other end is twisted through an angle of 30°. The angle of shear is (a) 18° (b) 0.18° (c) 36° (d) 0.36° 39. A steel wire of uniform cross-section of 1mm2 is heated upto 50°C and clamped rigidlyat itsends. If temperatureof wirefalls to40°C, changein tension in thewire is(coefficient of linear expansion of steel is 1.1 × 10–5/°C and Young's modulus of elasticity of steel is 2 × 1011 N/m2 ) (a) 22N (b) 44N (c) 88N (d) 88 × 106 N 40. In Searle'sexperiment tofindYoung'smodulus thediameter ofwire is measured as d = 0.05cm, length of wireis l = 125cm and when a weight, m = 20.0 kg is put, extension in wire wasfoundtobe0.100 cm. Find maximum permissible error in Young's modulus (Y). Use : 2 mg Y ( / 4) d x = p l . (a) 6.3% (b) 5.3% (c) 2.3% (d) 1% 41. If the ratio of lengths, radii and Young’s modulus of steel and brass wires shown in the figure are a, b, and c, respectively. The ratio between the increase in lengths of brass and steel wires would be (a) 2 2 b a c 4kg 2kg Steel Brass (b) 2 2 bc a (c) 2 2 ba c (d) 2 2 a b c 42. A uniform cube is subjected to volume compression. Ifeach side is decreased by 1%, then bulk strain is (a) 0.01 (b) 0.06 (c) 0.02 (d) 0.03 43. When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by5 cms will be (g = 9.8 m/sec2) (a) 4.900joule (b) 2.450joule (c) 0.495joule (d) 0.245joule 44. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? (a) Length = 100 cm, diameter = 1mm (b) Length = 200 cm, diameter = 2 mm (c) Length = 300 cm, diameter = 3 mm (d) Length = 50cm, diameter = 0.5mm 45. A steel rod of radius R = 10 mm and length L= 100 cm is stretched along its length bya force F = 6.28 × 104 N.If theYoung’smodulusofsteelisY= 2 ×1011 N/m2, the percentage elongation in the length of the rod is : (a) 0.100 (b) 0.314 (c) 2.015 (d) 1.549 DIRECTIONS for Qs. 46 to 50 : These areAssertion-Reason type questions. Each of these question contains twostatements: Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions fromthe following four options. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1 (b) Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1 (c) Statement-1 is True, Statement- 2 is False (d) Statement-1 is False, Statement -2 is True 46. Statement-1 Identical springsofsteel and copper areequally stretched. More work will be done on the steel spring. Statement-2 Steel is more elastic than copper. 47. Statement-1 Stress is the internal force per unit area of a body. Statement-2 Rubber is less elastic than steel. 48. Statement 1 : The stress-strain graphs are shown in the figurefor twomaterialsAand Bareshown in figure.Young's modulus ofA is greater than that of B. A B Stress Strain Statement 2 : TheYoung's modules for small strain is, stress Y strain = = slope of linear portion, of graph; and slope of A is more than slope that of B. 49. Statement 1:Young’s modulus for a perfectlyplastic body is zero. Statement 2: For a perfectlyplastic body, restoring force is zero. 50. Statement 1: Strain causes the stress in an elastic body. Statement 2:An elastic rubber is more plastic in nature.
- 256. 252 PHYSICS Exemplar Questions 1.Modulus of rigidityof ideal liquids is (a) infinity (b) zero (c) unity (d) some finite small non-zero constant value 2. The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will (a) be double (b) be half (c) be four times (d) remain same 3. The temperature of a wire is doubled. TheYoung's modulus of elasticity (a) will also double (b) will become four times (c) will remain same (d) will decrease 4. A spring is stretched byapplying a load to its free end. The strain produced in the spring is (a) volumetric (b) shear (c) longitudinal and shear (d) longitudinal 5. A rigid bar of mass M is supported symmetrically bythree wires each of length l. Those at each end are of copper and themiddle one is ofiron. The ratiooftheir diameters, ifeach is to have the same tension, is equal to (a) Ycopper / Yiron (b) iron copper Y Y (c) 2 iron 2 copper Y Y (d) iron copper Y Y 6. A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontallybetween two pillars (figure ). Amass m is suspended from the mid-point ofthe wire. Strain in the wire is m x 2L (a) 2 2 2 x L (b) x L (c) 2 x L (d) 2 2 x L 7. A rectangular frame is to be suspended symmetrically by two strings ofequal length on two supports (figure). It can be done in one of the following three ways; (a) (b) (c) The tension in the strings will be (a) the same in all cases (b) least in (a) (c) least in (b) (d) least in (c) 8. Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidlyat one end to the roof. A mass M is attached to each of the free ends at the centre of the rods. (a) Both the rods will elongate but there shall be no perceptible change in shape (b) The steel rod will elongate and change shape but the rubber rod will onlyelongate (c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse (d) The steel rod will elongate, without any perceptible changein shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre NEET/AIPMT (2013-2017) Questions 9. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [2013] (a) Length = 100 cm, diameter = 1mm (b) Length = 200 cm, diameter = 2 mm (c) Length = 300 cm, diameter = 3 mm (d) Length = 50cm, diameter = 0.5mm
- 257. 253 Mechanical Properties ofSolids 10. If the ratio of diameters, lengths and Young’s modulus of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be [NEET Kar. 2013] (a) 7 (5 ) q sp (b) 2 5 (7 ) q sp Steel 2m Copper 5m (c) 2 7 (5 ) q sp (d) 2 (5 ) q sp 11. Copper of fixed volume ‘V; is drawn into wire of length ‘l’. When this wire is subjected to a constant force ‘F’, the extension produced in the wire is ‘Dl’. Which of the following graphs is a straight line? [2014] (a) Dl versus 1 l (b) Dl versus l2 (c) Dl versus 2 1 l (d) Dl versus l 12. The approximate depth of an ocean is 2700 m. The compressibilityof water is 45.4 × 10–11 Pa–1 and densityof water is 103 kg/m3.What fractional compression of water will be obtained at the bottom of the ocean ? [2015] (a) 1.0 × 10–2 (b) 1.2 × 10–2 (c) 1.4 × 10–2 (d) 0.8 × 10–2 13. The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of : [2015 RS] (a) 2 :1 (b) 4 : 1 (c) 1 :1 (d) 1 : 2 14. Thebulkmodulus ofa sphericalobject is'B'. Ifit is subjected to uniform pressure 'p', the fractional decrease in radius is (a) B 3p (b) 3p B [2017] (c) p 3B (d) p B
- 258. 254 PHYSICS EXERCISE -1 1. (c) 2. (b) 3. (d) 4. (a) 5. (d) 6. (c) 7. (a) 8. (c) Couple per unit angle of twist, l 2 r C 4 h p = Couple t = l 2 r C 4 q h p = q Here h, l, C & t are same. So, r4q = constant ÷ ÷ ø ö ç ç è æ = q q 4 1 4 2 2 1 r r 9. (a) strain stress ) 90 tan( = q - 10. (c) If l isthe original length ofwire, then changein length offirst wire, ) ( 1 1 l l l - = D change in length of second wire, ) ( 2 2 l l l - = D Now, 2 2 1 1 A T A T Y l l l l D ´ = D ´ = or 2 2 1 1 T T l l D = D or l l l l - = - 2 2 1 1 T T or T1 l 2 – T1 l = T2 l 1 – l T2 or 1 2 2 1 1 2 T T T T - - = l l l 11. (c) F F mg Y A YA YA = Þ D = = D l l l l l 12. (d) The elasticity of a material depends upon the temperature of the material. Hammering & annealing reduces elastic property of a substance. 13. (b) The ratio of stress to strain is always constant. If stress is increased, strain will also increase so that their ratio remains constant. 14. (c) 1 FL l l YA A = Þ µ 15. (b) Stress = 2 Force 1 Stress Area πr µ 2 2 (2) 4 B A B A A B S r S S S r æ ö = = Þ = ç ÷ è ø 16. (c) 17. (d) According to Hooke's Law, within the elastic limits stress is directly proportional to strain. 18. (b) Let T be the tension in the ring, then ) r R ( A r T ) r R ( 2 . A r 2 . T Y - = - p p = r ) r R ( A Y T - = 19. (d) 0.5 2 (1 ) Y Y - h = h + s Þ s = h 20. (b) 1 / p V B V V B V D D = Þ µ D [Dp = constant] 21. (b) The wire starts behaving like a liquid at point b. It behaves like a viscous liquid in the region bc of the graph. 22. (b) Let V be the volume of the load and r its relative density So, a a A L g V A L F Y l l r = = .....(1) When the load is immersed in the liquid, then w w A L ) g 1 V g V ( A L F Y l l ´ ´ - r = ¢ = ....(2) (Q Now net weight = weight – upthrust) From eqs. (1) and (2), we get w a ) 1 ( l l - r = r or ) ( w a a l l l - = r 23. (c) Energy stored per unit volume = strain stress 2 1 ´ ´ = ) modulus s ' Young / stress ( stress 2 1 ´ ´ = ) modulus s ' Young /( ) stress ( 2 1 2 ´ = Y 2 S2 24. (c) 25. (c) For a perfectly rigid body strain produced is zero for the given force applied, so Y = stress/strain = ¥ EXERCISE - 2 1. (b) Let W newton be the load suspended. Then 1 1 1 1 A L W ) L / ( ) A / W ( Y l l = = ...(1) and 2 2 2 2 A L W ) L / ( ) A / W ( Y l l = = ....(2) Dividing equation (1) byequation (2), we get ÷ ø ö ç è æ ÷ ÷ ø ö ç ç è æ = ÷ ÷ ø ö ç ç è æ ÷ ÷ ø ö ç ç è æ = 1 2 A A 1 1 2 1 2 1 2 l l l l 1 2 2 1 = l l or l1 : l2 = 2 : 1 2. (c) Given that F/A = 4.8 × 107 Nm–2 F = 4.8 ×107 ×Aor 2 7 6 mv 4.8 10 10 48 r - = ´ ´ = or 48 r r m 2 2 = w or r m 48 2 = w sec / rad 4 16 3 . 0 10 48 = = ÷ ø ö ç è æ ´ = w Hints & Solutions
- 259. 255 Mechanical Properties ofSolids 3. (b) We know that Y= F L/pr2 l or r2 = F L/(Yp l) ) Y /( L F R B 2 B l p = and ) Y /( L F R S 2 S l p = or 2 10 10 2 Y Y R R 10 10 B S 2 S 2 B = ´ = = or 2 S 2 B R 2 R = or S B R 2 R = 2 / R R B S = 4. (d) F =YAat = (2.0 × 1011) (10–6) (1.1 × 10–5) (20) = 44 newton 5. (d) 6. (d) Elastic energy = x F 2 1 ´ ´ F=200 N, x=1 mm=10–3 m J 1 . 0 10 1 200 2 1 E 3 = ´ ´ ´ = - 7. (a) U/ volume = 2 strain Y 2 1 ´ = 3600 J m–3 [Strain = 0.06 × 10–2] 8. (b) 2 l F Y A F r L = ´ ´ Þ µ (Y, l and and Lare constant) Ifdiameter is madefour times then force required will be 16 times, i.e., 16 × 103 N 9. (b) 2 8 . 0 100 1 r ´ = q = f l = 0.004 radian 10. (d) Q Both wires are same materials so both will have sameYoung’s modulus, and let it beY. ( ) ΔL/L A. F strain stress Y = = , force applied F = A area of cross-section of wire = Now, 2 1 Y Y = ( )( ) ( )( ) 2 2 1 1 L A FL L A FL D = D Þ Since load and length are same for both 2 2 2 1 2 1 L r L r D = D Þ , 4 r r L L 2 1 2 2 1 = ÷ ÷ ø ö ç ç è æ = ÷ ÷ ø ö ç ç è æ D D 1 2 L : L 4:1 D D = 11. (d) 2 (1 ) Y = h + s 2.4 2 (1 ) 1.2 1 0.2 h = h + s Þ = + s Þ s = 12. (b) Bulk modulus V PV ) V / V ( P B D - = D - = ....(1) and T . V . 3 T V V a = D g = D or . T . 3 1 V V a = D - ...(2) From eqs. (1) and (2), ) T . 3 /( P B a = or B 3 P T a = 13. (a) P V / V B 1 K D = = . Here, P = 100 atm, K = 4 ×10–5 andV =100cm3. Hence, DV=0.4cm3 14. (a) I = d Y 3 W 3 l , where W= load, l= length of beam and I is geometrical moment ofinertia for rectangular beam, 3 bd Ι 12 = where b = breadth and d = depth For square beam b = d 4 1 b Ι 12 = For a beam of circular cross-section, 4 2 πr Ι 4 æ ö =ç ÷ è ø 4 3 4 3 1 b Y W 4 b Y 3 12 W l l = ´ = d (for sq. cross section) and ) r ( Y 3 W 4 ) 4 / r ( Y 3 W 4 3 4 3 2 p = p = d l l (forcircular cross-section) Now p = p p = p = d d 3 ) r ( r 3 b r 3 2 2 4 4 4 2 1 ( 2 2 b π r = Q i.e., theyhave same cross-sectional area) 15. (d) Weight of load Weight of puller Compressive stress area + = 5 6 20A d 10 5 10 1.6 10 area ´ ´ + ´ Þ = ´ Where d is the density. 3 5 6 20A 2.5 10 10 5 10 1.6 10 A ´ ´ ´ + ´ = ´ ie, 5× 105 = 1.1× 106A or A= 0.45m2 16. (a) l l 2 r C , 2 ) r r ( C 4 2 4 1 4 2 1 ph = - ph = Initial volume= Final volume r p = r - p l l 2 2 1 2 2 r ] r r [ ) r r )( r r ( r r r r 1 2 1 2 2 2 1 2 2 2 - + = Þ - = Þ ) 98 . 7 02 . 8 )( 98 . 7 02 . 8 ( r2 - + = Þ cm 8 . 0 r cm 64 . 0 04 . 0 16 r2 = Þ = ´ = Þ 4 4 4 4 4 1 4 2 2 1 ] 8 . 0 [ ] 98 . 7 [ ] 02 . 8 [ r r r C C - = - = 17. (a) Weights without and with liquid proportional to rb and rb – rl. 18. (b) 19. (c) d For a beam, the depression at the centre is given by, 3 f L 4Ybd æ ö d = ç ÷ è ø [f, L, b, d are constants for a particular beam] i.e. 1 Y d µ
- 260. 256 PHYSICS 20. (c) 6 3 2509.8 F/ A 50 10 Y / 0.5 10 2 - - ´ ´ = = D ´ l l 3 6 10 5 . 0 2 10 50 8 . 9 250 - - ´ ´ ´ ´ = Þ 2 10 m / N 10 6 . 19 ´ 21. (d) Work done by constant force in displacing the object by a distance l. = change in potential energy 1 2 = ´ stress × strain ×volume 1 2 2 = ´ ´ ´ ´ = l l F F A L A L 22. (c) Volume V = cross sectional A × length l or V = Al Strain = Elongation Original length Y l = Young’s modulus Y = Stress Strain Work done, W = 1 stress × strain × volume 2 ´ 2 1 (strain) 2 W Y Al = ´ ´ ´ 2 2 2 1 1 2 2 y YA Y Al y W y l l æ ö æ ö = ´ ´ ´ = Þ µ ç ÷ ç ÷ è ø è ø 23. (a) Energystored per unit volume = 1 – 2 2 F l Fl A L AL æ öæ ö = ç ÷ç ÷ è øè ø 24. (c) 6 100 10 0.01/100 K = = atm = 1011 N/m2 = 1012 dyne/cm2 25. (b) Stress = 1 kg wt/mm2 = 9.8 N/mm2 = 9.8 × 106 N/m2. 11 2 Y 1 10 N / m = ´ , 100 ? D ´ = l l Stress Stress Y Strain / = = Dl l 6 11 Stress 9.8 10 Y 1 10 D ´ = = ´ l l 11 6 100 9.8 10 100 10 - D ´ = ´ ´ ´ l l = 9.8 × 10–3 = 0.0098 % 26. (d) At extension l1, the stored energy = 2 1 1 2 Kl At extension l2, the stored energy = 2 2 1 2 Kl Work done in increasing its extension from l1 to l2 2 2 2 1 1 ( – ) 2 K l l = 27. (d) 3 9 200 10 10 2 10 / / 0.1/100 P h g K V V V V D r ´ ´ = = = = ´ D D 28. (a) When same stress is applied at two different temperatures, the increase in length is more at higher temperature. Thus T1 > T2. 29. (d) 30. (b) 1 1 2 2 W F l mgl = ´ ´ = 3 1 10 10 1 10 0.05 2 J - = ´ ´ ´ ´ = 31. (c) We know that Young's modulus l L r F Y 2 ´ p = SinceY, F are same for both the wires, we have, 2 2 2 2 1 1 2 1 L r 1 L r 1 l l = or, 2 2 1 1 2 2 2 1 L r L r ´ ´ = l l = 2 2 1 1 2 2 L ) 2 / D ( L ) 2 / D ( ´ ´ or, 8 1 L 2 L ) D 2 ( D L D L D 2 2 2 2 2 2 2 2 1 1 2 2 2 1 = ´ = ´ ´ = l l So, 8 : 1 : 2 1 = l l 32. (c) Y l A Wire (1) Wire (2) 3A Y l/3 As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of cross-section 3A will be l/3 (same volume as wire 1). For wire 1, / / = D l F A Y x ...(i) For wire 2 , '/ 3 /( /3) F A Y x = D l ...(ii) From (i) and (ii) , ' 3 3 ´ = ´ D D l l F F A x A x ' 9 Þ = F F 33. (c) Young’s modulus of rubber, rubber Y F F YA. A D = ´ Þ = D l l l l On putting the values from question, 2 2 6 8 10 10 10 5 10 25 10 5 F - - - ´ ´ ´ ´ ´ ´ = N 6250 10 25 25 1 2 = ´ ´ = - kinetic energy = potential energy of rubber 2 mv 2 1 l D = F 2 1 2 3 F 6250 5 10 v m 5 10 - - D ´ ´ = = ´ l = 62500 = 25 × 10 = 250 m/s
- 261. 257 Mechanical Properties ofSolids 34. (b) The atoms when brought from infinity are attracted due to interatomic electrostatic force of attraction. At point B, the potential energyis minimum and force of attraction is maximum. But if we bring atoms closer than x = B, forceof repulsion between two nuclei starts and P.E. increases. x B C P.E. A 35. (c) F x Fcom Fext From the figure, it is clear that Fcom < Fext. 36. (d) 4 . Constant 2 r r C L ph q = q = = 4 4 0 0 ( ) ( / 2) ( ') 2 2( / 2) r r l l ph q -q ph q -q Þ = l/2 l q0 q q¢ = 0 0 0 0 ( ) 8 2 16 9 q- q q Þ = Þ q = q 37. (a) 6 3 10 100 34 3 3 10 10 S L m dg = = = = ´ ´ 38. (b) r 6mm 30 r 0.18 1m q ´ ° q = f Þ f = = = ° l l 39. (a) F YA YA D = = aDq l l = 2 × 1011 × (1 × 10–6) × 1.1 × 10–5 × (10) = 22 N 40. (a) 2 mg Y ( / 4) d x = p l .........(1) max dY m d x 2 Y m d x D D D D æ ö = + + + ç ÷ è ø l l m = 20.0 kg Þ Dm= 0.1 kg l = 125cm Þ Dl = 1 cm. d =0.050 cm. Þ Dd = 0.001 cm x= 0.100 cm. Þ Dx=0.001cm. max 0.1kg 1cm dY 20.0kg 125cm 100% 6.3% 0.001cm 0.001cm Y 2 0.05cm 0.100cm æ ö + ç ÷ æ ö = ´ = ç ÷ ç ÷ è ø + ´ + ç ÷ è ø 41. (d) Given, 1 2 , l a l = 1 2 r b r = , 1 2 Y c Y = Brass T T 2g Steel T 2g Let Young’s modulus of steel beY1, and that of brass be Y2 1 1 1 1 1 F l Y A l = D …(i) and 2 2 2 2 2 F l Y A l = D …(ii) Dividing equation (i) byequation (ii), we get 1 1 2 1 2 2 2 1 2 1 Y F A l l Y F A l l D = D …(iii) Force on steel wire from free bodydiagram 1 (2 ) Newton T F g = = Force on brass wire from free bodydiagram 2 1 2 4 Newton F T T g g = = + = Now putting the value of F1, F2 in equation (iii), we get 2 1 2 1 2 2 2 2 2 2 1 1 1 2 1 1 . . . 4 2 Y r l l l g a Y g l l l r b æ ö é ù æ ö æ ö æ ö p D D æ ö = = ç ÷ ê ú ç ÷ ç ÷ ç ÷ ç ÷ è ø è ø D D è ø è ø p è ø ë û 42. (d) If side of the cube is L then 3 3 dV dL V L V L = Þ = % change in volume = 3 × (% change in length) = 3 × 1% = 3% Bulk strain 0.03 V V D = 43. (b) 2 2 4 9.8 19.6 10 2 10 F K x - ´ = = = ´ ´ Work done = 2 2 1 19.6 10 (0.05) 2.45 2 J ´ ´ = 44. (d) F = YA L × l So, extension, lµ L A µ 2 L D [Q F and Y are constant] l1 µ 2 100 1 µ 100 and l2 µ 2 200 2 µ 50 l3µ 2 300 3 µ 100 3 and l4 µ 50 1 4 µ 200 The ratio of 2 L D is maximum for case (d). Hence, option (d) is correct.
- 262. 258 PHYSICS 45. (a)Percentage elongation in the wire F AY = l 4 2 11 6.28 10 (1) 1 1000 (0.01) 2 10 ´ ´ = = p ´ ´ 46. (a) Work done 2 1 1 Stress Strain (Strain) 2 2 Y = ´ ´ = ´ ´ Since, elasticity of steel is more than copper, hence more work has to be done in order to stretch the steel. 47. (b) Stress is defined as internal force (restoring force) per unit area of a body. Also, rubber is less elastic than steel, because restoring force is less for rubber than steel. 48. (d) 49. (d) 50. (a) EXERCISE - 3 Exemplar Questions 1. (b) As liquid is ideal so no. frictional force exists hence, tangential forces are zero so there is no stress developed. 2. (d) As we know that, Breaking stress = Breaking force Area of cross-section ..(i) When length of the wire changes (or byreducing half) area of cross-section remains same. Hence, breaking force will be same because breaking stress does not depend on length. 3. (d) Aswe knowthat, length of awirewhen thetemperature increased 0 (1 ) t L L T = + aD where DT is change in the temperature. L0 is original length, a is coefficient oflinear expansion and Lt is length at temperature T. Now, 0 0 t L L L L T D = - = aD Now, Young's modulus (Y) Stress Strain = 0 0 0 1 FL FL A L AL T T = = µ ´D aD D As, 1 Y T µ D So, if temperature increass DT increases, hence Young's modulus of elasticity (Y) decreases. 4. (c) When a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. So strain produced when change in length corresponds to longitudinal strain andchange in shape corresponds to shearing strain. 5. (b) As we know that, The Young's modulus Stress / Strain / F A F L Y L L A L = = = ´ D D 2 2 4 ( / 2) F L FL L D D L = ´ = D p p D 2 4 4 FL FL D D LY LY = Þ = pD pD If F and L L D are constants. So, 1 D Y µ Hence, we can find ratio as copper iron iron copper D Y D Y = 6. (a) Consider the given diagram m 90°–q 90°–q q q x O B A C L L So, change in length 0 0 ( ) ( ) L A B AC CB D = + - + = 2BO – 2AC = 2 [BO – AC] ( AO = BO, AC = CB) 2 2 1/ 2 2[( ) ] x L L = + - 1/ 2 2 2 2 1 1 x L L é ù æ ö ê ú = + - ç ÷ ç ÷ ê ú è ø ë û 2 2 2 1 2 1 1 2 x x L L L L é ù D » + - = ê ú ë û [ ] x L << Q 2 2 2 / Strain = 2 2 2 L x L x L L L D = = 7. (c) Let us consider the free body diagram of the rectangular frame Tsinq Tsinq T Tcosq Tcosq q q m mg Net forces acting on frame will be zero. So, Balancing vertical forces 2T sinq – mg = 0 [T is tension in the string] 2T sinq = mg ...(i) Total horizontal force cos cos 0 T T = q - q =
- 263. 259 Mechanical Properties ofSolids Nowfrom Eq. (i), 2sin mg T = q max min 2sin mg T = q As mg is constant then 1 sin T µ q min min sin 0 0 q = Þ q = No option matches with q = 0° min max 2sin mg T = q (since, sinqmax = 1) max sin 1 90 q = Þ q = ° So tension is all three cases are different rejects option (a) For minimum tension q must be 90° i.e. sin q = 1 Hence, tension is the least for the case (b). 8. (d) A mass M is attached at the centre or midpoint of rod of rubber and steel. As the mass is attached to both the rods, both rod will be elongated as shown in figuresbut due todifferent elasticproperties ofmaterial rubber changes shape also. M M (Steel) (Rubber) As the Young's modulus of rigidity for steel, is larger than rubber, so strain L L D for rubber is larger than steel for same stress. NEET/AIPMT (2013-2017) Questions 9. (d) F = YA L × l! So, extension, lµ L A µ 2 L D [Q F and Y are constant] l1 µ 2 100 1 µ 100 and l2 µ 2 200 2 µ 50 l3µ 2 300 3 µ 100 3 and l4 µ 50 1 4 µ 200 The ratio of 2 L D is maximum for case (d). Hence, option (d) is correct. 10. (c) From formula, Increase in length 2 4 FL FL L AY D Y D = = p 2 S S C C S C C S S C L F D Y L L F D Y L æ ö D = ç ÷ D è ø = 2 7 1 1 5 q p s æ ö æ ö ´ç ÷ ç ÷ è ø è ø = 2 7 (5 ) q sp 11. (b) As Y= F F A AY Þ D = D l l l l But V = Al soA= V l Therefore Dl = 2 2 F VY µ l l Hence graph of Dl versus l2 will give a straight line. 12. (b) Compressibilityof water, K = 45.4 × 10–11 Pa–1 densityof water P = 103 kg/m3 depth of ocean, h = 2700 m We have to find V ? V D = As we know, compressibility, K= 1 ( V / V) (P gh) B P D = = r So, (DV/V)= Krgh = 45.4 × 10–11 × 103 ×10 ×2700 = 1.2258× 10–2 13. (a) Young's modulusY = W . A l l D 1 2 1 2 W W Y Y = [Q A, l, Dl same for both brass and steel] Brass A, , l Dl Steel l, A, Dl Y2 Y1 1 1 2 2 W Y 2 W Y = = [Ysteel/Ybrass = 2 given] 14. (c) Bulk modulus is given by B = P V V D æ ö ç ÷ è ø or V P V B D = R P 3 R B D = (here, R R D = fractional decreases in radius) R P R 3B D Þ =
- 264. 260 PHYSICS FLUIDS Fluid issomething that can flow. All liquids and gases are fluids. The force exerted normally at a unit area of the surface of a fluid is called fluid pressure. i.e., P = F a Its S.I. unit is Nm–2 or Pascal. Its dimensions are [ML–1T–2]. PASCAL'S LAW AND ITS APPLICATIONS Pascal's law : Pressure in a fluid in equilibrium is the same everywhere, if the effect of gravity is neglected. Another form of Pascal's law : The excess pressure, applied anywhere in a mass of confined incompressible fluid is transmitted by the fluid in all directions without being diminished in magnitude. Applications of Pascal's Law Hydraulic lift : Its working is based on Pascal's law. Apiston of small cross-sectional area (a) exerts a force (f) on the liquid. Applied force = f F = pA f = pa Area = a Area = A The pressure is transmitted undiminished to the larger cylinder of cross-sectional area A. = = f F P a A Þ = A F f a Hydraulic brakes also work on pascal's law. ATMOSPHERIC, HYDROSTATIC AND GUAGE PRESSURE Atmospheric pressure : The atmosphere exerts pressure on the earth's surface. The atmospheric pressure at sea level is given by P0 = 1.01 × 105 Pa Hydrostatic pressure : The hydrostatic pressure at a depth h below the surface of a fluid is given by P = hrg where r is the density of the fluid, g acceleration due to gravity and h is the depth of the liquid column. Gauge pressure : The pressure at anypoint in a fluid is equal to the sum of the atmospheric pressure acting on its surface and the hydrostatic pressure due to the weight of the fluid above that point which is at a depth h below the surface of the fluid. The gauge pressure is given by P = P0 + hrg or, P – P0 = hrg BUOYANCY AND ARCHIMEDES' PRINCIPLE Buoyancy If a body is partiallyor wholly immersed in a fluid, it experiences an upward force due tothe fluid surrounding it. The phenomenon of force exerted by fluid on the body called buoyancy and the force is called buoyant force or upthrust. A bodyexperiences buoyant force whether it floats or sinks, under its own weight or due to other forces applied on it. Archimedes' Principle When any body is immersed (totally or partially) in a liquid it appears to lose part of its weight and the apparent loss of weight is equal to the weight of liquid displaced. Let a bodyof weight W is immersed in a fluid and W' is upthrust on it then (i) if W > W', then body will sink. (ii) if W = W', then the bodyfloats with whole or some part of its volume inside the fluid. Let V be the volume of a body of density d and V' be the volume of liquid of density r displaced. If the body floats then Vd = V¢r V ' V d = r gives the fraction of the volume inside the liquid in which the body floats. Also, body immersed in a fluid experiences an upward buoyant force equivalent to the weight of the fluid displaced by it. The proof of this principle is very simple. Imagine a body of arbitrary shape completely immersed in a liquid of density r as 10 Mechanical Properties of Fluids
- 265. 261 Mechanical Properties ofFluids shown in the figure (a). Abody is being acted upon by the forces from all directions. Let us consider a vertical element of height h and cross-sectional area dA (as shown in the figure (b)). h2 dA h h1 F1 F2 (a) (b) The force acting on the upper surface of the element is F1 (downward) and that on the lower surface is F2 (upward). Since F2 > F1, therefore, the net upward force acting on the element is dF = F2 – F1 It can be easily seen from the figure (b), that 1 1 F ( gh )dA = r and 2 2 F ( gh )dA = r so dF g(h)dA = r Also, h2 – h1 = h and d (dA) = dV The net upward force is F gdV Vg = r = r ò Hence, for the entire body, the buoyant force is the weight of the volume of the fluid displaced. The buoyant force acts through the centre of gravity of the displaced fluid. Keep in Memory 1. The pressure is perpendicular to the surface of the fluid. 2. The upthrust on a body immersed in a liquid does not depend on the mass, density or shape of the body. It only depends on the volume of the body. 3. The weight of the plastic bag full of air is same as that of the empty bag because the upthrust is equal to the weight of the air enclosed. 4. The cross-section of the water stream from a top decreases as it goes down in accordance with the equation of continuity. 5. We cannot sip a drink with a straw on the moon, because there is no atmosphere on the moon. 6. The line joining the centre of gravity and centre of buoyancy is called central line. 7. Metacenter - is a point where the vertical line passing through the centre of buoyancy intersects the central line. 8. The floating body is in stable equilibrium when the metacenter is above the centre of gravity(centre of gravity is below the centre of buoyancy). 9. The floating bodyis in the unstable equilibrium when the metacenter lies below thecentre ofgravity(centre of gravity is above the centre of buoyancy). 10. The floating bodyis in the neutral equilibriumwhen centre of gravity coincides with the metacenter (centre of gravity coincides with the centre of buoyancy). 11. The wooden rod cannot float vertically in a pond of water because centre of gravity lies above the metacenter. 12. (i) If a body just floats in a liquid (density of the body is equal to the densityof liquid) then the bodysinks if it is pushed downwards. (ii) If two bodies have equal upthrust when just immersed in a liquid, both will have the same volume. (iii) If a person floats on his back on the surface of water, the apparent weight of person is zero. 13. The hydrometer can be used to measure density of the liquid or fluid. Relative Density (or Specific Gravity) Liquids may be treated as incompressible. Hence their density may be assumed to be constant throughout. Relative density = Weight of substance in air Weight of equal volume of water = Weight of substance in air Loss of weight in water 4º = Density of substance Density of water at C Density in SI system = 1000 × density in the cgs system. (i) The density of liquid of bulk modulus B at a depth h is given by 0 1 h gh B r æ ö r = r + ç ÷ è ø where 0 r is the density of liquid on its surface and r is the average densityof liquid. (ii) The density of liquid changes with pressure as 0 1 h P P P B D æ ö = + ç ÷ è ø where = DP change in pressure and B = bulk modulus of liquid. (iii) If two liquids of masses m1, m2 and densities r1, r2 are mixed together, then the density of the mixture is given by 1 2 1 2 1 2 m m m m + r = + r r And if m1 = m2 but different densities are mixed together, then the density of the mixture is harmonic mean of the densities. i.e., 2 1 2 1 2 r + r r r = r or ú û ù ê ë é r + r = r 1 2 1 1 1 2 1 (iv) If two drops of same volume but different densities are mixed together, then the density of the mixture is the arithmetic mean of the densities. i.e., 2 2 1 r + r = r (as 1 0 2 0 0 0 V V V V r + r r = + ) SURFACE TENSION It is defined as the force per unit length acting at right angles on either side of an imaginary line drawn on the free surface of the liquid. i.e., = l F S The surface tension is also defined as the work required to increase unit area of that liquid film. Its SI unit is N/m or J/m2 and dimensions are [ML0T–2].
- 266. 262 PHYSICS Keep inMemory 1. The liquid surface always acquires minimumsurface area due to surface tension (ST). So, the small droplet of any liquid is always spherical. 2. The ST is a molecular phenomenon as ST is due to 'cohesion' between the molecules of a liquid. 3. The force of attraction between the molecules of the same substance is called a cohesive force and that between molecules of different substance is called adhesive force. 4. Themolecular rangeis the maximum distance (10–9 m) upto which the molecules attract each other. 5. In general the ST of liquids decreases with increase in temperature but the ST of molten Cadmium and Copper increases with increase in temperature. 6. Ifthe impurityis completelysoluble then on mixing it in the liquid, its surface tension increases. For example on dissolving ionic salts in small quantities in a liquid, its surface tension increases. On dissolving salt in water, its surface tension increases. 7. Ifthe impurityis partiallysoluble in a liquid, then its surface tension decreases. For example on mixing detergent or phenol in water its surface tension decreases. 8. On increasing temperature surface tension decreases. At critical temperatureand boilingpoint it becomes zero. Surface tension of water is maximum at 4°C. S.T. 4°C Temp. S.T.=0 C of water ANGLE OF CONTACT The angle between the tangent to the liquid surface and the tangent to the solid surface at the point of contact (inside the liquid) is known as angle of contact. q q < 90º Water Glass Glass Mercury q q > 90° Some values of angle of contact of solid and liquid : Pair of surface Angle of contact Pure water and glass 0° Silver and glass 90° Alcohol and glass 138° Normal water & glass 8° Mercury & glass 135° Adhesion > cohesion Adhesion = cohesion Adhesion < cohesion 1. Liquid will wet the solid Critical Liquid will not wet the solid 2. Meniscus is concave Meniscus is plane Meniscus is convex 3. Angle of contact is acute (q < 90°) Angle of contact is 90º Angle of contact is obtuse (q > 90°) 4. Pressure below the menisucs is lesser Pressure below the meniscus Pressure below the meniscus is more than above it by (2T/r), i.e. p = p0 – 2T r is same as above it, i.e. p = p0 then above it by(2T/r), i.e., p = p0 + 2T r 5. In capillarytube liquid will ascend. No capillaryrise In capillarytube liquid will descend. Keep in Memory 1. The value of angle of contact lies between 0º and 180º. For pure water and glass it is 0º, for tap water and glass it is 8º and for mercuryand glass it is 135º. 2. For all those liquids which wet the solid surface and which rise up in a capillary tube, the angle of contact is an acute angle (q < 90°), e.g. water and glass. 3. For all those liquids which do not wet a solid surface and which depress in a capillary tube, the angle of contact is an obtuse angle (q > 90°), e.g. glass and mercury. 4. For all those liquids which neither rise nor get depressed in a capillary tube, the angleof contact is right angle (q = 90°), e.g. silver and water. 5. Angle of contact depends on impurities, water proofing agent, surface in contact and temperature. Angle of contact qC µ T where T is the temperature. Capillarity : The phenomenon of rise or fall of liquids in capillary tube is known as capillarity.
- 267. 263 Mechanical Properties ofFluids The rise or fall of a liquid in a capillary tube is given by 2 cos 2 T T h r g R g q = = r r 2T h g R Þ r = q q R r water where T = surface tension, q = angle of contact, r = density of liquid, r = radius of capillary tube, R = radius of meniscus. (i) Ifcapillarytube is ofinsufficient length l(i.e. l < h), then the liquidrises toa full height h with radiusR'such that hR= lR' (ii) When the capillary tube is tilted from vertical by an angle a, then the vertical height h of liquid column remains the same. The length of liquid in capillary increases such that h h cos ¢ = a or a = ¢ cos h h . According to Zurin's law capillary rise 1 h r µ where r is the radius of the capillary tube. Keep in Memory 1. Work done in forming a liquid drop of radius R, surface tension T is, W = 4pR2T. 2. Work done in forming a soap bubble of radius R, surface tension T is, W = 2×4pR2T = 8pR2T. 3. When n no. of smaller drops of liquid, each of radius r, surface tension T are combined to form a bigger drop of radius R then 1/3 R n r = . 4. The surface area of bigger drop = 4pR2 = 4pn2/3 r2. It is less than the area of n smaller drops. SHAPE OF LIQUID MENISCUS : The pressure on the concave side is always greater than the pressure on the convex side. q < 90º Water A P = Atmospheric 0 pressure Concave meniscus q > 90º Mercury A P = Atmospheric 0 pressure Convex meniscus PA=P0–2T/r PA=P0+2T/r (r is radius of meniscus) Excesspressure of liquiddropandsoapbubble : (i) Excess of pressure for spherical soap bubble is p = 4T/rand excessofpressure for liquid drop and air bubble in a liquid is p = 2T/r. (ii) (a) Excess of pressure within a cylindrical liquid drop p = T/R (b) Excess of pressure within a cylindrical soap bubble p = 2T/R where T = surface tension, R = radius of the cylindrical drop. Keep in Memory 1. Work done in breaking a liquid drop of radius R into n equal small drops = T ) 1 n ( R 4 3 / 2 - p 1 ; where T = surface tension. 2. Work done in breaking a soap bubble of radius R into n equal small drops = T ) 1 n ( R 8 3 / 2 - p 1 ; where T = surface tension. Example 1. A solid uniform ball having volume V and density rfloats at the interface of two unmixable liquids as shown in fig. The densities of the upper and the lower liquids are r1 and r2 respectively, such that r1 < r < r2. What fraction of the volume of the ball will be in the lower liquid ? r r1 r2 Solution : Let V1 and V2 be the volumes of the ball in the upper and lower liquids respectively. So V1 +V2 =V. As ball is floating in the two liquids ; weight of the ball = upthrust on ball due to two liquids i.e., V r g = V1 r1 g + V2 r2 g ; or V r = V1 r1 + (V – V1) r2 ; or V V 2 1 2 1 ÷ ÷ ø ö ç ç è æ r - r r - r = Fraction in the upper liquid = 2 1 2 1 V V r - r r - r = Fraction in the lower liquid = 1 V 1 V - 2 1 1 2 1 2 1 r - r r - r = - = r - r r - r Example2. A piece of cork is embedded inside of block of ice which floats on water. What will happen to the level of water when all the ice melts? Solution : Let , M = mass of the block ofice, m = mass ofpiece of cork and V = Volume of water displaced. Now(M+ m) =V × 1= V ...(1) When the ice melts, let it be converted into V' c.c. of water. Also M V' 1 V' = ´ = The piece of cork floats on the surface ofwater when all ice melts. Let the cork displaces a volume V'' c.c. of water. . Then m V" 1 V" = ´ = If V1 be the volume of water displaced by melted ice and cork, then ( ) 1 M m V' V" V + = + = ...(2)
- 268. 264 PHYSICS From eqns.(1) and (2), V = V1 Hence, no change in the level of water. Example3. Two substances of densities r1 and r2 are mixed in equal volume and the relative density of mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. Determine the values of r1 and r2 . Solution : When the substances are mixed in equal volumes, then 4 V 2 V V 2 1 ´ = r + r ... (1) When the two substances are mixed in equal masses, then 3 m 2 m m 2 1 = r + r ...(2) From eq. (1), 8 2 1 = r + r ... (3) From eqn. (2) 3 2 or 3 2 1 1 2 1 2 1 2 1 = r r r + r = r + r or 12 or 3 2 8 2 1 2 1 = r r = r r . ...(4) Now 2 / 1 2 1 2 2 1 2 1 ] 4 ) [( r r - r + r = r - r 4 ] 48 64 [ 2 / 1 = - = ...(5) Solving eqns. (3) and (5), we get 6 1 = r and 2 ρ 2 = ] Example4. A sealed tank containing a liquid of density r moves with a horizontal acceleration a, as shown in fig. Find the difference in pressure between the points A and B. C A B h a l Solution : Since points A and C are in the same horizontal line but separated by distance l and liquid tank is moving horizontallywith acceleration a, hence PC – PA = lra or PC = PA + lra Points B and C are vertically separated by h PB – PC = h r g or PB – (PA + l r a) = h r g or PB – PA = h r g + l ra Example5. Calculate the excess pressure within a bubble of air of radius 0.1 mm in water. If the bubble had been formed 10 cm below the water surface when the atmospheric pressure was 1.013 × 105 Pa, then what would have been the total pressure inside the bubble? Solution : Excess pressure within air bubble = 2T r = 3 3 2 73 10 0.1 10 - - ´ ´ ´ = 1460 Pa The pressure at a depth d, in liquid P = hdg. Therefore, the total pressure inside the air bubble is Pin = Patm + hdg + 2T r or Pin = 1.013× 105 + 10 ×10–2 × 103 × 9.8 + 1460 = 101300 + 980 + 1460 =103740 =1.037 ×105 Pa. Example6. A capillary of the shape as shown is dipped in a liquid. Contact angle between the liquid and the capillary is 0° and effect of liquid inside the meniscus is to be neglected. T is the surface tension of the liquid, r isradius of the meniscus, g is acceleration due to gravity and r is density of the liquid then determine the height h in equilibrium. h Solution : As weight ofliquidin capillaryisbalancedbysurfacetension, then 2 1 T 2 r r h g ´ p = p r (for a tube ofuniform radius r) 1 2T h r g = r r h1 But weight ofliquid in tapered tube ismore than uniform tube of radius r, then in order to balance h < h1. 2T h r g < r r Example7. A hydraulic automobile lift is designed to lift car with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximumpressure would the smaller piston have to bear ? Solution : Here mass ofcar = 3000 kg. Area of cross section of larger piston = 425cm2 = 425 ×10–4 m2. The maximum pressure that the smaller piston would have to bear = 4 Weight of car 3000 9.8 Area of cross-section 425 10- ´ = ´ 5 2 6.92 10 Nm- = ´
- 269. 265 Mechanical Properties ofFluids FLOW OF LIQUIDS The motion of fluids are of following four types : (i) Streamline motion : When fluid in motion, if fluid particles preceeding or succeeding a fluid particle follow the same path, then the path is called streamline and the motion of the fluid is called streamline motion. This type of motion takes place in non-viscous fluids having very small speed. Principle of continuity : Whenincompressible, non-viscous liquid flows in non-uniform tube then in streamline flow product of area and velocity at any section remains same. The mass of liquid flowing in equals the mass flowing out. i.e., m1 =m2 or, v1A1r1 Dt = v2 A2r2Dt .....(1) v1 A1 v2 A2 P Q As we have considered the fluid incompressible thus, v1A1 = v2A2 or Av = constant ....(2) (Since r1 = r2) Equations(1) and(2)are saidtobe asequationofcontinuity. (ii) Steady state motion : In a liquid in motion, when liquid particles, crossing a point, cross it with same velocity, then the motion of the liquid is called steady state motion. This type of motion takes place in non-viscous liquids having very small speed. (iii) Laminar motion : Viscous liquids flow in bounded region or in a pipe, in layers and when viscous liquid is in motion, different layers have different velocities. The layers in contact with the fixed surface has least velocity and the velocity of other parallel layers increases uniformly and continuously with the distance from the fixed surface to the free surface of the liquid. This is called laminar motion of the liquid. (iv) Turbulentmotion : Whenthe velocityofaliquid isirregular, haphazard and large, i.e. Beyond a limiting value called critical velocity the flow of liquid loses steadiness then the motion of the liquid is called turbulent motion. Critical velocity Vc K r h = r Here h is called coefficient of viscosity. VISCOSITY The internal friction of the fluid, which tends to oppose relative motion between different layers of the fluid is called viscosity. The viscous force between two layers of a fluid of area A having a velocity gradient dv dx is given by = -h dv F A dx where h is called the coefficient of viscosity. Its S.I. unit is poiseuille or decapoise. The C.G.S. unit is called poise. 1 decapoise =10 poise. Effects on Viscosity : (1) Effect of temperature : On increasing temperature viscosity of a liquid decreases. (2) Effect of pressure : On increasing pressure viscosity of a liquid increases but viscosity of water decreases. Keep in Memory 1. The viscosity of gases increases with increase of temperature, the rate of diffusion increases. 2. The viscosity of liquids decrease with increase of temperature, because the cohesive force between the liquid molecules decreases with increase of temperature. Critical Velocity : It is the maximum velocity of a fluid above which a stream line flow changes to a turbulent flow, i.e. it is the maximum velocity of a liquid below which its flow remains streamline. (i) Reynold's formula for critical velocityis η ρ = c N V r or r = h c V r N where N = Reynold's no., h = coefficient of viscosity, , r = densityofliquid; r = radius of tube, N = 1000 for narrow tube. (ii) (a) If0 < N < 2000 then the flow is laminar (b) If 2000 < N < 3000 then flow of liquid is unstable and maychangefrom laminar to turbulent (c) If N > 3000, then the flow is turbulent. (iii) When velocity of fluid is less than its critical velocitythen the flow of liquid is determined by its viscosity, its density has no effect on its flow. (iv) When the velocityof liquid is more than its critical velocity then its flow is determined by its density, where viscosity has little effect on its flow. For example lava from volcanois highly thick, despite that it comes out with high speed. (v) When c V V £ the flow of liquid is streamline and when V > Vc then the flow of liquid is turbulent. 1 c c r r V V 2 2 1 = for same liquid, which is flowing in twotubes of radii r1 and r2 respectively. (vi) The critical velocity of a liquid with high viscosity and smaller radius is higher than that of a liquid with low viscosity and greater radius. Reynold's Number (N) : It is pure number which determines the nature of flow of liquid through a pipe. 2 η η = = = æ ö ç ÷ è ø c c c V dr V Inertial force/area N V Viscous force/area r N is a dimensionless quantity and carries no unit.
- 270. 266 PHYSICS BERNOULLI'S THEOREM Fornon-viscous, incompressible, streamline flow of fluids the sum of pressure per unit volume, potential energy per unit volume and Kinetic energy per unit volume remain constant. i.e., r p + gh + 2 1 v2 = constant where r = density of fluid. When h = 0 then 2 p 1 v constant 2 + = r Bernoulli's theorem is strictly applicable for an ideal fluid. An ideal fluid is one which is (a) incompressible (b) streamline (c) irrotational and (d) non-viscous. Applications of Bernoulli’s Principle Dynamic lift : (i) Wings of aeroplane : The wings of the aeroplaneare having tapering. Due to this specific shape of wings when the aeroplane runs, air passes at higher speed over it ascompared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called 'dynamic lift' (= pressure difference area of wing) acts on the plane. Ifthis force becomes greater than the weight of the plane, the plane will riseup. v large, p small v small, p large (ii) Ball moving without spin: The velocityof fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball. (iii) Ball moving with spin: A ball which is spinning drags air along with it. Ifthesurfaceis rough moreair will bedragged. The streamlines ofair for a ball which ismoving and spinning at the same time. The ball is moving forward and relative toit the air is moving backwards. Therefore, the velocity of air above the ball relative to it is larger and below it is smaller. The stream lines thus get crowded above and rarified below. This difference in the velocities of air results in the pressure difference between the lower and upper faces and their is a net upward force on the ball. This dynamic lift due to spinning is called Magnuseffect. Some other applications of Bernoulli’s principle : (i) The action of carburator, sprayer or atomizer based on Bernoulli’s principle. (ii) The action of bunsen burner, exhaust pump etc. (iii) Air foil or lift on aircraft wingworkson Bernoulli’s principle. (iv) Motion of a spinning ball i.e., magnus effect. (v) Blowing of roofs by wind storms etc. based on Bernoullis principle. STOKE’S LAW When a solid moves through a viscous medium, its motion is opposed by a viscous force depending on the velocity and shape and size of the body. The energy of the body continuously decreases in overcoming the viscous resistance of the medium. Thisis whycars, aeroplanesetc. areshaped streamlinetominimize the viscous resistance on them. The viscous drag on a spherical body of radius r, moving with velocity v, in a viscous medium of viscosity h is given by 6 = ph viscous F rv This relation is called Stokes' law. . Importance ofStoke’slaw: (i) It is used in the determination of electronic charge with the help ofmilikan’sexperiment. (ii) It accounts the formation of clouds. (iii) It accounts why the speed of rain drops is less than that of a bodyfallingfreelywith a constant velocityfrom the height of clouds. (iv) It helps a man coming down with the help of a parachute. Terminal Velocity : When a spherical body is allowed to fall through viscous medium, its velocity increases till the viscous drag plus upthrust is equal to the weight of the body. After that body moves with constant velocity, called terminal velocity. The terminal velocity is given by 2 2 ( ) 9 = - s h R v d g where d = density of body, s = densityof medium, h = coefficient of viscosityof medium, R = radius of the spherical body. From terminal velocity, h a 1 V , i.e. greater the viscosity, smaller is the terminal speed. Flowof liquid throughtube /pipe : (i) Poiseuille's equation is 4 pr Q 8 p = hl , where p is the pressure difference between the two ends of the tubes, r is the radius, l is the length of the tube and h is the coefficient of viscosity, Q = rate of flow of liquid. Equivalent liquid resistance = 2 1 2 4 4 1 2 8 8 h h + p p l l r r , when tube are joined in series. It means that liquid flow through capillary tube is similar to flow of electric current through a conductor i.e., Q (rate of liquid flow) corresponds to I (rate of flow of charge), pressure difference similar to potential difference. (ii) When two tubes are joined in series then the volume of fluid flowing through the two tubes is the same but the pressure difference across the two tubes is different. The total pressure difference, P = P1 + P2. (iii) If two tubes are joined in parallel then the pressure difference across the two tubes is the same but the volume of fluid flowing through the two tubes is different. The total volume of the fluid flowing through the tube in one second is Q = Q1 + Q2.
- 271. 267 Mechanical Properties ofFluids VELOCITY OF EFFLUX AND TORRICELLI'S THEOREM (i) Torricelli's theorem : For liquid filled in a tank upto a height H having a hole O at a depth h from free level of liquid through which the liquid is coming out, velocity of efflux of liquid v = 2gh . (ii) Time taken by the liquid in falling from hole to ground level 2( ) - = H h t g H h Free surface (iii) The horizontal range covered by the liquid x = horizontal velocity (v) ×time (t) 2( ) 2 - = ´ H h gh g 2 ( ) = - h H h . For maximum horizontalrange, differentiating both sidew.r.t. h and we get h = H/2, the range is maximum. Keep in Memory 1. The cross-section of the water stream from a top decreases as it goes down in accordance with the equation of continuity. 2. When a hale blows over a roof, the force on the roof is upwards. 3. Sudden fall in atmospheric pressure predicts possibility of a storm. 4. Venturimeter is a device used for the measurement of the rate of flow of incompressible fluid through a tube. Its working is based on Bernoulli’s principle. Example8. The vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. Find the time taken to decrease the level from H1 to H2. Solution : The average velocity of efflux, 1 2 2g H 2g H v 2 + = Let t be the time taken to empty the tank from level H1 to H2. Then, ] H H [ A t a 2 H g 2 H g 2 2 1 2 1 - = ´ ´ + or t ( ) ( ) ( )( )ú ú û ù ê ê ë é - + - ´ - ´ ÷ ÷ ø ö ç ç è æ ´ = 2 1 2 1 2 1 2 1 H H H H H H H H g 2 a A ( ) 2 1 H H g 2 a A - ´ ÷ ÷ ø ö ç ç è æ ´ = ( ) 1 2 A 2 H H a g æ ö = ´ ´ - ç ÷ è ø Example9. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m/s and 63 m/s respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m–3. Solution : Let v1 be the speed and P1 be the pressure on the upper surface of the wing, and corresponding values on the lower surface be v2 and P2 respectively. v1 = 70 m/s, v2 = 63 m/s, A= 2.5 m2, P =1.3 kg m–3. According to Bernoulli’s theorem 2 2 1 1 2 2 1 1 P v P v 2 2 + r = + r 2 2 2 1 1 2 1 P P (v v ) 2 - = r - = 2 2 2 1 1 P P 3 (70 63 ) 2 - = ´ ´ - Force (lift) on the wing = 2 1 A (P P ) - = 2 2 1 2.5 1.3 (70 63 ) 2 ´ ´ ´ - = 1 2.5 1.3 133 7 2 ´ ´ ´ ´ 3 1.5 10 N = ´ Example10. In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air. Solution : Here, r = 2.0 × 10–5 m, r = 1.2 × 103 kg m–3, h = 1.8 × 10–5 Pa s. From formula, terminal velocity 2 2 r ( ) g V 9 r - s = h Þ 5 2 3 5 2 (2 10 ) (1.2 10 0) 9.8 V 9 1.8 10 - - ´ ´ ´ - ´ = ´ ´ 2 1 5.8 10 ms - - = ´ Now viscous force on the drop F= 6phrv Þ 5 5 2 22 F 6 (1.8 10 ) (2 10 ) 5.8 10 7 - - - = ´ ´ ´ ´ ´ ´ ´ = 3.93 × 10–19 N
- 272.
- 273. 269 Mechanical Properties ofFluids 1. The constant velocity attained by a body while falling through a viscous medium is termed as (a) critical velocity (b) terminal velocity (c) threshold velocity (d) None of these 2. The difference between viscosity and solid friction is/are (a) viscositydepends on area while solid friction does not (b) viscosity depends on nature of material but solid friction does not (c) both (a) and (b) (d) neither (a) nor (b) 3. Water is not used in thermometer because (a) it sticks to glass (b) its shows anamalous expansion (c) both (a) and (b) (d) neither (a) nor (b) 4. Toricelli’s theorem is used to find (a) the velocity of efflux through an orifice. (b) the velocity of flow of liquid through a pipe. (c) terminal velocity (d) critical velocity 5. Gases do not posses (a) density (b) surface tension (c) volume (d) viscosity 6. Paint-gun is based on (a) Bernoulli’stheorem (b) Archimedes’ principle (c) Boyle’s law (d) Pascal’s law 7. Water is flowing through a horizontal pipe having a restriction, then (a) pressure will be greater at the restriction. (b) pressure will be greater in the wider portion. (c) pressure will be samethroughout the length ofthe pipe. (d) None of these 8. Fevicol is added to paint to be painted on the walls, because (a) it increases adhesive force between paint and wall. (b) it decreases adhesive force between paint and wall molecules. (c) it decreases cohesive force between paint molecules. (d) None of these 9. A beaker containing a liquid of densityr moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is (a) h r g (b) h r (g – a) (c) h r (g + a) (d) ÷ ÷ ø ö ç ç è æ - + r a g a g g h 2 10. Thedensityof ice is x gram/litre and that ofwater is ygram/ litre. What is the change in volume when m gram of ice melts? (a) mx y(x – y) (b) m/(y– x) (c) ÷ ÷ ø ö ç ç è æ - x 1 y 1 m (d) (y– x)/x 11. Consider a 1 c.c. sample of air at absolute temperature T0 at sea level and another 1 c.c. sample of air at a height where the pressure is one-third atmosphere. The absolute temperature T of the sample at the height is (a) equal to T0/3 (b) equal to 3/T0 (c) equal to T0 (d) cannot bedetermined in termsofT0 from theabove data 12. Asmall ball (menu)falling under graivtyin a viscousmedium experiences a drag force proportional to the instantaneous speed u such that drag F Ku = . Then the terminal speed of the ball within the viscous medium is (a) mg K (b) K mg (c) K mg (d) 2 K mg ÷ ø ö ç è æ 13. A cylinder is filled with non-viscous liquid of densityd to a height h0 and a hole is made at a height h1 from the bottom of the cylinder. The velocity of the liquid issuing out of the hole is (a) 0 gh 2 (b) ) h h ( g 2 1 0 - (c) 1 dgh (d) 0 dgh 14. The terminal velocity depends upon (a) r 1 (b) 2 r 1 (c) 3 r 1 (d) 2 r 15. The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon (a) size of orifice (b) height of liquid (c) acceleration due to gravity (d) density of liquid 16. At the boiling point of a liquid, surface tension (a) is zero (b) is infinite (c) is same as that at any other temperature (d) cannot be determined 17. Thesurface energyof a liquid drop ofradius r is proportional to (a) r3 (b) r2 (c) r (d) r 1
- 274. 270 PHYSICS 18. Aliquid is contained in a vessel. The liquid-solid adhesive force is very weak as compared to the cohesive force in the liquid. The shape of the liquid surface will be (a) horizontal (b) vertical (c) concave (d) convex 19. Two liquids drops coalesce to form a large drop. Now, (a) energy is liberated (b) energy is neither liberated nor absorbed (c) some mass gets converted into energy (d) energy is absorbed 20. A man is sitting in a boat which is floating in pond. If the man drinks some water from the pond, the level of water in the pond will (a) rise a little (b) fall a little (c) remain stationary (d) none of these 21. A liquid is allowed to flow into a tube of truncated cone shape. Identify the correct statement from the following. (a) The speed is high at the wider end and high at the narrow end. (b) The speed is low at the wider end and high at the narrow end. (c) The speed is same at both ends in a stream line flow. (d) The liquid flows with uniform velocityin the tube. 22. The rain drops falling from the sky neither injure us nor make holes on the ground because they move with (a) constant acceleration (b) variable acceleration (c) variable speed (d) constant terminal velocity 23. The lift of an air plane is based on (a) Torricelli'stheorem (b) Bernoulli's theorem (c) Lawof gravitation (d) conservation of linear momentum 24. Surface tension may be defined as (a) the work done per unit area in increasing the surface area of a liquid under isothermal condition (b) the work done per unit area in increasing the surface area of a liquid under the adiabatic condition (c) the work done per unit area in increasing the surface area ofliquid under both isothermal and adiabatic condition (d) free surface energy per unit volume 25. A liquid does not wet the sides of a solid, if the angle of contact is (a) Zero (b) Obtuse (more than 90°) (c) Acute (less than 90°) (d) 90° (right angle) 1. The reading of spring balance when a block is suspended from it in air is 60 newton. This reading is changed to 40 newton when the block is submerged in water. The specific gravity of the block must be therefore (a) 3 (b) 2 (c) 6 (d) 3/2 2. An ice-berg floating partlyimmersed in sea water ofdensity 1.03 g/cm3. The densityoficeis 0.92 g/cm3. The fraction of the total volume of the iceberg above the level of sea water is (a) 8.1% (b) 11% (c) 34% (d) 0.8% 3. A boat having a length of 3 metres and breadth 2 metres is floating on a lake. The boat sinks by one cm when a man gets on it. The mass of the man is (a) 60kg (b) 62kg (c) 72kg (d) 128kg 4. The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume ofthe first bubble is n times the volume ofthe second where n is (a) 0.125 (b) 0.250 (c) 1 (d) 2 5. The level of water in a tank is 5m high. Aholeofarea 1 cm2 is made in the bottom of the tank. The rate of leakage of water from the hole(g = 10 m/s2) is (a) 10–2 m3/s (b) 10–3 m3/s (c) 10–4 m3/s (d) 103 m3/s 6. A spherical ball of iron of radius 2 mm is falling through a column of glycerine. If densities of glycerine and iron are respectively1.3 ×103 kg/m3 and8 ×103 kg/m3. hforglycerine = 0.83 Nm–2 sec, then the terminal velocityis (a) 0.7 m/s (b) 0.07m/s (c) 0.007m/s (d) 0.0007m/s 7. Acylinder ofheight 20m is completelyfilledwith water. The velocityof efflux of water (in ms–1) through a small hole on the side wall or the cylinder near its bottom is (a) 10m/s (b) 20m/s (c) 25.5m/s (d) 5 m/s 8. 8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of (a) 1 (b) 2 (c) 4 (d) 6 9. Water risesto aheight of10 cm in capillarytube and mercury falls to a depth of 3.1 cm in the same capillary tube. If the density of mercury is 13.6 and the angle of contact for mercuryis 135°, the approximate ratioof surface tensions of water and mercuryis (a) 1:0.15 (b) 1: 3 (c) 1: 6 (d) 1.5:1 10. A vessel with water is placed on a weighing pan and it reads 600 g. Now a ball of mass 40 g and density 0.80 g cm–3 is sunkinto the water with a pin ofnegligible volume, as shown in figure keeping it sunk. The weighting pan will show a reading (a) 600g (b) 550g (c) 650g Weighing pan (d) 632g
- 275. 271 Mechanical Properties ofFluids 11. Water flows in a stream line manner through a capillarytube ofradius a, the pressure difference being P and the rate flow Q. If the radius is reduced to 2 a and the pressureis increased to 2P, the rate of flow becomes (a) 4Q (b) Q (c) 2 Q (d) Q 8 12. A rain drop ofradius 0.3 mm has a terminal velocityin air = 1 m/s. The viscosity of air is 8 × 10–5 poise. The viscous force on it is (a) 45.2 × 10–4 dyne (b) 101.73×10–5 dyne (c) 16.95 × 10–4 dyne (d) 16.95 × 10–5 dyne 13. A big drop of radius R is formed by 1000 small droplets of water. The radius of small drop is (a) 10 R (b) 100 R (c) 500 R (d) 1000 R 14. 1 m3 water is brought inside the lake upto 200 metres depth from the surface of the lake. What will be change in the volume when the bulk modulus of elastically of water is 22000 atmosphere? (density of water is 1 × 103 kg/m3 atmosphere pressure = 105 N/m2 and g = 10 m/s2) (a) 8.9 × 10–3 m3 (b) 7.8 × 10–3 m3 (c) 9.1 × 10–4 m3 (d) 8.7 × 10–4 m3 15. Horizontal tube of non-uniform cross-section has radii of 0.1 m and 0.05 m respectively at M and N for a streamline flow ofliquid the rate of liquid flow is M N (a) continuously changes with time (b) greater at M than at N (c) greater at N than at M (d) same at M and N 16. There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm = 105 N/m2) then the velocityof water flowing from hole is (a) 400 m/s (b) 600 m/s (c) 60 m/s (d) None of these 17. In the figure, the velocityV3 will be A = 0.2 m 2 2 V = 4 ms 1 –1 V = 2 ms 2 –1 A = 0.2 m 1 2 A = 0.4 m 3 2 V3 (a) Zero (b) 4 ms–1 (c) 1 ms–1 (d) 3 ms–1 18. 1 centipoise is equal to (a) 1 kg m–1 s–1 (b) 1000 kg m–1 s–1 (c) 0.1 kg m–1 s–1 (d) 0.001 kg m–1 s–1 19. If a water drop is kept between two glass plates, then its shape is (a) (b) (c) (d) None of these 20. The fraction of a floating object of volume V0 and density d0 above the surface of liquid of density d will be (a) 0 0 d d d - (b) d d d 0 - (c) d d0 (d) 0 0 d d d d + 21. An open vessel containing water is given a constant acceleration a in the horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle q, given by (a) a g cos 1 - = q (b) g a tan 1 - = q (c) g a sin 1 - = q (d) a g tan 1 - = q 22. Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be (a) 10 cm per sec (b) 2.5 cm per sec (c) 5 × (4)1/3 cm per sec (d) 3 5´ cm per sec 23. Atank is filled with water upto a height H. Water is allowed tocome out of a hole P in oneof the walls at a depth h below the surface of water (seefig.) Express the horizontal distance X in terms Hand h. h P x H (a) ) h H ( h X - = (b) ) h H ( 2 h X - = (c) ) h H ( h 2 X - = (d) ) h H ( h 4 X - = 24. A bodyof density ' r is dropped from rest at a height h into a lake of density r where ' r > r neglecting all dissipative forces, calculatethe maximum depth towhich the bodysinks before returning to float on the surface : (a) ' h r - r (b) r r' h (c) ' ' h r - r r (d) ' h r - r r
- 276. 272 PHYSICS 25. Twocapillaryof length Land 2Land of radius R and 2R are connected in series. The net rate of flow of fluid through them will be (given rateto the flowthrough single capillary, X= L 8 PR4 h p ) (a) X 9 8 (b) X 8 9 (c) X 7 5 (d) X 5 7 26. One drop of soap bubble of diameter D breaks into 27 drops having surface tension s. The change in surface energy is (a) 2psD2 (b) 4psD2 (c) psD2 (d) 8psD2 27. In case A, when an 80 kg skydiver falls with arms and legs fully extended to maximize his surface area, his terminal velocity is 60 m/s. In Case B, when the same skydiver falls with arms and legs pulled in and bodyangled downward to minimize his surface area, his terminal velocityincreases to 80 m/s. In going from Case A to Case B, which of the following statements most accurately describes what the skydiver experiences? (a) Fair resistance increases and pressure P increases (b) Fair resistance increases and pressure P decreases (c) Fair resistance decreases and pressure P increases (d) Fair resistance remains the same and pressure P increases 28. A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontallyfrom the pan of a balance, so that it comes in contact with the water in a glass vessel. If an extra 3.97. If is required to pull it away from water, the surface tension of water is (a) 72 dyne cm–1 (b) 70.80 dyne cm–1 (c) 63.35 dyne cm–1 (d) 60 dyne cm–1 29. A capillary tube of radius r is immersed in a liquid. The liquid rises to a height h. The corresponding mass is m. What mass of water shall rise in the capillaryifthe radius of the tube is doubled? (a) m (b) 2m (c) 3m (d) 4m 30. In a satellite moving round any planet, the gravitational force is effectively balanced. If an ice cube exists there, and it melts with passageof time, its shape will (a) remain unchanged (b) change to spherical (c) become oval-shaped with long-axis along the orbit plane (d) become oval-shaped with long axis perpendicular to orbit plane 31. An egg when placed in ordinarywater sinks but floats when placed in brine. This is because (a) density of brine is less than that of ordinary water (b) density of brine is equal to that of ordinary water (c) density of brine is greater than that of ordinary water (d) None of these 32. Two pieces of metals are suspended from the arms of a balance and are found to be in equilibrium when kept immersed in water. The mass of one piece is 32 g and its density 8 g cm–3. The density of the other is 5 g per cm3. Then the mass of the other is (a) 28 g (b) 35 g (c) 21 g (d) 33.6g 33. Figure shows a weigh-bridge, with a beaker P with water on one pan and a balancing weight R on the other. A solid ball Q is hanging with a thread outside water. It has volume 40 cm3 and weighs 80 g. If this solid islowered to sink fullyin water, but not touching the beaker anywhere, the balancing weight R' will be Q P R (a) same as R (b) 40 g less than R (c) 40 g more than R (d) 80 g more than R 34. Figure here shows the vertical cross section of a vessel filled with a liquid of densityr. The normal thrust per unit area on the walls of the vessel at the point P, as shown, will be q h P H (a) hrg (b) Hrg (c) (H – h) rg (d) (H – h) rg cosq 35. Two vessels A and B of cross-sections as shown in figure contain a liquid up to the same height. As the temperature rises, the liquid pressure at the bottom (neglecting expansion of the vessels) will A B (a) increase in A, decrease in B (b) increase in B, decrease in A (c) increase in both A and B (d) decrease in both A and B
- 277. 273 Mechanical Properties ofFluids 36. A beaker with a liquid of density 1.4 g cm–3 is in balance over one pan of a weighing machine. If a solid of mass 10 g and density 8 g cm–3 is now hung from the top of that pan with a threadand sinking fullyin theliquid without touching the bottom, the extra weight to be put on the other pan for balance will be (a) 10.0g (b) 8.25g (c) 11.75g (d) –1.75g 37. The pressure energy per unit volume of a liquid is (a) r P (b) P (c) P × r (d) P r 38. A water tank of height 10m, completelyfilled with water is placed on a level ground. It has two holes one at 3 m and the other at 7 m from its base. The water ejecting from (a) both the holes will fall at the same spot (b) upper holewill fall farther than that from the lower hole (c) upper hole will fall closer than that from the lower hole (d) more information is required 39. A fast train goes past way side station platform at high speed. A person standing at the edge of the platform is (a) attracted to train (b) repelled from train (c) unaffected by outgoing train (d) affected onlyif the train's speed is more than the speed of sound 40. Three tubes X, Y and Z are connected to a horizontal pipe in which ideal liquid is flowing. The radii of the tubes X, Y and Z at the junction are respectively3 cm, 1 cm and 3 cm. It can be said X Y Z (a) the height of the liquid in the tube A is maximum. (b) the height of liquid in the tubes Aand B is same. (c) the height liquid in the tubes A,B and C is same. (d) the height of the liquid in the tubes A and C is the same. 41. A sphere of brass released in a long liquid column attains a terminial speed v0. If the terminal speed attained by sphere ofmarble ofthe same radius and released in the same liquid is nv0, then the value of n will be (Given : The specific gravities of brass, marbles and the liquid are 8.5, 2.5 and 0.8 respectively.) (a) 17 5 (b) 77 17 (c) 31 11 (d) 5 17 42. Figure shows a capillary rise H. If the air is blown through the horizontal tube in the direction as shown then rise in capillarytube will be H (a) =H (b) >H (c) <H (d) zero 43. Two soap bubbles (surface tension T) coalesce to form a big bubble under isothermal condition. Ifin this process the change in volume be V and the surface area be S, then the correct relation is (P is atmospheric pressure) (a) PV+TS =0 (b) 3PV+4TS=0 (c) 3PV+TS =0 (d) 4PV+3TS=0 44. Two liquids of densities d1 and d2 are flowing in identical capillary tubes uder the same pressure difference. If t1 and t2 are time taken for the flow of equal quantities (mass) of liquids, then the ratio of coefficient of viscosity of liquids must be (a) 2 2 1 1 t d t d (b) 2 1 t t (c) 1 2 1 2 t t d d (d) 2 2 1 1 t d t d 45. A tank has a small hole at its botom of area of cross-section a. Liquid is being poured in the tank at the rate Vm3/s, the maximum level ofliquid in the container will be (Area of tank A) (a) gaA V (b) gAa 2 V2 (c) gAa V2 (d) gaA 2 V Directions for Qs. (46 to 50) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethecorrectanswer (ONLYONEoptioniscorrect)fromthefollowing- (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 46. Statement 1 : Smaller the droplets of water, spherical they are. Statement 2 : Force of surface tension is equal, and opposite to force of gravity. 47. Statement 1 : If a bodyis floating in a liquid, the densityof liquid is always greater than the density of solid. Statement 2 : Surface tension is the property of liquid surface. 48. Statement 1 : The velocity of flow of a liquid is smaller when pressure is larger and vice-versa. Statement 2 : According to Bernoulli’s theorem, for the stream line flow of an ideal liquid, the total energy per unit mass remains constant.
- 278. 274 PHYSICS Exemplar Questions 1.A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicatethe one that represents the velocity (v) of the pebble as a function of time (t) . v t (a) v t (b) v t (c) v t (d) 2. Which of the following diagrams does not represent a streamline flow? (b) (a) (d) (c) 3. Along a streamline, (a) the velocityof a fluid particle remains constant (b) the velocity of all fluid particles crossing a given position is constant (c) the velocity of all fluid particles at a given instant is constant (d) the speed of a fluid particle remains constant 4. An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is (a) 9: 4 (b) 3: 2 (c) 3 : 2 (d) 2 : 3 5. The angle of contact at the interface of water-glass is 0°, ethyl alcohol-glass is 0°, mercury-glass is 140° and methyliodide-glass is 30°.Aglass capillaryis put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is (a) water (b) ethylalcohol (c) mercury (d) methyliodide NEET/AIPMT (2013-2017) Questions 6. The wetability ofa surface by a liquid depends primarilyon (a) surface tension [2013] (b) density (c) angle of contact between the surface and the liquid (d) viscosity 7. A fluid is in streamline flow across a horizontal pipe of variable area ofcross section. For this which of the following statements is correct? [NEET Kar. 2013] (a) The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe (b) The velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe (c) Velocity and pressure both are maximum at the narrowest part of the pipe (d) Velocityand pressure both are maximum at the widest part of the pipe 49. Statement 1 : Falling raindrops acquire a terminal velocity. Statement 2 : A constant force in the direction of motion and a velocity dependent force opposite to the direction of motion, always result in the acquisition of terminal velocity. 50. Statement 1 : Thebuoyant force on a submerged rigid object can be considered to be acting at the centre of mass of the object. Statement 2 : For a rigid body a force field distributed uniformlythrough its volume can be considered tobe acting at the centre of mass of the body.
- 279. 275 Mechanical Properties ofFluids 8. A certain number ofspherical drops of a liquid of radius ‘r’ coalesce to form a single drop ofradius ‘R’ and volume ‘V’. If ‘T’ is the surface tension of the liquid, then : [2014] (a) energy = 1 1 4VT r R æ ö - ç ÷ è ø is released (b) energy = 1 1 3VT r R æ ö + ç ÷ è ø is absorbed (c) energy = 1 1 3VT r R æ ö - ç ÷ è ø is released (d) energy is neither released nor absorbed 9. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the forcewill be (rair = 1.2 kg/m3) [2015] (a) 4.8 × 105 N, upwards (b) 2.4 × 105 N, upwards (c) 2.4 × 105 N, downwards (d) 4.8 × 105 N, downwards 10. The cylindrical tube ofa spray pump has radius, R, one end ofwhich has n fine holes, each ofradius r. Ifthe speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is : [2015 RS] (a) 2 2 VR nr (b) 2 3 2 VR n r (c) 2 V R nr (d) 2 2 2 VR n r 11. Water rises to a height 'h' in a capillarytube. If the length of capaillarytube above the surface of water is made less than 'h' then : [2015 RS] (a) water rises upto the top of capillary tube and stays there without overflowing (b) water rises upto a point a little below the top and stays there (c) water does not rise at all. (d) Water rises uptothe tip of capillarytube and then starts overflowing like fountain. 12. Two non-mixing liquids of densities r and nr(n > 1) are put in a container. Theheight ofeach liquid is h.Asolid cylinder oflength Land densityd is put in thiscontainer. Thecylinder floats with its axis vertical and length pL(p < 1) in the denser liquid. The density d is equal to : [2016] (a) {1 + (n + 1)p}r (b) {2 + (n + 1)p}r (c) {2 + (n – 1)p}r (d) {1 + (n – 1)p}r 13. A U tube with both ends open tothe atmosphere, is partially filledwith water.Oil,whichisimmisciblewithwater,ispoured intoone side until it stands at a distance of10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The densityof the oil is [2017] A Oil B C Water Initial water level Final water level 10 mm Pa Pa F E D 65 mm 65 mm (a) 425 kg m–3 (b) 800 kg m–3 (c) 928 kg m–3 (d) 650 kg m–3
- 280. 276 PHYSICS EXERCISE -1 1. (b) 2. (a) 3. (c) 4. (a) 5. (b) 6. (a) 7. (b) 8. (a) 9. (c) When a beaker containing a liquid of density r moves up with an acceleration a, it will work as a lift moving upward with acceleration a. The effective acceleration due to gravity in lift = (a + g) Pressure of liquid of height h = h r (a + g) 10. (c) Volume of m g of ice = m/x and volume of m g of water = m/y. So change in volume = ÷ ÷ ø ö ç ç è æ - = - x 1 y 1 m x m y m 11. (a) T P µ 12. (b) 13. (b) Velocityof liquid flowing out of hole = gh 2 . Here h = (h0 – h1) 14. (d) Terminal velocity, ( ) s - r h = 9 g r 2 v 2 T , 2 T r v µ 15. (a) = v velocityof efflux through an orifice gH 2 = 0 v = H v It is independent of the size of orifice. 16. (a) 17. (b) Surface energy µ surface area = pr2 18. (d) 19. (a) When liquid drops coalesce, there is a decrease of surface area and therefore decrease of surface energy. Hence, energyis liberated. 20. (c) 21. (b) The theorem of continuity is valid. A1v1r = A2v2r as the density of the liquid can be taken as uniform. A2 A1 A1v1 = A2v2 Þ Smaller the area, greater the velocity. 22. (b) 23. (b) Apply Bernoulli’s theorem. 24. (a) In Isothermal conditons W T A = D where, T = surface tension, W = work done, DA = change in area. 25. (b) EXERCISE - 2 1. (a) weight of block in air Specific gravity of block loss of weight in water = = 3 40 60 60 = - 2. (b) Let v be the volume of the ice-berg outside the sea water and V be the total volume of ice-berg. Then as per question 0.92 V=1.03(V –v) or v/V= 1–0.92/1.03 = 11/103 (v/V)× 100 = 11× 100 / 103 @ 11% 3. (a) Weight of a man = wt. of water displaced = volume × density = 3 10 100 1 2 3 ´ ´ ´ = 60 kg 4. (a) Given, 2 1 r T 4 2 r T 4 ´ = or r2 = 2r1 3 1 3 2 3 1 ) r 2 ( 3 4 n r 3 4 n r 3 4 p ´ = p ´ = p or 125 . 0 8 1 n = = 5. (b) Velocityof efflux, v 2gh = ; volume of liquid flowing out per sec = v A 2gh A ´ = ´ = 3 4 10 10 5 10 2 - - = ´ ´ ´ m3/s 6. (b) Terminal velocity, h r - r = 9 g ) ( r 2 v 0 2 0 = 83 . 0 9 8 . 9 10 ) 3 . 1 8 ( ) 10 2 ( 2 3 2 3 ´ ´ ´ - ´ ´ ´ - = 0.07 ms–1 7. (b) P.E.= K.E. 20 m = h v 2 1 mgh mv 2 = gh 2 v = = 20 10 2 ´ ´ (Hereg = 10 m/s2) = 20m/s Hints & Solutions
- 281. 277 Mechanical Properties ofFluids 8. (c) Surface energy = suface tension × area of surface For 1 drop, volume = 3 R 3 4 p if = R radius of drop. Total volume of 8 drops 3 R 3 4 . 8 p = ( )3 R 2 3 4 p = , R 2 ' R = new radius of big drop Newarea = 4p4R2 = 4 × old area Energyµ area E1 = 4pR2 .....(1) E2 =4.4pR2 .....(2) From equation (1) and (2) we get, E2 = 4E1 9. (c) g r cos 2 h r q s = h cos r Þ s µ q m m m w w w m w h cos cos h r q ´ q r = s s Þ 6 . 13 1 . 3 135 cos 0 cos 1 10 ´ - ° ´ ° ´ = 6 1 6 . 13 1 . 3 ) 707 . 0 ( 10 » ´ - - ´ = 10. (c) Volume ofball = 3 40 50 cm 0.8 = Downthrust on water = 50 g. Therefore reading is 650 g. 11. (d) 4 a (2P) Q 2 Q' 8 8 æ ö p ç ÷ è ø = = hl 4 Pa Q 8 é ù p = ê ú h ê ú ë û Q l 12. (a) F = 6 p h r n = 6 ×3.14× (8 ×10–5) × 0.03 ×100 = 4.52 ×10–3 dyne 13. (a) 14. (c) K= P V / V D V D = PV K P = h g r = 200 ×103 × 10N/m2 K=22000atm=22000 ×105 N/m2 V=1m3 3 4 3 5 200 10 10 1 V 9.1 10 m 22000 10 - ´ ´ ´ D = = ´ ´ 15. (d) According to principle of continuity, for a streamline flow of fluid through a tube of non-uniform cross- section the rate of flow of fluid (Q) is same at every point in the tube. i.e., Av = constant Þ A1v1 = A2v2 Therefore, the rate offlowof fluid is same at M and N. 16. (a) Pressure at the bottom of tank P = 5 2 3 10 N h g m r = ´ Pressure due to liquid column 5 5 5 1 3 10 1 10 2 10 P = ´ - ´ = ´ and velocity of water 2 v gh = 5 1 3 2 2 2 10 400 m/s 10 P v ´ ´ = = = r 17. (c) According to equation of continuity A1V1 = A2V2 + A3V3 Þ 4 × 0.2 = 2 × 0.2 +0.4 × V3 Þ V3 = 1 m/s. 18. (d) 1 centipoise = 10–2 gcm–1 s–1 = 0.001 kg m–1s–1 19. (c) Angle of contact is acute. 20. (b) Let x be the fraction of volume of object floating above the surface of the liquid. As weight of liquid displaced = weight of object (V0 – x V0)d g = V0 d0 g (1 – x)d = d0 or d d d d d 1 x 0 0 - = - = 21. (d) a g ma mg tan = = q or q = tan–1 g/a 22. (c) If R is radius of bigger drop formed, then 3 3 r 3 4 2 R 3 4 p ´ = p or R = 21/3 r As 2 0 r v µ 3 / 2 2 2 3 / 1 2 2 0 01 2 r ) r 2 ( r R v v = = = or 3 / 1 3 / 2 0 01 ) 4 ( 5 2 v v ´ = ´ = 23. (c) Vertical distance covered by water before striking ground = (H – h). Time taken is, g / ) h H ( 2 t - = : Horizontal velocity of water coming out of hole at P, gh 2 u = Horizontal range = ut 2gh 2(H h)/g = ´ - = ) h H ( h 2 - 24. (c) The effective acceleration of the body g ' 1 ' g ÷ ÷ ø ö ç ç è æ r r - = h Now, the depth to which the body sinks ' g 2 gh 2 ' g 2 u ' h 2 = ÷ ÷ ø ö ç ç è æ = gh h ' g' ' ´r = = r -r 25. (a) Fluid resistance is given by R = 4 8 L r h p When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is 2 1 S R R R + = = 4 4 8 L 8 2L R (2R) h h´ + p p = 8 9 R L 8 4 ´ ÷ ÷ ø ö ç ç è æ p h
- 282. 278 PHYSICS Effective weight= (32 – 4) gf = 28 gf If m be the mass of second body, volume of second body is 5 m Now, m 28 m m 35 g 5 = - Þ = 33. (c) Upthrust = weight of 40 cm3 of water = 40 g = down thrust on water 34. (c) Pressure is proportional to depth from the free surface and is same in all directions. 35. (a) As temperature rises, the density decreases, height increases. In A, the top cross-section is smaller. Therefore hA > hB. 36. (a) 10g is the force on water = extra wt. on other pan. 37. (b) Bernoulli’stheorem. 38. (a) Velocityof water from hole A = gh 2 v1 = Velocityof water from hole B 2 0 v 2g(H h) = = - Time of reaching the ground from hole B g / ) h H ( 2 t 0 1 - = = Time of reaching the ground from holeA g / h 2 t2 = = 39. (a) ApplyBernoulli’s theorem. 40. (d) Usethe equation ofcontinuityand Bernoulli’s theorem. 41. (b) For the same radius, terminal velocity is directly proportional to density difference. 42. (b) Due to increase in velocity, pressure will be low above the surface of water. 43. (b) 44. (a) 45. (b) 46. (c) 47. (d) 48. (a) 49. (a) 50. (d) EXERCISE - 3 Exemplar Questions 1. (c) When the pebble is dropped from the top of cylinder filled with viscous oil and pebble falls under gravity with constant acceleration, but as it is dropped it enter in oil. So dragging or viscous force is F = 6phrv where ris radius ofthe pebble, v isinstantaneousspeed, h is coefficient of viscosity. As the force is variable, hence acceleration is also variable so v-t graph will not be straight line due to viscosity of oil. First velocity increases and then becomes constant known as terminal velocity. 2. (d) In a streamline flow the velocity of fluid particles remaines constant across anycross-sectional area, then a point on the area cannot have different velocities at the same time, hence two streamlines flow layers do not cross each other. Rate of flow = 4 S P PR 8 R 8 L 9 p = ´ h = 9 8 X 4 PR as X 8 L é ù p = ê ú h ê ú ë û 26. (d) Volume of bigger bubble = volume of 27 smaller bubbles 3 3 d 3 4 27 D 3 4 p ´ = p Þ 3 D d = Þ Initial surface energy 2 i D 4 S p = s Final surface energy 2 f S 27 4 d = ´ p s i f S S S - = D and using 3 D d = ú ú û ù ê ê ë é - ´ p ´ s = D 2 2 D 9 D 27 4 S = 2 2 D 8 4 D 2 ps = s ´ p ´ 27. (d) For thefirst part ofthe question, remember that terminal velocity means the acceleration experienced becomes zero. Since a = 0 m/s2, then, 0 F F F w resistance air y = - = S mg F F w resistance air = For the second part of the question, while the velocity is higher, the acceleration is still zero. Therefore, the Fair resistance is still equal to the skydiver’s weight. F air resistance Case A = Fair resistance Case B What has changed is the surface area of the skydiver. Since pressure is P= F/A, asAdecreases, the pressure experienced increases. PAAA = PB AB = mg Since AA > AB , then PA < PB 28. (a) 1 2 (2 r 2 r ) mg p + p s = 980 97 . 3 2 5 . 8 2 2 7 . 8 2 ´ = s ú û ù ê ë é ´ p + ´ p 1 72 dyne cm- Þ s = 29. (b) Mass of liquid which rises in the capillary, r ´ r q s ´ p = r p = g r cos 2 r h ) r ( m 2 2 Þ m µr 30. (b) Because of surface tension. 31. (c) Brine due to its high density exerts an upthrust which can balance the weight of the egg. 32. (b) Volume of first piece ofmetal = 3 32 4cm 8 = Upthrust = 4 gf
- 283. 279 Mechanical Properties ofFluids 3. (b) In streamline flow, the speed of liquid of each particle at a point in a particular cross-section is constant, between two cross-section of a tube of flow because AV = constant (law of continuity). 4. (a) As given that, Diameter at 1st section (d1) = 2.5. Diameter at 2nd section (d2) = 3.75. According to equation of continuity, for cross-sections A1 and A2. A1v1 = A2v2 2 2 1 2 2 2 2 2 1 1 1 ( ) ( ) v A r r v A r r æ ö p = = = ç ÷ p è ø 2 2 2 2 2 1 1 2 2 d d d d æ ö ç ÷ æ ö è ø = = ç ÷ è ø æ ö ç ÷ è ø 2 1 2 1 , 2 2 d d r r é ù = = ê ú ë û Q 2 3.75 9 2.5 4 æ ö = = ç ÷ è ø 1 2 : 9 : 4 v v = 5. (c) We observed that meniscus of liquid is convex shape as shown in figure which is possible if only, the angle of contact is obtuse. Hence, the combination will be of case of mercury-glass (140°). Hence verifies the option (c). 140° convex mercury NEET/AIPMT (2013-2017) Questions 6. (c) Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid. If angle of contact is acute liquids wet the solid and vice-versa. 7. (b) According to Bernoulli’s theorem, 2 1 2 P v + r = constant and Av v = constant IfA isminimum,v ismaximum,Pisminimum. 8. (c) As surface area decreases so energy is released. Energy released = 4pR2T[n1/3 – 1] where R = n1/3r = 3 1 1 4 R T r R é ù p - ê ú ë û = 1 1 3VT r R é ù - ê ú ë û 9. (b) According to Bernoulli’s theorem, 1 2 2 0 P v P 0 + r = + So,DP = 1 2 rv2 F = DPA = 1 2 rv2A P0 P = 1 2 ×1.2×40 ×40×250 = 2.4 × 105 N (upwards) 10. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the liquid pR2V = npr2(v) Þ v= 2 2 2 2 R V VR n r nr p = p 11. (a) Water rises upto the top of capillary tube and stays there without overflowing. 12. (d) As we know, Pressure P = Vdg r d (1 – p)L nr pL Here, LA d g = (pL)A(nr)g + (1 – p)LAr g Þ d = (1 – p)r + pn r = [1 + (n – 1)p]r 13. (c) Here, oil oil water water h g h g ´r ´ = ´r ´ r0g × 140 × 10–3 = rwg × 130 × 10–3 roil = 3 3 130 10 928kg / m 140 ´ » [Q rw= 1 kgm–3]
- 284. 280 PHYSICS TEMPERATURE Thetemperature ofabodyis that physical quantitywhich indicates how much hot or cold the body is? So temperature indicates the hotness or coldness of a body. The flow of heat is always from higher temperature to lower temperature. Temperatureis a macroscopicconcept. Twobodies are said to be in thermal equilibrium with each other, when no heat flows from one bodyto the other, that is when both the bodies are at the same temperature. Temperature measuring device : Thermometer is a device to measure the temperature. Thermometers used for measuring very high temperatures are called pyrometers. Different types of temperature scales Name of the scale Symbol for each degree Lower fixed point Upper fixed point No. of divisions on the scale Reaumur ºR 0ºR 80ºR 80 Celsius ºC 0ºC 100ºC 100 Fahrenheit ºF 32ºF 212ºF 180 Rankine Ra 460Ra 672Ra 212 Kelvin K 273K 373K 100 * Lower fixed point (LFP) is the freezing point of water. * Upper fixed point (UFP) is the boiling point of water. TºK = (tºC + 273) or tºC = (TK – 273) Relationship betweendifferent temperature scales 32 460 273 80 100 180 212 100 - - - = = = = R C F Ra K The zero of the Kelvin scale is called absolute zero. The Kelvin scale is often termed as the absolute scale. In common use the absolute zero corresponds to –273ºC. However, its exact value is –273.16ºC. IDEAL GAS EQUATION AND ABSOLUTE TEMPERATURE The equation, PV = nRT where, n = number of moles in the sample of gas R = universal gas constant; (its value is 8.31 J mol–1 K–1), is known as ideal-gas equation It is the combination of following three laws : (i) Boyle'slaw : When temperature isheld constant, the pressure is inversely proportional to volume. i.e., 1 P V µ ( at constant temperature) (ii) Charle'slaw : When the pressureisheldconstant, thevolume ofthegasisdirectlyporportionalto theabsolutetemperature. i.e., V T µ (at constant pressure) (iii) Avogadro's law : When the pressure and temperature are kept constant, the volume is directly proportional to the number of moles of the ideal gas in the container. i.e., V n µ (at constant pressure and temperature) Absolute Temperature The lowest temperature of –273.16 °C at which a gas is supposed to have zero volume and zero pressure and at which entire molecular motion stops is called absolute zero temperature. A new scale of temperature starting with –273.16°C byLord Kelvin as zero. This is called Kelvin scale or absolutescale oftemperature. T(K)=t°C +273.16 Thermometric property : It is the propertyused to measure the temperature. Anyphysical quantitysuch as length ofmercury(in a glass capillary), pressure ofa gas (at constant volume), electrical resistance (of a metal wire), thermo emf (thermo-couple thermometer)which changes with a change in temperature can be used to measure temperature. The unknown temperature t on a scale by utilising a general property X of the substance is given by 0 100 0 100º æ ö - = ´ ç ÷ - è ø t X X t C X X Keep in Memory 1. The mercury thermometer with cylindrical bulb are more sensitive than those with spherical bulb. 2. Alcoholthermometer ispreferredtothe mercurythermometer dueto thelarger valueof the coefficient ofcubical expansion. 3. Properties that make mercury the ideal thermometric substance are : (i) It does not stick to the glass walls (ii) Availablein pureform (iii) Low thermal conductivityand low specific heat (iv) Vapour Pressure is low 1 1 Thermal Properties of Matter
- 285. 281 Thermal Properties ofMatter (v) Coefficient ofexpansion is uniform (vi) It shines 4. Gas thermometer is more sensitive than the mercury thermometer, because its coefficient of cubical expansion is much larger. 5. At the following temperatures different temperature scales havethesame reading: (i) Fahreinheit and Reaumur at –25.6ºF = –25.6ºR (ii) Reaumur and Celsius at 0ºR = OºC (iii) Celsius and Fahrenheit at –40ºC = –40ºF (iv) Fahrenheit and Kelvin at 574.25ºF = 574.25K 6. The Celsius scale and the Kelvin cannot have the same reading at anytemperature. 7. Six’s thermometer measures minimum and maximum temperature during a day. It uses mercuryas well as alcohol as the thermometric substances. 8. Normal human bodytemperatureis 37ºC = 98.6ºF. 9. Clinical thermometers are much shorter than the laboratory thermometers because they are used to measure a limited range (96º F to 100º F) oftemperature. 10. Rapidlychanging temperature is measured byusing thermo electricthermometer. 11. The temperature inside a motor engine is measured using platinum resistance thermometer. 12. The radiation thermometers can measure temperature from a distance. 13. Adiabatic demagnetisation can be used to measure temperature very near to the absolute zero. 14. TheRankine is the smallest temperature range among all the scales of temperature. 15. When a substance is heated it expands along each dimension in the same proportion. (ifit is uniform) THERMALEXPANSION Thermal Expansion of Substances Linear expansion : i.e., increase in length with increase in temperature T 0D a = D l l or ) T 1 ( T 0 0 0 D a + = Þ D a = - l l l l l where l0 is the initial length, l = final length, DT = change in temperature, a = coefficient of linear expansion. Areal expansion : i.e., increase in area with increase in temperature. 0 D = b D A A T where b = coefficient of area expansion Volume expansion : i.e. increase in volume with increase in temperature. 0 D = g D V V T where g = coefficient of volume expansion. The units of a, b and g is (°C)–1 or K–1. Graph of av versus T: 500 T (K) 250 (For copper) a ® v (10 K ) –5 –1 Keep in Memory 1. During change in temperature, the mass does not change but the density decreases due to increase in volume. Therefore 0 1 r r = + gDT 2. Relation between a, b andg a = g 3 and a = b 2 3. Water has negative coefficient of volume expansion for temperature range (0°C – 4°C). Therefore water contracts when the temperaturerisesfrom 0°C to4°Cand then expands as the temperature increases further. Densityofwater reaches a maximum valueof1000 kg m–3 at 4°C.This is the anomalous behaviour of water. Vo lume Temperature 4°C Volume Temperature 4°C Thermal ExpansionofLiquid(whichisheated inavessel) T ) 3 ( V Vapp D a - g = D where V D app = apparent change in volume V= volume of theliquid initially g = coefficient ofvolume expansion ofliquid a = coefficient of linear expansion of vessel Itisclear from theformula that if a = g 3 , 0 Vapp = D butif a > g 3 , app V D is positive and the liquid rises in the vessel and if a < g 3 , app V D is negative and liquid falls in vessel. Change in density of solidsandliquidswith temperature. The density at t°C is given by r r = + g 0 (1 ) t t where r0 is the density at 0°C. If r1 and r2 are the densities at temperature t1 and t2 respectively, then r -r g = r - 1 2 2 2 1 ( ) t t
- 286. 282 PHYSICS Isothermal andadiabatic elasticities: (i) Isothermal elasticity, . P V dV dP Eiso = ÷ ø ö ç è æ - º (since PV = RT = constant, V P dV dP - = ) (ii) Adiabatic elasticity, P V dV dP Eadia g = ÷ ø ö ç è æ - = ; where v p C C = g (iii) iso adia E E g = Expansionof Gases There are two coefficients of expansion of gases. (a) Pressure coefficient of expansion of gas (gP) 0 0 - g = t P P P P t where P0 = Pressure at 0°C, Pt = Pressure at t°C (b) Volume coefficient of expansion of gas (gv) 0 0 - g = t v V V V t where V0 = Volume at 0°C, Vt =Volume at t°C HEAT Heat is the form of energy which is transferred from one body to another body due to temperature difference between two bodies. Thermal energyof a bodyis the sum of kinetic energy of random motion of the molecules/atoms and the potential energy due to the interatomic forces. Calorimeter is an instrument used to measure the heat. The SIunitofheat isjouleand itsdimensionsare[ML2 T–2]. Other units of heat are : (i) Calorie: It is the amount of heat required to raise the temperatureof1gofwater from14.5ºCto15.5ºC. 1 calorie= 4.2joule. Calorieis theunit ofheat in CGSsystem. (ii) Kilocalorie,1Kcal= 1000cal (iii) British Thermal Unit (BTU): It is the fps unit of heat = 252 cal = amount of heat required to raise the temperature of1 lb ofwater through 1ºF (from 63°F to 64ºF). SPECIFIC HEAT CAPACITY It is the amount of heat required to raise the temperature of unit mass of substance through 1 degree. Specific heat capacity, = ´ D Q c m t Its SI unit is J kg–1K–1 CGS unit cal g–1°C–1 It dimensions are [M0L2T–2q–1] For gases : During the isothermal process Dt = 0 Þ c = ¥ (infinite) Thermal Stress If a rod fixed at two ends is heated or cooled, then stress are produced in the rod. Thermal stress = T Y D a ´ 0 T strain æ ö D = aD = ç ÷ è ø l l Effectof Temperature on PendulumClock A Pendulum clock consists of a metal bar attached with a bob at oneend and fixed at the other end. Let pendulum clock read correct time when its length is l0. The time period t0 = l0 g 2π Suppose that the temperatureis raised byDT, then newtime period is g t 2 = p l = Þ l l 0 0 t t .....(i) We have = + aD l l0(1 t) Þ = + aD l l0 (1 t) .....(ii) From equations (i) and (ii), = + aD 0 1 2 t (1 t) t = + aD 1 2 1 t (since a is verysmall) Þ - = aD 0 t 1 1 t 2 t - Þ = aD 0 0 t t 1 t 2 t so, fractional change in time D = aD 0 t 1 t 2 T Loss of time per day = aD ´ 1 2 T 86400 second . (Q no. of seconds in one day = 86400 sec.) When temperature increases (during summer), the length of the pendulum increases due to which the time period increases and the clock loses time. On the other hand, when temperature decreases (during winter season), the length decreases and the time period decrease, the clock in this case, gains time. Error in scale reading : As the temperature changes, the length ofthe metal scale changes so is the difference in graduation a m R R (1 T) = + aD where Ra = actual reading, Rm = measured reading
- 287. 283 Thermal Properties ofMatter and during the adiabatic process Q = 0 Þ c = 0 (zero) Specific heat of gas : The specificheat ofa gas varies from zeroto infinity. It may have any positive or negative value. Following of two specific heats of gas are more significant. (a) Molar specific heat at constant volume (CV) : It is the amount of heat required to raise the temperature of 1 mole of a gas through 1K or 1°C at constant volume. 1 Q T æ ö = ç ÷ è ø V d C n d (b) Molar specific heat at constant pressure (CP) : It is the amount of heat required toraise the temperature of1 mole of a gas through 1K or 1°C at constant pressure. 1 C æ ö = ç ÷ è ø P dQ n dT Mayer’s relation : CP – CV = R [where R is a gas constant. (R = 8.314 Jmol–1K–1)] Why CP is greater thanCV ? When heat is supplied to a gas at constant volume entirely used to raise its temperature. When a gas is heated at constant pressure, some work is done in expansion of gas which is in addition to raise its temperature and hence CP > CV. THERMAL CAPACITY OR HEAT CAPACITY It is the amount of heat required to raise the temperature of substance through 1ºC or 1K. Thermal capacity = mass × specific heat. Thisis the relationbetween sp. heat capacity and heat capacity. Its SI unit is JK–1 CGS unit cal/°C. Its dimensions are [ML2T– 2q–1] Water Equivalent It is the mass of water in gram which would require the same amount of heat to raise its temperature through 1ºC as the body when heated through the same temperature. It is measured in gram. Hence Water equivalent of a body = mass of the body × specific heat Principle of Heat (or Calorimetry) When two bodies at different temperatures are placed in contact with each other then heat will flow from the body at higher temperature to the bodyat lower temperature until both reach toa common temperature. i.e. Heat lost by hot body = heat gained by cold body. It follows the law of conservation of energy. LATENT HEAT It is defined as the amount of heat absorbed or given out by a body during the change of state while its temperature remaining constant. It is of two types : (i) Latent heat of fusion : It is the quantityof heat required to change unit mass ofthe solid intoliquid at its melting point. Latent heat of fusion of ice is 80 cal/g. (ii) Latentheat ofvaporisation:Itisthequantityofheatrequired to convert unit mass of liquid into vapour at its boiling point. Latent heat of vaporisation of water is 536 cal/g. Latent heat increases the intermolecular potential energy of the molecules while kinetic energyremains constant. Different Processes of Phase Change or State Change (i) Sublimation : It is the process of conversion of solids to gaseous state on heating. On cooling the vapours get converted back into solids. (ii) Condensation : It is the process of conversion of gases/ vapours toliquid state on cooling. On heating, these liquids are converted into vapours/gases. (iii) Boiling : It is the process of conversion of liquid to gaseous state on heating. It is a cooling process. (iv) Melting : It is the process in which solid gets converted intoliquid on heating. (v) Evaporation : Conversion of liquid intogaseous state at all temperatures is called evaporation. It is a phenomenon that occurs at the surface of the liquid. Melting and boiling occur at definite temperature called melting point and boiling point respectively. The liquid boils at a temperature, at which its vapour pressure is equal to the atmospheric pressure. Howevaporationisdifferentfromboiling? Evaporation is a slow process occurring at all the temperature at the surface ofliquid while boiling is a fast process involving whole ofthe liquid heated toa particular temperature called boiling point ofthe liquid. Keep in Memory 1. The heat capacity or thermal capacitydepends on nature of the substance (specific heat capacity) and mass of quantity of matter of the body. 2. In case of melting and boiling, the temperature does not change, only intermolecular potential energyof the system changes. 3. System are in thermal equilibrium when their temperature are same or average kinetic energyper molecule is same. 4. Internal energyis the sum of all energies in a system which includes energy of translational, rotational as well as vibrational motion of the molecules. 5. The molar specific heat capacity depends on the molecular weight of the substance. 6. Thetemperature, volume or pressure ofa system mayremain constant when it absorbs heat. 7. When we press two block of ice together such that after releasing the pressure two block join this phenomenon is called regelation.
- 288. 284 PHYSICS PhaseDiagram D B C P E F T O A TC Ttr P Liquid Vapour Solid 4.58 mm ofmercury 273.16ºK water (i) The diagram shows vaporisation curve (AB), fusion curve (CD), sublimation curve (EF). (ii) For water, AB is called steam line, CD is called ice line and EF is called hoar-frost line. (iii) The temperature at which all the three states of matter (solid, liquid, gas) are mutuallyin thermal equilibrium is called the triple point (point P in diagram) of the substance. (iv) The temperature upto which a gas can be liquified under pressure alone is called the critical temperature (Tc). Example1. A solid material is supplied with heat at a constant rate. The temperature of the material is changing with the heat input as shown in Fig. Study the graph carefully and answer the following : (i) What is represented by horizontal regions AB & CD? (ii) If CD = 2 AB. What do you infer? Solution : A B C D E Heat Supplied Temp. Y O X 0ºC(ice) (0ºC liquid) (100ºC liquid) (100ºC Vapour) Q1 Q2 Q3 Q4 (i) In the horizontal regions AB and CD heat is supplied but temp. remains constant. Therefore, theyrepresent change of state. The graphAB represents change from solid state to liquid state, at constant temp. = melting point of solid. AB = latent heat of fusion. Again CD represents change of state from liquid to vapour at boiling point of the liquid. CD = latent heat of vaporisation. (ii) As CD = 2AB Latent heat of vaporisation is twice the latent heat of fusion. Example2. A clock which keeps correct time at 25ºC has a pendulum made of brass whose coefficient of linear expansion is 0.000019. How many seconds a day will it gain if the temperature fall to 0ºC? Solution : Let L0 and L25 be the length of pendulum at 0ºC and 25ºC respectively. We know that L25 = L0(1+ aT)= L0(1+ 0.000019×25)= 1.000475L0 IfT25 andT0 bethe timeperiodsat 25ºC and 0ºC respectively, then T25 = 2p 25 L g æ ö ç ÷ è ø and T0 = 2p 0 L g æ ö ç ÷ è ø 25 0 T T = 25 0 L L æ ö ç ÷ è ø = 0 0 1.000475L L æ ö ç ÷ è ø = (1.000475) = 1.000237 Now 25 0 0 T T T - =0.000237 Gain in time for one vibration = 2 × 0.000237 sec. Number of vibration in one day= 24 60 60 2 ´ ´ (T = 2 sec.) Hence the gain in time in one day =2 ×0.000237 × 24 60 60 2 ´ ´ =20.52sec. Example3. A circular hole in an aluminium plate is 2.54 cm in diameter at 0ºC. What is the diameter when the temperature of the plate is raised to 100ºC ? Given aA = 2.3 × 10– 5 (ºC) – 1 Solution : Let D0 and Dt be diameters of hole at 0ºC and tºC respectively. Circumference ofhole at 0ºC l0 = 2pr0 = pD0 Circumference of holeat t = 100ºC lt = 2prt = pDt From relation lt = l0 (1 + a.t), we get pDt = pD0 (1 + 2.3 × 10–5 ×100) Dt =2.54(1+0.0023)= 2.5458cm Example4. If the volume of a block of a metal changes by 0.12%when it is heated through 20ºC, what is the coefficient of linear expansion of metal ? Solution : Coefficient of cubical expansion of metal is given by g = V Vt D Here g = V 0.12 Vt 100t D = , t = 20ºC g = 0.12 100 20 ´ = 6.0 × 10–5 per ºC Coefficient of linear expansion a = 5 6.0 10 3 3 - g ´ = = 2.0 × 10–5 Per ºC
- 289. 285 Thermal Properties ofMatter Example5. From what height must a block of ice be dropped into a well of water so that 5% of it may melt? Given : both ice and water are at 0°C, L = 80 cal g–1, J = 4.2 J cal–1 and g = 980 cm s–2. Solution : Let m be the mass of ice. Let h be the height from which block of ice is dropped. Work done, W = mgh = m× 980 × h erg Mass of ice to be melted = 5 m 100 ´ Heat required, 5 Q 100 = × m × 80 cal or Q = 4 m cal [Q L = 80 cal g–1] [Q W= JQ] Now, m × 980 ×h = J× 4m 7 4.2 10 4m h m 980 ´ ´ = ´ cm= 1714.3m [Q J = 4.2 × 107 erg cal–1] SATURATED AND UNSATURATED AIR The air is said tobe saturatedwhen the maximum possibleamount of water vapours are present in it. The pressure of the water vapours in the saturated air is called saturation vapour pressure. Ifthe air contains vapours less than the maximum possible amount possible in the air, then it (air) is said to be unsaturated. Vapour Pressure A vapour differs from a gas in that the former can be liqufied by pressure alone, wheras the latter cannot be liqufied unless it is first cooled. If a liquid is kept in a closed vessel, the space above the liquid becomes saturated with vapour. The pressure exerted by this vapour is called saturated vapour pressure. Its value depends only on temperature. It is independent of any external pressure. Humidity The humidity shows the presence of water vapours in the atmosphere. The amount ofwater vapours present per unit volume oftheair is called absolute humidity. Relative humidity (RH) is defined as the ratio of the mass of water vapour (m) actually present in the given volume of air to the mass of water vapour (M) required to saturate the same volume at the same temperature. Normally RH is expressed in percentage i.e., % 100% = ´ RH M m RH is also defined as the ratio of actual pressure (p) of water vapour to the saturated vapour pressure (P) of the water at the same temperature i.e., % ×100% = p RH P Also, RH = Saturated vapour pressure of water at dew point Saturated vapour pressure of water at room temperature DewPoint : It is the temperature at which the amount of water vapours actuallypresent in a certain volume of the air which is sufficient to saturate that volume of air. At the dew point the actual vapour pressure becomes the saturated vapour pressure. It means that vapour pressure at room temperature = saturated vapour pressure(S.V.P) at dew point If room teperature = dewpoint then Relative humidity(R.H.) = 100% Keep in Memory 1. Hygrometry is the branch of thermal physics that deals with the measurement ofamount of water vapours present in the atmosphere. 2. Hygrometer is an instrument used todetermine dew point and relative humidity. 3. Relative humidityis low when the air is dry. 4. Saturated vapours do not obey (except Dalton’s law) gas laws, while unsaturated vapours obey them. The S.V.P. varies linearlywith temperature (keeping volume constant) and equal to one atmospheric pressure at boiling point.(shown in fig.(a)) S.V.P. constant at constant temperature on increasing or decreasing volume fig. (b). V = constant T Boiling point 1 atm (S.V.P.) (a) P T = constant V (S.V.P.) (b) P 5. At 0ºC, the saturation vapour pessure is 4.6 mm of Hg. Therefore, water vapours are always present around the ice. It increases with the increase in temperature. 6. The relative humidity is 100% when thetemperatureofthe atmosphere is equal to the dew point. 7. Cryogenics is the study of low temperature. 8. As RH is expressed in percentage, hence the maximum value of RH maybe 100%. 9. The saturation vapour pressure of water at 100ºC is 760 mm ofHg at sea level. 10. As the temperature rises, the absolute humidity may increase and the relative humiditymayremain constant or may even decrease. 11. Ifthe relative humidityis high then at the same temperatue one feels hotter. 12. If the air is absolutley dry, the dew point is not observed. 13. The relative humidity decreases with the increase in temperature. 14. Ifwater is sprinkled in the room, both relative humidityas well as the dew point increase. 15. Dew appears on the leaves of the trees etc. due to the condensation of saturated vapours. 16. Themist or fog is thesmall droplets of water formed due to the condensation of water vapours near the surface of earth. 17. Mist or fog formed much higher above the surface of earth is called cloud. 11.1
- 290. 286 PHYSICS 18. TheRelative humidity between 50% and 60% gives comfortable feeling to human beings. 19. When the relative humidityis 100%, the reading of the dry and wet bulbs is same. 20. Smaller the difference in the temperature of dry and wet bulbs, larger is the relative humidity. 21. Water vapours are always seen around the ice, because the temperature around it is less than the dew point. HEAT TRANSFER Three modes of transmission of heat : (a)Conduction (b) Convection and(c)Radiation (a) Conduction : In thermal conduction particles of body at higher temperature transmit heat to the particles at lower temperature by mutual contact (collision) only and not by the movement of the particles. All solids are heated by conduction. Conduction cannot take place in vacuum. Dx D D Q t D D Q t A T1 T2 T > T 1 2 Figure shows a solid of cross-section area A and thickness Dx. Thefaceareatdifferent temperatureT1 and T2 (T1 > T2) The rate of heat flow ÷ ø ö ç è æ D D t Q as found experimentally is given by ÷ ø ö ç è æ D D µ D D x T A t Q where 2 1 T T T - = D Þ Δ Δ Δ Δ æ ö = ç ÷ è ø Q T KA t x …(1) where K is proportionality constant called thermal conductivity. It is a measure of how quickly heat energy can conduct through the substance. Coefficientof thermalconductivity(K): Thecoefficientofthermal conductivity, K, of a material is defined as the amount of heat that flowing per second through a rod of that material having unit length and unit area of cross-section in the steady state, when the difference of temperature between two ends of the rod is 1 ºC and flow of heat is perpendicular to the end faces of the rod. Unit of coefficient of thermal conductivity in SI system is watt/m-K Dimensions : [M L T–3 q–1] For a perfect conductor thermal conductivity K is infinite and for a perfect insulator K is zero. In general, solids are better conductors than liquids and liquids are better conductors than gases. (Heat transfer through the process of conduction is possible in liquids and gases also, if theyare heated from the top.) Metals are much better conductors than non-metals. A good conductor of heat is also a good conductor of electricity The conduction ofboth heat and electricityis due to the movement of free electrons. (b) Convection : It is the process by which heat is transferred from one place to another in a medium by the movement of particles of the medium. It occurs due to densitydifference. This phenomenon occurs in fluids. (c) Radiation : It is the process by which heat is transferred from one place to another without any intervening medium. The light reaches from Sun to Earth byradiation process. Keep in Memory 1. The equation Δ Δ Δ Δ æ ö = ç ÷ è ø Q T KA t x is valid for steady state condition. The condition is said to occur when no part of heat is used up in raising the temperature of any part of cross-section of the solid. 2. On comparing equation (1)with the following equation used for flowing of charge on account of potential difference. V1 Dq V2 s l We find : l V A t q D s = D D The role of resistance (thermal resistance) is played by Dx KA 3. Series combination of conductors l1 l2 K , A 1 1 T1 T T2 K , A 2 2 Equivalent thermal conductivity: 1 2 1 2 2 = + eq K K K K K 1 2 1 2 1 2 ( – ) A T T H l l K K = + where H = heat flow per second 4. Parallel combination of conductors l K , A 1 1 T1 T2 K , A 2 2 Equivalent thermal conductivity: Keq = 1 1 2 2 2 2 + + K A K A A A 1 2 1 2 1 2 1 2 1 2 K K T + T l l H = K K + l l
- 291. 287 Thermal Properties ofMatter 5. Davy’s safety lamp is based on the conduction. It is used in mines toknow the ignition temperature of gases. Danger of explosion can be avoided. 6. (i) Principle of chimney used in a kitchen or a factory is based on the convection. (ii) Land and sea breezes are due ot the convection. (iii) Temperature of theupper part of theflameis more than the temperature on the sides, because the currents of air carrythe heat upwards. (iv) Radiation can be detected by differential air thermometer, Bolometer, thermopile, etc. Example6. Two plates of same area are placed in contact. Their thickness as well as thermal conductivities are in the ratio 2 : 3. The outer surface of one plate is maintained at 10ºC and that of other at 0ºC. What is the temp. of the common surface? Solution : Let q be the temp. of common surface. As 2 1 t Q t Q ÷ ÷ ø ö ç ç è æ D D = ÷ ÷ ø ö ç ç è æ D D 2 2 2 1 1 1 x ) 0 ( A K x ) 10 ( A K - q = q - or 1 1 2 2 K x (10 θ) θ K x - = K1 x1 x2 K2 Plate 1 Plate 2 10°C 8°C 0°C Now, 3 2 K K x x ; A A 2 1 2 1 2 1 = = = ; . C º 5 or 3 2 ) 10 ( 3 2 = q q = q - Example7. A slab consists of two parallel layers of two different materials of same thickness and thermal conductivities K1 and K2. Find the equivalent thermal conductivity of the slab. Solution : Rate of flow of heat through each layer of slab is same, because slabs are in series. x x DT1 DT2 l l l 2 A ) T T ( K T A K T A K 2 1 2 2 1 1 D + D = D = D or C 2 ) T T ( K T K T K 2 1 2 2 1 1 = D + D = D = D , say K C 2 T T , K C T , K C T 2 1 2 2 1 1 = D + D = D = D or K C 2 K C K C 2 1 = + or 2 1 2 1 K K K K 2 K + = Example8. Two walls of thickness d1 and d2 and thermal conductivities K1 and K2 are in contact. In the steady state, if the temperatures at the outer surfaces are T1 and T2 then find the temperature at the common wall. Solution : K1 T1 T2 T d1 d2 K2 Let the temperature of common wall be T. Then 2 2 2 1 1 1 d t ) T T ( A K d t ) T T ( A K - = - or, 2 2 2 1 1 1 d t ) T T ( K d t ) T T ( K - = - or, )] T T ( K [ d )] T T ( K [ d 2 2 1 1 1 2 - = - or, 2 2 1 2 1 2 1 2 1 1 T K d T K d T d K d T K - = - or, ] K d d K [ T T d K d T K 2 1 2 1 2 1 2 2 1 1 + = - ú û ù ê ë é + - = 1 2 2 1 2 1 2 2 1 1 d K d K T d K d T K T HEAT TRANSFER BY RADIATION AND NEWTON’S LAW OF COOLING Stefan’s Law: This law is also called Stefan Boltzmann law. This law states that the power radiated for overall wavelength from a black body is proportional to the fourth power of thermodynamic temperature T. 4 4 E T E T µ Þ = s for perfectly black body ....(1) where E is energyemitted per second from unit sruface area of the black body, T is temperature and s is Stefan’s constant, and s = 5.67×10–8 Wm–2 K–4. If body is not perfectly black body, then E = es T4 ....(2) where e is emissivity. Then energyradiated per second by a body of area A is E1 = es T4 ....(3) If the body temperature is T and surrounding temperature is T0, then net rate of loss of energy by body through radiation from equation (3) is ' 4 4 1 0 ( ) E eA T T = s - ....(4) Now rate of loss of energy in terms of specific heat c is ' 1 dQ dT mc E dt dt = = ....(5) and 4 4 0 ( ) dT eA T T dt mc s = - ....(6) (T- temperature, t - time in second)
- 292. 288 PHYSICS So bodycools by radiation and rate of cooling depends on e (emissivity of body), A (area of cross-section of body), m (mass of body) and c (specific heat capacity of body). A black body is defined as a body which completely absorbs all the heat radiation falling on it without reflecting and transmitting any of it. Weins Law : lmT = b, where b is the Weins constant and b = 2.898 × 10–3mK and T = temperature. Graphof lm versus T T 2 1 1 2 l = l m m T T Hence on increasing temperature of a body, its colour changes gradually from red to orange ® yellow ® green ® blue ® violet. Thus the temperature of violet star is maximum and temperature of red star is minimum. Sun is a medium categorystar with lm = 4753Aº (yellow colour) and temperature6000K. Kirchoff’s Law According to this law, the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at the same temperature. i.e. e E A a = where e = emissive power of a given surface a = absorptive power of a given surface E = emissive power of a perfect black body A = absorptive power of a perfect black body (i) For a perfect black body, A = 1, = e E a (ii) If emissive and absorptive power are considered for a particular wavelenth l then l l l = e E a (iii) SinceEl is constant at a given temperature, according to this law if a surface is good absorber of a particular wavelength then it is also a good emitter of that wavelength. Fraunhoffer's Line (i) Fraunhoffer lines are the dark lines in the spectrum of sun and are explained on the basis ofKirchoff’s law. When white light emitted from central core of sun (photosphere) passes through its atmosphere (chromosphere) radiation of those wavelength will be absorbed by the gases present there, which theyusuallyemit (as good emitter is a good absorber) resulting dark lines in the spectrum of sun. (ii) On the basis ofKirchoff’s law, Fraunhoffer identified some of the elements present in the chromosphere. They are hydrogen, helium, sodium, iron, calcium, etc. Fraunhoffer had observed about 600 darklines in the spectrum of sun. Newton's Law of Cooling The rate of loss of heat of a body is directly proportional to the temperature difference between the body and the surroundings. i.e. 0 ( ) dT K T T dt = - - where T = temperature of body, T0 = temperature ofsurrounding. and mc eA K s = where, e = emissivity of body, A = area of surface of body, s = Stefan’s constant, m = mass ofbody, c = specific heat of body. It is a special case of Stefan’s lawand above relation is applicable onlywhen the temperature ofbody is not much different from the temperature of surroundings. Solar constant : Solar constant is the solar radiation incident normally per second on one square metre at the mean distance of the earth from the sun in free space. It is given byS = 1.34 × 109 Jm–2s–1. Temperature of sun is given T4 = S/s(R/r)2, where R is mean distance of the earth from the sun and r is the radius of the sun. Keep in Memory 1. Good absorber is a good emitter. 2. Cooking utensils are provided with wooden or ebonite handles, since wood or ebonite is a bad conductor of heat. 3. Good conductor of heat are good conductors of electricity, Mica is an exception which being a good conductor ofheat is a bad conductor of electricity. Example9. Compare the rate of loss of heat from a metal sphere of temperature 827°C, with the rate of loss of heat from the same sphere at 427 °C, if the temperature of surroundings is 27°C. Solution : Given : T1 = 827 °C = 1100 K, T2 = 427 °C = 700 K and T0 = 27 °C = 300 K According to Steafan Boltzmann law ofradiation, dQ dt = s A Ae (T4 – T0 4) 4 4 4 4 1 0 1 4 4 4 4 2 0 2 dQ dt (T T ) [(1100) (300) ] dQ (T T ) [(700) (300) ] dt æ ö ç ÷ è ø - - = = æ ö - - ç ÷ è ø =6.276 or 1 2 dQ dQ : 6.276 :1 dt dt æ ö æ ö = ç ÷ ç ÷ è ø è ø Example10. The filament of an evacuated light bulb has a length 10 cm, diameter 0.2 mm and emissivity 0.2, calculate the power it radiates at 2000 K. (s = 5.67 × 10–8 W/m2 K4) Solution : l= 10cm= 0.1 m, d= 0.2mm =2 × 10–4 m r = 0.1mm=1×10–4 m, e = 0.2, T = 2000 K, s= 5.67 ×10–8 W/m2 K4 According to stefan's law of radiation, rate of emission of heat for an ordinary body, E = sAeT4 = s(2 pr l) eT4 [QA= 2prl] = 5.67 ×10–8 × 2× 3.14 ×1 ×10–4 × 0.1 × 0.2 × (2000)4 = 11.4 W Power radiated bythe filament = 11.4 W
- 293.
- 294. 290 PHYSICS 1. Thecoefficient of thermal conductivity depends upon (a) temperature difference between the two surfaces. (b) area of the plate (c) material of the plate (d) All of these 2. The wavelength of radiation emitted by a body depends upon (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface (d) All of the above 3. A vessel completelyfilled with a liquid is heated. If a and g represent coefficient of linear expansion ofmaterial ofvessel and coefficient of cubical expansion of liquid respectively, then the liquid will not overflowif (a) g = 3 a (b) g > 3 a (c) g < 3 a (d) g £ 3 a 4. The fastest mode of transfer of heat is (a) conduction (b) convection (c) radiation (d) None of these 5. In order that the heat flows from one part of a solid to another part, what is required? (a) Uniform density (b) Temperature gradient (c) Density gradient (d) Uniform temperature 6. The sprinkling of water slightly reduces the temperature of a closed room because (a) temperature of water is less than that of the room (b) specific heat of water is high (c) water has large latent heat of vaporisation (d) water is a bad conductor of heat 7. Which of the following is not close to a black body? (a) Black board paint (b) Green leaves (c) Black holes (d) Red roses 8. According to Newton’s law ofcooling, the rate of cooling of a body is proportional to (Dq)n, where Dq is the difference of the temperature of the body and the surroundings, and n is equal to (a) two (b) three (c) four (d) one 9. A metallic rod l cm long, A square cm in cross-section is heated through tºC. If Young’s modulus of elasticity of the metal is Eandthemean coefficientoflinear expansion isa per degree celsius, then the compressional force required to prevent the rod from expanding along itslength is (a) E A a t (b) E A a t/(1 + at) (c) E A a t/(1 – a t) (d) E l a t 10. A black body rediates energyat the rate ofE watt per metre2 at a high temperature T K. when the temperature is reduced to (T/2) K, the radiant energy will be (a) E/16 (b) E/4 (c) E/2 (d) 2E 11. The tempearture of an isolated black body falls from T1 to T2 in time t, then t is (Let c be a constant) (a) 2 1 1 1 t c T T æ ö = - ç ÷ è ø (b) 2 2 2 1 1 1 t c T T æ ö = - ç ÷ è ø (c) 3 3 2 1 1 1 t c T T æ ö = - ç ÷ è ø (d) 4 4 2 1 1 1 t c T T æ ö = - ç ÷ è ø 12. Twostraight metallic strips each of thickness t and length l are rivetted together. Their coefficients of linear expansions are a1 and a2 . If they are heated through temperature DT, the bimetallic strip will bend to form an arc ofradius (a) 1 2 t /{α α )ΔT} + (b) 2 1 t /{(α α )ΔT} - (c) 1 2 t (α α )ΔT - (d) 2 1 t (α α )ΔT - 13. Two solid spheres, of radii R1 and R2 are made of the same material and have similar surfaces. The spheres are raised to the same temperature and then allowed to cool under identical conditions. Assuming spheres to be perfect conductors of heat, their initial rates of loss of heat are (a) 2 2 2 1 / R R (b) 2 1 / R R (c) 1 2 / R R (d) 2 1 2 2 / R R 14. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (a) Ec (b) 2E/c (c) E/c (d) 2 E/c 15. Which of the following graph correctly represents the relation between E ln and T ln where E is the amount of radiation emitted per unit time from unit area of bodyand T is the absolute temperature (a) E ln T ln (b) E ln T ln (c) E ln T ln (d) E ln T ln
- 295. 291 Thermal Properties ofMatter 1. The resistance of a resistance thermometer has values 2.71 and 3.70 ohms at 10ºC and 100ºC respectively. The temperature at which the resistance is 3.26 ohm is (a) 40ºC (b) 50ºC (c) 60ºC (d) 70ºC 2. The temperature of an iron block is 140ºF. Its temperature on the Celsius scale is (a) 60º (b) 160º (c) 140º (d) 132º 3. A pendulum clock is 5 seconds fast at temperature of 15ºC and 10 seconds slow at a temperature of 30ºC. At what temperaturedoes it givethecorrect time?(taketime interval = 24 hours) (a) 18ºC (b) 20ºC (c) 22ºC (d) 25ºC 4. If a bar is made of copper whose coefficient of linear expansion is one and a half times that of iron, the ratio of force developed in the copper bar to the iron bar of identical lengths and cross-sections, when heated through the same temperature range (Young’s modulus ofcopper maybetaken to be equal to that of iron) is (a) 3/2 (b) 2/3 (c) 9/4 (d) 4/9 5. A metallic bar is heated from 0ºC to100ºC. The coeficient of linear expansion is 10–5 K–1. What will be the percentage increase in length? (a) 0.01% (b) 0.1% (c) 1% (d) 10% 16. For the construction of a thermometer, one of the essential requirements is a thermometric substance which (a) remains liquid over the entire rangeof temperatures to be meaured . (b) has property that varies linearly with temperature (c) has a property that varies with temperature (d) obey Boyle's law 17. The temperature of stars is determined by (a) Stefan’s law (b) Wien’s displacement law (c) Kirchhoff’s law (d) Ohm’slaw 18. The temperature of the Sun is measured with (a) platinum thermometer (b) gas thermometer (c) pyrometer (d) vapour pressure thermometer. 19. Which of the following circular rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature will conduct most heat? (a) r = 2r0; l = 2l0 (b) r = 2r0; l = l0 (c) r = r0; l = 2l0 (d) r = r0; l = l0 20. At a certain temperature for given wave length , the ratio of emissive power of a body to emissive power of black body in same circumstances is known as (a) relative emissivity (b) emissivity (c) absorption coefficient (d) coefficient of reflection 21. Ifa liquid is heated in weightlessness, the heat is transmitted through (a) conduction (b) convection (c) radiation (d) None of these, because the liquid cannot be heated in weightlessness. 22 The rate of heat flow through the cross-section of the rod shown in figure is (T2 > T1 and thermal conductivity of the material of the rod isK) r2 r1 L T1 T2 (a) 1 2 2 1 K r r (T T ) L p - (b) 2 1 2 2 1 K (r r ) (T T ) 4L p + - (c) 2 1 1 2 1 K (r r ) (T T ) L p + - (d) 2 1 1 2 1 K (r r ) (T T ) 2L p + - 23. When a body is heated, which colour corresponds to high temperature (a) red (b) yellow (c) white (d) orange 24. The two ends of a rod of length L and a uniform cross- sectional area A are kept at two temperatures T1 and T2 (T1 > T2). The rate of heat transfer, dQ dt through the rod in a steady state is given by (a) 1 2 k (T T ) dQ dt LA - = (b) 1 2 dQ kLA (T T ) dt = - (c) 1 2 kA (T T ) dQ dt L - = (d) 1 2 kL (T T ) dQ dt A - = 25. Two rods of the same length and areas of cross-section A1 and A2 have their ends at the same temperature. K1 and K2 are the thremal conductivities of the two rods. The rate of flow of heat is same in both rods if (a) 1 1 2 2 A K A K = (b) 1 2 2 1 A K A K = (c) A1A2 = K1K2 (d) A1 K1 2 = A2 K2 2
- 296. 292 PHYSICS 6. Steamis passed into 22 gm of water at 20ºC. The mass of water that will be present when the water acquires a temperatueof 90ºC (Latent heat of steam is 540 cal/g) is (a) 24.83gm (b) 24gm (c) 36.6gm (d) 30gm 7. Acylindrical rodofaluminiumis oflength 20 cms andradius 2 cms. The two ends are maintained at temperatures of 0ºC and 50ºC the coefficient of thermal conductivity is 0.5 cal cm×sec׺C . Then the thermal resistance of the rod in cal sec ׺C is (a) 318 (b) 31.8 (c) 3.18 (d) 0.318 8. Ametal ballofsurfacearea 200squarecm, temperature527ºC is surrounded by a vessel at 27ºC. If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is approximately 8 2 2 joule 5.67 10 m sec K - s = ´ ´ ´ (a) 108 joule (b) 168 joule (c) 182 joule (d) 192 joule 9. The rate ofradiation ofa black bodyat 0ºC is E joule per sec. Then the rate ofradiation of this black bodyat 273ºc will be (a) 16E (b) 8E (c) 4E (d) E 10. Solar radiation emitted by sun resembles that emitted by a blackbodyat a temperature of6000 K.Maximum intensityis emitted at wavelength of about 4800 Å. If the sun were to cool down from 6000 K to 3000 K, then the peak intensity would occur at a wavelength (a) 4800 Å (b) 9600Å (c) 2400 Å (d) 19200Å 11. A bucket full of hot water is kept in a room and it cools from 75ºCto70ºC in T1 minutes, from 70ºC to65ºC in T2 minutes and from 65ºC to 60ºC in T3 minutes. Then (a) T1 = T2 = T3 (b) T1 < T2 < T3 (c) T1 > T2 > T3 (d) T1 < T2 > T3 12. The maximum energy in the thermal radiation from a hot source occurs at a wavelength of 11 ×10–5 cm. According to Wien’s law, the temperature of this source (on Kelvin scale) will be n timesthe temperature of another source (on Kelvin scale) for which the wavelength at maximum energy is 5.5 × 10–5 cm. The value n is (a) 2 (b) 4 (c) 1/2 (d) 1 13. The temperature of a furnace is 2324ºC and the intensityis maximum in its radiation spectrum nearlyat 12000 Å. If the intensity in the spectrum of a star is maximum nearly at 4800Å, then the surface temperature of the star is (a) 8400ºC (b) 7200ºC (c) 6492.5ºK (d) 5900ºC 14. A mountain climber finds that water boils at 80ºC. The temperature of this boiling water is...... Fahrenheit (a) 50º (b) 150º (c) 176º (d) 200º 15. A metal cube of length 10.0 mm at 0°C (273K) is heated to 200°C (473K). Given : its coefficient of linear expansion is 2× 10–5 K–1. The percent change of its volume is (a) 0.1 (b) 0.2 (c) 0.4 (d) 1.2 16. No other thermometer is suitable as a platinum resistance thermometer tomeasure temperatures in the entire range of (a) –50ºC to + 350ºC (b) –200ºC to + 600ºC (c) 0ºC to100ºC (b) 100ºCto 1500ºC 17. Calculate the surface temperature ofthe planet, ifthe energy radiated byunit area in unit time is 5.67 ×104 watt. (a) 1273°C (b) 1000°C (c) 727°C (d) 727K 18. In a thermocouple, the temperature ofthe cold junction and theneutral temperature are –40°C and 275°C respectively. If the cold junction temperature is increased by 60°C, the neutral temperature and temperature of inversion respectively become (a) 275°C,530°C (b) 355°C,530°C (c) 375°C,590°C (d) 355°C,590°C 19. In a surrounding medium oftemperature 10°C, a bodytakes 7min for afalloftemperature from60°Cto40°C. In whattime the temperature ofthe bodywill fall from 40°C to28°C? (a) 7min (b) 11min (c) 14min (d) 21min 20. Tworods P and Q of same length and same diameter having thermal conductivity ratio 2 : 3 joined end to end. If temperature at one end of P is 100°C and at one and of Q 0°C, then the temperature ofthe interface is (a) 40°C (b) 50°C (c) 60°C (d) 70°C 21. 100 g ofice is mixed with 100g ofwater at 100ºC. What will be the final temperature of the mixture ? (a) 13.33ºC (b) 20ºC (c) 23.33ºC (d) 40ºC 22. A crystal has a coefficient of expansion 13×10–7 in one direction and 231 × 10–7 in everydirection at right angles to it. Then the cubical coefficient of expansion is (a) 462 × 10–7 (b) 244 × 10–7 (c) 475 × 10–7 (d) 257 × 10–7 23. A glass flask of volume 1 litre is fullyfilled with mercuryat 0ºC. Both the flask and mercuryare now heated to 100ºC. If the coefficient ofvolume expansion of mercuryis 1.82 × 10– 4/ºC, volume coefficient of linear expansion of glass is 10 × 10–6/ºC, the amount of mercury which is spilted out is (a) 15.2ml (b) 17.2ml (c) 19.2ml (d) 21.2ml
- 297. 293 Thermal Properties ofMatter 24. A rectangular block is heated from 0ºC to 100ºC. The percentage increase in its length is 0.10%. What will be the percentage increase in its volume? (a) 0.03% (b) 0.10% (c) 0.30% (d) None of these 25. The coefficient of apparent expansion ofmercury in a glass vessel is 153 × 10–6/ºC and in a steel vessel is 144 × 10–6/ºC. If a for steel is 12 × 10–6/ºC, then, that of glass is (a) 9 × 10–6/ºC (b) 6 × 10–6/ºC (c) 36 × 10–6/ºC (d) 27 × 10–6/ºC 26. A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperatue of water in the beaker is 20ºC. If 440 gm of hot water at 92ºCispouredin it, thefinal temperature, neglecting radiation loss, will be nearest to (a) 58ºC (b) 68ºC (c) 73ºC (d) 78ºC 27. Steam at 100ºC is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15ºC till the temperature of the calorimeter and its contents rises to 80ºC.The mass of the steam condensed in kg is (a) 0.130 (b) 0.065 (c) 0.260 (d) 0.135 28. A body of mass 5 kg falls from a height of 20 metres on the ground and it rebounds to a height of 0.2 m. If the loss in potential energy is used up bythe body, then what will be the temperaturerise? (specificheat ofmaterial = 0.09 cal gm–1 ºC–1) (a) 0ºC (b) 4ºC (c) 8ºC (d) None of these 29. 80 g of water at 30ºC are poured on a large block of ice at 0ºC. The mass of ice that melts is (a) 1600g (b) 30 g (c) 150g (d) 80 g 30. Ice starts forming in a lake with water at 0ºC when the atomspheric temperature is –10ºC. If the time taken for the first 1 cm of ice to be formed is 7 hours, then the timetaken for the thickness of ice to change from 1 cm to 2 cm is (a) 7 hours (b) 14 hours (c) 21 hours (d) 3.5 hours 31. Two identical rods of copper and iron are coated with wax uniformly. When one end of each is kept at temperature of boiling water, the length upto which wax melts are 8.4 cm amd 4.2 cm, respectively. Ifthermal conductivity of copper is 0.92, then thermal conductivityof iron is (a) 0.23 (b) 0.46 (c) 0.115 (d) 0.69 32. Twovessels ofdifferent materials are similar in size in every respect. The same quantityofice filled in them gets melted in 20 min and 35 min, respectivley. The ratio ofcoefficients of thermal conduction of the metals is (a) 4: 7 (b) 7: 4 (c) 25: 16 (d) 16:25 33. Apartition wall hastwolayersA and B, in contact, each made of a different material. Theyhave the same thickness but the thermal conductivityof layer A is twice that of layer B. If the steady state temperature difference across the wall is 60 K, then the corresponding difference across the layer A is (a) 10K (b) 20K (c) 30K (d) 40K 34. The spectral energy distribution of the sun (temperature 6050K) is maximum at 4753Å. The temperatureofa star for which this maximum is at 9506 Å is (a) 6050K (b) 3025K (c) 12100K (d) 24200K 35. Two bodies A and B are placed in an evacuated vessel maintained at a temperature of27ºC. ThetemperatureofA is 327ºC and that of B is 227ºC. The ratio of heat loss from A and B is about (a) 2: 1 (b) 1: 2 (c) 4: 1 (d) 1: 4 36. The thermal capacity of 40g of aluminium (specific heat = 0.2 cal g–1 °C–1) is (a) 40 cal °C–1 (b) 160 cal °C–1 (c) 200 cal °C–1 (d) 8 cal °C–1 37. Which of the following temperatures is the highest? (a) 100K (b) –13°F (c) –20°C (d) –23°C. 38. 5 kg of water at 10°C is added to 10 kg of water at 40°C. Neglecting heat capacity of vessel and other losses, the equilibrium temperature will be (a) 30°C (b) 25°C (c) 35°C (d) 33°C 39. Abeaker contains200g of water.Theheat capacityofbeaker is equal to that 20 gofwater. The initial temperature ofwater in the beaker is 20°C . If440 g of hot water at 92°C is poured in, the final temperature, neglecting radiation loss, will be (a) 58°C (b) 68°C (c) 73°C (d) 78°C 40. The rectangular surface of area 8 cm × 4 cm of a black body at temperature 127°Cemits energyE per second. Ifthelength and breadth are reduced to half of the initial value and the temperature is raised to327°C, the rateofemission ofenergy becomes (a) 8 3 E (b) 16 81 E (c) 16 9 E (d) 64 81 E
- 298. 294 PHYSICS 41. Tworods of same length and transfer a given amount of heat 12 second, when theyare joined as shown in figure (i). But when they are joined as shwon in figure (ii), then they will transfer same heat in same conditions in l l l Fig. (i) Fig. (ii) (a) 24 s (b) 13 s (c) 15 s (d) 48 s 42. A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1 : 4. Ifthe free face of brass is at 100°C and that of copper at 0°C, the temperature of interface is (a) 80°C (b) 20°C (c) 60°C (d) 40°C 43. 540 g ofice at 0°C is mixed with 540 g ofwater at 80°C. The finaltemperatureofmixture is (a) 0°C (b) 40°C (c) 80°C (d) less than 0°C 44. Three identical rods A, B and C of equal lengths and equal diameters are joined in series as shown in figure. Their thermal conductivities are 2k, k and k/2 respectively. The temperatures at two junction points are T1 A B C 0°C 0.5°C k 2k 100°C T2 (a) 85.7,57.1°C (b) 80.85,50.3°C (c) 77.3,48.3°C (d) 75.8,49.3°C 45. Which one of the following graphs best represents the ways in which the total power P radiated bya black bodydepends upon the thermodynamic temperature T of the body? (a) T O P (b) T O P (c) T O P (d) T O P 46. A black body at 227°C radiates heat at the rate of 7 cals/ cm2s.At a temperature of727°C, the rate ofheat radiated in the same units will be (a) 50 (b) 112 (c) 80 (d) 60 47. The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167 WK–1m–1 and thickness 1 cm. Thetemperature is maintained bycirculating oil as shown in figure. Find the radiation loss to the surrounding in Jm–2s–1 if temperature of the upper surface of the disc is 27°C and temperature of the surrounding is 27°C. Oil out Oil in (a) 595 Jm–2s–1 (b) 545 Jm–2s–1 (c) 495 Jm–2s–1 (d) None of these 48. Twomarks on a glass rod 10 cm apart are found to increase their distanceby0.08 mm when the rod isheatedfrom 0°C to 100°C. Aflask made ofthe same glassas that of rod measures a volumeof 1000 cc at 0°C. The volumeit measures at 100°C in cc is (a) 1002.4 (b) 1004.2 (c) 1006.4 (d) 1008.2 DIRECTIONS for Qs. 49 to 50 : These are Assertion-Reason type questions. Each of these questioncontains twostatements: Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions fromthe following four options. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1 (b) Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1 (c) Statement-1 is True, Statement- 2 is False (d) Statement-1 is False, Statement -2 is True 49. Statement 1 : The equivalent thermal conductivity of two plates of same thickness in series is less than the smaller value of thermal conductivity. Statement 2 : For two plates ofequal thickness in series the equivalent thermal conductivity is given by 2 1 K 1 K 1 K 1 + = 50. Statement 1 : As the temperature of the black body increases, the wavelength at which the spectral intensity ( El ) ismaximum decreases. Statement 2 : The wavelength at which the spectral intensity will be maximum for a black bodyisproportional tothefourth power of its absolute temperature.
- 299. 295 Thermal Properties ofMatter Exemplar Questions 1. A bimetallic strip is made of aluminium and steel (aAl > asteel). On heating, the strip will (a) remain straight (b) get twisted (c) will bend with aluminium on concave side (d) will bend with steel on concave side 2. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. Ifit isheated uniformly to raise its temperature slightly (a) its speed of rotation increases (b) its speed of rotation decreases (c) its speed of rotation remains same (d) its speed increases because its moment of inertia increases 3. The graph between two temperature scales A and B isshown in figure between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by O 100 180 DtB = 100° DtA = 150° Temperature (°B) Temperature (°A) (a) 180 100 150 A B t t - = (b) 30 150 100 A B t t - = (c) 180 150 100 B A t t - = (d) 40 100 180 B A t t - = 4. An aluminium sphere is dipped into water. Which of the following is true? (a) Buoyancywill beless in water at 0°C than that in water at4°C (b) Buoyancywill bemorein water at 0°C than that in water at4°C (c) Buoyancyin water at 0°C will be same as that in water at4°C (d) Buoyancymaybemoreor less in water at 4°Cdepending on the radius of the sphere 5. As the temperature is increased, the period of a pendulum (a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob (b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob (c) increases as its effectivelength increases due toshifting to centre of mass below the centre of the bob (d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob 6. Heat is associated with (a) kinetic energy of random motion of molecules (b) kinetic energy of orderlymotion of molecules (c) total kinetic energy of random and orderly motion of molecules (d) kinetic energy of random motion in some cases and kinetic energy of orderlymotion in other 7. The radius ofa metal sphere at room temperature T is R and the coefficient of linear expansion of the metal is a. The sphere heated a little by a temperature DT so that its new temperature is T + DT. The increase in the volume of the sphere is approximately. (a) 2 R T p aD (b) 2 R T p aD (c) 3 4 / 3 R T p aD (d) 3 4 R T p aD 8. Asphere, a cubeanda thin circular plate, all of samematerial andsame mass are initiallyheatedtosame high temperature. (a) Plate will cool fastest and cube the slowest (b) Sphere will cool fastest and cube the slowest (c) Plate will cool fastest and sphere the slowest (d) Cube will cool fastest and plate the slowest PastYears (2013-2017) NEET/AIPMTQuestions 9. A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finallyturns to white hot. The correct explanation for the above observation is possible by using [2013] (a) Wien’s displacement law (b) Kirchoff’s law (c) Newton’s law of cooling (d) Stefan’s law 10. The density of water at 20°C is 998 kg/m3 and at 40°C 992 kg/m3. The coefficient of volume expansion of water is [NEET Kar. 2013] (a) 10–4/°C (b) 3 × 10–4/°C (c) 2 × 10–4/°C (d) 6 × 10–4/°C
- 300. 296 PHYSICS 11. Twometal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivitiesareK1 and K2 and cross sectional areas A1 and A2, respectively. If the rate of heat conduction in rod 1 is four times that in rod 2, then [NEET Kar. 2013] (a) K1A1 = K2A2 (b) K1A1 = 4K2A2 (c) K1A1 = 2K2A2 (d) 4K1A1 = K2A2 12. Certain quantityofwater cools from 70°Cto60°C in the first 5minutes and to54°C in thenext5minutes. Thetemperature of the surroundings is: [2014] (a) 45°C (b) 20°C (c) 42°C (d) 10°C 13. Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be: [2014] [Take specific heat ofwater = 1 cal g– 1 °C– 1 and latent heat of steam = 540 cal g– 1] (a) 24 g (b) 31.5g (c) 42.5g (d) 22.5g 14. On observing light from three different stars P, Q and R, it was found that intensityof violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in thespectrum of Rand the intensityof red colour ismaximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperature ofP, Q and R, then it can beconcluded from the above observations that [2015] (a) TP > TR > TQ (b) TP < TR < TQ (c) TP < TQ < TR (d) TP > TQ > TR 15. The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flowin the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be [2015] (a) 16.8 J/s (b) 8.0 J/s (c) 4.0 J/s (d) 44.0 J/s 16. The value of coefficient of volume expansion of glycerine is 5 × 10-4 K-1. Thefractional changein thedensityofglycerine for a rise of 40°C in its temperature, is: [2015 RS] (a) 0.020 (b) 0.025 (c) 0.010 (d) 0.015 17. Apiece of ice falls from a height h sothat it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value ofh is : [2016] [Latent heat ofice is 3.4 × 105 J/kg and g = 10 N/kg] (a) 34km (b) 544km (c) 136km (d) 68km 18. Coefficient of linear expansion of brass and steel rods are a1 and a2. Lengths of brass and steel rods are 1 l and 2 l respectively. If 2 1 ( ) l l - is maintained same at all temperatures, which one of the following relations holds good ? [2016] (a) 2 2 1 2 2 1 l l a =a (b) 1 2 2 1 l l a =a (c) 1 2 2 1 l l a =a (d) 1 1 2 2 a = a l l 19. A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U2 and that at 1000nmisU3.Wien's constant, b = 2.88 × 106 nmK. Which of the following is correct ? [2016] (a) U1 = 0 (b) U3 = 0 (c) U1 > U2 (d) U2 > U1 20. A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be : (a) 450 (b) 1000 [2017] (c) 1800 (d) 225 21. Two rods Aand B of different materials are welded together as shown in figure. Their thermal conductivities are K1 and K2. The thermal conductivityof the composite rod will be : A K1 B K2 T1 T2 d [2017] (a) 1 2 3(K K ) 2 + (b) K1 + K2 (c) 2 (K1 + K2) (d) 1 2 K K 2 +
- 301. 297 Thermal Properties ofMatter EXERCISE - 1 1. (c) 2. (d) 3. (d) 4. (c) 5. (b) 6. (c) 7. (a) 8. (d) 9. (a) strain stress / A / F E = D = l l where Dl=(l'–l) = lat so F = EAat 10. (a) 11. (c) 12. (b) Let the angle subtended by the arc formed be q. Then 2 1 1 2 r r r or r - - = D D = q = q l l l l t T ) ( 1 2 D a - a = q l t T ) ( r or 1 2 D a - a = l l So, T ) ( t r 1 2 D a - a = 13. (a) Initial rate of loss of heat 2 2 2 1 2 4 1 4 R R e A T e A T = ´ ´ s ´ ´ s = 14. (b) Momentum of photon E c = Change in momentum 2E c = = momentum transferred to the surface (the photon will reflect with same magnitude of momentum in opposite direction). 15. (c) We know 4 eT E = T ln 4 e ln E ln + = Þ negative is e ln 1 e because < E ln T ln 16. (c) The temperature is measured by the value of the thermodynamic property of a substance i.e., the propertywhich varies linearlywith the temperature. 17. (b) Temperature of stars can be determined by Wein’s displacement law = lm constant 18. (c) 19. (b) We know that Q = T T R H L - Also Thermal resistance, R= l l KA K r = p 2 Heat flowwill be maximum when thermal resistance is minimum. From given option (i) r = 2r0, l = 2l0 = = R K r K r 2 2 2 0 0 2 0 0 2 l l p p ( ) (ii) r = 2r0, l = l0 = = R K r K r l l 0 0 2 0 0 2 2 4 p p ( ) (iii) r = r0, l = 2l0 = = R K r K r 2 2 0 0 2 0 0 2 l l p p (iv) r = r0, l = l0 = = R K r K r l l 0 0 2 0 0 2 p p It is clear that for option (b) resistance is minimum, hence heat flowwill bemaximum. 20. (c) 21. (a) In condition of weightlessness, convection is not possible. 22. (a) eff 1 2 r r r = 2 1 1 2 2 1 KA(T T ) K r r (T T ) dQ dt L L - p - = = 23. (c) When a body is heated then relation between colours and temperature is according to Prevost’s theory of radiation which states that everybody emitting radiant energyin all directions at a rate depending only on the nature of its surface and its temperature e.g., when a bodyis placed in an enclosure(furnace) it would acquire the temperature of furnace and seem white means radiate white light. So it becomes first dark and then white. 24. (c) 1 2 kA (T T ) dQ dt L - = [(T1–T2) is the temperature difference] 25. (b) EXERCISE - 2 1. (c) ) t 1 ( R R 0 t a + = ) 10 1 ( R 71 . 2 0 ´ a + = ...(1) ) 100 1 ( R 70 . 3 0 ´ a + = ...(2) ) t 1 ( R 26 . 3 0 a + = ...(3) Solve these equations to obtain the value of t. 2. (a) 5 C 9 32 140 = - or C 9 160 700 = - or C = 60ºC 3. (c) t T 2 1 t ´ D a = D Hints & Solutions
- 302. 298 PHYSICS 86400 ) 15 T ( 2 1 5 ´ - a = and86400 ) T 30 ( 2 1 10 ´ - a = 4. (a) a µ a = F or A t Y F (Q Y t A is same for both copper and iron) or C C Ι Ι F α and F α µ µ 2 3 1 2 / 3 F FC = = I 5. (b) Δ αΔT = l l 5 3 10 100 10 - - = ´ = 3 Δ 100% 10 100 - ´ = ´ l l 1 10 0.1% - = = 6. (a) Let m be the mass of steam condensed. Then m×540+m×10/2=22 ×70 m= 2.83gm Now, total mass = 22 + 2.83 = 24.83 gm 7. (d) Thermal resistance R =l/KA Where l =20cm. & A(cylindrical rod) =pr2 =40pcm2 So = = ´ p ´ 20 cal R 0.318 0.5 40 sec º C 8. (c) According to Stefan’s Law, the rate of loss of heat is e ) T T ( A t Q 4 2 4 1 ´ - s = here s = 5.67 × 10–8J/m2 × sec.K2, T1=527+273=800K, T2 = 27+ 273 = 300K& A= 200×10–4m2 So. 8 2 4 4 Q 5.67 10 2 10 t [(800) (300) ] 0.4 - - = ´ ´ ´ - ´ @ 182 joule 9. (a) According to Stefan’s Law energy radiated per sec E = sAT4 (here e = 1 for black body) for first case E = sA(273)4 for second case E1=sA(546)4 so E1=16E 10. (b) According to Wein's displacement law, λ T = b. m where b=2.884 × 10–3 mK. So l = l m m 1 2 1 2 T T or 4800 × 6000 = l ´ m2 3000 or l = ° m2 9600A 11. (b) The time of cooling increases as the difference between the temperature of body& surrounding is reduced. So T1 < T2 < T3 (according to Newton’s Law of cooling). 12. (c) 5 5 1 2 11.0 10 T 5.5 10 T - - ´ = ´ 2 1 T T 2 1 = 13. (c) 14. (c) 5 80 9 32 F = - F – 32 = 144 or F = 176ºF. 15. (d) Percentage change in volume = gt × 100 t 300 100 t 3 a = ´ a = 2 . 1 200 10 2 300 5 = ´ ´ ´ = - 16. (b) 17. (c) According to Stefan's Boltzmann law, the energy radiated per unit time E = sAT4. 5.67 × 104 = 5.67 × 10–8 × 1 × T4 ( A = 1 m2) or, 1/ 4 4 –8 5.67 10 T 1000K 5.67 10 æ ö ´ = = ç ÷ ´ è ø or , T = 1000 – 273 = 727°C. 18. (a) When the temperature of cold junction is increased, the neutral temperature remains constant for a given thermocouple and it is independent of the temperature of the cold junction. qn = 275°C, qi – qn = qn – qc qi = 2qn – qc = 530°C. 19. (a) According to Newton's law of cooling, 1 2 1 2 0 K t 2 q - q q + q é ù = - q ê ú ë û where q0 is the surrounding temperature. 60 40 60 40 K 10 7 2 - + æ ö = - ç ÷ è ø Þ 20 40K 7 = Þ 1 K 14 = 40 28 40 28 K 10 t 2 - + é ù = - ê ú ë û Þ 12 24K t = or 12 12 14 t 7min 24K 24 ´ = = = 20. (a) Let q be temperature of interface. Q P K A( 0) K A(100 ) q - - q = l l P Q K K 100 q = - q Þ 2 3 100 q = -q or, 200 – 2q = 3q or, 5q= 200 or, q = 40°C 21. (a) Let the final temperature of mixture be T. Then 100 × 80 + 100 (T – 0) ×½ (as specific heat of ice is 0.5 cal/g C° and specific heat ofwater is 1 cal/gC°) = 100 × 1 ×(100 – T) Solving, we get T = 13.33ºC. 22. (a) 3 2 1 a + a + a = g 7 7 7 10 231 10 231 10 13 - - - ´ + ´ + ´ = 7 10 475 - ´ =
- 303. 299 Thermal Properties ofMatter 23. (a) T ) ( V V g m 0 D g - g = D 100 )] 10 10 ( 3 10 82 . 1 [ 1 6 4 - - ´ ´ - ´ = 100 )] 10 3 . 0 10 82 . 1 [ 1 4 4 - - ´ - ´ = =15.2ml 24. (c) Given Δ / 0.10% 0.001 and ΔT 100º C = = = l l Now Δ αΔT = l l or 0.001 α 100 = ´ or 5 α 10 /º C - = Further 5 γ 3α 3 10 /ºC - = = ´ 3 Δ V 100 (3 10 ) (100) 0.30% V - ´ = ´ = 25. (a) We know that real apparent vessel γ γ γ = + So, app vessel glass app vessel steel (γ γ ) (γ γ ) + = + (Q real g is same in both cases) or 6 vessel glass 153 10 (γ ) - ´ + steel vessel 6 ) ( 10 144 g + ´ = - Further vessel steel (γ ) 6 6 3 3 (12 10 ) 36 10 /ºC - - = a = ´ ´ = ´ 6 vessel glass 153 10 (γ ) - ´ + 6 6 10 36 10 144 - - ´ + ´ = Solving we get 6 vessel glass ( ) 27 10 /ºC - g = ´ or glass 6 9 10 /ºC 3 - g a = = ´ 26. (b) Let the final temperature be T. Then 200 × 1 × (T – 20)+ 20× (T– 20) = 440 (92 – T) Solving it, we get T = 68ºC. 27. (a) ) 80 100 ( m mL - + ) 15 80 ( 02 . 0 ) 15 – 80 ( 1 1 . 1 - ´ + ´ ´ = 30 . 1 5 . 71 m 20 540 m + = + ´ 80 . 72 m 560 = 130 . 0 m = 28. (d) h mg mgh W W W 2 1 ¢ - = - = ) h h ( mg ¢ - = 8 . 19 10 5 ) 2 . 0 20 ( 10 5 ´ ´ = - ´ = joule 990 198 5 = ´ = This energyis converted intoheat when the ball strikes the earth. Heat produced is calorie 2 . 4 990 Q = C º 32 11 09 . 0 5000 42 100 99 mc Q T = ´ ´ ´ = = D 29. (b) 80× 1 × 30= m × 80 Þ m = 30gm. 30. (c) 31. (a) Use 2 2 2 1 2 1 K K l l = 32. (b) Here flow of heat, area of vessels & temperature gradient are same so 1 1 2 2 dT dT Q K A t K A t dx dx = = so 4 7 20 35 t t K K 1 2 2 1 = = = 33. (b) 34. (b) 1 1 2 2 λ T λ T = 2 T . 9506 4753 6050 = ´ K 3025 T2 = 35. (a) ) T T ( ) T T ( E E 4 0 4 2 4 0 4 1 2 1 - s - s = 4 4 4 4 ) 300 ( ) 500 ( ) 300 ( ) 600 ( - - = 36. (d) Thermal capacity= 40 × 0.2 = 8 cal °C–1 37. (b) –13°F is (13 + 32)° below ice point on F scale. 38 (a) Mass ratio 1 : 2; hence DT ratio 2 : 1. Therefore equilibriumtemp is30°C. 39. (b) Heat capacityof cold : hot =1 : 2 Sofinal temp. is 3 1 × (2 × 92 + 1 × 20) = 68°C. 40. (d) 4 T area E ´ ´ s = ; T increases by a factor 2 3 . Area increases by a factor 4 1 . 41. (d) 2 t , t ' A A / 2 µ µ l l 42. (a) q = q - Þ - q = q - 4 400 ) 0 ( K ) 100 ( K 4 80 C Þ q = ° 43. (a) 540 ×80 +540q = 540 (80–q) ° = q Þ = q Þ q - = q + Þ 0 0 2 80 80 44. (a) 45. (c) The total power radiated by a black body of area A at temperature TK is given byP = AsT4 Where s = Stefan's constant = 5.7 × 10–8 W m–2 K–4 Which is best represented in graph. (c) 46. (b) According to Stefan’s law 4 E T , = s T1 = 500K T2 = 1000 K 4 4 2 2 1 1 E T 1000 16 E T 500 æ ö æ ö = = = ç ÷ ç ÷ è ø è ø E2 = 16 × 7 = 112 cal / cm2s 47. (a) The rate of heat loss per unit area due to radiation = Îs (T4–T0 4) = 0.6 × 5.67 × 10–8 [(400)4–(300)4] = 595 Jm–2s–1.
- 304. 300 PHYSICS 48. (a) 21 1 2 1 V V ; V (T T ) - g = - 2 1 1 2 1 (T T ) - a = - l l l T1 = 0°C, T2 =100°C 2 1 2 1 1 1 V V ; 100V 100 - - g = a = l l l l 1 = 10 cm; l 2 – l 1 = 0.08 mm = 0.008 cm 6 0.008 8 10 / C; 10 100 - a = = ´ ° ´ g= 3a = 24 × 10–6 / °C 24 × 10–6 2 V 1000 1000 100 - = ´ V2 – 1000 = 24 × 10–6 × 105 = 2.4 V2 = 1002.4 cc 49. (a) For equivalent thermal conductivity, the relation is 2 1 R K 1 K 1 K 1 + = ; If K K K 2 1 = = 2 K K K 2 K 1 K 1 K 1 R R = Þ = + = Which is less than K. If 2 1 K K > suppose x K K 2 1 + = 2 1 1 2 1 2 K K 1 1 1 K K K K K + = + = 2 2 2 2 K K x 1 K (K x)K + + Þ = + Þ x K 2 x K K K 2 2 2 2 + + = Now, K K2 - = x K 2 x K K K 2 2 2 2 2 + + - ) x K 2 ( x K K x K K 2 2 2 2 2 2 2 2 + - - + = 2 2 2 K positive 2K x = = + So, K2 > K, sothevalue ofK is smaller than K2 and K1. 50. (c) From Wein's law lmT = constant i.e. peak emission wavelength lm µ 1 T . Hence as T increases lm decreases. EXERCISE - 3 Exemplar Questions 1. (d) If strips of aluminium and steel are fixed together on metallic strip and both are heated then (aAl > asteel) aluminium will exapnd more because the metallic strip with higher coefficient of linear expansion (aAl) will expand more Thus it should have larger radius of curvature. Hence, aluminium will beon convex side. q O Aluminium Steel 2. (b) On heating a uniform metallicrodits length will increase so moment of inertia of rod increased from I1 to I2. No external torque is acting on the system so angular momentum should be conserved. L =Angular momentum = Iw= constant 1 1 2 2 I I w = w Rod So, 2 1 1 2 1 I I w = < w w2 < w1 (Q Due to expansion of the rod I2 > I1) So, angular veloctiy decreases. 3. (b) In the given graph shows lowest fixed point for scale A is 30° and lowest point for scale B is 0°. Upper fixed point for the scale A is 180° and upper fixed point for scale B is 100°. Hence, formula is (LEP) (LEP) (UFP) (LFP) (UFP) (LFP) A A B B A A B B t t - - = - - where , LFP - Lower fixed point, UFP - Upper fixed point. O 100 180 DtB = 100° Temperature (°B) Temperature (°A) DtA = 150° 30° 90°–q q C B (+A) O (+ B) So, 30 0 180 30 100 0 A B t t - - = - - 30 150 100 - = A B t t 4. (a) As we know that, the Buoyant force (F) on a body volume (V) and densityof(r), when immersed in liquid of density (rl) is ' l V g = r where V ' = volume of displaced liquid bydipped body (V). Let volume of the sphere is V and r is its density, then we can write buoyant force s F V G = r l F µ r (for liquid) 4 4 0 0 1 C C C C F F ° ° ° ° r = > r 4 0 ( ) C C ° ° r > r Q 4 0 C C F F ° ° > Hence, buoyancy will be less in water at 0°C than that in water at 4°C .
- 305. 301 Thermal Properties ofMatter 5. (a) As the temperature increased the length (L) of the pendulum increases due to expansion i.e., linear. Pendulum L So, Time period of pendulum 2 L T g = p T L µ Hence on increasing temperature, time period (T) also increases. 6. (a) As we know that when the temperature increases vibration of molecules about their mean position increases hence the kinetic energy associated with random motion of molecules increases. 7. (d) Let the radius of the sphere is R. As the temperature increases radius of the sphere increases as shown. R dV Original volume 3 4 3 V R = p Coefficient of linear expansion = a Coefficient of volume expansion = 3a As we know that V Y V t D = D By putting the value of V, increase in the volume 3 V V t Þ D = aD 3 4 R t = p aD 8. (c) Loss of heat temperature on cooling temperature increase depend on material of object surface area exposed to surrounding and temperature difference between body and surrounding. Let us consider the diagram where all the three objects are heated to same temperature T. As we know that density mass volume r = where r is same for all the three objects hence, volume will also be same. m T Plate Cube Sphere T T m m As thickness of the plate is least so, surface area of the plateismaximum. We know that, according to Stefan's law of heat loss 4 H AT µ where, A is surface area of object and T is temperature. So, Hsphere : Hcube : Hplate = Asphere : Acube : Aplate Soarea ofcircular plate ismaximum. For sphere, as the sphere is having minimum surface area. Hence, the sphere cools slowest and circular platewill cool faster. PastYears (2013-2017) NEET/AIPMTQuestions 9. (a) Wein’s displacement law According to this law lmax µ 1 T or, lmax × T = constant So, as the temperature increases l decreases. 10. (b) From question, Dr = (998– 992) kg/m3 = 6 kg/m3 r = 3 3 998 992 kg/m 995 kg/m 2 + = r = m V Þ V V Dr D = - r Þ V V Dr D = r Coefficient of volume expansion of water, 1 1 V V t t D Dr = D r D = 4 6 3 10 / C 995 20 - » ´ ° ´ 11. (b) Q1 = 4Q2 (Given) Þ 1 1 2 2 4 K A t K A t L L D D = Þ K1A1 = 4K2A2. 12. (a) Let the temperature of surroundings be q0 By Newton's law of cooling 1 2 1 2 0 k t 2 q - q q + q é ù = - q ê ú ë û Þ 0 70 60 70 60 k 5 2 - + é ù = - q ê ú ë û Þ 2 = k [65 – q0] ...(i)
- 306. 302 PHYSICS Similarly, 0 6054 60 54 k 5 2 - + é ù = - q ê ú ë û Þ 6 5 = k [57 – q0] ...(ii) Bydividing (i) by(ii) we have 0 0 65 10 6 57 - q = - q Þ q0 = 45º 13. (d) According to the principle of calorimetry. Heat lost = Heat gained mLv + mswDq = mwswDq Þ m×540+m×1×(100–80) =20 ×1×(80 –10) Þ m= 2.5g Therefore total mass of water at 80°C =(20 +2.5) g= 22.5g 14. (a) From Wein’s displacement law lm × T = constant P – max. intensityis at violet Þlm isminimumÞtempmaximum R – max. intensityis at green Þ lm is moderate Þ temp moderate Q – max. intensityis at red Þ lm is maximum Þtemp. minimum i.e., Tp > TR > TQ 15. (c) As the temperature difference DT = 10°C as well as the thermal resistance is same for both the cases, sothermal current or rate of heat flow will also be same for both the cases. 16. (a) From question, Rise in temperature Dt = 40°C Fractional change in the density 0 Dr r = ? Coefficient of volume expansion g = 5 × 10–4K–1 r = r0 (1 –gDt) Þ 0 Dr r = gDT = (5 ×10–4) (40) = 0.02 17. (c) According to question only one-quarter of the heat produced by falling piece of ice is absorbed in the meltingof ice. i.e., mgh 4 =mL Þ h = 5 4L 4 3.4 10 136 km g 10 ´ ´ = = . 18. (d) From question, (l2 – l1) is maintained same at all temperatures hence change in length for both rods should be same i.e., Dl1 = Dl2 As we know, coefficient of linear expansion, a = 0 T D D l l l1a1DT = l2a2DT l1a1 = l2a2 19. (d) Accordingtowein'sdisplacementlaw, maximum amount of emitted radiation corresponding to lm = b T lm = 6 2.88 10 nmK 5760K ´ = 500 nm U2 U Emitted (radiation) 250 nm 500 nm 1000 nm wave length K From the graph U1 < U2 > U3 20. (c) Given r1 = 12 cm, r2 =6 cm T1 = 500 K and T2 = 2 × 500 = 1000 K P1 = 450 watt Rate of power loss 2 4 P r T µ 2 4 1 1 1 2 4 2 2 2 P r T P r T = 2 4 2 2 2 1 2 4 1 1 r T P P r T = Solving we get, P2 = 1800 watt 21. (d) Heat current H = H1 + H2 = 1 1 2 2 1 2 K A(T T ) K A(T T ) d d - - + EQ 1 2 1 2 1 2 K 2A(T T ) A(T T ) [K K ] d d - - = + Hence equivalent thermal conductivities for two rods of equal area is given by 1 2 2 EQ k k K + =
- 307. THERMAL EQUILIBRIUM ANDZEROTH LAW OF THERMODYNAMICS Thermal Equilibrium Twosystems are said tobe in thermal equilibrium with each other if theyhave the same temperature. Zeroth Law of Thermodynamics If objects A and B are separately in thermal equilibrium with a third object C then objects A and B are in thermal equilibrium with each other. FIRST LAW OF THERMODYNAMICS First law of thermodynamics gives a relationship between heat, work and internal energy. (a) Heat : It is the energy which is transferred from a system to surrounding or vice-versa due to temperature difference between system and surroundings. (i) It is a macroscopic quantity. (ii) It is path dependent i.e., it is not point function. (iii) If system liberates heat, then by sign convention it is taken negative, If system absorbs heat, it is positive. (b) Work : It is the energy that is transmitted from one system to another by a force moving its points of application. The expression of work done on a gas or by a gas is 2 1 V V W dW PdV = = ò ò where V1 is volume of gas in initial state and V2 in final state. (i) It is also macroscopic and path dependent function. (ii) Bysign convention it is +ive if system does work (i.e., expands against surrounding) and it is – ive, if work is done on system (i.e., contracts). (iii) In cyclic process the work done is equal to area under the cycle and is negative if cycle is anti-clockwise and +ive if cycle is clockwise (shown in fig.(a) and (b)). (c) Internal energy : The internal energy of a gas is sum of internal energy due to moleculer motion (called internal kinetic energy UK) and internal energy due to molecular configuration (called internal potential energy UP.E.) i.e., U = UK + UP.E. ……(1) (i) In ideal gas, as there is no intermolecular attraction, hence 3 2 K n U U RT = = ……(2) (for n mole ofideal gas) (ii) Internal energyispath independent i.e., point function. (iii) In cyclic process, there is no change in internal energy (shown in fig.) i.e., dU= Uf – Ui = 0 Þ Uf = Ui (iv) Internal energy of an ideal gas depends only on temperature eq.(2). Firstlawof thermodynamics isa generalisation of the law of conservation of energy that includes possible change in internal energy. First law of thermodynamics “If certain quantity of heat dQ is added to a system, a part of it is used in increasing the internal energy by dU and a part is use in performing external work done dW i.e., dQ dU dW dU dQ dW = + Þ = - The quantity dU (i.e., dQ – dW) is path independent but dQ and dW individually are not path independent. Applications of First Law of Thermodynamics (i) In isobaric process P is constant so ò - = = 2 1 V V 1 2 ) V V ( P PdV dW so dQ = dU + dW = n CP dT 1 2 Thermodynamics
- 308. 304 PHYSICS For idealgas, dQ = 0 dU = mCVdT (for any process) 2 2 1 1 V V V V K dW PdV dV Vg = = ò ò (where PVg = K = constant) 2 2 1 1 1 1 2 1 ( ) 1 1 1 1 PV PV K V V g g g g - - æ ö - = - = ç ÷ - - è ø where PVg = constant isapplicable onlyin adiabaticprocess. Adiabatic process is called isoentropic process (in these process entropy is constant). (iii) Isobaric process : A process taking place at constant pressure is called an isobaric process. In this process dQ = n CpdT, dU = n CVdT and dW = P(V2–V1) (iv) Isochoric process : A process taking place at constant volume is called isochoric process. In this process, dQ = dU =n CVdT and dW = 0 (v) Cyclic process : In this process the inital state and final state after traversing a cycle (shown in fig.) are same. In cyclic process, dU = 0 = Uf – Ui and dW = area of cycle = area (abcd) Slope of adiabaticandisothermal curve: For isothermal process PV = constant On differentiating, we get PdV + VdP = 0 And slope of isothermal curves V P dV dP isothermal - = ÷ ø ö ç è æ ... (1) Isothermal Adiabatic V P For adiabatic process PVg = constant On differentiation, we get slope of adiabatic curve æ ö = -g ç ÷ è øadiabatic P (P/V) V d d ....(2) It is clear from equation (1) and (2) that the slope of adiabatic curve is more steeper than isothermal curve as shown by fig by g time (g = CP/CV) (ii) In cyclic process heat given to the system is equal to work done (area of cycle). (iii) In isothermal process temperature T is constant and work done is 1 2 V V e V V Log nRT PdV dW 2 1 ò = = Since, T = constant so for ideal gas dU = 0 Hence, 1 2 e V V Log nRT dW dQ = = (for ideal gas) (iv) In isochoric process W = 0 as V = constant It means that heat given to system is used in increasing internal energy of the gas. (v) In adiabatic process heat given or taken by system from surroundingiszeroi.e.,dQ=0 ( ) 1 2 nR dU dW T T 1 é ù = - = - - ê ú g - ë û 1 1 2 2 (P V P V ) 1 é ù - = ê ú g - ë û It means that if system expands dW is +ive and dU is –ive (i.e., temperature decrease) and if system contracts dW is –ive and dUis +ive (i.e., temperature increase). THERMODYNAMIC PROCESSES (i) Isothermal process : If a thermodynamic systemisperfectly conducting to surroundings and undergoes a physical change in such a way that temperature remains constant throughout, then process is said to be isothermal process. T = constant V P For isothermal process, the equation of state is PV = nRT = constant, where n is no. of moles. For ideal gas, since internal energy depends only on temperature. 2 2 1 1 0 V V V V dV dU dQ dW PdV nRT V = Þ = = = ò ò or 2 2 10 1 1 log 2.303 log e V V dQ nRT nRT V V = = (ii) Adiabatic process : If system is completely isolated from the surroundings so that no heat flows in or out of it, then any change that the system undergoes is called an adiabatic process. V P
- 309. 305 Thermodynamics Graphs of thermodynamicprocesses : 1. In the figure (i) P–V graph the process ab is isothermal, bc is isobaric and ca is isochoric. V P a c b Fig (i) The fig (ii) is the P–T diagram offig (i) T P a c b Fig.(ii) 2. Figure below shows P – V diagrams for two processes. P I II V The heat absorbed in process I is more than that in II. Because, area under process I is also more than area under process II. The work done in the process I is more than that in II. Also, the change in internal energy is same in both cases. 3. The P–V and corresponding V–T diagram for a cyclic process abca on a sample of constant mass of ideal gas are shown below: V P a c b T V a c b 4. For isochoric process, the P–V, V–T and P–T graphs : P 2 1 V 2 1 V T 2 1 P T 5. For isobaric process, the P – V, P – T and V – T graphs : V P P T V T 6. For isothermal process, the P – V, V – T and P– T graphs : P V V T P T Keep in Memory 1. In thermodynamics heat and work are not state variables, whereas internal energyis a state variable. 2. For ideal-gas (i) relation between P and V is PVg = constant (ii) relation between V and T is TV g–1 = constant (iii) relation between P and T is TgP1–g = constant 3. A quasi-static process is an infinitely slow process such that system remains in thermal and mechanical equilibrium with the surroundings throughout. 4. Pressure, volume, temperature and mass are state variables. Heat and work are not state variables. 5. A graphical representation ofthe state of a system with the help of two thermodynamical variables is called indicator diagram.
- 310. 306 PHYSICS REVERSIBLE ANDIRREVERSIBLE PROCESS Reversible Process : A process which can proceed in opposite direction in such a way that the system passes through the same states as in direct process and finally the system and the surroundings acquire the intial conditions. Conditions for a processtobe reversible : (a) The process must be extremely slow. (b) There should no loss of energy due to conduction, or radiation. The dissipating forcesshould not be in thesystem. (c) The system must always be in thermal and chemical equilibrium with the surroundings. Examples : Fusion of ice, vaporisation of water, etc. Irreversible Process : The process which cannot be traced back in the opposite direction is defined as irreversible process. Examples : Work done against friction, magnetic hysteresis. • In nature all process are irreversible, because no natural process can fulfil the requirement of a reversible process. HEAT ENGINE A heat engine is a device which converts heat energy into mechanical energy. Hot Reservoir T1 Cold Reservoir T2 Working Substance Work (W) = Q – Q 1 2 Q1 Q2 Efficiency of heat engine is given by 1 2 2 1 1 ( ) η ( ) 1 1 Work done W Efficiency Heat taken from source T Q T Q T = = - = - where Q2 = amount of heat rejected per cycle to the sink (of tempT2) Q1 = amount of heat energy absorbed per cycle from the source (of tempT1). The efficiency of heat engine h is never greater than unity, h =1 onlyfor ideal engine & for practical heat engine h < 1. REFRIGERATOR AND HEAT PUMP : Refrigerator or heat pump is a heat engine running in backward direction i.e. working substance (a gas) takes heat from a cold body and gives out to a hotter body with the use of external energy i.e. electrical energy. A heat pump is the same as a refrigerator. Hot Reservoir T1 Cold Reservoir T2 Working Substance Work (W) = Q – Q 1 2 Q1 Q2 The coefficient of performance of refrigerator or heat pump is Heat extractedfromcoldreservoir Work doneon refrigerator b = = = - - 2 2 1 2 1 2 Q T Q Q T T , whereT2 is temperature of cold body and T1 is temperature of hot body. CARNOT ENGINE Carnot devised an ideal engine which is based on a reversible cycle of four operations in succession : isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. A D B Q1 Q2 C P Adia b a tic expa n s io n Isothermal expansion Isothermal compression Ad ia b a tic c o mp re s s io n T1 T2 V V1 V2 Efficiency of Carnot engine, 1 W Q h = = 2 4 1 2 1 3 2 1 1 V V µRT In µRT In V V V µRT In V æ ö æ ö + ç ÷ ç ÷ è ø è ø æ ö ç ÷ è ø The points B and C are connected by an adiabatic path as are the points D and A. Hence, using this eqn. and the adiabatic gas eqn. T1V2 (g – 1) = T2 V3 (g – 1) and T1V1 (g – 1) = T2 V4 (g – 1). Combination of the above eqns. gives 3 2 1 4 V V V V = , and, 1 2 1 T T T - h = = 1 2 1 Q Q Q - or, 2 2 1 1 1 1 Q T Q T h = - = - . The percentage efficiency of Carnot’s engine, 1 2 1 100% T T T - h = ´ or, 1 2 1 100% Q Q Q - h = ´ The efficiency of a Carnot engine is never 100% because it is100% onlyiftemperatureofsinkT2 =0 which isimpossible. In a Carnot cycle, 2 2 1 1 Q T Q T = or 1 2 1 2 Q Q T T = .
- 311. 307 Thermodynamics Carnot Theorem :No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same hot and cold reservoirs. i.e., R I h > h or 2 2 1 1 1 1 T Q T Q - > - SECOND LAW OF THERMODYNAMICS It states that it is impossible for a self acting machine unaided by any external agency, to transfer heat from a body at a lower temperature to a body at higher temperature. It is deduced from this lawthat the efficiencyof any heat engine can never be 100%. Entropy : Entropy is a measure of disorder of the molecular motion of a system. The greater the disorder, the greater is the entropy. The change in entropy is given by ( ) ( ) Heat absorbed by the system dQ S Absolute temperature T D = 1 2 dQ S S T - = ò (here T is not differentiable) Clausius inequality dQ 0 T £ ò Ñ or, dQ dS T ³ ò or, dQ = TdS ³ dU+ PdV Also, S = K logew 2 e 1 S Klog w D = w is the microscopic form of entropy, where K is Boltzmann's constant and wrespresents the number of possible microscopic states. Energy entering a body increases disorder. Energy leaving a body decreases disorder. When a hot body is brought into thermal contact with a coldbody for a short time, then : (i) Each bodywill experience a change in the entropy of its particle. (ii) The hot body experiences a decrease in entropy (a negative change) of magnitude 1 1 Q S T D D = (iii) The cold body experiences an increase in entropy (a positive change) of magnitude 2 2 Q S T D D = (iv) The net change in entropy 1 2 S S S D = D + D The effect of naturally occurring processes is always to increase the total entropy (or disorder) of the universe. Example1. A cyclic process is shown in fig. Work done during isobaric expansion is 1 2 3 B A D C 2 × 10 2 P (N/m ) 2 10 2 V(m ) 3 (a) 1600J (b) 100J (c) 400J (d) 600J Solution : (c) Isobaric expansion is represented by curve AB; Work done = area under AB = 2 × 102 × (3 –1) =4 × 102 = 400 J. Example2. An ideal gas heat engine operates in carnot cycle between 227ºC and 127ºC. It absorbs 6 × 104 cal of heat at higher temp. Amount of heat converted into work is (a) 1.2 × 104 cal (b) 2.4 × 104 cal (c) 6 × 104 cal (d) 4.8 × 104 cal Solution : (a) As 1 2 1 2 T T Q Q = ; 2 4 Q 127 273 400 227 273 500 6 10 + = = + ´ (As 1 2 1 2 T T 1 Q Q 1 - = - = h ) 4 4 2 4 6 10 4.8 10 5 Q cal = ´ ´ = ´ = - = ´ - ´ = ´ 4 4 4 1 2 W Q Q 6 10 4.8 10 1.2 10 cal. Example3. An ideal carnot engine whose efficiency is 40% receives heat at 500 K. If its efficiency were 50%, then what would be intake temp. for same exhaust temp ? Solution : From, h - = - = h 1 T T ; T T 1 1 2 1 2 5 3 100 40 1 = - = ; K 300 500 5 3 T 5 3 T 1 2 = ´ = = Again 2 1 T 1 η' T ' = - or 1 300 50 1 1 T ' 100 2 = - = or 1 T ' 600K =
- 312. 308 PHYSICS Example4. When asystem is taken from state a to state b, in fig. along the path a ® c ® b, 60 J of heat flow into the system, and 30 J of work are done : (i) How much heat flows into the system along the path a ® d ® b if the work is 10 J. (ii) When the system is returned from b to a along the curved path, the work done by the system is –20 J. Does the system absorb or liberate heat, and how much? (iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the process a ® d and d ® b. P V c a b d Solution : For the path a ® c ® b dU = dQ – dW = 60 – 30 = 30 J or Ub – Ua = 30 J (i) Along the path a ® d ® b dQ = dU+ dW = 30 + 10 = 40 J (ii) Along the curved path b – a dQ = (Ua – Ub) + W = (–30) + (–20) = –50 J, heat flows out the system (iii) Qad = 32 J; Qdb = 8 J Example5. Two samples of a gas initially at same temperature and pressure are compressed from avolume V to V/2. One sample is compressed isothermally and the other adiabatically. In which sample is the pressure greater? Solution : Let initial volume, V1 = Vand final volume, V2 = V/2 Initial pressure, P1 = P ; final pressure, P2 = ? For isothermal compression P2V2 = P1V1 or 1 1 2 2 P V PV P 2P V V / 2 = = = For adiabatic compression P2' V2 g = P1V1 g or P2' = P1 1 2 V V g æ ö ç ÷ è ø = V P V / 2 g æ ö ç ÷ è ø or P2´ = 2gP Since g > 1 2g > 2 P2' > P2 Pressure during adiabatic compression is greater than the pressure during isothermal compression. Example6. A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle? Solution : W = 800 J, T1 = 600 K, T2 = 300 K h = 1 – 1 2 1 T W T Q = = 1 300 800 1 600 Q - = or 0.5 = 1 800 Q Heat energy supplied by source, 800 Q 0.5 = = 1600 joule per cycle Example7. The temperatures T1 and T2 of the two heat reservoirs in an ideal Carnot engine are 1500°C and 500°C respectively. Which of the following : increasing T1 by 100°C or decreasing T2 by 100°C would result in a greater improvement in the efficiency of the engine? Solution : The efficiency of a Carnot's engine is given by 2 1 T 1 T h = - Given T1 =1500°C =1500+ 273= 1773Kand T2 = 500°C = 500 + 273 = 773 K. When the temperature of the source is increased by100°C, keeping T2 unchanged, the new temperature of the source is T´1 =1500+100=1600°C =1873K.Theefficiencybecomes 2 1 T 773 ´ 1 1 0.59 T´ 1873 h = - = - = On the other hand, if the temperature ofthe sink isdecreased by 100°C, keeping T1 unchanged, the new temperature of the sink is T´2 = 500 – 100 = 400°C = 673 K. The efficiency now becomes 2 1 T´ 673 ´´ 1 1 0.62 T 1773 h = - = - = Since h´´ is greater than h´, decreasing the temperature of the sink by 100°C results in a greater efficiency than increasing the temperature of the source by100°C. Example8. Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105 N/m2 and 6 litre respectively. The final volume of the gas is 2 litres. Molar specific heat of the gas at constant volume is 3R/2. [(3)5/3 = 6.19] Solution : For an adiabatic change PVg = constant P1V1 g = P2V2 g
- 313. 309 Thermodynamics As molar specificheat of gas at constant volume v 3 C R 2 = CP = CV + R = 3 5 R R R 2 2 + = ; P V C (5 / 2) R 5 C (3/ 2) R 3 g = = = From eqn. (1) 1 2 1 2 V P P V g æ ö = ç ÷ è ø = 5/3 5 2 6 10 N / m 2 æ ö ´ ç ÷ è ø = (3)5/3 × 105 = 6.19 ×105 N/m2 Work done = 5 3 5 3 1 [6.19 10 2 10 10 6 10 ] 5 1 3 - - - ´ ´ ´ - ´ ´ æ ö - ç ÷ è ø 2 2 10 3 (6.19 3) 2 é ù ´ ´ = - - ê ú ê ú ë û = – 3 × 102 × 3.19 = – 957 joule [–ve sign shows external work done on the gas] Example9. A refrigerator is to maintain eatables kept inside at 90C. If room temperature is 360C, calculate the coefficient of performance. Solution : Here, T1 =36°C =36 +273= 309K, T2 =10°C =10 +273 =283K 2 1 2 T 283 283 COP 10.9 T T 309 283 26 = = = = - - Example10. One mole of an ideal gas at pressure P0 and temperature T0 is expanded isothermally to twice its volume and then compressed at constant pressure to (V0/2) and the gas is brought back to original state by a process in which P µ V (Pressure is directly proportional to volume). The correct representation of process is (a) V P V /2 0 V0 2V0 P0 (b) V P V /2 0 V0 2V0 P0 (c) P0 T P P0/2 T0 T /4 0 (d) T V T0 V /2 0 V0 Solution : (c) ProcessAB is isothermal expansion, BC is isobaric compression and in process CA 2 nRT P P T P µ Þ µ P0 A P P0/2 T0 T /4 0 B C Example11. A Carnot’s heat engine works with an ideal monatomic gas, and an adiabatic expansion ratio 2. Determine its efficiency. Solution : Given, 3 2 V 2 V r = = and g for a monatomicgas = 5/3. Using, 1 1 1 g - æ ö h = -ç ÷ r è ø we have, the required efficiency 5 1 3 1 1 1 0.63 0.37 or 37% 2 - æ ö h = - = - = ç ÷ è ø
- 314.
- 315. 311 Thermodynamics 1. Which ofthe following is incorrect regarding first law of thermodynamics? (a) It is a restatement of principle of conservation of energy. (b) It is applicable to cyclic processes (c) It introduces the concept of entropy (d) It introduces the concept of internal energy 2. Choose the incorrect statement related to an isobaric process. (a) constant V T = (b) W= PDV (c) Heat given to a system is used up in raising the temperature only. (d) DQ > W 3. The internal energy of an ideal gas does not depend upon (a) temperature of the gas (b) pressure of the gas (c) atomicity of the gas (d) number of moles of the gas. 4. During isothermal expansion, the slope of P-V graph (a) decreases (b) increases (c) remains same (d) mayincrease or decrease 5. During melting of ice, its entropy (a) increases (b) decreases (c) remains same (d) cannot say 6. Which of the following processes is adiabatic ? (a) Melting of ice (b) Bursting of tyre (c) Motion of piston of an engine with constant speed (d) None of these 7. At a given temperature the internal energy of a substance (a) in liquid state is equal to that in gaseous state. (b) in liquid state is less than that in gaseous state. (c) in liquid state is more than that in gaseous state. (d) is equal for the three states of matter. 8. Air conditioner is based on the principle of (a) Carnot cycle (b) refrigerator (c) first low of thermodynamics (d) None of these 9. A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabaticallytoits original volume. Assuming g to be 1.5, the new pressure of the gas is (a) 2P (b) P (c) 4P (d) P/2 10. One mole of an ideal gas at temperature T was cooled isochorically till the gas pressure fell from P to n P . Then, by an isobaric process, the gas was restored to the initial temperature. The net amount of heat absorbed by the gas in the process is (a) nRT (b) n RT (c) RT (1 – n–1) (d) RT (n – 1) 11. Ice contained in a beaker starts melting when (a) the specific heat of the system is zero (b) internal energy of the system remains constant (c) temperature remains constant (d) entropy remains constant 12. A uniform sphere is supplied heat electrically at the centre at a constant rate. In the steady state, steady temperatures are established at all radial locations r, heat flows outwards radial and is ultimately radiated out by the outer surface isotropically. In this steadystate, the temperature gradient varies with radial distance r according to (a) r–1 (b) r–2 (c) r–3 (d) r–3/2 13. For an ideal gas graph is shown for three processes. Process 1, 2 and 3 are respectively. 1 2 3 Work done (magnitude) Temperature change DT (a) Isobaric, adiabatic, isochoric (b) Adiabatic, isobaric, isochoric (c) Isochoric, adiabatic, isobaric (d) Isochoric, isobaric, adiabatic 14. The efficiency of carnot engine when source temperature is T1 and sink temperature is T2 will be (a) 1 2 1 T T T - (b) 2 1 2 T T T - (c) 1 2 2 T T T - (d) 1 2 T T 15. In the equation PVg = constant, the value ofg is unity. Then the process is (a) isothermal (b) adiabatic (c) isobaric (d) irreversible
- 316. 312 PHYSICS 16. Foradiabatic processes (Letters have usual meanings) (a) PgV = constant (b) TgV = constant (c) TVg–1 = constant (d) TVg = constant 17. The gas law T PV = constant is true for (a) isothermal changes only (b) adiabatic changes only (c) both isothermal and adiabatic changes (d) neither isothermal nor adiabatic change 18. When heat is given to a gas in an isothermal change, the result will be (a) external work done (b) rise in temperature (c) increase in internal energy (d) external work done and also rise in temperature 19. Volume of one mole gas changes according to the V = a/T. If temperature change is DT, then work done will be (a) RDT (b) – RDT (c) R 1 g - DT (d) R (g – 1)DT 20. In changing the state ofthermodynamics fromAto Bstate, the heat required is Q and the work done by the system is W. The change in its internal energy is (a) Q + W (b) Q – W (c) Q (d) 2 Q W - 21. If DQ and DW represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as (a) Q U W D = D + D (b) Q U W D = D - D (c) Q W U D = D - D (d) – Q W U D = D - D 22. The work done in which of the following processes is equal to the internal energy of the system? (a) Adiabatic process (b) Isothermal process (c) Isochoric process (d) None of these 23. Which of the following processes is reversible? (a) Transfer of heat by conduction (b) Transfer of heat by radiation (c) Isothermal compression (d) Electrical heating of a nichrome wire 24. In thermodynamic processes which of the following statements is not true? (a) In an isochoric process pressure remains constant (b) In an isothermal process the temperature remains constant (c) In an adiabatic process PVg = constant (d) In an adiabatic process the system is insulated from the surroundings 25. Monatomic, diatomic and polyatomic ideal gases each undergo slow adiabatic expansions from the same initial volumeand same initial pressure tothe same final volume. The magnitude of the work done by the environment on the gas is (a) the greatest for the polyatomic gas (b) the greatest for the monatomic gas (c) the greatest for the diatomic gas (d) the question is irrelevant, there is no meaning ofslow adiabatic expansion 1. Agas at 27ºC and pressureof 30 atm. is allowed toexpand to atmospheric pressure and volume 15 times larger. The final temperature of the gas is (a) –123ºC (b) +123ºC (c) 273ºC (d) 373ºC 2. A system changes from the state (P1, V1) to (P2, V2) as shown in the figure. What is the work done by the system? Pressure i n N/m 2 6×10 5 5×10 5 4×10 5 3×10 5 2×10 5 1×10 5 1 2 3 4 5 Volume in metre3 (P , V ) 2 2 (P , V ) 1 1 (a) 7.5 × 105 joule (b) 7.5 × 105 erg (c) 12 × 105 joule (d) 6 × 105 joule 3. A refrigerator works between 0ºC and 27ºC. Heat is to be removed from the refrigerated space at the rate of 50 kcal/ minute, the power of the motor of the refrigerator is (a) 0.346kW (b) 3.46kW (c) 34.6kW (d) 346kW 4. A perfect gas goes from a state A to another state B by absorbing 8 × 105 J ofheat and doing 6.5 × 105 J ofexternal work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process (a) work done by gas is 105 J (b) work done on gas is 105 J (c) work done by gas is 0.5 × 105 J (d) work done on the gas is 0.5 × 105 J
- 317. 313 Thermodynamics 5. The temperatureof 5 moles of a gas which was held at constant volume was changed from 100º to 120ºC. The change in the internal energyofthe gas was found to be 80 joule, the total heat capacity of the gas at constant volume will be equal to (a) 8 joule per K (b) 0.8 joule per K (c) 4.0 joule per K (d) 0.4 joule per K 6. A polyatomic gas (g = 4/3) is compressed to 1/8th of its volume adiabatically. If its initial pressure is P0, its new pressure will be (a) 8P0 (b) 16P0 (c) 6P0 (d) 2P0 7. The efficiency of a Carnot engine operating with reservoir temperatures of 100ºC and –23ºC will be (a) 100 23 100 + (b) 100 23 100 - (c) 373 23 100 + (d) 373 23 100 - 8. Bywhat percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature? (a) 8.1% (b) 9.1% (c) 10.1% (d) 11.1% 9. A gas has pressure P and volume V. It is now compressed adiabaticallyto1/32 times the original volume. Given that (32)1.4 = 128, the final pressure is (g= 1.4) (a) P/128 (b) P/32 (c) 32P (d) 128P 10. At 27ºC a gas is compressed suddenlysuch that its pressure becomes (1/8) of original pressure. Final temperature will be (g= 5/3) (a) 450K (b) 300K (c) –142ºC (d) 327ºC 11. A diatomic gas initallyat 18ºC is compressed adiabatically to one eighth of its original volume. The temperature after compression will be (a) 18ºC (b) 887ºC (c) 327ºC (d) None of these 12. Absolute zero is obtained from (a) P–V graph (b) V 1 P - graph (c) P–T graph (d) V–Tgraph 13. An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 104 cals of heat at higher temperature. Amount of heat converted to work is (a) 4.8 × 104 cal (b) 6 × 104 cal (c) 2.4 × 104 cal (d) 1.2 × 104 cal 14. Three moles ofan ideal gas kept at a constant temperature at 300 K are compressed from a volume of 4 litre to 1 litre. The work done in the process is (a) – 10368 J (b) –110368J (c) 12000J (d) 120368J 15. The temperature at which speed of sound in air becomes double of its value at 27° C is (a) 54°C (b) 327°C (c) 927°C (d) None of these 16. 1 gm ofwater at a pressure of1.01 × 105 Pa is converted into steam without anychange of temperature. The volume of1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is (a) »167cal (b) »500cal (c) 540cal (d) 581cal 17. An ideal refrigerator has a freezer at a temperature of 13ºC. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is (a) 320ºC (b) 39ºC (c) 325K (d) 325ºC 18. Onemole ofan ideal monoatomic gasisheated at a constant pressure of one atmosphere from 0ºC to 100ºC. Then the work done by the gas is (a) 6.56 joule (b) 8.32 ×102 joule (c) 12.48× 102 joule (d) 20.8× 102 joule 19. The pressure inside a tyreis4 timesthat ofatmosphere. Ifthe tyre bursts suddenly at temperature 300 K, what will be the newtemperature? (a) 300(4)7/2 (b) 300(4)2/7 (c) 300(2)7/2 (d) 300(4)–2/7 20. A monatomic ideal gas expands at constant pressure, with heat Q supplied. The fraction of Q which goes as work done by the gas is (a) 1 (b) 3 2 (c) 5 3 (d) 5 2 21. A carnot’s engine takes 300 calories of heat at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is (a) 1000K (b) 750K (c) 250K (d) 125K 22. The source and sink temperatures of a Carnot engine are 400 K and 300 K, respectively. What is its efficiency? (a) 100% (b) 75% (c) 33.3% (d) 25% 23. The volume of a gas is reduced adiabatically to 1/4 of its volume at 27ºC. If g = 1.4 the new temperature is (a) (300)20.4 K (b) (300)21.4 K (c) 300 (4)0.4 K (d) 300 (2)1.4 K 24. In pressure-volume diagram, the isochoric, isothermal, isobaric and iso-entropic parts respectively, are A B C D P V (a) BA,AD, DC,CB (b) DC, CB,BA,AD (c) AB,BC, CD, DA (d) CD, DA,AB, BC
- 318. 314 PHYSICS 25. Twocylinders fitted with pistons contain equal amount of an ideal diatomic gas at 300 K. The piston of A is free to move, whilethat ofBis held fixed. Thesameamount ofheat is given to the gas in each cylinder. Ifthe rise in temperature ofthe gas inAis 30 K, then the rise in temperature of gas in B is (a) 30K (b) 18K (c) 50K (d) 42K 26. A Carnot engine works first between 200°C and 0°C and then between 0°C and –200°C. The ratio of its efficiencyin the two cases is (a) 1.0 (b) 0.577 (c) 0.34 (d) 0.68 27. A Carnot’s engine works as a refrigerator between 250 K and300K. Ifitreceives750caloriesofheatfrom thereservoir at the lower temperature, the amount of heat rejected at the higher temperature is (a) 900 calories (b) 625calories (c) 750calories (d) 1000calories 28. A Carnot engine is working between 127°C and 27°C. The increasein efficiencywill bemaximumwhen thetemperature of (a) the source is increased by 50°C (b) the sink is decreased by 50°C (c) source is increased by 25°C and that of sink is decreased by 25°C (d) both source and sink are decreased by 25°C each. 29. During an adiabatic process an object does 100J of work and its temperature decreases by 5K. During another process it does 25J of work and its temperature decreases by 5K. Its heat capacity for 2nd process is (a) 20J/K (b) 24J/K (c) 15J/K (d) 100J/K 30. The internal energychange in a system that has absorbed 2 kcals of heat and done 500 J of work is (a) 6400J (b) 5400J (c) 7900J (d) 8900J 31. In an adiabatic process, the pressure is increased by 2 % 3 . If g = 3 2 , then the volume decreases by nearly (a) 4 % 9 (b) 2 % 3 (c) 1% (d) 9 % 4 32. A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume of gas in two parts respectivelyare(P, 5V) and (10P,V). Ifnowthe piston is left free and the system undergoes isothermal process, then the volumes of the gas in two parts respectively are (a) 2V,4V (b) 3V,3V (c) 5V ,V (d) 10 20 V, V 11 11 33. Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperatureof gas at cand d are 300 K and 500 K. Calculate the heat absorbed by the gas during the process. P U a d b c P0 2P0 (a) 400Rln 2 (b) 200R ln 2 (c) 100R ln 2 (d) 300R ln 2 34. The figure shows the P-V plot ofan ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, D P A B C V 3 2 1 0 1 2 3 (a) the process during the path A ® B is isothermal (b) heat flows out of the gas during the path B ® C ® D (c) work done during the path A ® B ® C is zero (d) positive work is done by the gas in the cycle ABCDA 35. A thermodynamic system goes from states (i) P1, V to 2P1, V (ii) P, V1 to P, 2V1. Then work done in the two cases is (a) zero, zero (b) zero, PV1 (c) PV1, zero (d) PV1, P1V1 36. For an isothermal expansion of a perfect gas, the value of P P D is equal to (a) 1/2 – V V D g (b) – V V D (c) – V V D g (d) 2 – V V D g 37. One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be (a) (T – 4) K (b) (T + 2.4)K (c) (T – 2.4) K (d) (T + 4) K
- 319. 315 Thermodynamics 38. If U Dand W D represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true? (a) , U W D = -D in an adiabatic process (b) , U W D =D in an isothermal process (c) , U W D =D in an adiabatic process (d) , U W D =- D in an isothermal process 39. During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies that (a) 150 J heat has been removed from the gas (b) 300 J of heat has been added to the gas (c) noheat is transferredbecause the process is isothermal (d) 150 J of heat has been added to the gas 40. When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C, is (a) 273cal/K (b) 8 × 104 cal/K (c) 80cal/K (d) 293cal/K 41. A mass of diatomic gas (g = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in final state is (a) 28atm (b) 68.7atm (c) 256atm (d) 8atm 42. A thermodynamicsystem is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is D C B A 3V V P 2P Volume Pressure (a) 2PV (b) 4PV (c) 1 2 PV (d) PV 43. An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram : P V A 1 2 B 3 If Q1, Q2, Q3 indicate the heat a absorbed bythe gas along the three processes and DU1, DU2, DU3 indicate the change in internal energy along the three processes respectively, then (a) Q1 > Q2 > Q3 and DU1 = DU2 = DU3 (b) Q3 > Q2 > Q1 and DU1= DU2 = DU3 (c) Q1 = Q2 = Q3 and DU1 > DU2 > DU3 (d) Q3 > Q2 > Q1 and DU1> DU2 > DU3 44. Choose the correct relation between efficiency h of a Carnot engine and the heat absorbed (q1) and released by the working substance (q2). (a) h = 2 1 1 q + q (b) h = 1 2 1 q + q (c) h = 1 2 1 q - q (d) 2 1 1 q h = - q 45. In the given (V – T) diagram, what is the relation between pressure P1 and P2 ? T V P2 P1 q1 q2 (a) P2 > P1 (b) P2 < P1 (c) P2 = P1 (d) Cannot be predicted 46. A system goes from A to B via two processes I and II as shown in figure. If DU1 and DU2 are the changes in internal energies in the processes I and II respectively, then II I B A p v (a) relation between DU1 and DU2 can not be determined (b) DU1 = DU2 (c) DU1 < DU2 (d) DU1 > DU2 47. Which of thefollowing statements about a thermodynamic process is wrong ? (a) For an adiabatic process DEint = – W (b) For a constant volume process DEint = + Q (c) For a cyclic process DEint = 0 (d) For free expansion of a gas DEint > 0
- 320. 316 PHYSICS 48. Ina Carnot engine efficiency is 40% at hot reservoir temperature T. For efficiency 50%, what will be the temperature of hot reservoir? (a) T (b) 2 T 3 (c) 4 T 5 (d) 6 T 5 Directions for Qs. (49 to 50) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethecorrectanswer (ONLYONE option is correct ) from the following. (a) Statement-1 is false, Statement-2 is true Exemplar Questions 1. An ideal gas undergoes four different processes from the same initial state (figure). Four processes are adiabatic, isothermal, isobaric and isochoric.Out of 1, 2, 3 and 4 which one is adiabatic? p V 1 2 3 4 (a) 4 (b) 3 (c) 2 (d) 1 2. If an average person jogs, he produces 14.5 × 103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaporation) is (a) 0.025kg (b) 2.25kg (c) 0.05kg (d) 0.20kg 3. Consider p-V diagram for an ideal gas shown in figure. p V 1 2 p V constant = Out of the following diagrams, which figure represents the T-pdiagram? (i) T p 1 2 (ii) T p 1 2 (iii) T p 1 2 (iv) T p 1 2 (a) (iv) (b) (ii) (c) (iii) (d) (i) 4. An ideal gas undergoes cyclic process ABCDA as shown in given p-V diagram. The amount of work done bythe gas is D C B A p V V0 3V0 2p0 p0 (a) 6p0V0 (b) –2p0V0 (c) +2p0V0 (d) +4p0V0 5. Consider twocontainers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermallywhile the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is (a) 1 2g- (b) 1 1 2 g- æ ö ç ÷ è ø (c) 2 1 1 æ ö ç ÷ - g è ø (d) 2 1 1 æ ö ç ÷ g - è ø (b) Statement-1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement-1 istrue, Statement-2 is false 49. Statement-1 : At a given temperature the specific heat ofa gas at constant volume is always greater than its specific heat at constant pressure. Statement-2 : When a gas is heated at constant volume some extra heat is needed compared to that at constant pressure for doing work in expansion. 50. Statement -1 : If an ideal gas expands in vacuum in an insulated chamber, DQ, DUand DWall are zero. Statement-2 : Temperature of the gas remains constant.
- 321. 317 Thermodynamics 6. Three copperblocks of masses M1, M2 and M3 kg respectivelyare brought intothermal contact till theyreach equilibrium. Before contact, theywere at T1, T2, T3 (T1 > T2 > T3). Assuming there is no heat loss to the surroundings, theequilibrium temperature Tis (s is specific heatofcopper) (a) 1 2 3 3 T T T T + + = (b) 1 1 2 2 3 3 1 2 3 M T M T M T T M M M + + = + + (c) 1 1 2 2 3 3 1 2 3 3( ) M T M T M T T M M M + + = + + (d) 1 1 2 2 3 3 1 2 3 M T s M T s M T s T M M M + + = + + NEET/AIPMT (2013-2017) Questions 7. A gas is taken through the cycle A ® B® C ® A, as shown in figure. What is the net work done by the gas ? P (10 Pa) 5 V (10 m ) 3 3 A B C 7 6 5 4 3 2 1 0 2 6 4 8 [2013] (a) 1000J (b) zero (c) – 2000 J (d) 2000J 8. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of p v C C for the gas is [2013] (a) 2 (b) 5 3 (c) 3 2 (d) 4 3 9. A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is Ua = 10 J. Along the path adc the amount of heat absorbed dQ1 = 50 J and the work donedW1 = 20 J whereas along the path abc the heat absorbed dQ2 = 36 J. Theamount ofwork done along the path abc is [NEET Kar. 2013] a b c d p V (a) 6 J (b) 10J (c) 12J (d) 36J 10. Which ofthe following relations does not give the equation of an adiabatic process, where terms have their usual meaning? (a) PgT1–g = constant [NEET Kar. 2013] (b) P1–g Tg = constant (c) PVg = constant (d) TVg–1 = constant 11. Two Carnot engines A and B are operated in series. The engineA receives heat from the source at temperature T1 and rejects the heat to the sink at temperature T. Thesecond engine B receives the heat at temperature T and rejects to its sink at temperature T2. For what value of T the efficienciesofthe two enginesare equal? [NEET Kar. 2013] (a) 1 2 2 T T + (b) 1 2 2 T T - (c) T1T2 (d) 1 2 T T 12. Athermodynamic system undergoes cyclic processABCDA as shown in fig. The work done by the system in the cycle is : [2014] (a) P0V0 (b) 2P0V0 (c) 0 0 P V 2 P C B D A V0 2V0 V P0 2P0 3P0 (d) Zero 13. A monoatomic gas at a pressure P, having a volume V expands isothermallyto a volume2V and then adiabatically to a volume 16V. The final pressure of the gas is :(take g = 5 3 ) [2014] (a) 64P (b) 32P (c) P 64 (d) 16P 14. Figure below shows two paths that may be taken by a gas to go from a state A to a state C. 2 × 10 m 4 × 10 m –3 3 –3 3 2×10 Pa 4 6×10 Pa 4 P B C A V
- 322. 318 PHYSICS In processAB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be [2015] (a) 500J (b) 460J (c) 300J (d) 380J 15. A carnot engine having an efficiency of 1 10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is :- [ 2015, 2017] (a) 90 J (b) 99 J (c) 100 J (d) 1 J 16. The coefficient of performance of a refrigerator is 5. If the insidetemperatureof freezer is–20°C, then thetemperature of the surroundings to which it rejects heat is [2015 RS] (a) 41°C (b) 11°C (c) 21°C (d) 31°C 17. An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? [2015 RS] (a) Isobaric (b) Isochoric (c) Isothermal (d) Adiabatic 18. A gas is compressed isothermallyto half its initial volume. Thesame gas is compressedseparatelythrough an adiabatic process until its volume is again reduced to half. Then : [2016] (a) Compressing the gas isothermally will require more work to be done. (b) Compressing the gas through adiabatic process will require more work to be done. (c) Compressing the gas isothermallyor adiabaticallywill require the same amount of work. (d) Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas. 19. A refrigerator worksbetween 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 joules) [2016] (a) 2.365W (b) 23.65W (c) 236.5W (d) 2365W 20. Thermodynamic processes are indicated in the following diagram : [2017] P i IV f III f f f I 700k 500k 300k V II Match the following Column-1 Column-2 P. Process I A. Adiabatic Q. Process II B. Isobaric R. Process III C. Isochoric S. Process IV D. Isothermal (a) P ® C, Q ® A, R ® D, S ® B (b) P ® C, Q ® D, R ® B, S ® A (c) P ® D, Q ® B, R ® A, S ® C (d) P ® A, Q ® C, R ® D, S ® B
- 323. 319 Thermodynamics EXERCISE - 1 1.(c) 2. (c) 3. (b) 4. (a) 5. (a) 6. (b) 7. (b) 8. (b) 9. (a) Let P and V be the initial pressure and volume of ideal gas.Afterisothermalexpansion,pressureisP/4.Sovolume is4V. Let P1 be the pressure after adiabatic compression. Then g g = ) V 4 ( ) 4 / P ( V P1 P 2 ) 4 ( ) 4 / P ( P 2 / 3 1 = = 10. (c) The temperature remains unchanged therefore i f U U = . Also, W Q D = D . In the first step which is isochoric, 0 W = D . In second step, pressure = n P . Volume V is increased fromVon nV. 1 P W (nV V) n n 1 PV n RT(1 n ) - = - - æ ö = ç ÷ è ø = - 11. (c) During melting temperature remains constant 12. (b) Flow rate µ gradient × r2. When flow rate is constant, gradient µ r–2. 13. (d) Isochoric proceess dV = 0 W = 0 proceess 1 Isobaric: W = P DV = nRDT Adiabatic | W | = nR T 1 D g - 0 < g – 1 < 1 As workdone in case of adiabatic process is more so process 3 is adiabatic and process 2 is isobaric. 14. (a) Efficiencyof carnot engine = 1 2 1 T T T - = h where 1 T = source temperature 2 T = sink temperature. 15. (a) PV = constant represents isothermal process. 16. (c) 17. (c) 18. (a) 19. (b) PV= RT ; 2 RT RT P V a = = a V T = 2 a dV dT T = - 2 2 W P dV RT a dT a T = æ ö = - ç ÷ è ø ò ò W = – R DT 20. (b) Q U W D = D + D U Q W Q W Þ D = D - D = - (using proper sign) 21. (b) From FLOT Q U W D = D + D Q Heat supplied to the system so DQ ® Positive and work is done on the system so DW® Negative Hence Q U W +D = D - D 22. (a) In adiabatic process DQ= 0 DW = – DU 23. (c) For process to be reversible it must be quasi-static. For quasi static process all changes take place infinitelyslowly. Isothermal process occur veryslowly so it is quasi-static and hence it is reversible. 24. (a) In an isochoric process volume remains constant whereas pressure remains constant in isobaric process. 25. (a) nRdT W 1 = g - g is minimum for a polyatomic gas Hence, W is greatest for polyatomic gas EXERCISE - 2 1. (a) We know that 2 2 2 1 1 1 T V P T V P = 1 1 1 2 2 2 V P T V P T = Here P1 = 30 atm., P2 = 1 atm, V1 = V(say), V2 = 15V, T1 = 27ºC = 27 + 273 = 300ºK and T2 =? Substituting these values, we get T2 = 150ºK T2 = 150ºK = 150 – 273 = –123ºC. 2. (c) 2 ) 1 5 ( ) 10 6 10 1 ( W 5 5 - ´ + ´ = 5 5 6 10 4 12 10 joule 2 ´ ´ = = ´ 3. (a) W Q T T T 2 2 1 2 = - Hints & Solutions
- 324. 320 PHYSICS W 000 , 50 273 300 273 = - 27 50,000 cal/min 273 W ´ = 4.227 50,000 Joule/sec 60 273 W P t ´ ´ = = ´ = 346 watt = 0.346 kW 4. (d) ) 10 5 . 6 10 8 ( dW dQ dU 5 5 ´ - ´ = - = J 10 5 . 1 5 ´ = 5 5 10 5 . 1 10 dU dQ dW ´ - == - = 5 0.5 10 J = - ´ – ve sign indicates that work done on the gas is J 10 5 . 0 5 ´ . 5. (c) v v dU nC dT or 80 5 C (120 100) = = ´ - Cv = 4.0 joule/K 6. (b) 0 0 4/3 V 4/3 4/3 P V = P P = P 8 1 1 8 æ ö Þ ç ÷ è ø = 16Po 7. (c) 373 23 100 373 123 373 250 1 T T 1 1 2 + = = - = h Þ - = h 8. (d) T V ) 100 / 90 ( P T V P ¢ = or 90 10 1 90 100 P P + = = ¢ or % 1 . 11 90 10 P P P = = - ¢ 9. (d) PVg = g ÷ ø ö ç è æ 32 V P1 or g = = = 1.4 1 P (32) , P (32) , P 128P 10. (c) g - g g - g = 1 2 2 1 1 1 P T P T 11. (d) 1 T 18ºC (273 18) 291 K = = + = and 8 / V V 1 2 = We know that 1 V T - g = constant or , 1 1 1 1 2 2 V T V T - g - g = 1 4 . 1 1 2 1 1 2 ) 8 ( 291 V V T T - - g ´ = ÷ ÷ ø ö ç ç è æ = C º 5 . 395 K 5 . 668 = = 12. (c) T Pµ if V is constant, where P is pressure of certain amount of gas and T is absolute zero temperature. 13. (d) We know that efficiency of carnot engine = I T T L H - Also, Efficiency of Heat engine = Work output Heat input 1 1 - = æ ö Þ = - ç ÷ è ø L H S S H T W T Q TL W Q T 4 4 127 273 6 10 1 227 273 1.2 10 cal + æ ö = ´ - ç ÷ è ø + = ´ 14. (a) Work done in an isothermal process is given by 2 10 1 V W 2.3026nRTlog V = Here, n = 3, R = 8.31 J mol–1 degree–1 T = 300 K, V1 =4 litre, V2 = 1litre. Hence, W = 2.3026 × 3 × 8.31 × 300 × 10 1 log 4 =17221.15(–2log102) =–17221.15×2×0.3010=– 10368J. 15. (c) The speed of sound in air, v T µ 1 1 2 2 v T v T = Þ 1 1 2 v (27 273) 2v T + = 2 1 300 2 T = T2 =1200 K=(1200– 273)°C=927°C 16. (b) Joule 10 ] 1 1671 [ 10 01 . 1 V P dW 6 5 - ´ - ´ = D = 1.01 167 cal. 4.2 40cal. nearly ´ = = ΔQ mL 1 540, ΔQ ΔW ΔU = = ´ = + or . cal 500 40 540 U = - = D 17. (b) 2 2 1 2 T 273 13 260, T K T T = - = = - 1 1 260 5 T 260 or T 260 52 = - - = 1 2 T 312K, T 312 273 39ºC = = - =
- 325. 321 Thermodynamics 18. (b) p 5 dQn C dT 1 R 100 2 = = ´ ´ J 10 8 . 20 J 5 . 2077 2 ´ = = v dU n C dT 3 1 R 100 2 1246.5J = = ´ ´ = 2 dU dQ dU 8.31 10 J = - = ´ 19. (d) Under adiabatic change g g - ÷ ÷ ø ö ç ç è æ = 1 2 1 1 2 P P T T or g g - = 1 2 1 1 2 ) P / P ( T T air for 5 / 7 4 . 1 ; ) 1 / 4 ( 300 T ) 5 / 7 ( ) 5 / 7 ( 1 2 = = g = - or 7 / 2 2 ) 4 ( 300 T - = 20. (d) p Q nC T = D and W PΔV nRΔT. = = For monatomic gas, 2 R 5 Cp = . W 2 Q 5 Þ = 21. (c) 1 2 1 2 1 2 1 2 T T Q Q T T 1 Q Q 1 = Þ - = - = h So K 250 300 500 150 Q T Q T 1 1 2 2 = ´ = ´ = 22. (d) Efficiency, T2 = 1 – T 1 η T1 (source temp.) = 400 K T2 (sink temp.) = 300 K 300 1 25% 400 4 = 1 – = = h 23. (c) ) 1 4 . 1 ( 2 ) 1 4 . 1 ( 4 V . T V ) 300 ( - - ÷ ø ö ç è æ = 4 . 0 2 ) 4 ( ) 300 ( T = 24. (d) From C to D, V is constant. So process is isochoric. From D to A, the curve represents constant temperature. So the process is isothermal. From A to B, pressure is constant . So, the process is isobaric. BC represents constant entropy. 25. (d) In cylinder A, heat is supplied at constant pressure while in cylinder Bheat issupplied at constant volume. (DQ)A =nCP(DT)A and (DQ)B = nCV (DT)B Given : (DQ)A =(DQ)B (DT)B P A V C ( T) C 1.4 30 42K = D = ´ = [Q for diatomic gas P V C 1.4 C = ] 26. (b) 2 1 T 1 T 273 0 1 273 200 200 473 h = - + = - + = 2 1 T ' 1 T (273 200) 1 273 0 200 473 h = - - = - + = 200 273 ' 473 200 0.577 h = ´ h = 27. (a) 250 300 250 W 750 - = Heat rejected = 750 + 150 = 900 cal. 28. (b) 1 2 1 T T T - is maximum in case(b). 29. (c) For adiabatic process, dU = – 100 J which remains same for other processes also. Let C be the heat capacity of 2nd process then –(C) 5 = dU + dW = – 100+ 25 = – 75 C = 15J/K 30. (c) According to first law of thermodynamics Q = DU + W DU = Q – W = 2 × 4.2 × 1000 – 500 = 8400 –500 = 7900 J 31. (a) PV3/2 = K
- 326. 322 PHYSICS 3 log Plog V log K 2 + = P 3 V 0 P 2 V D D + = V 2 P V 3 P D D = - or V 2 2 V 3 3 4 9 D æ ö æ ö = - ç ÷ ç ÷ è ø è ø = - 32. (a) The volume on both sides will be so adjusted that the original pressure × volume is kept constant as the piston moves slowly (isothermal change) P5V=P'V' ...........(1) 10PV=P'V'' ...........(2) From(1) and(2), V'' =2V' andfrom V'+V'' =6V V'=2V, V''=4V 33. (a) Change in internal energy for cyclic process (DU) = 0. For process a ® b, (P-constant) 400 ® = D = D = - a b W P V nR T R For process b ® c, (T-constant) b c W 2R (300) ln 2 ® = - For process c ® d, (P-constant) c d W 400R ® = + For process d ® a, (T-constant) d a W 2R (500) ln 2 ® = + Net work a b b c c d d a ( W) W W W W ® ® ® ® D = + + + W 400R ln 2 D = dQ = dU+ dW, first law of thermodynamics dQ = 400 R ln 2. 34. (b,d)(a) Process is not isothermal. (b) Volume decreases and temperature decreases DU = negative, So, DQ = negative (c) Work done in process A ® B ® C is positve (d) Cycle is clockwise, so work done bythe gas is positive 35. (b) (i) Case ® Volume = constant Þ 0 PdV = ò (ii) Case ® P = constant Þ 2 2 1 1 1 1 1 V V V V PdV P dV PV = = ò ò 36. (b) Differentiate PV = constant w.r.t V 0 – Þ D + D = D D Þ = P V V P P V P V 37. (a) 1 nRT nRT 1 V P V P W 2 1 2 2 1 1 - g - = - g - = 1 ) T T ( nR 2 1 - g - = n = 1, T1 = T R 6 1 3 / 5 ) T T ( R 2 = - - Þ Þ Tf = ( T – 4)K 38. (a) Byfirst law of thermodynamics, Q U W D =D + D In adiabatic process, Q D = 0 U W D = -D In isothermal process, U D = 0 Q W D = D 39. (a, d) If a process is expansion then work done is positive so answer will be (a). But in question work done by gas is given –150J so that according to it answer will be (d). 40. (d) Change in entropy is given by dS = dQ T or f mL Q S T 273 D D = = 1000 80 S 293cal/ K. 273 ´ D = = 41. (c) T1 = 273+27 =300K T2 = 273 + 927 = 1200K For adiabatic process, P1–g Tg = constant Þ P1 1–g T1 g = P2 1–g T2 g 1 2 1 P P -g æ ö Þ ç ÷ è ø = 1 2 T T g æ ö ç ÷ è ø 1 1 2 P T -g æ ö Þ ç ÷ è ø = 2 1 T T g æ ö ç ÷ è ø 1 1.4 1.4 1 2 P 1200 P 300 - æ ö æ ö = ç ÷ ç ÷ è ø è ø 0.4 1.4 1 2 P (4) P - æ ö = ç ÷ è ø 0.4 1.4 2 1 P 4 P æ ö = ç ÷ è ø 1.4 7 0.4 2 2 1 1 P P 4 P 4 æ ö æ ö ç ÷ ç ÷ è ø è ø = = = P1 (27) = 2 × 128 = 256 atm 42. (a) Q Internal energy is the state function. In cyclie process; DU = 0 According to 1st law of thermodynamics D = D + Q U W So heat absorbed DQ = W = Area under the curve = – (2V)(P)= – 2PV So heat rejected = 2PV
- 327. 323 Thermodynamics 43. (a) Initialand final condition is same for all process DU1 = DU2 = DU3 from first law of thermodynamics DQ= DU + DW Work done DW1 > DW2 > DW3 (Area of P.V. graph) So DQ1 > DQ2 > DQ3 44. (d) Efficiency 1 W h = q and W = q1 – q2 1 2 1 – q q h = q = 2 1 1 q - q 45. (b) P1 > P2 P1 P2 T V q2 q1 As V = constant Þ Pµ T Hence from V–T graph P1 > P2 46. (b) Change in internal energy do not depend upon the path followed by the process. It onlydepends on initial and final states i.e., DU1 = DU2 47. (a) For adiabatic process Q = 0. Byfirst law of thermodynamics, Q = DE + W Þ DEint = – W. 48. (d) The efficiency of carnot's heat engine 2 1 T 1– T h = where T2 is temperature ofsink, and T1 is temperature of hot reservoir or source. When efficiencyis 40% i.e. h = 40/100 = 2 1 T 1– T or 2 1 T 2 1– [T T(given)] 5 T = = T2= 3 T 5 Now, when efficiency is 50% 1 1 3 T 50 6 5 1– T T 100 5 T h = = = 49. (a) 50. (c) EXERCISE - 3 Exemplar Questions 1. (c) For the straight line in the graph denoted by 4, that shows pressure is constant, so curve 4 represents an isobaric process. p V 1 2 3 4 For the straight line in graph denoted by1, that shows volume is constant, so curve 1 represents isochoric process. Out of curves 3 and 2, curve 2 is steeper. Hence, curve 2 is adiabatic and curve 3 is isothermal. 2. (a) As we know that amount of sweat evaporated/minute Sweat produced/ minute Number of calories required for evaporation/kg = Amount of heat produced per minute in jogging Latent heat (in cal/kg) = 580 × 103 calories are needed to convert 1 kg H2O into stream. 1 cal. will produce sweat = 1 kg/ 580 × 103 14.5 × 103 cal will produce (sweat) 3 3 14.5 10 145 kg= kg/m 580 580 10 ´ = ´ =0.025kg. 3. (c) According to given P–V diagram that pV = constant So we can say that the gas is going through an isothermal process. Ifpressure (P)increase at constant temperature volume V decreases, the graph (iii) shows that pressure (P) is smaller at point 2 and larger at point 1 point sothe gas expands and pressure decreases. Hence verifiesoption (c). 4. (b) According to the given p-V diagram. Work done in the process ABCD = (AB) × BC = (3V0 – V0) × (2p0 – p0) = 2V0 × p0 = 2p0V0 Here the direction of arrow is anti-clockwise, sowork done is negative. Hence, work done by the gas = –2p0V0 That shows external work done on the system. 5. (a) Let us consider the p-V diagram for container A (isothermal) and for container B (adiabatic).
- 328. 324 PHYSICS 2 2 11 p p p0 p0 V0 V0 2V0 2V0 V V Container (Isothermal) A Container (Adiabatic) B In both process compression of the gas. For isothermal process (gas A) during 1®2 1 1 2 2 p V p V = 1 0 2 0 ( 2 , ) V V V V = = Q 0 0 2 0 (2 ) ( ) p V p V = 2 0 2 p p = For adiabatic process, (gas B)during (1®2) 1 2 1 2 p V p V g g = 1 0 2 0 ( 2 , ) V V V V = = Q 0 0 2 0 (2 ) ( ) p V p V g g = 0 2 0 0 0 2 (2) V p p p V g g æ ö = = ç ÷ è ø So, ratio of final pressure 1 0 2 1 0 (2) ( ) 2 ( ) 2 A p p p p g g - B æ ö = = = ç ÷ è ø where, g is ratio of specific heat capacities for the gas. Hence, verifies the option (a). 6. (b) Consider the equilibrium temperature ofthe system is T. Let us consider, T1, T2 < T < T3. As given that, there is no net loss to the surroundings. Heat lost by M3 = Heat gained by M1 + Heat gained by M2 3 3 1 1 2 2 ( ) ( ) ( ) M s T T M s T T M s T T - = - + - 3 3 3 1 1 1 M sT M sT M sT M sT - = - 2 2 2 M sT M sT + - (where, s is specified heat ofthe copper material) 1 2 3 3 3 1 1 2 2 [ ] T M M M M T M T M T + + = + + 1 1 2 2 3 3 1 2 3 M T M T M T T M M M + + = + + NEET/AIPMT (2013-2017) Questions 7. (a) Wnet = Area of triangleABC = 1 2 AC × BC = 1 2 × 5 × 10–3 × 4 × 105 = 1000 J 8. (c) According to question P µ T3 But as we know for an adiabatic process the pressure P µ 1 T g g - . So, 1 g g - = 3 Þ g = 3 2 or, , p v C C = 3 2 9. (a) From first law ofthermodynamics Qadc = DUadc + Wadc 50J = DUadc + 20 J DUadc = 30J Again, Qabc = DUabc + Wabc Wabc = Qabc – DUabc = Qabc – DUadc = 36 J – 30 J = 6 J 10. (a) Adiabatic equations of state are PVg = constant TVg–1 = constant P1–gTg = constant. 11. (d) Efficiency of engine A, 1 1 1 , T T h = - Efficiencyof engine B, 2 2 1 T T h = - Here, h1 = h2 2 1 T T T T = Þ 1 2 T T T = 12. (d) Work done by the system in the cycle =Area under P-V curve andV-axis = 0 0 0 0 1 (2P P )(2V V ) 2 - - + 0 0 0 0 1 (3P 2P )(2V V ) 2 é ù æ ö - - - ç ÷ ê ú è ø ë û = 0 0 0 0 P V P V 0 2 2 - = 13. (c) For isothermal process P1V1 = P2V2 Þ PV= P2(2V) Þ P2 = P 2 For adiabatic process 2 3 2 3 P V P V g g = Þ 3 P (2v) P 16v) 2 g g æ ö = ç ÷ è ø Þ P3 = 5/3 3 1 P 2 8 64 æ ö = ç ÷ è ø
- 329. 325 Thermodynamics 14. (b) Incyclic process ABCA Qcycle = Wcycle QAB + QBC + QCA = ar. of DABC + 400 + 100 + QC®A = 1 2 (2× 10–3)(4 × 104) Þ QC ® A = – 460 J Þ QA ® C = + 460 J 15. (a) Given, efficiency of engine, h = 1 10 work done on system W = 10J Coefficient of performance ofrefrigerator 2 Q 1 W - h b = = h = 1 9 1 10 10 1 1 10 10 - = = 9 Energyabsorbed from reservoir Q2 = bw Q2 = 9×10= 90J 16. (d) Coefficient of performance, Cop = 2 1 2 T T T - 5= 1 1 273 20 253 T (273 20) T 253 - = - - - 5T1 – (5 × 253) = 253 5T1 =253 +(5×253) =1518 T1 = 1518 5 =303.6 or, T1 = 303.6 – 273= 30.6 @ 31°C 17. (d) Since area under the curve is maximum for adiabatic process so, work done (W = PdV) on the gas will be maximum for adiabatic process P V Adiabatic Isobaric Isothermal 18. (b) Wext = negative of area with volume-axis W(adiabatic) > W(isothermal) P O V0 2V0 V Adiabatic Isothermal 19. (c) Coefficient of performance of a refrigerator, b = 2 2 1 2 Q T W T T = - (Where Q2 is heat removed) Given: T2 =4°C=4 +273=277k T1 =30°C =30+273 =303k b = 600 4.2 277 W 303 277 ´ = - Þ W= 236.5 joule Power P = W t = 236.5joule 1sec = 236.5 watt. 20. (a) Process I volume is constant hence, it is isochoric In process IV, pressure is constant hence, it is isobaric
- 330. 326 PHYSICS BEHAVIOUR OFGASES Ideal gas : In an ideal gas, we assume that molecules are point masses and there is nomutual attraction between them. The ideal gas obeys following laws : (i) Boyle’s law :According to Boyle’s law for a given mass of ideal gas, the pressureofa ideal gas is invereslyproportional to the volume at constant temperature i.e., 1 PV = constant µ Þ P V (ii) Charle’s law : For a given mass, the volume of a ideal gas is proportional to temperature at a constant pressure i.e., constant µ Þ = V V T T (iii) Gay-Lussac’s law : For a given mass of ideal gas, the pressure is proportional to temperature at constant volume i.e., constant µ = P P T T (iv) Avogadro’s law :According toAvogadro’s law, the number of molecules of all gases are same at same temperature, pressure and volume i.e., 1 2 = A A N N for same P, V and T. The value ofAvagadronumber is 6.02 × 1023 molecules. (v) Graham’slaw: (a) At constant temperature and pressure, the rms speed of diffusion of two gases is inversely proportional to the square root of the relative density i.e., vrms 1 a d Þ ( ) ( ) 1 2 1 2 = rms rms v d d v (b) According to Graham’s law, the rate of diffusion of a gas is inversely proportional to the square root of its density, provided pressure and temperature are constant 1 (rate of diffusion) ρ r µ (vi) Dalton’s law: The pressure exerted bya gaseous mixture is equal to sum of partial pressure of each component gases present in the mixture, i.e., P= P1 + P2 + P3+.................. Pn Arelation connecting macroscopic properties P, Vand T of a gas describing the state of the system is called equation of state. The equation of state for an ideal gas of n mole is PV nR T = where R is universal gas constant whose value is R = 8.31 J/mol K and R = NAk, where k is Boltzmann’s constant (NA isAvagadronumber). nisthe number ofmoles of a gas and = = A m N n M N , NA isAvagadro's number where m is the mass of a gas, N is the number of molecules and M is the molecular weight of a gas. EquationofReal Gas: The real gas follows Vander Wall’s law. According to this: 2 2 ( ) æ ö + - = ç ÷ è ø n a p V nb nRT V ; here a and b are Vander Wall’s constant Critical Temperature, Volume and Pressure : (a) The temperature at or below which a gas can be liquefied byapplying pressure alone is called critical temperature. It is given by 8 27 = c a T Rb (b) The volume of gas at a critical temperature TC is called critical volume VC, where VC = 3b (c) The pressure of gas at a critical temperature TC is called critical pressure PC, where 2 27 = C a P b Gas Equation (i) The gases found in nature are real gases. (ii) The real gas do not obey ideal gas equation but they obey vander wall's gas equation ( ) 2 P V RT V a b æ ö + - = ç ÷ è ø 1 3 Kinetic Theory
- 331. 327 Kinetic Theory kT T N R C m 3 1 2 = =( N R k = is Boltzmann constant) or kT 2 3 C m 2 1 2 = ....(vi) or 2 . . 1 3 2 2 = r m s mv kT or 2 . . 3 = r m s kT v m ....(vii) or . . 3 3 = = r m s nkT RT v nm M (a) It is clear from eq.(vi) that ata given temperature, theaverage translational kinetic energy ofanygas molecules are equal i.e., it depends only on temperature. (b) From eqn. (vii) It is clear that (a) r.m.s v T µ (b) r.m.s 1 v M µ , where M is moleculer mass ofthe gas. DEGREE OF FREEDOM The degree of freedom of a particle is the number of independent modes of exchanging energy or the number of idependent motion, which the particle can undergo. For monatomic (such as helium, argon, neon etc.) gas the molecules can have three independent motion i.e., it has 3 degree of freedom, all translational. Z Y X For a diatomic gas molecules such as H2, O2, N2, etc. it has two independent rotational motion besides of three independent translational motion, so it has 5 degree of freedom. Z Z Y Y X X Diatomic Triatomic In polyatomic gas molecules such as CO2, it can rotate about any of three coordinate axes. It has six degree (three translational +three rotational) of freedom. At high temperature the molecule can vibrate also and degree of freedom due to vibration also arises, but we neglect it. (a) 'a' depends upon the intermolecular force and the nature of gas. (b) 'b' depends upon the size of the gas molecules and represents the volume occupied by the molecules of the gas. (iii) The molecules of real gas have potential energyas well as kinetic energy. (iv) The real gas can be liquefied and solidified. (v) The real gases like CO2, NH3, SO2 etc. obeyVander Wall's equn at high pressure and low temperature. KINETIC THEORY OF AN IDEAL GAS The basic assumptions of kinetic theory are : (i) A gas consist of particles called molecules which move randomlyin all directions. (ii) The volume of molecule is very small in comparison tothe volume occupied by gas i.e., the size of molecule is infinitesimelysmall. (iii) The collision between twomolecules or between a molecule and wall are perfectelyelastic and collision time (duration ofcollision) is verysmall. (iv) The molecules exert no force on each other or on the walls of containers except during collision. (v) The total number of molecules are large and they obey Newtonian mechanics. 2 1 3 = pV mn C ....(i) where 2 C is called mean squarevelocityand 2 C iscalled root mean square velocity i.e, 2 2 2 2 1 2 ............... + + = n C C C C n ....(ii) and 2 . s . m . r C = n ....(iii) where m = mass of one molecule and n = number of molecules 2 2 r.m.s M 1 P C ρv 3V 3 Þ = = ....(iv) or ú û ù ê ë é r = 2 s . m . r v 2 1 3 2 P where and æ ö = ´ r = ç ÷ è ø M M m n V or P 2 E 3 = ....(v) where E is translational kinetic energy per unit volume of the gas. It is clear that pressure of ideal gas is equal to 2/3 of translational kinetic energyper unit volume. Kinetic interpretation of temperature : From eqn. (iv), we get 2 C M 3 1 PV = For 1 mole of a gas at temperature T : PV = RT so RT C m 3 1 2 =
- 332. 328 PHYSICS MEAN FREEPATH The distance covered by the molecules between two successive collisions is called the free path. The average distance covered by the molecules between two successive collisions is called the mean free path i.e., 2 1 2 . nd l = p 2 2 B K T d P = p where, n = number of molecules per unit volume d = diameter of each molecule KB = Boltzmann’s constant T = temperature P = pressure Mean free path depends on the diameter of molecule (d) and the number of molecules per unit volume n. At N.T.P., lfor air molecules is 0.01µm. LAW OF EQUIPARTITION OF ENERGY If we dealing with a large number of particles in thermal equilibrium to which we can apply Newtonian mechanics, the energy associated with each degree of freedom has the same average value (i.e., 1 T 2 k ), and this average value depends on temperature. From the kinetic theoryof monatomic ideal gas, we have 2 1 3 2 2 = mC kT ....(i) or Energy of molecule = (number of degree of freedom) × kT 2 1 ....(ii) So it is clear from equation (i) and (ii) that monatomic gas has three degree of freedom and energy associated per degree of freedom is kT 2 1 (where k is Boltzmann’s constant) Since we know that internal energy of an ideal gas depends only on temperature and it is purelykinetic energy. Let us consider an 1 mole ideal gas, which has N molecules and fdegree offreedom, then total internal energy U = (total number ofmolecules) × (degree of freedom of one molecule) × (energy associated with each degree of freedom) =(N) (f) × ( kT 2 1 ) U = Nf kT 2 1 ....(iii) Differentiating eq.(iii) w.r.t T, at constant volume, we get, 2 NfK T U V = ÷ ø ö ç è æ ¶ ¶ ....(iv) Now molar, specific heat at constant volume is defined as V V V T U T Q C ÷ ø ö ç è æ ¶ ¶ = ÷ ø ö ç è æ ¶ ¶ = ....(v) so V N K R C 2 2 f f = = ....(vi) where R is universal gas constant. NowbyMayer’s formula, 1 2 æ ö - = Þ = + ç ÷ è ø P V P f C C R C R ... (vii) So ratio of specific heat V P C C = g is V P C C f 2 1 = + = g ...(viii) where f is degree of freedom of one molecule. (a) For monatomic gas, f = 3, 3 2 = v C R , 5 2 = p C R so 2 1 1.67 3 g = + = (b) For diatomic gas, f = 5, 5 2 = v C R , 7 2 = p C R 4 . 1 5 2 1 = + = g (c) For Polyatomic gas, f = 6, 3 = v C R , 4 = p C R 33 . 1 6 2 1 = + = g Keep in Memory 1. Real gases behave like perfect gas at high temperature and low pressure. In real gas, we assume that the molecules have finite size ant intermolecular attraction acts between them. 2. Realgases deviatemost from the perfect gasat high pressure and lowtemperature. 3. Gaseous state of matter below critical temperature is called vapours. Belowcritical temperaturegas is vapour and above critical temperature vapour is gas. 4. Random motion of the consituents of the system involving exchange of energy due to mutual collisions is called thermalmotion. 5. Total kinetic energy or internal energyor total energydoes not depend on the direction of flowofheat. It isdetermined by the temperature alone. 6. The internal energyof a perfect gas consists onlyofkinetic energy of the molecules. But in case of the real gas it consists of both the kinetic energy and potential energy of inter molecular configuration. Distribution of Molecular Speeds : The speed of all molecules in a gas is not same but speeds of individual molecules vary over a wide range of magnitude . Maxwell derived the molecular distribution law (by which we can find distribution ofmolecules in different speeds) for sample
- 333. 329 Kinetic Theory of agas containing N molecules, which is kT 2 / mv 2 2 / 3 2 e v kT 2 m N 4 ) v ( N - ÷ ø ö ç è æ p p = ...(1) whereN(v)dv is the number ofmolecules in thegas samplehaving speeds between v and v + dv. where T = absolute temperature of the gas m = mass ofmolecule k = Boltzmann’s constant The total number of molecules N in the gas can be find out by integrating eqation (1) from 0 to¥ i.e., 0 ( ) ¥ = ò N N v dv ....(2) T2 T1 T2 > T1 vrms v Speed (m/sec) No. of molecules per unit speed interval v vp Figure shows Maxwell distribution law for molecules at two different temperature T1 and T2(T2 > T1). The number of molecules between v1 and v2 equals the area under the curve between the vertical lines at v1 and v2 and the total number of molecules as given by equation (2) is equal to area under the distribution curve. The distribution curve is not symmetrical about most probable speed, vP,(vP is the speed, which is possesed in a gas by a large number of molecules) because the lowest speed must be zero, whereas thereis nolimit tothe upper speed a molecule can attain. It is clear from fig.1 that vrms(root mean square) > v (average speed of molecules) > (most probable speed) vP AVERAGE, ROOTMEANSQUAREAND MOSTPROBABLE SPEED Average speed Tofind the average speed v , we multiplythe number ofparticles in each speed interval by speed v characteristic of that interval. We sum these products over all speed intervals and divide by total number of particles i.e., ( ) ¥ = òo N v vdv v N (where summation is replaced by integration because N is large) 8 1.59 = = p kT kT hv m m ....(1) Root Mean Square Speed In this case we multiply the number of particles in each speed interval byv2 characterstic of that interval; sum of these products over all speed interval and divide by N i.e., 2 2 ( ) ¥ = òo N v v dv v N ....(2) Root mean square speed is defined as 2 . . 3 1.73 = = = r m s kT kT v v m m Most Probable Speed It is the speed at which N(v) has its maximum value (or possessed by large number of molecules), so Ζ ∴ p d 2kT kT N(v) 0 v 1.41 du m m < Þ < < ....(3) It is clear from eqn. (1), (2) and (3) . . > > r m s p v v v The root mean square velocity of a particle in thermal system is given by 3 3 = = rms RT kT C M m where R - universal gas constant T - temperature of gas m - mass of the gas M - molecular weight of gas or mass of one mole of a gas. The average speed of the gas molecules is given by 8 8 = = p p av kT RT C m M where M is the mass of one mole and m is the mass of one particle. Most probablespeed isthat with which themaximum number of molecules move. It is given by 2 2RT 2 T V 3 M = = = p rms k c m The most probable speed, the average speed as well as root mean square speed increases with temperature. Example1. The root mean square velocity of the molecules in a sample of helium is 5/7th that of the molecules in a samzple of hydrogen. If the temperature of the hydrogen gas is 0ºc, then find the temperature of helium sample. Solution : ( ) ( ) = = H H He He H s . m . r He s . m . r m / KT 3 ) m / KT 3 ( v v He H H He m m T T ´ 273 T 2 1 7 5 4 273 1 T He He = Þ ´ ´ = or C º 14 . 284 K º 14 . 557 THe = =
- 334. 330 PHYSICS Example2. Let rmsP v, v and v respectively denote the mean speed, r.m.s. speed, and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m, then which of the following is correct ? (a) No molecule can have a speed greater than max 2 v (b) No molecule can have a speed less than p v / 2 (c) p r.m.s. v v v < < (d) None of these Solution : (c) According to kinetic theory of gases, a molecule of a gas can have speed such that 0< v < ¥ , so the alternatives (a) and (b) can never be correct, Since . . . 3 r m s k T v m = ; 8 3 av k T v m = ; and 2 mp k T v m = so . . . r m s av p v v v > > i.e. . . . p r m s v v v < < ; Example3. A gas cylinder of 50 litres capacity contained helium at 80 atmospheric pressure. Due to slowleakage it was found after a while that the pressure had dropped to 40 atmosphere. Find the proportion of the gas that has escaped and also the volume the escaped gas would occupy at atmospheric pressure. Solution : Volume of the remaining gas = volume of cylinder = 50 l at 40 atmospheric pressure The volume of this gas at 80 atmospheric pressure, using Boyle’s law p1V1 = p2V2 = 40 50 80 ´ = 25 l Volume of the gas (at 80 atmospheric pressure) which has escaped = 50 – 25 = 25 l Mass of gas escaped 25 1 Original mass of gas 50 2 = = Again using Boyle’s law, 1 ×V =80 ×25 or V = 2000 litre Example4. In Vander Wal’s equation the critical pressure Pc is given by (a) 3b (b) 2 a 27 b (c) 2 27a b (d) 2 b a Solution : (b) The Vander Wall’s equation of state is RT ) b V ( V a P 2 = - ÷ ÷ ø ö ç ç è æ + ; or 2 V a b V RT P - - = At the critical point, P = Pc, V = Vc, and T = Tc ; 2 c c c c V a b V RT P - - = ...(i) At the critical point on the isothermal, 0 dV dP c c = 2 3 2 0 ( ) c c c RT a V b V - = - - ; or 3 c 2 c c V a 2 ) b V ( RT = - ...(ii) Also at critical point, 0 dV P d 2 c c 2 = 4 c 3 c c V a 6 ) b V ( RT 2 0 - - = ; or 4 c 3 a c V a 6 ) b V ( RT 2 = - ...(iii) Dividing eqn. (iii) by(ii), we get, c c V 3 1 ) b V ( 2 1 = - or b 3 Vc = ...(iv) Putting this value in eqn. (ii), we get, 3 2 b 27 a 2 b 4 RT = or R b 27 a 8 Tc = ...(v) Putting the values of Vc and Tc in eqn. (i), we get, 2 2 c b 27 a b 9 a R b 27 a 8 b 2 R P = - ÷ ÷ ø ö ç ç è æ = Keep in Memory 1. Brownian motion, provides a direct evidence for the existance ofmolecules and their motion. The zig-zag motion of gas molecules is Brownian motion. 2. Average speed 8 8 8 π π π = = = B k T RT PV v m M M 3. Root mean square speed, 3 3 3 = = = B rms k T RT PV v m M M 4. Most probable speed 2 2 2 = = = B mp k T RT PV V m M M 5. Vrms : V : Vmp = 1.73 : 1.60 : 1.41 Vrms > V > Vmp.
- 335.
- 336. 332 PHYSICS 1. Thekinetic theory of gases (a) explains the behaviour of an ideal gas. (b) describes the motion of a single atom or molecule. (c) relates the temperature of the gas with K.E. of atoms of the gas (d) All of the above 2. The ratio of principal molar heat capacities of a gas is maximumfor (a) a diatomic gas (b) a monatomic gas (c) a polyatomic gas having linear molecules. (d) a polyatomic gas having non-linear molecules. 3. The correct statement of the law of equipartition of energy is (a) the total energy of a gas is equally divided among all the molecules. (b) The gas possess equal energies in all the three directions x,y and z-axis. (c) the total energy of a gas is equally divided between kinetic and potential energies. (d) the total kinetic energy of a gas molecules is equally divided among translational and rotational kinetic energies. 4. The internal energy of an ideal gas is (a) the sum of total kinetic and potential energies. (b) the total translational kinetic energy. (c) the total kineticenergyof randomlymoving molecules. (d) the total kinetic energy of gas molecules. 5. A fixed mass ofgas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocityof its molecules is doubled. The new volume will be (a) V/2 (b) 2 / V (c) 2V (d) 4V 6. In kinetic theory of gases, it is assumed that molecules (a) have same mass but can have different volume (b) have same volume but mass can be different (c) have different mass as well as volume (d) have same mass but negligible volume. 7. Gases exert pressure on the walls of the container because the gas molecules (a) possess momentum (b) collide with each other (c) have finite volume (d) obey gas laws. 8. Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T. Themass ofa molecule is‘m’ Neglecting vibrational energy terms, the false statement is (a) a moleculecan havea speedgreater than 2v (b) v is proportional to T (c) the averagerotational K.E. ofa molecule is 2 1 4 mv (d) the average K.E. of a molecule is 2 5 6 mv 9. The adjoining figure shows graph of pressure and volume ofa gasat twotemperturesT1 and T2. Which ofthefollowing inferences is correct? T2 T1 V P (a) T1 > T2 (b) T1 = T2 (c) T1 < T2 (d) None of these 10. At constant pressure, the ratio of increase in volume of an ideal gas per degreerisein kelvin temperatureto its original volume is (T = absolute temperature of the gas) is (a) T2 (b) T (c) 1/T (d) 1/T2 11. The K.E. ofonemoleof an ideal gasis E= (3/2)RT. Then Cp will be (a) 0.5R (b) 0.1R (c) 1.5R (d) 2.5R 12. The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is (a) 2v (b) v (c) 2v (d) 4v 13. Which of the following formula is wrong? (a) –1 V R C = g (b) –1 P R C g = g (c) Cp / CV = g (d) Cp – CV = 2R 14. For a gas if ratio of specific heats at constant pressure and volume is g then value of degrees of freedom is (a) 3 –1 2 –1 g g (b) 2 –1 g (c) 9 ( –1) 2 g (d) 25 ( –1) 2 g 15. In the kinetic theory of gases, which of these statements is/are true ? (i) The pressure ofa gas is proportional tothe mean speed of the molecules. (ii) The root mean square speed of the molecules is proportional to the pressure. (iii) The rate of diffusion is proportional to the mean speed of the molecules. (iv) The mean translational kinetic energy of a gas is proportional to its kelvin temperature. (a) (ii)and (iii)only (b) (i),(ii)and (iv) only (c) (i)and (iii) only (d) (iii) and (iv) only
- 337. 333 Kinetic Theory 16. Agas mixture consists of molecules of type 1, 2 and 3, with molar masses m1 > m2 > m3. vrms and K are the r.m.s. speed and average kinetic energy of the gases. Which of the following is true? (a) (vrms)1 < (vrms)2 < (vrms)3 and 1 2 3 ( ) ( ) ( ) K K K = = (b) (vrms)1 = (vrms)2 = (vrms)3 and 1 2 3 ( ) ( ) ( ) K K K = > (c) (vrms)1 > (vrms)2 > (vrms)3 and 1 2 3 ( ) ( ) ( ) K K K < > (d) (vrms)1 > (vrms)2 > (vrms)3 and 1 2 3 ( ) ( ) ( ) K K K < < 17. Asampleof an idealgasoccupies a volume ofV ata pressure P and absolute temperature. T. The mass of each molecule is m. The equation for density is (a) mkT (b) P/k T (c) P/(kT V) (d) Pm/k T 18. The value ofcritical temperature in terms of van der Waal’s constant a and b is given by (a) c 8a T = 27 R b (b) c a T = 27 bR (c) c a T = 2R b (d) c –a T = R b 19. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is (a) H2 (b) F2 (c) O2 (d) Cl2 20. Two gases A and B having the same temperature T, same pressurePand samevolumeVaremixed. Ifthemixtureis at the same temperature T and occupies a volume V, the pressure of the mixture is (a) 2P (b) P (c) P/2 (d) 4P 21. The perfect gas equation for 4 gram of hydrogen gas is (a) PV=RT (b) PV=2RT (c) PV= 2 1 RT T (d) PV=4RT 22. Maxwell's laws of distribution of velocities shows that (a) the number of molecules with most probable velocity is very large (b) the number of molecules with most probable velocity is very small (c) the number of molecules with most probable velocity is zero (d) the number of molecules with most probable velocity is exactly equal to 1 23. According to kinetic theory of gases, which one of the following statement(s) is/are correct? (a) Real gas behave as ideal gas at high temperature and low pressure. (b) Liquid state of ideal gas is impossible (c) At any temerature and pressure, ideal gas obeys Boyle's lawand charles' law (d) The molecules of real gas do not exert any force on one another. 24. For hydrogen gas Cp – Cv = a and for oxygen gas Cp – Cv = b. So, the relation between a and b is given by (a) a = 16 b (b) 16 a = b (c) a = 4 b (d) a = b 25. The relation between the gas pressure P and average kinetic energy per unit volume E is (a) 1 2 P E = (b) P = E (c) 3 2 P E = (d) 2 3 P E = 1. If the critical temperature of a gas is 100ºC, its Boyle temperature will be approximately (a) 337.5ºC (b) 500ºC (c) 33.3ºC (d) 1000ºC 2. The r.m.s. velocity of oxygen molecule at 16ºC is 474 m/sec. The r.m.s. velocityin m/s ofhydrogen molecule at 127ºC is (a) 1603 (b) 1896 (c) 2230.59 (d) 2730 3. The gases are at absolute temperature 300ºK and 350ºK respectively. The ratio of average kinetic energy of their molecules is (a) 7: 6 (b) 6: 7 (c) 36: 49 (d) 49:36 4. The total degree of freedom of a CO2 gas molecule is (a) 3 (b) 6 (c) 5 (d) 4 5. If one mole of a monatomic gas (g = 5/3) is mixed with one mole ofa diatomic gas(g= 7/3), the value ofgfor the mixture is (a) 1.40 (b) 1.50 (c) 1.53 (d) 3.07 6. The molecules of a given mass of gas have a root mean square velocity of 200m s–1 at 27°C and 1.0 × 105 N m–2 pressure. When the temperature is 127°C and the pressure 0.5 × 105 Nm–2, the root mean square velocityin ms–1, is (a) 3 400 (b) 2 100 (c) 3 2 100 (d) 3 100
- 338. 334 PHYSICS 7. Ifmassesof all molecule ofa gas are halved and their speed doubled then the ratio of initial and final pressure will be (a) 2: 1 (b) 1: 2 (c) 4: 1 (d) 1: 4 8. On a particular day, the relative humidity is 100% and the room temperature is 30ºC, then the dew point is (a) 70ºC (b) 30ºC (d) 100ºC (d) 0ºC 9. The velocity of the molecules of a gas at temperature 120 K is v. At what temperature will the velocitybe 2 v? (a) 120K (b) 240K (c) 480K (d) 1120K 10. The density of a gas is 6 × 10–2 kg/m3 and the root mean square velocity of the gas molecules is 500 m/s. The pressure exerted by the gas on the walls of the vessel is (a) 3 2 5×10 N/m (b) –4 2 1.2×10 N/m (c) –4 2 0.83×10 N/m (d) 2 30N/m 11. The temperature of the mixture of one mole of helium and one mole of hydrogen is increased from 0°C to 100°C at constant pressure. The amount of heat delivered will be (a) 600cal (b) 1200cal (c) 1800cal (d) 3600cal 12. Helium gas is filled in a closed vessel (having negligible thermal expansion coefficient) when it is heated from 300 K to600 K, then averagekinetic energyofhelium atom will be (a) 2 times (b) 2 times (c) unchanged (d) half 13. One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas. J= 4200 J/kcal. (a) 1.979k cal/kmol K (b) 2.378k cal/kmolK (c) 4.569kcal/kmolK (d) 3.028kcal/ kmolK 14. Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec and 5 km/sec. The root mean square speed of these molecules (in km/sec) is (a) 4 / 54 (b) 2 / 54 (c) 3.5 (d) 3 3 15. The density of air at pressure of 105 Nm–2 is 1.2 kg m–3. Under these conditions, the root mean square velocity of the air molecules in ms–1 is (a) 500 (b) 1000 (c) 1500 (d) 3000 16. Howmanydegrees offreedom are associated with 2 grams of He at NTP ? (a) 3 (b) 3.01 × 1023 (c) 9.03 × 1023 (d) 6 17. If 2 mole of an ideal monatomic gas at temperature T0 is mixed with 4 moles of another ideal monatomic gas at temperature 2T0, then the temperature ofthe mixture is (a) 0 5 T 3 (b) 0 3 T 2 (c) 0 4 T 3 (d) 0 5 T 4 18. The temperature at which the root mean square velocityof the gas molecules would become twice of its value at 0°C is (a) 819°C (b) 1092°C (c) 1100°C (d) 1400°C 19. At what temperature is the r.m.s. velocity of a hydrogen moleculeequal to that ofan oxygen molecule at 47ºC (a) 80K (b) –73K (c) 3K (d) 20K 20. One litre of oxygen at a pressure of 1 atm, and 2 litres of nitrogen at a pressure of0.5 atm. areintroduced in the vessel of 1 litre capacity, without anychange in temperature. The total pressure would be (a) 1.5atm. (b) 0.5atm. (c) 2.0atm. (d) 1.0atm. 21. The air in a room has 15 gm of water vapours per cubic metre ofits volume. However for saturation one cubic metre of volume requires 20 gm of water vapour then relative humidityis (a) 50% (b) 75% (c) 20% (d) 25% 22. A vessel contains air at a temperature of15ºC and 60% R.H. What will betheR.H. if it isheated to 20ºC?(S.V.P. at 15ºCis 12.67 &at 20ºC is 17.36mm of Hg respectively) (a) 262% (b) 26.2% (c) 44.5% (d) 46.2% 23. To what temperature should be the hydrogen at 327°C be cooked at constant pressure so that the root mean square velocityof its molecules becomes half of its previous value (a) –123°C (b) 120°C (c) –100°C (d) 0°C 24. Ifpressure ofa gas contained in a closed vessel is increased by 0.4% when heated by 1ºC, the initial temperature must be (a) 250K (b) 250ºC (c) 2500K (d) 25ºC 25. N molecules, each of mass m, of gas Aand 2 N molecules, each ofmass 2 m, of gas Bare contained in the same vessel which is maintainedat a temperature T. The mean square of the velocity of molecules of B type is denoted by v2 and the mean square of the X component of the velocity of A type is dentoed by w2, w2/v2 is (a) 2 (b) 1 (c) 1/3 (d) 2/3 26. At what temperature, pressure remaining constant, will the r.m.s. velocity of a gas be half of its value at 0ºC? (a) 0ºC (b) –273ºC (c) 32ºC (d) –204ºC
- 339. 335 Kinetic Theory 27. Graphof specific heat at constant volume for a monatomic gas is (a) Y X O cv T (b) Y X O cv T 3R (c) Y X O cv T 3 — 2 R (d) Y X O cv T 28. Theorder ofmagnitudeof thenumber ofnitrogen molecules inan airbubbleofdiameter 2mmunder ordinaryconditionsis (a) 105 (b) 109 (c) 1013 (d) 1017 29. At identical temperatures, the rms speed of hydrogen molecules is 4 times that for oxygen molecules. In a mixture oftheseinmassratioH2:O2=1:8,thermsspeedofallmolecules is n times the rms speed for O2 molecules, where n is (a) 3 (b) 4/3 (c) (8/3)1/2 (d) (11)1/2 30. InVander Waal’s equation the critical pressurePc is given by (a) 3 b (b) 2 b 27 a (c) 2 b a 27 (d) a b2 31. 1moleofagaswith g =7/5ismixedwith 1moleofagaswith g = 5/3, then the value of g for the resultingmixtureis (a) 7/5 (b) 2/5 (c) 24/16 (d) 12/7 32. N molecules, each of mass m, of gas Aand 2 N molecules, each ofmass 2 m, of gas Bare contained in the same vessel which is maintained at a temperature T. The mean square velocity of molecules of B type is denoted by V2 and the mean square velocityofA type is denoted byV1, then 2 1 V V is (a) 2 (b) 1 (c) 1/3 (d) 2/3 33. An ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V.When itexpandstoa volume2V, the temperaturebecomes (a) 2 / T (b) 2T (c) 2 T 2 (d) 4T 34. Two monatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in gas 2 is given by (a) 2 1 m m (b) 1 2 m m (c) 2 1 m m (d) 1 2 m m 35. If the potential energy of a gas molecule is U = M/r6 – N/r12, M and N being positive constants, then the potential energyat equilibrium must be (a) zero (b) M2/4N (c) N2/4M (d) MN2/4 36. Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air temperature is 22°C. During the day, temperature rises to 42°C and the tube expands by 2%. The pressure of the air in the tube at this temperature, will be approximately (a) 212kPa(b) 209kPa (c) 206kPa (d) 200kPa 37. Work done by a system under isothermal change from a volume V1 to V2 for a gas which obeys Vander Waal's equation 2 ( ) n V n P nRT V æ ö a -b + = ç ÷ ç ÷ è ø is (a) 2 2 1 2 1 1 2 loge V n V V nRT n V n V V æ ö æ ö - b - + a ç ÷ ç ÷ - b è ø è ø (b) 2 2 1 2 10 1 1 2 log V n V V nRT n V n V V æ ö æ ö - b - + a ç ÷ ç ÷ - b è ø è ø (c) 2 2 1 2 1 1 2 loge V n V V nRT n V n V V æ ö æ ö - b - +b ç ÷ ç ÷ - b è ø è ø (d) 2 1 1 2 2 1 2 log – e V n V V nRT n V n V V æ ö æ ö - b + a ç ÷ ç ÷ - b è ø è ø 38. A gas mixture consists of2 moles ofoxygen and 4 moles of Argon at temperature T. Neglecting all vibrational moles, the total internal energy of the system is (a) 4RT (b) 15RT (c) 9RT (d) 11RT 39. The molar heat capacities of a mixture of two gases at constant volume is 13R/6. The ratio of number of moles of the first gas to the second is 1 : 2. The respective gases may be (a) O2 and N2 (b) He and Ne (c) He and N2 (d) N2 and He
- 340. 336 PHYSICS 40. Agraph is plotted with PV/T on y-axis and mass of the gas along x-axis for different gases. The graph is (a) a straight line parallel to x-axis for all the gases (b) a straight line passing through origin with a slope having a constant value for all the gases (c) a straight line passing through origin with a slope having different values for different gases (d) a straight line parallel to y-axis for all the gases 41. Four mole of hydrogen, twomole ofhelium and one mole of water vapour form an ideal gas mixture. What is the molar specific heat at constant pressure of mixture? (a) 16 R 7 (b) 7 R 16 (c) R (d) 23 R 7 42. An ideal gas is taken through a process in which thepressure and the volume are changed according to the equation P = kV. Molar heat capacity of the gas for the process is (a) C = Cv + R (b) C = Cv + R 2 (c) C = Cv – R 2 (d) C = Cv + 2R 43. A vessel has 6g of hydrogen at pressure P and temperature 500K. Asmall hole is made in it sothat hydrogen leaks out. How much hydrogen leaks out if the final pressure is P/2 and temperature falls to300 K ? (a) 2g (b) 3g (c) 4g (d) 1g 44. Figure shows a parabolic graph between T and 1/V for a mixture of a gas undergoing an adiabatic process. What is the ratio ofVrms ofmolecules and speed ofsound in mixture? (a) 3 / 2 (b) 2 (c) 2 / 3 T 1/V 2T0 T0 1/V0 4/V0 (d) 3 45. Which one the following graphs represents the behaviour of an ideal gas ? (a) V PV (b) V PV (c) V PV (d) V PV 46. The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant pressure to that at constant volume is (a) 5/7 (b) 9/7 (c) 7/5 (d) 8/7 DIRECTIONS for Qs. 47 to 50 : Each of these question contains two statements: STATEMENT-1 (Assertion) and STATEMENT-2 (Reason).Answer these questions from the following four options. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1 (b) Statement-1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1 (c) Statement-1 is True, Statement- 2 is False (d) Statement-1 is False, Statement -2 is True 47. Statement 1 : The ratio V P C C for a monatomic gas is less than for a diatomic gas. Statement 2 : Themolecules ofa monatomic gas have more degrees of freedom than those of a diatomic gas. 48. Statement 1 : Air pressure in a car tyre increases during driving. Statement 2 : Absolutezero temperature is not zero energy temperature. 49. Statement 1 : The total translational kinetic energy of all themolecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume. Statement 2 : The molecules of a gas collidewith each other and the velocities of the molecules change due to the collision. 50. Statement 1 : Mean free path of a gas molecules varies inversely as density of the gas. Statement 2 : Mean free path varies inversely as pressure of the gas.
- 341. 337 Kinetic Theory Exemplar Questions 1.A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of500 m s–1 in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground (a) remains the same because 500 ms–1 is very much smaller than vrms of the gas (b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls (c) will increase bya factor equal to 2 2 2 rms rms ( (500) ) / v v + where vrms was the original mean square velocity of the gas (d) will be different on the top wall and bottom wall ofthe vessel 2. 1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it.At anygiven time, C D G H F E A B (a) the pressure on EFGH would be zero (b) the pressure on all the faces will the equal (c) the pressure of EFGH would be double the pressure on ABCD (d) the pressure on EFGH would be half that on ABCD 3. Boyle's law is applicable for an (a) adiabatic process (b) isothermal process (c) isobaric process (d) isochoric process 4. A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction (figure). Ifthe temperature is increased M (a) both p and V of the gas will change (b) only p will increase according to Charles' law (c) V will change but not p (d) p will change but not V 5. Volume versus temperature graphs for a given mass of an ideal gas are shown in figure. At two different values of constant pressure. What can be inferred about relation between p1 and p2? 40 30 20 10 100 200 300 400 500 V l ( ) T K ( ) p1 p2 (a) p1 > p2 (b) p1 = p2 (c) p1 < p2 (d) Data is insufficient 6. 1 mole of H2 gas iscontained in a box of volume V= 1.00 m3 at T =300 K. The gas is heated toa temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal) (a) same as the pressure initially (b) 2 times the pressure initially (c) 10 times the pressure initially (d) 20 times the pressure initially 7. A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let f1(v)dv, denote the fraction of molecules with speed between v and (v + dv ) with f2(v)dv, similarlyfor oxygen. Then, (a) 1 2 ( ) ( ) ( ) f v f v f v + = obeys the Maxwell's distribution law (b) f1(v), f2(v) will obey the Maxwell's distribution law separately (c) neither f1(v), nor f2(v) will obey the Maxwell's distribution law (d) f2(v) and f1(v) will be the same 8. An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be (a) 1.1 p (b) p (c) less than p (d) between p and 1.1
- 342. 338 PHYSICS NEET/AIPMT (2013-2017)Questions 9. In the given (V – T) diagram, what is the relation between pressure P1 and P2 ? [2013] (a) P2 > P1 (b) P2 < P1 T V P2 P1 q1 q2 (c) Cannot be predicted (d) P2 = P1 10. The amount of heat energyrequired to raisethe temperature of 1g ofHelium at NTP, from T1K toT2K is [2013] (a) 3 2 NakB(T2 – T1) (b) 3 4 NakB(T2 – T1) (c) 3 4 NakB 2 1 T T (d) 3 8 NakB(T2 – T1) 11. In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be [NEET Kar. 2013] (a) 4P (b) 2P (c) P (d) P/2 12. The mean free path of molecules of a gas, (radius ‘r’) is inversely proportional to : [2014] (a) r3 (b) r2 (c) r (d) r 13. The ratio of the specific heats p v C C = g in terms of degrees of freedom (n) is given by [2015] (a) n 1 3 æ ö + ç ÷ è ø (b) 2 1 n æ ö + ç ÷ è ø (c) n 1 2 æ ö + ç ÷ è ø (d) 1 1 n æ ö + ç ÷ è ø 14. One mole of an ideal diatomic gas undergoes a transition from Ato B along a path AB as shown in the figure. P(in kPa) V(in m ) 3 2 5 A B 4 6 The change in internal energy of the gas during the transition is: [2015] (a) – 20 kJ (b) 20J (c) –12kJ (d) 20kJ 15. Two vessels separately contain two ideal gases A and B at the same temperature. The pressure ofAbeing twice that of B. Under such conditions, the density of A is found to be 1.5 times the densityofB. The ratio of molecular weight of A and B is : [2015 RS] (a) 3 4 (b) 2 (c) 1 2 (d) 2 3 16. 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0JK–1. If the speed of sound in this gas at NTP is 952 ms–1, then the heat capacity at constant pressure is (Take gas constant R = 8.3 JK–1 mol–1) [2015 RS] (a) 7.5 JK–1 mol–1 (b) 7.0 JK–1 mol–1 (c) 8.5 JK–1 mol–1 (d) 8.0 JK–1 mol–1 17. The molecules of a given mass of a gas have r.m.s. velocity of 200 ms–1 at 27°C and 1.0 × 105 Nm–2 pressure. When the temperature and pressure ofthe gas are respectively, 127°C and 0.05 × 105 Nm–2, the r.m.s. velocity ofits molecules in ms–1 is : [2016] (a) 100 2 (b) 400 3 (c) 100 2 3 (d) 100 3 18. Agas mixture consists of 2 moles ofO2 and 4 moles ofAr at temperature T. Neglecting all vibrational modes, the total internal energy of the system is :- [2017] (a) 15RT (b) 9 RT (c) 11 RT (d) 4 RT
- 343. 339 Kinetic Theory EXERCISE -1 1. (d) 2. (b) 3. (b) 4. (d) 5. (d) Sincevrms is doubled by increasing the temp. so by rms 3KT v m = , the temp. increase by four times. Now for constant pressure 1 2 1 2 V V T T = V1=V, T1=TºK, T2= 4TºK, V2=? V2 =4V 6. (d) 7. (a) 8. (c) 9. (c) For a given pressure, volume will be more if temperature ismore (Charle's law) Constant pressure T1 V P V1 V2 From the graph it is clear that V2 > V1 Þ T1 > T2 10. (c) At constant pressure V T V T V T D D µ Þ = Hence ratio of increase in volume per degree rise in kelvin temperature to it’s origianl volume ( / ) 1 D D = = V T V T 11. (d) v 3 C dE/dT R 2 = = p 3 5 C R R R 2.5R 2 2 = + = = 12. (c) rms 3RT v M = . According to problem T will become 2T and M will becomes M/2 so the value of vrms will increase by 4 2 = times i.e. new root mean square velocitywill be 2v. 13. (d) The difference of CP and CV is equal to R, not 2R. 14. (b) 2 2 1 2 1 , –1 2 –1 –1 f f f f g = + Þ g = Þ = Þ = g g 15. (d) 16. (a) 1 rms v M µ Þ (vrms)1 < (vrms)2 < (vrms)3 also in mixture temperature of each gas will be same, hence kinetic energyalso remains same. 17. (d) We know that PV = n RT = (m/M) RT where M = Molecular weight. Now T R M m d m P ÷ ø ö ç è æ = ÷ ø ö ç è æ ´ where d = density of the gas M T N k M T R d P A = = ...(2) where R= k NA, k is Boltzmann constant. But A N M = m = mass of each molecule so P m d kT ´ = 18. (a) 19. (a) For one mole of gas M RT 3 vrms = 3 8.31 300 1930 M ´ ´ = Þ M = 2×10–3 kg = 2 g So that di-atomic gas is hydrogen. 20. (a) Here, initiallyP1 = P, V1 = V +V= 2 V; Finally, P2 = P; V2 = V As P1V1 = P2V2 or V V P P 1 1 2 = = P 2 V V 2 P = ´ 21. (b) 4 n 2 2 = = 22. (a) Based on Maxwell's velocity distribution curve. 23. (a, b, c) 24. (d) Both are diatomic gases and Cp – Cv = R for all gases. 25. (d) 2 3 P E = EXERCISE - 2 1. (a) Boyle temperature is defined as B c c a 27 8a 27 T T 3.375 T Rb 8 27Rb 8 æ ö = = = = ç ÷ è ø = 3.375 × 100 = 337.5ºC 2. (c) 32 289 R 3 v . oxg ´ = æ ö ç ÷ è ø 3RT v = rms M 2 400 R 3 vH ´ = so vH = 2230.59 m/sec 3. (b) 4. (c) Hints & Solutions
- 344. 340 PHYSICS 5. (c)Let g1 and g2 denote the ratio of two specific heats of gas 1 and gas 2 respectively when they are mixed in the ratio of n1 and n2 mole. The resultant value of g is the weighted mean of g1 and g2 i.e., 1 1 ) 5 / 7 ( 1 ) 3 / 5 ( 1 n n n n 2 1 2 2 1 1 + + = + g + g = g 53 . 1 30 46 2 15 21 25 = = ´ + = 6. (a) 2 1 c 400 2 c 300 3 = = Þ 1 2 2 400 c 200 ms 3 3 - = ´ = 7. (b) Since V mNv 3 1 P 2 rms = since m is halved & speed is doubled so pressure become twice. 8. (b) Relative humidity (R.H) S.V.P at dew point S.V.P at room temp. = since here R.H is 100% Þ room temp. = dewpoint = 30ºC 9. (c) v T µ v' T ' v T = Given ' v =2 v or, 2 T' 1 T = ' T = 4 T=4 ×120K= 480K 10. (a) 2 2 2 ) 500 ( ) 10 6 ( 3 1 3 1 P ´ ´ ´ = u r = - 2 3 m / N 10 5´ = 11. (b) 1 2 1 2 1 2 ( ) p p p mix C C C m +m = m +m 1 5 ( ( ) 2 p C He R = and 2 2 7 ( ) ) 2 p C H R = (Cp)mix 5 7 1 1 2 2 1 1 R R ´ + ´ = + = 3R= 3 × 2 = 6cal/mol.°C Amount of heat needed to raise the temperature from0°C to100°C ( ) 2 6 100 1200 p p Q C T cal D =m D = ´ ´ = 12. (b) Average K.E. = 2 1 3 mc kT T 2 2 = µ 600K 300K (Av.K.E.) 600 2 (Av.K.E.) 300 = = At 600K it will be 2 times than that at 300K. 13. (a) V=22.4litre=22.4 ×10–3 m3,J= 4200J/kcal by ideal gas equation for one mole of a gas, 5 3 PV 1.013 10 22.4 10 R T 273 - ´ ´ ´ = = 5 p v R 1.013 10 22.4 C C J 273 4200 ´ ´ - = = ´ =1.979kcal/kmolK 14. (a) 1/ 2 2 2 2 2 rms (2) (3) (4) (5) v 4 é ù + + + = ê ú ê ú ë û ú û ù ê ë é = 4 54 15. (a) 5 1 3P 3 10 c 500 ms 1.2 - ´ = = = r 16. (c) Moles of He = 2 1 4 2 = Molecules = 23 1 6.02 10 2 ´ ´ = 3.01 × 1023 As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas. Total degrees of freedom = 3 × 3.01 × 1023 = 9.03 × 1023 17. (a) Let T be the temperature of the mixture, then U = U1 + U2 Þ 1 2 f (n n ) RT 2 + = 1 0 2 0 f f (n ) (R) (T ) (n ) (R) (2T ) 2 2 + Þ (2 + 4)T = 2T0 + 8T0 (Q n1 = 2, n2 = 4) T = 0 5 T 3 18. (a) t 0 c 273 t c 273 + = t 273 273 4 = - ´ Þ t 819 C Þ = ° 19. (d) The r.m.s. velocityofthemolecule ofa gas ofmolecular weight M at Kelvin temperature T is given by, M RT 3 v . s . m . r = Let MO and MH are molecular weight of oxygen and hydrogen and TO and TH the corresponding Kelvin temperature at which vr.m.s. is same for both, H H O O . s . m . r M RT 3 M RT 3 v = = Hence, H H O O M T M T = To = 273º + 47º = 320 K, Mo = 32, MH = 2 K 20 320 32 2 TH = ´ =
- 345. 341 Kinetic Theory 20. (c)According to Dalton’s Law P= P1 +P2..................... here 1 litre of orygen at a pressure of1atm, & 2 litre of nitrogen at a pressure of 5 atm are introduced in a vessel of 1 litre capacity. so 1 1 2 2 V P P V 1 1 2 .5 P 2atm volumeof vessel 1 + ´ + ´ = = = 21. (b) Relative humidity % 100 M m ´ = where m is the mass of vapour actually present in certain volume of air & M is the mass required to saturate the air fully of that volume so % 75 100 20 15 H . R = ´ = 22. (c) Bydefinition Þ 100 60 ; 100 . temp room at P . V . S . temp room at P . V H . R ´ = 15 (V.P.) 12.67 = (V.P)15 = 7.6mm of Hg. Now since unsaturated vapour obeys gas equation & mass of water remains constant so nRT P PαT V æ ö = Þ ç ÷ è ø 15 20 20 (V.P) 273 15 (V.P) (V.P) 273 20 + = Þ + =7.73 mm ofHg So 20 20 20 (V.P) (R.H) 100 44.5% (S.V.P) = ´ = 23. (a) 273 327 273 t ) v ( ) v ( 1 rms 2 rms + + = or 1 273 2 600 t + = or t = – 123ºC 24. (a) 1 1 1 T V P = constant for one mole & 1 1 T P = constant for constant volume so 1 1 1 1 1 P 1.004P T 250K T (T 1) = Þ = + 25. (d) Total K.E. ofAtype molecules 2 m 2 3 w = Total K.E. ofAtype of molecule is [ ] 2 z ms . r 2 y ms . r 2 x ms . r A ) v ( ) v ( ) v ( m 2 1 E . K + + = but w = x ms . r ) v ( so w = = z ms . r y ms . r ) v ( ) v ( Total K.E. of B type molecules 2 2 1 2 m v m v 2 = ´ = Now, 2 2 3 m ω m v 2 ´ = or 2 2 2 ( ω / v ) 3 = 26. (d) rms v T µ At 0ºC, rms v 273 µ At temp. T, rms v T 2 µ K 69 ~ K 2 . 68 4 273 T or 273 T 2 1 - = = = C º 204 273 69 T - = - = 27. (c) For a monatomic gas R 2 3 Cv = So correct graph is v C R 2 / 3 ® T 28. (d) Molar volume = 22400 cm3 Bubble volume = 3 3 4 d 4 10 3 8 3 - p = p No. of molecules 23 3 17 6 10 4 10 10 22400 3 - ´ ´ p = » ´ 29. (d) Molecule number ratio is 3 1 : 3 2 O : H 2 2 = . That gives 2 1 2 (c ) 16 1 rms 3 3 = + æ ö æ ö è ø è ø times the value for O2. 30. (b) The Vander waal’s equation of state is RT ) b V ( V a P 2 = - ÷ ÷ ø ö ç ç è æ + or 2 V a b V RT P - - = At the critical point, P = Pc, V = Vc, and T = Tc 2 c c c c V a b V RT P - - = ....(i)
- 346. 342 PHYSICS At thecritical point on the isothermal, 0 dV dP c c = c 2 3 c c RT 2a 0 (V b) V - = + - or 3 c 2 c c V a 2 ) b V ( RT = - ....(ii) Also at critical point, 0 dV P d 2 c c 2 = 4 c 3 c c V a 6 ) b V ( RT 2 0 - - = or c 3 4 c c 2RT 6a (V b) V = - ..(iii) Dividing (ii) by (iii), weget c c V 3 1 ) b V ( 2 1 = - or Vc = 3 b ....(iv) Putting this value in (ii), we get 2 3 RTc 2a 4b 27 b = or R b 27 a 8 Tc = ...(v) Putting the values of Vc and Tc in (i), we get 2 2 c b 27 a b 9 a R b 27 a 8 b 2 R P = - ÷ ÷ ø ö ç ç è æ = 31. (c) g for resulting mixture should be in between 7/5 and 5/3. 32 (b) For 1 molecule of a gas, vrms = 3KT m where m is the mass of one molecule For N molecule of a gas, v1= 3KT N m ´ For 2N molecule of a gas v2 = 3KI N N (2m) ´ ´ 1 2 v 1 v = 33. (a) Since VP2 = constant, 2 '2 VP 2VP = P P' 2 = As P constant or Tα P, T = thusT becomes T/Ö2 34. (a) We know that, ÷ ø ö ç è æ g = M RT Crms Here 1 2 rms 2 rms rms 1 C m 1 C ; M C m æ ö µ = ç ÷ è ø 35. (b) 3 12 2 13 dU d M N 6M 12N F dr dr r R r r - é ù é ù = = - - = - + ê ú ê ú ë û ë û In equilibrium position, F = 0 2 13 6M 12N 0 r r - = or, 6 2N r M = Potential energyat equilibrium position 2 2 2 2 M N M M M U (2N / M) 2N 4N 4N (2N / M) = = = - = 36. (b) 1 1 2 2 1 2 P V P V T T = Here, P1 =200kPa T1 =22°C =295K T2 =42°C =315K V2 1 1 1 2 V V 1.02V 100 = + = 1 2 1 200 315V P 295 1.02V ´ = ´ 209.37kPa = 37. (a) According to given Vander Waal's equation 2 2 nRT n P V n V a = - - b Workdone, 2 2 2 1 1 1 2 2 V V V V V V dV dV W PdV nRT n V n V = = - a - b ò ò ò [ ] 2 2 1 1 2 1 log ( ) V V e V V nRT V n n V é ù = - b + a ê ú ë û 2 2 1 2 1 1 2 loge V n V V nRT n V n V V æ ö é ù - b - = + a ç ÷ ê ú - b è ø ë û 38. (d) Internal energyof 2 moles of oxygen 2 5 5 Uo µ RT 2. RT 5RT 2 2 æ ö = = = ç ÷ è ø Internal energyof 4 moles ofArgon. Ar 3 3 U µ RT 4. RT 6RT 2 2 æ ö = = = ç ÷ è ø Total internal energy 2 O Ar U U U 11RT = + = 39. (c) 1 v 2 v 1 2 vmix 1 2 n C n C C n n + = + Þ 1 v 1 v 1 2 1 1 n C 2n C 13R 6 n 2n + = + 1 2 n 1 n 2 é ù = ê ú ë û Q Þ v v 1 2 13R C 2C 2 = +
- 347. 343 Kinetic Theory Possible valuesare, v v 1 2 3R 5R C , C 2 2 = = Gases are monatomic (like He) and diatomic (likeN2) 40. (c) PV m nR R T M æ ö = = ç ÷ è ø or PV R m T M æ ö = ç ÷ è ø i.e. PV T versus m graph is straight line passing through origin with slope R/M, i.e. the slopedepends on molecular mass of the gas M and is different for different gases. 41. (d) Cv for hydrogen = 5 R 2 Cv for helium = 3R 2 Cv for water vapour = 6R 2 = 3R (Cv)mix = 5 3 4 R 2 R 1 3R 16 2 2 R 4 2 1 7 ´ + ´ + ´ = + + Cp = Cv + R p 16 C R R 7 = + or p 23 C R 7 = 42. (b) P=kV Þ PV–1 =k It is polytropic process (PVn = constant) So n = – 1 So, C = CV + V R R C 1 n 2 = + - 43. (d) m PV RT M = Initially, PV= 6 R 500 M ´ Finally, P (6 x) V R 300 2 M - = ´ (if x g gas leaks out) Hence, 6 5 2 6 x 3 = ´ - x = 1 gram 44. (b) From graph, T2V = const. .....(1) As we know that TVg–1 = const Þ 1 1 VT cons. g- = ....(2) On comparing (1) and (2), we get Þ g = 3/2 Also rms 3P v = r and sound P v g = r Þ rms sound v 3 2 v = = g 45. (b) For an ideal gas PV = constant i.e. PV doesn’t vary withV. 46. (c) Molar specific heat at constant pressure 7 2 P C R = Since, CP – CV = R Þ CV – CP – R 7 5 – 2 2 R R R = = (7 / 2) 7 . (5/ 2) 5 P V C R C R = = 47. (c) For monatomic gas, f= 3, v p 3R 5R C , C 2 2 = = ; v p C 3 C 5 = For diatomic gas, f= 5 v p 5R 7R C , C 2 2 = = ; v p C 5 C 7 = 48. (b) When a person is driving a car then the temperature of air inside the tyre is increased because of motion. From theGayLussac’s law, P µ T Hence, when temperature increases the pressure also increase. 49. (b) Total translational kinetic energy 3 3 2 2 nRT PV = = In an ideal gas all molecules moving randomly in all direction collide and their velocity changes after collision. 50. (a) The mean free path of a gas molecule is the averge distance between two successive collisions. It is represented by l. 2 1 2 kT P l = ps and 2 2 m d l = ×ps Here, s = 0 diameter ofmoleculeand k = Boltzmann’s constant. Þ l µ 1 / d, l µ T and l µ 1 / P. Hence, mean free path varies inversely as density of the gas. It can easily proved that the mean free path varies directlyas the temperature and inversely as the pressure of the gas.
- 348. 344 PHYSICS EXERCISE -3 Exemplar Questions 1. (b) As the relative velocity of molecule with respect to the walls of container does not change in rocket, due to the mass ofa molecule is negligible with respect to the mass of whole system and system of gas moves as a whole and (g = 0) on molecule energy where. Hence pressure of the gas inside the vessel, as observed by us, on the ground remain the same. 2. (d) Pressure on the wall due to force exerted bymolecule on walls due to its rate of transfer of momentum to wall. In an ideal gas, when a molecule collides elastically with a wall, themomentum transferredto each molecule will betwice themagnitude ofits normal momentum is 2 mv. For the wall EFGH, absorbs those molecules. Which strike toit so rate of change in momentum toit become only mv so the pressure of EFGH would half ofABCD. 3. (b) Boyle's law is applicable at constant temperature and temperature remains constant in isothermal process. For ideal gas, pV = nRT = constant So, pV = constant (at constant temperature) 1 p V µ So, this process can be called as isothermal process. 4. (c) Let us consider the given diagram where an ideal gas is contained in a cylinder, having a piston of mass M. The pressure on gas does not change. M p pa p pa Mg/A A The pressure inside the gas will be p = pa + Mg /A where, pa = atmospheric pressure A = area of cross-section of the piston. Mg = weight of piston Hence, p = constant. As the piston and cylinder is frictionless so the equation for ideal gas pV = nRT, volume (V) increases at constant pressure. as p, R, n are constant so, V T µ so on increasing temperature of system its volume increased but pwill remain constant. 5. (a) As we know that an ideal gas equation, as the pressure and quantity of gas in system are constant nR pV nRT V T p æ ö = Þ = ç ÷ è ø T V p µ as n, R are constant 1 V T p æ ö µ ç ÷ è ø Slope of the V – T graph, dV nR m dT p = = [m = slope of V – t graph] 1 dV dT p µ or 1 m p µ [Q nR = constant] So, 1 p m µ hence, 1 2 2 1 = p m p m where, m1 is slope of the graph corresponding to p1 and similarlym2 is slope corresponding to p2. Soslope of p1 is smaller than p2 . Hence, (p1 > p2 ). 6. (d) Pressure exerted by gas is due to rate of change of momentum (p) imparted byparticles to wall. When the molecules breaks into atoms, the number of moles would become twice. From ideal gas equation, pV = nRT where, p = Pressure of gas, n = Number of moles R = Gas constant, T = temperature As gases breaks number of moles becomes twice of initial, so n2 = 2n1 So, p nT µ or 2 2 2 1 1 1 1 1 (2 )(3000) 20 (300) p n T n p n T n = = = So, p2 = 20p1 Hence, final pressure of the gas would be 20 times to the initial pressure. 7. (b) For a function f1(v) thenumber ofmolecules (n) which will have their speeds between v and v + dv. For each function f1(v) and f2(v) number of molecules remain same 1 mole each but due to mass difference their speed will be different. Hence both gases of each function f1(v) and f2(v) will obeythe Maxwell's distribution law separately. 8. (d) As we know that an ideal gas equation, pV = nRT where, n = Number of moles, p = Pressure, V = Volume, R = Gas constant, T = Temperature pV n RT = If n, R are constant for the system or as number of moles of the gas remains fixed, hence, we can write pV T = constant or 1 1 2 2 1 2 p V p V RT RT =
- 349. 345 Kinetic Theory 2 2 11 2 1 ( ) T p p V V T æ ö = ç ÷ è ø ( )( )(1.1 ) (1.05) ( ) p V T V T = [p1 = p, V2 = 1.05 V and T2 = 1.1 T] 1.1 1.05 p æ ö = ´ç ÷ è ø = p(1.0476) So, final pressure p2 will lies between p and 1.1p. NEET/AIPMT (2013-2017) Questions 9. (b) P1 > P2 P1 P2 T V q2 q1 As V = constant Þ Pµ T Hence from V–T graph P1 > P2 10. (d) From first law ofthermodynamics DQ = DU+ DW = 3 2 . 1 4 R (T2 – T1) + 0 = 3 8 NaKB (T2 – T1) [Q K = R N ] 11. (b) Q 2 rms 1 3 mn P V V = When mass is halved and speed is doubled then Resultant pressure, Pt = 2 rms 1 (2 ) 3 2 m n v V ´ ´ = 2 P. 12. (b) Mean free path lm = 2 1 2 d n p where d = diameter of molecule and d = 2r lm µ 2 1 r 13. (b) Let ‘n’ be the degree of freedom p v n 1 R C 2 2 1 n C n R 2 æ ö + ç ÷ è ø æ ö g = = = + ç ÷ è ø æ ö ç ÷ è ø 14. (a) Change in internal energyfrom A® B DU= f 2 nRDT = f 2 nR (Tf – Ti) = 5 2 {PfVf –PiVi} (As gas is diatomic f = 5) = 5 2 {2 × 103 × 6 – 5 × 103 × 4} = 5 2 {12 – 20} × 103 J= 5 × (–4) × 103 J DU= –20 KJ 15. (a) FromPV=nRT PA = A A M RT r and B B B M P RT r = From question, A A A A B B B B P M M 3 2 P M M 2 r = = = r So, A B M 3 M 4 = 16. (d) Molar mass of the gas = 4g/mol Speed of sound V = RT m g Þ 952= 3 3.3 273 4 10- g´ ´ ´ Þ g = 1.6 = 6 8 10 5 1 = Also, g = P V C 8 C 5 = So, CP = 8 5 5 ´ = 8JK–1mol–1 [CV = 5.0 JK–1 given] 17. (b) Here v1 = 200 m/s; temperature T1 = 27°C = 27 + 273 = 300 k temperature T2 =127°C=127+273=400 k,V=? R.M.S. Velocity, Vµ T Þ v 400 200 300 = Þ v = 200 2 m / s 3 ´ Þ v = 400 m / s 3 18. (c) Internal energy of the system is given by U= f nRT 2 Degree of freedom Fdiatomic = 5 fmonoatomic = 3 and, number of moles n(O2)= 2 n(Ar) = 4 Utotal = 5 3 (2)RT (4)RT 2 2 + = 11RT T
- 350. 346 PHYSICS F ¬ B OC A x ® i.e., F µ – x or, F kx = - r r where k is called force constant or spring constant or, 2 2 0 d x kx m m dt + = r r (b) Angular S.H.M. The restoring torque is proporational to the angular displacement from the mean position. C t = - q r r where C is called torsional rigidity or, I 2 2 d C dt q = - q r r or, 2 2 0. d C I dt q + q = r r Terms Related to S.H.M. (i) Amplitude : The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude. It is denoted byA. (ii) Time period : The time taken by a oscillating particle to complete one oscillation is called its time period. It is denoted by T. (iii) Frequency : It is the number of oscillations completed in one second. It is denoted by u. 1 T u = The S.I. unit of frequency is s–1 or Hz. (iv) Angular frequency 2 2 T p w = pu = The S.I. unit of angular frequencyis rad/sec. (v) Phase : The parameter, by which the position of particle from its mean position is represented, is known as phase. The phase at any instant tells the state of position & direction of motion at that instant. The phase at time t = 0 is known as the initial phase or epoch (e). PERIODIC AND OSCILLATORY MOTION Periodic Motion When a body repeats its motion after regular interval of time, it is said to be in periodic motion. The path ofperiodic motion may be rectilinear, open/ closed curvilinear. Example: (i) Motion of moon around earth. (ii) Motion of a piston in a cylinder. (iii) Motion of a simple pendulum etc. Oscillatory Motion If during a periodic motion, the particle moves to and fro on the same path, the motion is vibratory or oscillatory. Example:(i) The motion of a ball in bowl Bowl Ball (ii) The needle of a sewing machine (iii) Vibrations of prongs of tuning fork etc. (i)All oscillatorymotion areperiodic but all periodic motioniiare not oscillatorymotion. (ii) The oscillatory motion which can be expressed in terms of sine and cosine function, is said to be harmonic motion. SIMPLE HARMONIC MOTION (S.H.M.) “If a particle moves up and down (back and forth) about a mean position (also called equilibrium position) in such a way that a restoring force/ torque acts on particle, which is proportional to displacement from mean position, but in opposite direction from displacement, then motion of the particle is called simple harmonic motion. If displacement is linear, it is called linear S.H.M. andifdisplacement isangular, it is called an angularS.H.M. Example: (i) Motion of a body suspended by spring. (ii) Oscillations of simple pendulum. Equations of S.H.M. (a) Linear S.H.M. The restoring force is proportional tothe displacement from mean position. 14 Oscillations
- 351. 347 Oscillations (ii) Potential energy:Aparticle in S.H.M. possesses potential energy due to its displacement from the mean position. 2 2 2 1 1 . . 2 2 P E ky m y = = w (iii) Total mechanical energy E=K.E. +P.E. 2 2 2 2 2 1 1 m (A y ) m y 2 2 = w - + w 2 2 1 2 E m A = w 2 1 2 kA = (iv) The curves representing KE, PE and total energyare shown in figure. O K.E. P.E. Displacement – A + A Total energy E = K.E. + P.E. Energy Keep in Memory 1. Restoring force F = – Mw2x 2. Kinetic energy = (1/2) Mw2(A2 – x2) 3. Potential energy= 1/2 Mw2x2 4. Total energy of SHM = 1/2 Mw2A2 Equation a = – w2y shows that if body perform S.H.M. then acceleration of the body is proportional to displacement, but in the opposite direction of displacement. It is an essential requirement for any motion to be S.H.M. 5. The kinetic and potential energyofSHM varies sinusoidally with a frequency twice that of SHM. 6. Total energy 2 2 2 2 2 1 m A 2 mA n 2 = w = p where n = frequency of vibration. 7. x a 2 w - = where w is constant 2 2 x A v - w = a x x v 8. Geometrically the projection of the body undergoing uniform circular motion on the diameter of the circle is SHM. 9. In a non-inertial frame. eff g 2 T l p = x 2 2 y eff a ) a g ( g + - = (vi) Total phase angle : Thetotal angle (wt + q) isknown as total phase angle. Characteristics of S.H.M. (i) Displacement : The displacement ofa particle in S.H.M. is given by sin( ) = + y A t w f whereAis amplitude, wis angular frequencyand (wt + f) is called the phase of the particle at any instant t. T/4 T/2 3T/4 T +A –A (0,0) Time (t) Displacement (ii) Velocity : The velocity of a particle in S.H.M. is given by cos( ) dy A t dt = w w + f or, 2 2 v A y = w - At y= 0 (at mean position), max v A = w T/4 T/2 3T/4 T (0) Time (t) Velocity (iii) Acceleration : The acceleration of a particle in S.H.M. is given by 2 sin( ) dv a A t dt = = -w w + f or, a = – w2y The negativesign indicates that the acceleration is directed towards the mean position At y= A(at extreme position), amax = – w2A T/4 T/2 3T/4 T Time (t) Acceleration Energy in S.H.M. : (i) Kinetic energy : A particle in S.H.M. possesses kinetic energy by virtue of its motion. 2 2 2 2 1 1 . . ( ) 2 2 K E mv m A y = = w -
- 352. 348 PHYSICS SOME SYSTEMSEXECUTING S.H.M. Case 1 Spring mass system : (i) When two springs having force constants k1 and k2 connected in parallel, then k2 k1 M The force constant of the combination is k = k1 + k2 and hence T = 2p[M/(k1 + k2)]1/2 (ii) When two springs of force constants k1 and k2 are connected in series, then k2 k1 M The force constant of the combination is 1/k = 1/k1 + 1/k2. i.e., k = k1k2/(k1 + k2) Hence 2 m T K = p (iii) If two mass M1 and M2 are connected at the two ends of the spring, then their period of oscillation is given by M2 M1 T = 2p[m/k)]1/2 where 1 2 1 2 M M M M m = + is the reduced mass. (iv) When the length of spring increases, spring constant decreases. If thelength ofspring becomes n times, its spring constant becomes n 1 times and therefore time period will be increased by n times. (v) If we divide the spring into n equal parts, the spring constant of each part becomes n k. Hence time period when the same mass is suspended from each part is: 1/2 2 M T nk é ù = pê ú ë û Case 2 Simple pendulum : A simple pendulum consists of a point mass suspended by a weightless inextensible cord from a rigid support. q x =lq mg sinq mg mg cosq l l Let a bob of mass m is displaced from its, equilibrium position and released, then it oscillates in a vertical plane under gravity. Let q be the angular displacement at any time t, then corresponding linear displacement along the arc is x = lq. It is clear from the diagram that mg sinq, is the restoring force acting on m tending to return it to mean position. So from Newton’s second law q - = = sin mg dt x md F 2 2 ...(i) where negative sign indicates that restoring force mg sin q (= F) is opposite to displacement q. If q is very small, then sin q » q, so from equation (i) l x mg sin mg dt x md 2 2 - = q - = or ...(ii) where w2 = g/l. This is the equation of S.H.M. of the bob with time period 2 2 T g æ ö p = = p ç ÷ w è ø l OM How to find the time period of a body undergoing S.H.M.? Step 1 : First, find the equilibrium position. Equilibrium position will be one for which 0 F = S and 0 St = Step 2 :Displacethebody, from the equilibrium position byx. Find therestoring force acting on the bodyF = –kx (for translation) Find the restoring torque acting on the body q - = t k (for rotational) Step 3 : Since 2 2 dt x md ma F = = Use kx dt x md 2 2 - = ... (i) for translational q - = k dt x md 2 2 … (ii) for rotational Step 4 : k m 2 T p = (for translational) k I 2 T p = (for rotational) where, I = moment ofinertia COMMON DEFAULT Incorrect. The time period of spring mass-system is dependent on the value of g. Correct. Time period ofspring-mass system shifts onlythe equilibrium position. It does not change the time period. Because of this reason, time period of spring mass system remains same on plains / mountains / in satellites. Incorrect. The time taken tocover halfthe amplitude form equilibrium position is 8 T .
- 353. 349 Oscillations Correct. The actualtime taken is 12 T . Incorrect. In a spring mass system, mass oscillate about the end of a spring when the spring is in unstretched condition. Correct. Themass oscillates about the equilibrium position which may or may not be at the unstretched length. Case 3 Liquid in U-tube :AU-tube ofuniform cross-sectional area A has been set up vertically with open ends facing up. x x [Restoring force = –2Adg x] If m gm of a liquid of densityd is poured into it then time period of oscillation. Adg 2 m 2 T p = Case 4 Rectangular block in liquid : d h' h d' Rectangular block floating in a liquid, gd ' d ' h 2 T p = ; where d = density of liquid, d¢ = density of block, h = height of block Case 5 : Vibration of gas system in a cylinder with frictionless piston. A m p h Time period, AP mh 2 T p = where m = mass of gas, A = cross sectional area of piston P = pressure exerted by gas on the piston, h = height of piston Case 6: If a tunnel is dig in the earth diametrically or along a chordthen timeperiod, sec 36 min 84 g R 2 T = p = for a particle released in the tunnel. Case 7: The time period of a ball oscillating in the neck of a chamber B mV A 2 T p = Case 8 : Ifa dipole of dipole moment p is suspended in a uniform electric field E then time period of oscillation PE I T p 2 = Keep in Memory 1. In S.H.M. the phase relationship between displacement, velocity and accleration, is as follows : (a) The velocity is leading the displacement by a phase 2 p radian (b) The acceleration is leading the displacement by a phase p radian (c) The acceleration is leading the velocity bya phase 2 p radian. 2. (a) When 2 A x = , then velocityV = 0.86Vmax. (b) When V = Vmax/2, the displacement x = 0.87A. (c) When 2 A x = , the kineticenergyof S.H.M. is 75% of the total energy and potential energy25% of the total energy. (d) When the kinetic energyof S.H.M. is 50% of the total energy, the displacement is 71% of the amplitude. 3. The time period of a simple pendulum of length l which is comparable with radius of earth. 1/2 1/2 R R T 2 2 R ( R)g 1 g é ù é ù = p = p ê ú ê ú + æ ö ë û ê ú + ç ÷ è ø ê ú ë û l l l where R = radius of the earth and g is the acceleration due to gravity on the surface of the earth. (a) When l << R, then g 2 T l p = (b) When l = R, we find hour 1 g 2 R 2 T 2 / 1 » ú ú û ù ê ê ë é p = (c) When l = ¥ , then 2 / 1 g R 2 T ú ú û ù ê ê ë é p = = 84.6 minutes. Thus maximum ofT is 84.6 minutes. (d) Under weightlessness or in the freely falling lift ¥ = ú û ù ê ë é p = 2 / 1 0 2 T l . This means, the pendulum does not oscillate at all.
- 354. 350 PHYSICS 4. Timeperiod ofa simple pendulum in a train accelerating or retarding at the rate a is given by 2 / 1 2 2 a g 2 T ú ú û ù ê ê ë é + p = l 5. Ifa simple pendulum whose bob is of densitydo is made to oscillate in a liquid of density d, then its time period of vibration in liquid will increase and is given by g d d 1 2 T o ÷ ÷ ø ö ç ç è æ - p = l (where d0 > d) 6. The time period of a simple pendulum in a vehicle moving along a circular path of radius r and with constant velocity V is given by, 1/ 2 4 2 2 2 T V g r é ù = pê ú ê ú + ê ú ë û l 7. If T1 and T2 are the time periods ofa bodyoscillating under the restoring force F1 and F2 then the time period of the bodyunder the influence of the resultant force 1 2 = + r r r F F F will be 2 2 2 1 2 1 T T T T T + = 8. (a) The percentage change in time period of simple pendulum when its length changes is % 100 2 1 100 T T ´ ÷ ø ö ç è æ D = ´ D l l (b) The percentage change in time period of simple pendulum when g changes but l remains constant is 100 g g 2 1 100 T T ´ ÷ ÷ ø ö ç ç è æ D = ´ D % (c) The percentage change in time peirod of simple pendulum when both l and g change is % 100 g g 2 1 100 T T ´ ÷ ÷ ø ö ç ç è æ D + D = ´ D l l 9. If a wire of length l , area of cross-section A, Young’s s modulus Y is stretched by suspending a mass m, then the mass can oscillate with time period: l m TA 2 T p = 10. If a simple pendulum is suspended from the roof of compartment of a train moving down an inclined plane of inclination q, then the time period ofoscillations 2 / 1 cos g 2 T ú û ù ê ë é q p = l 11. If a ball of radius r oscillates in a bowl of radius R, then its time period is given by : 2 / 1 g r R 2 T ú û ù ê ë é - p = 12. Ifa disc ofradius r oscillates about a point at its rim, then its time period is given by: 2 / 1 g r 2 T ú û ù ê ë é p = It behaves as a simple pendulum of length r. 13. The graph between the length of a simple pendulum and its timeperiod is a parabola. 14. The graph between the length of a simple pendulum and the square of its time period is a straight line. 15. The graph between l & T and between l & T2 intersect at T = 1 second. 16. The time period of the mass attached to spring does not change with the change in acceleration due to gravity. 17. Ifthemass mattached toa spring oscillates in a non viscous liquid density s, then its time period is given by 2 / 1 1 k m 2 T - ú û ù ê ë é ÷ ÷ ø ö ç ç è æ r s - p = where k = force constant and r is density of the mass suspended from the spring. 18. The length of second pendulum (T = 2 sec) is 99 cm Physical Pendulum Trestoring = – mgd sin q If q is small, sin q » q cm mg d q Trestoring = – mgdq And = 2 I T mgd p Let a test-tube of radius r, carrying lead shots of mass m is held verticallywhen partlyimmersed in liquid ofdensityr. On pushing the tube little into liquid and let it executes S.H.M. of time period 2 2 = m T r g p p r Conical Pendulum When the bob of a simple pendulum moves in a horizontal circle it is called as conical pendulum. r T S q If l is the length of the pendulum and the string makes an angle q with vertical then time period, g cos 2 T q p = l
- 355. 351 Oscillations In this caseelectric force qE and gravity force are opposite. 5. A pendulum clock slows down in summer and goes faster in winter. 6. Potential energyof a particle executing S.H.M. is equal to average force × displacement. i.e., 2 2 2 P x m 2 1 x 2 x m 0 x 2 F 0 U w = ÷ ÷ ø ö ç ç è æ w + = ú û ù ê ë é + = . 7. If the total energyof a particle executing S.H.M. is E, then its potential energyat displacement x is E A x U 2 2 P = and kinetic energy E A x 1 U 2 2 k ÷ ÷ ø ö ç ç è æ - = FREE, DAMPED, FORCED OSCILLATIONS AND RESONANCE Free Oscillation If a system oscillates on its own and without any external influence then it is called as free oscillation. Frequency of free oscillation is called natural frequency. The equation for free S.H.M. oscillation 2 2 md x dt = Frestoring force = –kx, where k is constant. The differential equation of harmonic motion in absence of damping and external force is 2 2 0 2 0 d x x dt + w = , wherew0 isnaturalfrequencyofbody. . The time period is 0 2 2 = = m T k p p w Damped Oscillation Oscillation performed under the influence of frictional force is called as damped oscillation. (a) In case of damped oscillations the amplitude goes on decreasing and ultimately the system comes to a rest. t x (b) The damping force (Fdamping µ – v Þ Fdamping = – bv) is proportional to the speed of particle. Hence the equation of motion dV m kx bv dt = - - where b is positive constant and is called damping coefficient. Then the differential equation of a damped harmonic oscillation is Torsional Pendulum It is an arrangement which consists of a heavy mass suspended from a long thin wirewhoseother andisclampedtoa rigid support. Timeperiod 2 = T C p I where I= moment of inertia of body about the suspension wire as axis ofrotation. C= restoring couple per unit thirst. Keep in Memory 1. The displacement, velocityand acceleration of S.H.M. vary simple harmonically with the same time period and frequency. 2. The kinetic energy and potential energy vary periodically but not simple harmonically. The time period of kinetic energy or potential energy is half that of displacement, velocity and acceleration. 3. The graph between displacement, velocity or acceleration and t is a sine curve. But the graph between P.E. or K.E. of S.H.M. and time tisparabola. Disp laceme nt t x=A sin t w Veloc ity t t cos A dt dx v w w = = Acce lerat ion x dt x d a 2 2 2 w - = = t +A x O –A K.E., P.E. E 2 2 x m 2 1 . E . P w = ) x A ( m 2 1 . E . K 2 2 2 - w = ½E E=½ m A w 2 2 4. (a) If the bob of simple pendulum is -vely charged and a +vely charged plate is placed below it, then the effective acceleration on bob increases and consequently time period decreases. Time period, m qE g 2 T + p = l In this case electric force q E and gravity force act in same direction. (b) If the bob of a simple pendulum is -vely charged and is made to oscillate above the -vely charged plate, then the effective acceleration on bob decreases and the time period increases. T 2 qE g m = p - l
- 356. 352 PHYSICS 2 2 0 2 2 0 dx dx C x dt dt + + w = ...(i) where 2 C = b/m (C is damping constant) & 2 0 m k w = , the natural frequency of oscillating particle i.e., its frequency in absence of damping. In case of over damping the displacement of the particle is 0 exp( / 2 ) x A t = - t sin (wt + f) ...(ii) Þ x=Asin (wt + f)whereA0 = max. amplitudeoftheoscillator. 2 2 0 c - w = w and b m = t (relaxation time) It is clear from the fig & eqn. (ii) that the amplitude of damped harmonic oscillator decreases with time. In this case, the motion does not repeat itself & is not periodic in usual sense of term. However it has still a time period, 2 2 0 c 2 2 T - w p = w p = , which is the time interval between its successive passage in same direction passing the equilibrium point. A= –A e 0 –ct A=A e 0 –ct t Displa ceme nt Forced Oscillation and Resonance : The oscillation of a system under the action of external periodic force is called forced oscillation. External force can maintain the amplitudeof damped oscillation. When the frequency of the external periodic force is equal to the natural frequency of the system, resonance takes place. The amplitude of resonant oscillations is very very large. In the absence of damping, it may tend to infinity. At resonance, the oscillating system continuously absorbs energyfrom the agent applying external periodic force. In case offorced oscillations, the total force acting on the system is 0 Restorting Ext periodic force Damping force force bdx F kx F sin pt dt = - - + ……(i) Then by Newton’s second law : Þ pt sin F dt dx b kx dt x d m 0 2 2 + - - = or t sin f x dt dx C 2 dt x d 0 2 0 2 2 b = w + + ……(ii) where m F f , m k , m b C 2 0 0 2 0 = = w = The equation(ii)isthedifferential equationofmotionof forced harmonic oscillator. The amplitude at any time t is 0 2 2 2 2 0 f x sin(pt ) ( p ) 4c p = - q é ù w - + ë û where 2 2 p cp 2 tan - w = q & p is the frequencyof external periodic force. Example1. A spring is stretched by 0.20 metre when a mass of 0.50 kg is suspended. Calculate the period of the spring when a mass of 0.25 kg is suspended and put to oscillation. (g = 10 m/sec2). Solution : The force constant K of the spring is given by F 0.5 kg Wt K y 0.20m 0.5 10 newton 0.20 m 25 newton / metre = = ´ = = ÷ ø ö ç è æ p = K m 2 T Now 0.25 2 3.14 0.628 second. 25 æ ö = ´ = ç ÷ è ø Example2. What will be the force constant of the spring system shown in fig? (a) 1 2 k k 2 + (b) 1 1 2 1 1 2k k - é ù + ê ú ë û k2 k1 k1 (c) 1 2 1 1 2k k é ù + ê ú ë û (d) 1 1 2 2 1 k k - é ù + ê ú ë û Solution : (b) Two springs of force constants k1 and k2 are in parallel. Hence 1 1 1 k 2 k k k = + = ¢ The third spring is in series with spring of force constant k¢.
- 357. 353 Oscillations 1 2 1 2 1 k 1 k 2 1 k or k 1 k 2 1 k 1 - ú û ù ê ë é + = ú û ù ê ë é + = Example3. A particlestarts with S.H.M. from the mean position as shown in figure below. Its amplitude isA and its time period is T. At one time, its speed is half that of the maximum speed. What is this displacement at that time ? (a) 2 A 3 (b) 3 A 2 (c) 2A 3 (d) 3A 2 Solution : (b) We know that 2 2 1/ 2 v ω[A x ] = - 1/ 2 2 2 2 v x A ω é ù = - ê ú ê ú ë û Given that max v Aω v 2 2 = = . so, A . 2 3 4 A A 2 / 1 2 2 2 2 = ú ú û ù ê ê ë é w w - Example4. A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be P A M h (a) Mh T 2π PA æ ö = ç ÷ è ø (b) MA T 2π Ph æ ö = ç ÷ è ø (c) M T 2π PAh æ ö = ç ÷ è ø (d) T 2π (MPhA) = Solution : (a) We know that PV = R T ) V / V ( P P or 0 V P P V D = D = D + D or ÷ ø ö ç è æ - = ÷ ÷ ø ö ç ç è æ - = h x P h A x A P A F ÷ ø ö ç è æ - = h x A P F x h M A P M 1 h x A P a ÷ ÷ ø ö ç ç è æ - = ÷ ø ö ç è æ - = This gives, 2 2 2 æ ö = = = ç ÷ è ø P A M h or T M h P A p w p w Example5. A simple harmonic oscillator has an amplitude A and time period T. Determine the time required by it to travel from x = A to 2 A x = . Solution : For S.H.M., ÷ ø ö ç è æ p = t T 2 sin A x When x =A, ÷ ø ö ç è æ p = t . T 2 sin A A 1 t . T 2 sin = ÷ ø ö ç è æ p Þ ÷ ø ö ç è æ p = ÷ ø ö ç è æ p 2 sin t . T 2 sin Þ t = (T/4) When 2 A x = and ÷ ø ö ç è æ p = t . T 2 sin A 2 A or ÷ ø ö ç è æ p = p t T 2 sin 6 sin or t = (T/12) Now, time taken to travel from x =A to x =A/2 = T/4 –T/12 = T/6 Example6. Calculate the increase in velocity of sound for 1ºC rise of temperature, if the velocity of sound at 0ºC is 332 m/sec. Solution : t 1 0 0 1/2 1 0 0 0 0 0 t 0 0 T v 273 t ; v T 273 t t v v 1 v 1 273 2 273 t t v 1 v v 546 546 t v v v = 0.61 m/sec. 546 æ ö + æ ö = = ç ÷ ç ÷ è ø è ø æ ö æ ö = + = + ç ÷ ç ÷ è ø è ø ´ æ ö æ ö = + = + ç ÷ ç ÷ è ø è ø æ ö - = ç ÷ è ø (As v0 = 332 m/sec and Dt = 1ºC) Example7. Which one of the following equations does not represent S.H.M.; where x = displacement and t = time. Parameters a, b and c are the constants of motion (a) x = a sin bt (b) x = a cos bt + c (c) x = a sin bt + c cos bt (d) x = a sec bt + c cosec bt Solution : (d) Sec bt is not defined for bt = p/2.
- 358. 354 PHYSICS x =a sec bt + c cosec bt = bt cos bt sin bt cos c bt sin a + This equation cannot be modified in the form of simple equation of S.H.M. i.e. x = a sin (wt + f). so, it cannot represent S.H.M. Example8. If length of pendulum is increased, time period T increases. So, frequency decreases or angular velocity decreases. As maximum velocity max v Aω = , so max. velocity will also decrease. (where A is amplitude of bob). A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. What will be the speed of the bob, when the length makes an angle of 60º to the vertical? (g = 10 m/s2) Solution : Here, r w = 3 m/s; l = 0.5 m; K.E. at the lowest point m 2 9 ) 3 ( m 2 1 r m 2 1 2 2 2 = = w = Letv bethevelocityofthe bobatB when ÐOAB= 60° then, A O C B 60º l OC = h = (l – l cos q) = (l – l cos 60º) = l /2 If v is the velocity of bob at position B, then using law of conservation of energy, we have, 2 1 1 m 9 m v mg (1 cos θ) 2 2 ´ = + - l or 2 1 1 m v m 9 mg (1 cos θ) 2 2 = ´ - - l 1/2 1/2 1/2 [9 2 (1 cos )] [9 2 10 0.5 (1 1/ 2)] (9 – 5) 2 / = - - = - ´ ´ - = = l v g m s q Example9. A small spherical steel ball is placed a little away from the centre of a large concave mirror whose radius of curvature R = 2.5 cm. When the ball is released, it begins to oscillate about the centre. Show that the motion of the ball is simple harmonic and find the period of motion. Neglect friction and take g = 10 m/sec2. Solution : F = – mg sin q = – mg q (Q q issmall) = – mg (x/R) F= – kx, wherek = (mg/R) Now, a = F/m = – kx/m = – w 2x Hence, motion is simpleharmonic motion and T = 2p w = R 2 g æ ö p ç ÷ è ø = 2 × 3.14 × 2.5 10 æ ö ç ÷ è ø = 3.142 sec. Example10. A spring of stiffness constant k and natural lengthl is cut into two parts of length 3l / 4 and l / 4 respectively, and an arrangement is made as shown in the figure. If the mass is slightly displaced, find the time period of oscillation. m (3/4)l l/4 Solution : The stiffness of a spring is inversely proportional to its length. Therefore the stiffness of each part is 1 4 k k 3 = and k2 = 4k m k k 1 = (4/3) k k 2 = 4 Timeperiod, 1 2 m T 2 k k = p + or 3m 3m T 2 16k 2 k p = p =
- 359.
- 360. 356 PHYSICS 1. Achild swinging on swing in sitting position stands up. The time period of the swing will (a) increase (b) decrease (c) remain same (d) increase if the child is tall and decrease if the child is short. 2. The graph oftime period (T) of simple pendulum versus its length (l) is (a) T O l (b) T l O (c) O l T (d) O l T 3. Which ofthe following is a simple harmonic motion? (a) Particle moving through a string fixed at both ends. (b) Wave moving through a string fixed at both ends. (c) Earth spinning about its axis. (d) Ball bouncing between tworigid vertical walls. 4. Arodis hinged verticallyat one endandis forced tooscillate in a vertical plane with hinged end at the top, the motion of the rod: (a) is simple harmonic (b) is oscillatorybut not simple harmonic (c) is pericolic but not oscillatory (d) maybe simple harmonic 5. The graph plotted between the velocity and displacement from mean position ofa particle executing SHM is (a) circle (b) ellipse (c) parabola (d) straight line 6. Acceleration of a particle executing SHM, at it’s mean position is (a) infinity (b) variable (c) maximum (d) zero 7. A vertical mass-spring system executes simple harmonic oscillations with a period of 2s. A quantity of this system which exhibits simple harmonic variation with a period of 1 s is (a) velocity (b) potential energy (c) phasedifference between acceleration anddisplacement (d) difference between kinetic energyand potential energy 8. A simple pendulum has a metal bob, which is negatively charged. Ifit isallowedtooscillate above a positivelycharged metallic plate, then its time period will (a) increase (b) decrease (c) become zero (d) remain the same 9. In the fig. S1 and S2 are identical springs. The oscillation frequency of the mass m is f. If one spring is removed, the frequencywill become S1 S2 m (a) f (b) f× 2 (c) 2 f ´ (d) 2 / f 10. A pendulum is undergoing S.H.M. The velocityof the bob in the mean position is v. If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be (a) v/2 (b) v (c) 2 v (d) 4 v 11. The potential energyofa particle (Ux) executing S.H.M. is given by (a) 2 x k U (x a) 2 = - (b) 2 3 x 1 2 3 U k x k x k x = + + (c) bx x e A U - = (d) Ux = a constant 12. A particle moves such that its acceleration ‘a’ is given bya = – bxwhere xis the displacement from equilibrium position and b is constant. The period of oscillation is (a) 2 p/b (b) 2π / b (c) b / 2 p (d) b / 2 p 13. A particle executes S.H.M. having time period T, then the time period with which the potential energychanges is (a) T (b) 2T (c) T/2 (d) ¥ 14. An instantaneous displacement of a simple harmonic oscillator is x =Acos (wt + p/4). Its speed will be maximum at time (a) p/4w (b) p/2w (c) p/w (d) 2p/w 15. The tension in the string of a simple pendulum is (a) constant (b) maximum in the extremeposition (c) zero in the mean position (d) None of these
- 361. 357 Oscillations 16. Arectangular blockof mass m and area of cross-section A floats in a liquid of density r . If it is given a small vertical displacement from equilibrium. It undergoes oscillations with a time period T then (a) m Tµ (b) r µ T (c) A / 1 Tµ (d) r µ / 1 T 17. If the magnitude of displacement is numerically equal to that of acceleration, then the time period is (a) 1 second (b) p second (c) 2p second (d) 4p second 18. The graph shown in figure represents Time Velocity (a) motion of a simple pendulum starting from mean position (b) motion of a simple pendulum starting from extreme position (c) simple pendulum describing a horizontal circle (d) None of these 19. A particle executing simple harmonic motion along y-axis has its motion described bythe equation sin( ) y A t B w = + . The amplitude of the simple harmonic motion is (a) A (b) B (c) A+ B (d) A B + 20. Resonance is an example of (a) tuning fork (b) forced vibration (c) free vibration (d) damped vibration 21. Which one of the following equations of motion represents simple harmonic motion? (a) Acceleration = – k(x + a) (b) Acceleration = k(x + a) (c) Acceleration = kx (d) Acceleration = – k0x + k1x2 where k, k0, k1 and a are all postive. 22. Out of the following functions, representing motion of a particle, which represents SHM? (A) y sin t cos t = w - w (B) y = sin3 wt (C) 3 y 5cos 3 t 4 p æ ö = - w ç ÷ è ø (D) 2 2 y 1 t t = + w + w (a) Only(A) (b) Only (D) does not represent SHM (c) Only(A) and (C) (d) Only(A) and (B) 23. The total mechanical energy of a spring-mass system in simple harmonic motion is 2 2 A m 2 1 E w = . Suppose the oscillating particle is replaced byanother particle ofdouble themass while theamplitudeA remains the same. The new mechanical energywill (a) become 2E (b) become E/2 (c) become E 2 (d) remainE 24. Thedisplacement ofaparticle along the x-axis is given byx = a sin2 . t w The motion of the particle corresponds to: (a) simple harmonic motion of frequency / w p (b) simple harmonic motion of frequency 3 / 2 w p (c) non simple harmonic motion (d) simple harmonic motion of frequency / 2 w p 25. The period of oscillation of a mass M suspended from a spring of negligible mass isT. Ifalong with it another mass M is also suspended, the period of oscillation will now be (a) T (b) / 2 T (c) 2T (c) 2 T 1. A simple pendulum performs S.H.M. about x = 0 with an amplitude a, and time periodT. The speed of the pendulum at x = a/2 will be (a) T 3 a p (b) T 2 3 a p (c) T a p (d) T a 3 2 p 2. A body of mass 5 gram is executing S.H.M. about a fixed point O. With an amplitudeof 10 cm, its maximum velocity is 100 cm/s. Its velocity will be 50 cm s–1 at a distance (in cm) (a) 5 (b) 2 5 (c) 3 5 (d) 2 10 3. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the massexecutes SHMoftimeperiod T. Ifthemass is increased bym, thetime period becomes 3 T 5 . Then theratioof M m is (a) 9 25 (b) 9 16 (c) 3 5 (d) 5 3
- 362. 358 PHYSICS 4. Atunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period (a) 42 minutes (b) 1 day (c) 1 hour (d) 84.6minutes 5. A particle of mass 1 kg is moving in S.H.M. with an amplitude 0.02 and a frequency of 60 Hz. The maximum force acting on the particle is (a) 144p2 (b) 188p2 (c) 288p2 (d) None of these 6. The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface is (a) 1/6m (b) 6m (c) 1/36m (d) 36m 7. The displacement of a S.H.M. doing particle when K.E. = P.E. (amplitude= 4 cm)is (a) cm 2 2 (b) 2cm (c) cm 2 1 (d) cm 2 8. The equation of SHM of a particle isA+ 4p2x = 0 wherea is instantaneous linear acceleration at displacement x. The frequency of motion is (a) 1Hz (b) 4pHz (c) Hz 4 1 (d) 4Hz 9. y= 2 (cm)sin ú û ù ê ë é f + p 2 t what is the maximum acceleration of the particle doing the S.H.M. (a) 2 p cm/s2 (b) 2 2 p cm/s2 (c) 4 2 p cm/s2 (d) 4 p cm/s2 10. A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration 3 ms–2 is (a) 8 . 12 8 . 9 t (b) 8 . 9 8 . 12 t (c) 8 . 6 8 . 9 t (d) 8 . 9 8 . 6 t 11. Twoparticlesare oscillatingalong twoclose parallel straight lines side by side, with the same frequencyand amplitudes. Theypass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is (a) 0 (b) 2p/3 (c) p (d) p/6 12. If a simple pendulum of length l has maximum angular displacement q, then the maximum K.E. ofbobof massm is (a) 1 m / g 2 l (b) l 2 / mg (c) ) cos 1 ( mg q - l (d) /2 sin mg q l 13. Three masses of 500 g, 300 g and 100 g are suspended at the end ofa spring as shown, and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 2 second. When the 300 g mass is also removed, it will oscillate with a period of (a) 2 s (b) 4 s (c) 8 s 500 g 300 g 100 g (d) 1 s 14. If the mass shown in figure is slightly displaced and then let go, then the system shall oscillate with a time period of (a) k 3 m 2p (b) k 2 m 3 2p (c) k 3 m 2 2p m k k k (d) m k 3 2p 15. Two oscillators are started simultaneously in same phase. After 50 oscillations of one, theyget out ofphase byp, that is half oscillation. The percentage difference offrequencies of the two oscillators is nearest to (a) 2% (b) 1% (c) 0.5% (d) 0.25% 16. Two wires are kept tight between the same pair ofsupports. Thetensions in the wiresare in theratio2 : 1, the radii are in theratio3 : 1 and the densities are in the ratio1: 2. The ratio of their fundamental frequencies is (a) 2: 3 (b) 2: 4 (c) 2: 5 (d) 2: 6 17. The time period of the oscillating system (see figure) is (a) 1 2 m T 2 k k = p (b) 1 2 m T 2 k k = p + M k1 k2 (c) 1 2 T 2 mk k = p (d) None of these 18. Aparticle ofmass10gm isdescribingS.H.M. along astraight line with period of2 sec and amplitude of10 cm. Its kinetic energywhen it is at 5 cm from its equilibrium position is (a) 37.5p2 erg (b) 3.75p2 erg (c) 375p2 erg (d) 0.375p2 erg
- 363. 359 Oscillations 19. A boyis executing simple Harmonic Motion. At a displacement x its potential energy is E1 and at a displacement y its potential energy is E2. The potential energyE at displacement (x + y) is (a) 1 2 E E E = - (b) 1 2 E E E = + (c) 1 2 E E E = + (d) 1 2 E E E = - 20. A particle undergoes simpleharmonic motion having time period T. The time taken in 3/8th oscillation is (a) T 8 3 (b) T 8 5 (c) T 12 5 (d) T 12 7 21. Twoparticles are executing S.H.M. of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them ? (a) 5p/6 (b) 2p/3 (c) p/3 (d) p/6 22. Lissajous figure obtained by combining x = a sin wt and y= a sin (wt + p/4) will be a/an (a) ellipse (b) straight line (c) circle (d) parabola 23. A simple spring has length l and force constant K. It is cut into two springs of lengths l1 and l2 such that l1 = n l2 (n = an integer). The force constant ofspring of length l1 is (a) K(1 + n) (b) (K/n)(1 + n) (c) K (d) K/(n + 1) 24. The spring constant from the adjoining combination of springs is m 2K K K (a) K (b) 2K (c) 4K (d) 5K/2 25. A Second’s pendulum is placed in a space laboratory orbiting around the earth at a height 3 R from the earth’s surface where R is earth’s radius. The time period of the pendulum will be (a) zero (b) 2 3 (c) 4 sec (d) infinite 26. A simple pendulum attached to the roof of a lift has a time period of 2s in a stationary lift. If the lift is allowed to fall freely the frequency of oscillations of pendulum will be (a) zero (b) 2Hz (c) 0.5Hz (d) infinity 27. Twosimple pendulumsoflength 0.5 m and 20 mrespectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed oscillations [nT1=(n–1)T2, where T1 is timeperiod ofshorter length & T2 be time period of longer wavelength and n are no. of oscillations completed] (a) 5 (b) 1 (c) 2 (d) 3 28. A clock which keeps correct time at 20ºC, is subjected to 40ºC. If coefficient oflinear expansion of the pendulum is 12×10–6 per ºC. Howmuch will it gain or loose in time ? (a) 10.3 seconds/day (b) 20.6 seconds/day (c) 5 seconds/day (d) 20 minutes/day 29. A mass is suspended separately by two different springs in successive order then time periods is t1 and t2 respectively. It is connected by both springs as shown in fig. then time period is t0, thecorrect relation is k1 k2 m (a) 2 2 2 1 2 0 t t t + = (b) 2 2 2 1 2 0 t t t - - - + = (c) 1 2 1 1 1 0 t t t - - - + = (d) 2 1 0 t t t + = 30. When an oscillator completes 100 oscillations itsamplitude reduces to 3 1 ofits initial value. What will be its amplitude, when it completes 200 oscillations ? (a) 8 1 (b) 3 2 (c) 6 1 (d) 9 1 31. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity w, in gravity free space. The increase in length of the spring will be k m w (a) k m 2 l w (b) 2 2 m k m w - w l (c) 2 2 m k m w + w l (d) None of these 32. Frequency of oscillation is proportional to 2k k m (a) m k 3 (b) m k (c) m k 2 (d) k 3 m
- 364. 360 PHYSICS 33. Apendulum bob is raised to a height h and released from rest.At what height will it attain halfofits maximum speed? (a) 4 h 3 (b) 2 h (c) 4 h (d) h 707 . 0 34. A mass m fall on spring of spring constant k and negligible mass from a height h. Assuming it sticks to the pan and executessimple harmonicmotion, themaximum height upto which the pan will rise is (a) mg k (b) mg 2kh 1 1 k mg é ù + - ê ú ë û (c) mg 2kh 1 1 k mg é ù + + ê ú ë û k m h (d) mg kh 1 1 k mg é ù + - ê ú ë û 35. The graph shown in figure represents (a) S.H.M. Velocity Displacement +a –a O (b) circular motion (c) rectillinear motion (d) uniform circular motion 36. The time period of a simple pendulum of infinite length is (Re = radius ofEarth) (a) g R 2 T e p = (b) g R 2 2 T e p = (c) g 2 R 2 T e p = (d) ¥ = T 37. A block rests on a horizontal table which is executing SHM in the horizontal planewith an amplitude'a'. If thecoefficient of friction is 'm', then the block just starts to slip when the frequency of oscillation is (a) a g 2 1 m p (b) a g m (c) g a 2 m p (d) g a m 38. On Earth, a bodysuspended on a spring of negligible mass causes extension L and undergoes oscillations along length of the spring with frequency f. On the Moon, the same quantities are L/n and f ' respectively. The ratio f '/f is (a) n (b) n 1 (c) n–1/2 (d) 1 39. Auniform pole oflength l= 2 Lis laid on smooth horizontal table as shown in figure. The mass of pole is M and it is connected to a frictionless axis at O.A spring with force constant k is connected to the other end. The pole is displaced bya small angle q0 from equilibrium position and released such that it performs small oscillations. Then x =0 M O 2L (a) 0 M 3k w = (b) 0 k 3M w = (c) 0 3k M w = (d) 0 k 2M w = 40. A particle ofmass is executing oscillations about the origin on the x-axis. Its potential energyisV(x) = k | x |3, wherek is a positive constant. Ifthe amplitude ofoscillation is a, then its time period T is (a) proportional to 1 a (b) proportional to a (c) independent 3 2 a (d) None of these 41. A circular hoop of radius R is hung over a knife edge. The period of oscillation is equal to that of a simple pendulum of length (a) R (b) 2R (c) 3R (d) 3R 2 42. In the figure shown, the spring is light and has a force constant k. The pulley is light and smooth and the strring is light . The suspended block has a mass m. On giving a slight displacement vartically to the block in the downward direction from its equilibrium position the block executes S.H.M. on being released with time period T. Then (a) m T 2 k = p (b) m T 2 2k = p m (c) 2m T 2 k = p (d) m T 4 k = p
- 365. 361 Oscillations 43. A bodyof mass m falls from a height h onto a pan (of negligible mass) of a spring balance as shown. The spring also possesses negligible mass and has spring constant k. Just after striking thepan, thebodystarts socillatorymotion in vertical directioin of amplitudeAand energyE. Then (a) mg A k = (b) mg 2kh A 1 k mg = + m h (c) 2 1 E mgh kA 2 = + (d) 2 2mg E mgh 2k æ ö = +ç ÷ è ø 44. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency w. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time (a) at the mean position of the platform (b) for an amplitude of 2 g w (c) for an amplitude of 2 2 g w (d) at the highest position of the platform 45. A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10–3 Joule. Obtain theequation ofmotion ofthis particle if this initial phase of oscillation is 45º. (a) y 0.1sin 4t 4 p æ ö = ± + ç ÷ è ø (b) y 0.2sin 4t 4 p æ ö = ± + ç ÷ è ø (c) y 0.1sin 2t 4 p æ ö = ± + ç ÷ è ø (d) y 0.2sin 2t 4 p æ ö = ± + ç ÷ è ø 46. A bodyof mass 0.01 kg executes simple harmonic motion about x = 0 under the influence of a force as shown in figure. The time period of S.H.M. is 80 –80 0.2 –0.2 F(N) x(m) (a) 1.05s (b) 0.52s (c) 0.25s (d) 0.03s 47. A forced oscillator is acted upon by a force F = F0 sin wt. The amplitude of oscillation is given by 2 55 2 36 9 w - w + . The resonant angular frequency is (a) 2 unit (b) 9 unit (c) 18 unit (d) 36 unit 48. Astraight rod ofnegligiblemass is mountedon a frictionless pivot and masses 2.5 kg and 1 kg are suspended at distances 40 cm and 100 cm respectivelyfrom the pivot as shown. The rod is held at an angle q with the horizontal and released. Then ///////////////////////////////////////////////////////////// q 2.5kg 1 kg 40cm 100cm (a) the rod executes periodic motion about horizontal position after the release (b) the rod remains stationary after the release. (c) the rod comes to rest in vertical position with 2.5 kg mass at the lowest point (d) the rod executes periodic motion about vertical position after the release DIRECTIONS for Qs. (49-50) : Each question contains STATEMENT-1andSTATEMENT-2.Choosethecorrectanswer (ONLYONE option is correct ) from the following- (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 istrue,Statement-2 is true; Statement -2 isnot a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 49. Statement 1 : Thegraph between velocityanddisplacement for a harmonic oscillator is an ellipse. Statement -2 : Velocity does not change uniformly with displacement in harmonic motion. 50. Statement -1 : If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Statement -2 : The total energy is directlyproportional to the square of the amplitude of vibration of the harmonic oscillator.
- 366. 362 PHYSICS Exemplar Questions 1.The displacement of a particle is represented by the equation 3cos 2 . 4 y t p æ ö = - w ç ÷ è ø The motion of the particle is (a) simple harmonic with period 2p/w (b) simple harmonic with period p/w (c) periodicbut not simpleharmonic (d) non-periodic 2. The displacement of a particle is represented by the equation y = sin3 wt. The motion is (a) non-periodic (b) periodicbut not simpleharmonic (c) simple harmonic with period 2p/w (d) simple harmonic with period p/w 3. The relation between acceleration and displacement offour particles are given below (a) ax = +2x (b) ax = +2x2 (c) ax = –2x2 (d) ax = –2x Which, one of the particle is exempting simple harmonic motion? 4. Motion of an oscillating liquid column in a U-tube is (a) periodicbut not simpleharmonic (b) non-periodic (c) simple harmonic and time period is independent ofthe density of the liquid (d) simpleharmonicand timeperiod isdirectlyproportional to the density of the liquid 5. Aparticle isacted simultaneouslybymutuallyperpendicular simple harmonic motion x = a cos wt and y = a sin wt. The trajectoryof motion of the particle will be (a) an ellipse (b) a parabola (c) a circle (d) a straight line 6. The displacement of a particle varies with time according to the relation sin cos . y a t b t = w + w (a) The motion is oscillatory but not SHM (b) The motion is SHM with amplitude a + b (c) The motion is SHM with amplitude a2 + b2 (d) The motion is SHM with amplitude 2 2 a b + 7. Four pendulums A, B, C and D are suspended from the same G G D C B A elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. IfA is given a transverse displacement, (a) Dwill vibrate with maximum amplitude (b) C will vibrate with maximum amplitude (c) B will vibrate with maximum amplitude (d) All the four will oscillate with equal amplitude 8. Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection ofthe radius vector of the rotating particle P is (a) 2 ( ) sin 30 t x t B p æ ö = ç ÷ è ø y x B p t T ( = 0) = 30s (b) ( ) cos 15 t x t B p æ ö = ç ÷ è ø (c) ( ) sin 15 2 t x t B p p æ ö = + ç ÷ è ø (d) ( ) cos 15 2 t x t B p p æ ö = + ç ÷ è ø 9. The equation of motion of a particle is x = a cos(at)2. The motion is (a) periodic but not oscillatory (b) periodic and oscillatory (c) oscillatory but not periodic (d) neither periodic nor oscillatory 10. A particle executing SHMmaximum speed of30 cm/s and a maximum acceleration of 60cm/s2. Theperiod ofoscillation is (a) p sec (b) 2 p sec (c) 2p sec (d) t p sec PastYears (2013-2017) NEET/AIPMTQuestions 11. A particle of mass m oscillates along x-axis according to equation x = a sin wt. The nature of the graph between momentum and displacement ofthe particle is [NEET Kar. 2013] (a) straight line passing through origin (b) circle (c) hyperbola (d) ellipse
- 367. 363 Oscillations 12. The oscillationof a body on a smooth horizontal surface is represented by the equation, X = A cos (wt) where X = displacement at time t w = frequency of oscillation Which one of the following graphs shows correctly the variation of‘a’ with ‘t’? [2014] (a) a O T t (b) a O T t (c) a O T t (d) a O T t 13. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is [2015] (a) 2 2 2 1 2 2 1 2 x – x 2 V – V p (b) 2 2 1 2 2 2 1 2 V V 2 x x + p + (c) 2 2 1 2 2 2 1 2 V – V 2 x – x p (d) 2 2 1 2 2 2 1 2 x – x 2 V – V p 14. When two displacements represented by y1 = asin(wt) and y2 = b cos(wt) are superimposed the motion is: [2015] (a) simple harmonic with amplitude a b (b) simple harmonic with amplitude 2 2 a b + (c) simple harmonic with amplitude (a b) 2 + (d) not a simpleharmonic 15. Aparticleisexecutingasimpleharmonicmotion.Itsmaximum acceleration is a and maximum velocityis b. Then its time period of vibration will be : [2015 RS] (a) a b (b) 2 b a (c) 2pb a (d) 2 2 b a 16. A particleexecutes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocityis equal to that ofits acceleration. Then its time period in seconds is [2017] (a) 5 2p (b) 4 5 p (c) 2 3 p (d) 5 p 17. A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. Theyare connected in series and the new force constant is k'. Then theyare connected in parallel and force constant is k¢¢ . Then k' : k¢¢ is [2017] (a) 1 :9 (b) 1 : 11 (c) 1 : 14 (d) 1 : 6
- 368. 364 PHYSICS EXERCISE -1 1. (b) 2. (a) 3. (b) 4. (d) 5. (b) 6. (d) 7. (b) 8. (b) 9. (d) Here effective spring factor, keff = 2 k ; Frequency of oscillation, m k 2 2 1 f p = ; when one spring is removed, then spring factor = k ; New frequency of oscillation 2 f m k 2 1 f = p = ¢ 10. (c) max v a ω = ; max max v 2a 2 v 2v = w = = ¢ max (sincev v) = 12. (b) displacement T 2 π acceleration = = b / 2 x b x 2 p = p 11. (a) P.E. of bodyin S.H.M. at an instant, 2 2 2 1 1 U m y ky 2 2 = w = If the displacement, y= (a – x) then 2 2 1 1 U k(a x) k(x a) 2 2 = - = - 13. (c) P.E. changes from zero to maximum twice in each vibration so its time period is T/2 14. (a) Velocity, ) 4 / t ( sin A dt dx v p + w w - = = Velocitywill be maximum, when wt + p/4 = p/2 or wt = p/2 – p/4 = p/4 or t = p/4w 15. (d) Tension is maximum at the mean position. 16. (a) A rectangular block of mass m and area of cross- section A floats in a liquid of density r. If it is given a small vertical displacement from equilibrium. It undergoes oscillations with a time period T then : T µ m. 17. (c) 2 2 2 2 y y, 1, 1 T 4p = w w = = && or 2 2 4 T p = or T = 2p second. 18. (a) At displacement ± a, the velocity is zero. At zero displacement, velocityis maximum. 19. (a) The amplitude is a maximum displacement from the mean position. 20. (b) 21. (a) a = – kX, X = x + a. In simple harmonic motion acceleration is directly proportional to the displacement from the mean position. Also the acceleration is in the opposite direction of displacement. 22. (c) Only functions given in (A) & (C) represent SHM. 23. (d) 2 2 kA 2 1 E A m k m 2 1 E = Þ = Þ E does not depend on m 24. (a) x = a sin2 wt = 2 a (1 –cos 2wt) dx dt = 2 a 2 sin 2 t w w Þ 2 2 d x dt = 2 4 cos2 2 a t w × w This represents an S. H. M. of frequency = w p 25. (d) T = 2 p m K 1 2 T T = 1 2 M M T2 = T1 2 1 M M = T1 2M M T2 = T1 2 = 2 T (where T1 =T) EXERCISE - 2 1. (a) 2 2 y a v - w = At x= 0, v 2 2 a o a. = w - = w At x = 2 2 2 a a 3a ,v' a 2 2 4 æ ö = w - = w ç ÷ è ø v' 3 v 2 = or v' 3 a 2 = w 3 a 3 a a 2 T p p = w = 2 T p æ ö w = ç ÷ è ø Q 2. (c) w = = a 100 vmax ; w = 100/a = 100/10 = 10 rad/s v2 = w2 (a2 – y2) or 502 = 102 (102 – y2) or 25 = 100 – y2 or 3 5 75 y = = cm. Hints & Solutions
- 369. 365 Oscillations 3. (b) k M 2 T p = Mm T 2 k + = p ¢ k M Þ 5 M M m 2 2 3 k k + p = p Þ 25 M M m 9 k k + = Þ 9 16 M m = 4. (d) 6 R 64 10 T 2 2 g 9.8 ´ = p = p = 2 7 10 8 7 22 2 3 ´ ´ ´ ´ = 60 49 1000 8 22 2 ´ ´ ´ ´ min=84.6min 5. (c) Max. force = mass× max. acceleration =m4p2 n2 a = 1 × 4 ×p2 × (60)2 × 0.02 = 288 p2 6. (a) g 2 T l p = ; ) 6 / g ( 2 g 2 2 l l ¢ p = p = Time period will remain constant if on moon, l' = l/6 = 1/6 m 7. (a) Kinetic energy, ( ) 2 2 2 2 x A m 2 1 mv 2 1 E . K - w = = Potential energy, 2 2 2 1 1 P.E kx m x 2 2 = = w putting, E . P E . K = ( ) 2 2 2 2 2 x m 2 1 x A m 2 1 w = - w 2 2 x 2 A = Þ . cm 2 2 2 A x = = 8. (a) Comparing A+ w2x=0 p = w Þ p = w 2 4 2 2 , Hz 1 , 2 2 = n p = pn 9. (b) ÷ ø ö ç è æ f + p = 2 t sin 2 y velocity of particle ÷ ø ö ç è æ f + p p ´ = 2 t cos 2 2 dt dy acceleration ÷ ø ö ç è æ f + p p - = 2 t sin 2 dt y d 2 2 Thus max a = 2 2 p 10. (a) 1 1 t , t' 9.8 12.8 µ µ ( ) g' 9.8 3 12.8 = + = Q t' 9.8 9.8 t' t t 12.8 12.8 = Þ = 11. (b) Equation of SHM is given by x =Asin (wt + d) (wt + d) is called phase. When x = A 2 , then sin (wt + d) = 1 2 t 6 p Þ w + d = or 1 6 p f = For second particle, A A x = A/2 2 1 x=0 2 5 6 6 p p f = p - = 2 1 f = f - f = 4 2 6 3 p p = 12. (c) When the bob moves from maximum angular displacement q to mean position, then the loss of gravitational potential energy is mgh where ) cos 1 ( h q - = l 13. (d) k 100 2 ' T , k 400 2 T p = p = T ' 100 1 T 2 T' 1 s T 400 2 2 2 = = Þ = = = 14. (b) The equivalent situation is a series combination of two springs of spring constants k and 2k. If k' is the equivalent spring constant, then 3 k 2 k 3 ) k 2 )( k ( ' k = = k 2 m 3 2 T p = Þ 15. (b) Phase change p in 50 oscillations. Phase change 2p in 100 oscillations. Sofrequency different ~ 1 in 100.
- 370. 366 PHYSICS 16. (a)) 2 ( T D 1 ' , T 2 ) D 3 ( 1 r p = n pr = n l l 3 2 2 3 2 ' = ´ = n n 17. (b) 18. (c) Kinetic energy 2 2 2 1 ( ) 2 K m a y w = - 2 2 2 2 1 2 10 [10 5 ] 375 2 2 æ ö = ´ ´ - = ç ÷ è ø erg p p 19. (b) 2 2 1 2 1 2 2 2 1 1 , 2 2 = Þ = = Þ = E E E kx x E ky y k k and 2 2 ( ) 2 1 = + Þ + = E E k x y x y k 1 2 1 2 2 2 2 E E E E E E k k k Þ + = Þ + = 20. (c) Time to complete 1/4th oscillation is 4 T s. Time to complete 8 1 th vibration from extreme position is obtained from t T 2 cos a t cos a 2 a y p = w = = or s 6 T t = Sotime to complete 3/8th oscillation = 12 T 5 6 T 4 T = + 21. (b) ) t ( sin a y f + w = ; when y = a/2, then ) t ( sin a 2 a f + w = or 6 sin 2 1 ) t ( sin p = = f + w or 6 5 sin p So phase of two particles is p/6 and 5 p/6 radians Hence phase difference = (5 p/6) – p/6 = 2 p/3 22. (a) sin wt = x/a and 2 2 2 a / x 1 t sin 1 t cos - = w - = w y= sin (wt + p/4) = sin wt cos p/4 + cos wt sin p/4 = 2 1 a / x 1 2 1 a x 2 2 ´ ÷ ø ö ç è æ - + ´ or 2 2 x a x y 2 - + = or 2 2 2 2 2 2 x a x 2 ) x a ( x y 2 - + - + = = 2 2 2 x a x 2 a - + It is an equation of an ellipse. 23. (b) Let k be the force constant of spring of length l2. Since l1 = n l2, where n is an integer, so the spring is made of (n + 1) equal parts in length each of length l2. k ) 1 n ( K 1 + = or k = (n + 1) K The spring of length l1 (= n l2) will be equivalent to n springs connected in series where spring constant n / K ) 1 n ( n k k + = = ¢ & spring constant of length l2 isK(n+1). 24. (c) Here all the three springs are connected in parallel to mass m. Hence equivalent spring constant k = K + K+ 2 K= 4 K. 25. (d) The second pendulum placed in a space laboratory orbiting around the earth is in a weightlessness state. Hence g = 0 so T = ¥ 26. (a) When lift is falling freely, the effective acceleration due to gravity inside the lift is zero i.e. g' = g – g = 0. Therefore time period will be infinityand frequencyis zero 27. (b) Let T1, T2 be the time period of shorter length and longer length pendulums respectively.As per question, n T1 = (n – 1) T2 ; so g 20 2 ) 1 n ( g 5 . 0 2 n p - = p or 6 ) 1 n ( 40 ) 1 n ( n - » - = Hence n = 6/5 » 1 28. (a) g 2 T l p = ; 1/ 2 T 1 2 ( / g) / T 2 - D = p ´ ´ D l l l ) g 2 T T ( l l Q D + p = D + q D a = D = D 2 1 2 1 T T l l = 5 6 10 12 ) 20 40 ( 10 12 2 1 - - ´ = - ´ ´ ´ 5 10 12 T T - ´ ´ = D = 24 ×60× 60× 12 ×10–5 = 10.3 s/day 29. (b) 1 1 k m 2 t p = or 1 2 2 1 k m 4 t p = or 2 1 2 1 t m 4 k p = Similarly, 2 2 2 2 t m 4 k p = and 2 0 2 2 1 t m 4 ) k k ( p = + 2 2 2 2 1 2 2 0 2 t m 4 t m 4 t m 4 p + p = p or 2 2 2 1 2 0 t 1 t 1 t 1 + = 30. (d) It is a damped oscillation, where amplitude of oscillation at time t is given by γ t 0 A a e- = where a0 = initial amplitude of oscillation g = damping constant.
- 371. 367 Oscillations As per question,γ 100 / ν 0 0 a a e 3 - = ……(i) (where n is the frequency of oscillation) and γ 200 / ν 0 A a e- = ……(ii) From (i) ; γ 100/ ν o 0 a a e 3 - ´ = ……(iii) Dividing equation (ii) by(iii), wehave γ 200 / ν γ 100 / ν 0 A e a (1/ 3) e - - ´ = = γ 100 / ν 1 e 3 - ´ = or 0 0 a 9 1 3 1 3 1 a A = ´ ´ = 31. (b) Let x = the increase in the length of spring. Then the particle moves along a circular path of radius (l + x), and the spring force = kx = centripetal force mw2 (l+ x) =kx or 2 2 m k m x w - w = l 32. (a) k k 2 x Let mass is displaced towards left by x then force on mass = – kx –2kx = – 3kx [negative sign is taken because force is opposite to the direction of motion] F 3kx Þ = - x m 2 w - = 3k . m Þ w = 1 3k f . 2 2 m w = = p p Thus it is propotional to m / k 3 33. (a) Maximum velocity can be found out by energy conservation. 2 mv 2 1 mgh = gh 2 v or max = ; 2 gh ' v 2 vmax = = ' mgh ) ' v ( m 2 1 mgh 2 + = ; ' mgh 2 mgh 2 1 mgh + = Þ ' mgh 4 mgh 3 = 4 h 3 ' h or = 34. (b) Loss in PE of mass = gain in PE of the spring mg(h + x) = 2 kx 2 1 2 kx – 2mgx – 2mgh = 0 Þ x= k 2 ) mgh 2 ( k 4 g m 4 mg 2 2 2 - - ± or x= k mgh 2 k g m k mg 2 2 2 + ± x = A– k mg = k mg k mgh 2 k g m 2 2 2 - + = k mg ú ú û ù ê ê ë é - + 1 mg kh 2 1 35. (a) t = 0, v maximum. The motion begins from mean position. So it represents S.H.M. 36. (a) The both of the pendulum will more along a straight lineAB. q q Re mg x A B The direction ofthe Earth's gravitational field is radial Now, mg R m GM F 2 e e = = x R m GM R x F cos F F 3 e e e x - = - = q - = =–kx where e 3 e GM m k R = Time period ofa simple harmonic oscillator, 3 e e R / m GM m 2 k m 2 T p = p = or g R 2 R GM R 2 T e 2 e e e p = p = 37. (a) For the block is about to slip, a g 2 w = m or g a m Þ w = g 1 g 2 ν ν a 2 a m m Þ p = Þ = p 38. (d) Oscillations along spring length are independent of gravitation.
- 372. 368 PHYSICS 39. (c) 2 2 1 4 I M(2L) ML 3 3 = = Force applied by the spring is F = – kx Þ F = – k (2Lq) (q is the angular displacement from the equilibrium position). Further 2 2 | F | 4L ksin 4L k t = ´ = q = - q r r l Also, .. 2 I I 4L k t = a = q = - q Þ .. 3k 0 M q+ q = Þ 0 3k M w = 40. (a) V(x)=k |x|3 since, 2 dV(x) F 3k | x | dx = - = - ……(i) x = a sin (wt) This equation always fits to the differential equation 2 2 2 d x x dt = -w or 2 2 d x m m x dt = - w Þ F = – mw2x ……(ii) Equations (i) and (ii) give 2 2 3k | x | m x - = - w Þ 1/ 2 3kx 3ka [sin( t)] m m w = = w Þ a wµ Þ 1 T a µ 41. (b) 2R T 2 2 g g = p = p l 42. (d) Let the extension in the spring be x0 at equilibrium. If F0 be the tension in the string then F0 = kx0. Further if T0 is the tension in the thread then T0 = mg and 2T0 =kx0. Let the mass m be displaced through a slight displacement x downwards. Let the the new tension in the string and spring be T and F respectively. Þ 0 x F x 2 æ ö = ç ÷ è ø and F = 2T Þ 0 x 2T k x 2 æ ö = ç ÷ è ø Þ 0 kx (T T ) 4 - = Þ Time period m m 2 4 k k 4 = p = p 43. (b) 2 1 mg(h y) ky 2 + = Þ mg mg 2kh y 1 k k mg = ± + At equilibrium mg = ky0 Þ 0 mg y k = Þ Amplitude A = y – y0 mg 2kh 1 k mg = + Energyof oscillation is 2 2 1 mg E kA mgh 2 2k æ ö = = + ç ÷ è ø 44. (b) For block Ato move in S.H.M. mg x N A mean position mg–N =mw2x where x is the distance from mean position For block to leave contact N = 0 2 2 g x x m mg w = Þ w = Þ 45. (a) The displacement of a particle in S.H.M. is given by y= a sin (wt + f) velocity= dy dt = wa cos (wt + f) The velocity is maximum when the particle passes through the mean position i.e., max dy dt æ ö ç ÷ è ø = w a
- 373. 369 Oscillations The kinetic energyat this instant is given by 1 2 m 2 max dy dt æ ö ç ÷ è ø = 1 2 mw2 a2 = 8 × 10–3 joule or 1 2 × (0.1) w2 × (0.1)2 = 8 × 10–3 Solving we get w= ±4 Substituting the values of a, w and f in the equation of S.H.M., we get y= 0.1 sin (± 4t + p /4) metre. 46. (d) Slope of F - x curve = – k = 80 0.2 - Þ k = 400 N/m, Timeperiod, T = 2p m k = 0.0314 sec. 47. (b) At resonance, amplitude ofoscillation is maximum Þ 2w2 –36w+ 9 isminimum Þ 4w– 36 = 0 (derivative is zero) Þ w= 9 48. (b) Torque about hinge 2.5 g × 0.40 cos q – 1g × 1 cos q = 0 49. (c) 50. (b) EXERCISE - 3 Exemplar Questions 1. (b) As given that, 3cos 2 4 y t p æ ö = - w ç ÷ è ø Velocityof the particle 3cos 2 4 dy d v t dt dt é p ù æ ö = = - w ç ÷ ê ú è ø ë û 3( 2 ) sin 2 4 t é p ù æ ö = - w - - w ç ÷ ê ú è ø ë û 6 sin 2 4 t p æ ö = w - w ç ÷ è ø So, acceleration, 6 sin 2 4 dv d a t dt dt é p ù æ ö = = w - w ç ÷ ê ú è ø ë û (6 ) ( 2 )cos 2 4 t p æ ö = w ´ - w - w ç ÷ è ø 2 4 3cos 2 4 t é p ù æ ö = - w - w ç ÷ ê ú è ø ë û 2 4 a y = - w In simple harmonic motion acceleration (or force) is directly proportional to the negative of displacement ofparticle Þ as acceleration, a y µ - Hence, due to negativesign motion is simple harmonic motion (SHM.) Asimpleharmonic motion isalwaysperiodic. Somotion is periodic simpleharmonic. From the given equation, 3cos 2 4 y t p æ ö = - w ç ÷ è ø Compare it by standard equation cos( ) y a t = w + f So, ' 2 w = w 2 2 2 ' ' 2 T T p p p = w Þ = = w w Hence, the motion is SHM with period p w . 2. (b) A motion will be harmonic if a µ displacement and a simple harmonic motion is always periodic but all simple harmonic motion are periodic but all periodic are not harmonic. As given equation of motion is 3 sin y t = w (3sin 4sin3 )/ 4 t t = w - w 3 [ sin3 3sin 4sin ] q = q- q Q Differentiating both side w.r.t. t So, (3sin ) (4sin3 ) / 4 dy d d v t t dt dt dt é ù = = w - w ê ú ë û or 3 (3sin sin3 ) sin 4 q - q é ù q = ê ú ë û 4 3 cos 4 [3 cos3 ] dy t t dt = w w - ´ w w Again, differentiating both side w.r.t. t 2 2 2 4 3 sin 36 sin3 d y t t dt ´ = - w w + w w 2 2 2 3 sin 36 sin3 4 d y t t a dt - w w + w w = = So, 2 2 d y dt æ ö ç ÷ ç ÷ è ø is not directly proportional to y. Somotion is not harmonic. 3 ( ) sin , y t t = w 3 ( ) sin ( ) y t T t T + = w + 3 2 sin ( ) t T T p é ù = + ê ú ë û 3 3 sin (2 ) sin t t = p + w = w
- 374. 370 PHYSICS 3. (d)For motion to be SHM acceleration of the particle must be opposite of restoring force and proportional to negative of displacement. So, F = ma = m(–2x) i.e., F = –2mx, so F x µ - Hence, ax = –2x We should be clear that x has to be linear. 4. (c) Consider a U-tube filled in which a liquid column oscillates. When liquid column lifted upto height y from A toB in arm Q. The liquidlevel in arm Pbecomes at C', so the difference between the height of two columns are = AB +A'C = y + y = 2y. In this case, restoring force acts on the liquid due to gravity. Acceleration of the liquid column, can be calculated in terms of restoring force. P Q B A C B' A' C' h y Equilibrium level Restoring force, f = Weight of liquid column of height 2y ( 2 ) 2 f A y g A gy = - ´ ´r ´ = - r [ ] v h =r Q As restoring force at A opposite to gravitational force as liquid is lifted against ( ) 2 t mg k A g w m = r Then f y µ - so motion is SHM. Timeperiod (2 ) 2 2 2 2 m A h h T k A g g ´ ´r = p = p = p r 2 h T g = p Time period is independent ofthe density of the liquid and motion is harmonic. 5. (c) The resultant-displacement can be find by adding x and y-components. According to variation of x and y, trajectory will be predicted, so resultant displacement is y' = (x + y) As given that, cos x a t = w ...(i) sin y a t = w ...(ii) So, ' ( cos sin ) y a t a t = w + w ' (cos sin ) y a t t Þ = w + w cos sin ' 2 2 2 t t y a w w é ù = + ê ú ë û ' 2(cos cos 45 sin sin 45 ) y a t t = w °+ w ° 2 cos( 45 ) a t = w - ° So, the displacement is not straight line and not parabola also. Now, squaring and adding eqs. (i) and (ii), 2 2 2 x y a + = 2 2 [ cos sin 1] t t w + w = Q This is the equation of a circle, so motion in circular (independent of time). Clearly, the locus is a circle of constant radius a. 6. (d) As given that, the displacement is sin cos y a t b t = w + w Let sin a A = f ...(i) and cos b A = f ...(ii) Squaring and adding (i) and (ii) 2 2 2 2 2 2 sin cos a b A A + = f+ f 2 2 A a b = + (amplitude) sin sin cos cos y A t A t = f× w + f w cos( ) y A t = w - f sin( ) dy A t dt = - w w - f 2 2 2 2 cos( ) d y A t y dt = - w w - f = -w Thus, 2 2 ( ) d y y dt µ - so, motion is SHM. Hence, it is an equation of SHM with amplitude 2 2 . A a b = + 7. (b) When pendulum vibrate with transverse vibration then 2 l T g = p l = length of pendulum. G G Elastic support A B C D e Through the elastic rigid support the disturbance is transferred to all the pendulum A and C are having length and hence same frequency. They will be in resonance, because their time period of oscillation. So, a periodic force of period (T) produces resonance in AandCandtheywillvibratewith maximumamplitude as in resonance.
- 375. 371 Oscillations 8. (a) Asthe particle (P) is executing circular motion with radius B. Let particle P is at Q at instant any (t), foot of perpendicular on x axis is at R vector OQ makes Ðq , with its zero position not P displacement of particles for O to R. Consider angular velocity of the particle executing circular motion is w and when it is at Q makes and angle q as shown in the diagram. BR x y p t ( =0) Q q q r 90–q O Clearly, t q = w Now, we can write cos(90 – ) OR OQ = q ( ) OR X = Q sin sin x OQ OQ t = q = w sin r t = w [ ] OQ r = Q 2 2 sin sin 30 x B t B t T p p æ ö = = ç ÷ è ø 2 sin 30 x B t p æ ö = ç ÷ è ø Hence, this equation represents SHM. 9. (c) According to the question, 2 cos( ) x a t = a is a cosine function, so it is an oscillatory motion. Now, at, t = (t + T). The equation of motion of particle 2 ( ) cos[ ( )] x t T a t T + = a + 2 [ ( ) cos( ) ] x t a t = a Q 2 2 cos[ 2 ] ( ) a t T tT x t = a + a + a ¹ where, T is supposed as period of the function w(t). Hence, it is oscillatory but not periodic. 10. (a) Let us consider a equation of an SHM is represented by y = a sin wt cos dy v a t dt = = w w max ( ) 30 cm/sec v a = w = (given)...(i) Acceleration 2 ( ) sin = = - w w dv a a t dt 2 max 60 = w = a a (given)...(ii) Eqs. (i) and (ii), we get ( ) 60 (30) 60 a w w = Þ w = w= 2 rad/s 2 2 rad/s T p = sec = p T PastYears (2013-2017) NEET/AIPMTQuestions 11. (d) As 2 2 2 2 2 1 v y a a + = w This is the equation of ellipse. Hence the graph is an ellipse. P versus x graph is similar to V versus x graph. 12. (c) Displacement, x = Acos (wt) (given) Velocity, v = dx A sin ( t) dt = - w w Acceleration, a = 2 dv A cos t dt = - w w Hence graph (c) correctly dipicts the variation of a with t. 13. (a) As we know, for particle undergoing SHM, 2 2 V A – X = w 2 2 2 2 1 1 V (A – x ) =w 2 2 2 2 2 2 V (A – x ) =w Substracting we get, 2 2 2 2 1 2 1 2 2 2 V V x x + = + w w Þ 2 2 2 2 1 2 2 1 2 V – V x – x = w Þ 2 2 1 2 2 2 2 1 V – V w x – x = Þ 2 2 2 1 2 2 1 2 x – x T 2 V – V = p 14. (b) The two displacements equations are y1 = a sin(wt) and y2 = b cos(wt) = b sin t 2 p æ ö w + ç ÷ è ø yeq = y1 + y2 = a sinwt + b coswt = a sinwt + b sin t 2 p æ ö w + ç ÷ è ø Since the frequencies for both SHMs are same, resultant motion will be SHM. NowAeq = 2 2 a b 2ab cos 2 p + + b a 2 2 eq A a b Þ = +
- 376. 372 PHYSICS 15. (c)As, we know, in SHM Maximum acceleration ofthe particle, a =Aw2 Maximum velocity, b =Aw Þ w= a b Þ T = 2 2 p pb = w a 2 T p é ù w = ê ú ë û Q 16. (b) Given,AmplitudeA= 3 cm Which particle is at x = 2 cm According to question, magnitude of velocity = acceleration 2 2 2 A x x w - = w 2 2 2 (3) (2) 2 T p æ ö - = ç ÷ è ø 4 5 T p = Þ T = 4 5 p 17. (b) Let l be the complete length of the spring. Length when cut in ratio, 1 : 2 : 3 are , and 6 3 2 l l l Spring constant (k) 1 length ( ) µ l Spring constant for given segments k1 = 6k, k2 = 3k and k3 = 2k When they are connected in series 1 1 1 1 k ' 6k 3k 2k = + + Þ 1 6 k ' 6k = Force constant k' = k And when they are connected in parallel k"= 6k+ 3k + 2k Þ k" = 11k Then the ratios k ' 1 k" 11 = i.e., k' : k"= 1 : 11
- 377. (Compression) l C CC R R R l (Rarefaction) C Logitudinal waves propagate through medium with the help of compressions and rarefactions. Equation of a Harmonic Wave Harmonic waves are generated by sources that execute simple harmonicmotion. A harmonic wave travelling along the positive direction ofx-axis is represented by sin( ) y A t kx = w - sin 2 t x A T ì ü æ ö = p - í ý ç ÷ l è ø î þ ( ) 2 sin A vt x p ì ü = - í ý l î þ where, y = displacement ofthe particle ofthe medium at a location xat timet A = amplitude of the wave l = wavelength T = time period v = , ul wave velocityin the medium w = 2 , T p angular frequency k = 2 , p l angular wave number or propagation constant. If the wave is travelling along the negative direction of x-axisthen sin( ) y A t Kx = w + sin 2 . t x A T ì ü æ ö = p + í ý ç ÷ è ø l î þ WAVE MOTION Wave motion is a type of motion in which the disturbance travels from one point of the medium to another but the particles of the medium do not travel from one point to another. For the propagation of wave, medium must have inertia and elasticity. These two properties of medium decide the speed of wave. Thereare twotypes of waves 1. Mechanical waves: These waves require material medium for their propagation. For example : sound waves, waves in stretched string etc. 2. Non-mechanical waves or electromagnetic waves : These waves do not require any material medium for their propagation. For example : light waves, x-rays etc. Thereare twotypesof mechanical waves (i) Transverse waves : In the transverse wave, the particles of medium oscillate in a direction perpendicular to the direction of wave propagation. Waves in stretched string, waves on the water surface are transverse in nature. Transverse wave can travel onlyin solids and surface ofliquids. Transverse waves propagate in the form of crests and troughs. Wavelength Trough Wave motion l Crest All electromagnetic waves are transverse in nature. (ii) Longitudinal waves : In longitudinal waves particles of medium oscillate about their mean position along the direction of wave propagation. Sound waves in air are longitudinal. These waves can travel in solids, liquids and gases. 15 Waves
- 378. 374 PHYSICS (ii) Thespeedoftransverse waves onstretchedstring isgiven by T v = m where T is the tension in the string and m is the mass per unit length of the string. Speed of Longitudinal Waves : The speed of longitudinal waves in a medium of elasticity E and density r is given by E v = r For solids, E is replaced byYoung's modulus (Y) solid Y v = r For liquids and gases, E is replaced by bulk modulus of elasticity(B) liquid/gas B v = r The density of a solid is much larger than that of a gas but the elasticityis larger bya greater factor. vsolid > vliquid > vgas Speed of Sound in a Gas : Newton'sformula. P v = r where P is the atmospheric pressure and r is the density of air at STP. Laplace'scorrection P v g = r where g is the ratio of two specific heats Cp and Cv Power and Intensity of Wave Motion : If a wave is travelling in a stretched string, energy is transmitted along the string. Power of the wave is given by 2 2 1 2 P A v = mw where m is mass per unit length. Intensity is flow of energy per unit area of cross section of the string per unit time. Intensity 2 2 1 2 I A v = rw Principle of Superposition of Waves : If two or more waves arrive at a point simultaneously then the net displacement at that point is the algebraic sum of the displacement due to individual waves. y = y1 + y2 + ............... + yn. where y1, y2 .......... yn are the displacement due to individual waves and y is the resultant displacement. Differential equation of wave motion : 2 2 2 2 2 1 d y d y dx v dt = Relation between wave velocity and particle velocity : The equation of a plane progressive wave is sin( ) y A t kx = w - ...(i) The particle velocity p dy v A cos( t kx) dt = = w w - ...(ii) Slope of displacement curve or strain cos( ) dy Ak t kx dx = - w - ...(iii) Dividing eqn. (ii) by (iii), we get dy dt dy dx k w = - p v v since v, wave velocity dy dx k w æ ö Þ = = ç ÷ è ø - . p dy v v dx Þ = - i.e., Particle velocity = – wave velocity × strain. Particle velocity changes with the time but the wave velocity is constant in a medium. Relation between phase difference, path difference and time difference : • Phase difference of 2p radian is equivalent to a path difference l and a time difference of period T. • Phase difference = 2p l × path difference 2 x p f = ´ l 2 x l Þ = ´f p • Phase difference = 2 T p × time difference 2 t T p f = ´ 2 T t Þ = ´ f p • Time difference = T l × path difference T t x = ´ l x t T l Þ = ´ Speed of Transverse Waves : (i) The speed of transverse waves in solid is given by v h = r where h is the modulus of rigidity of the solid and r is the densityof material.
- 379. 375 Waves INTERFERENCE OF WAVES Whentwo waves of equal frequency and nearly equal amplitude travelling in same direction having same state of polarisation in medium superimpose, then intensity is different at different points. At some points intensityis large, whereas at other points it is nearlyzero. Consider two waves y1 = A1sin (wt – kx) and y2 =A2 sin (wt – kx + f) By principle of superposition y= y1 + y2 = Asin (wt – kx + d) where, A2 = A1 2 + A2 2 + 2A1A2 cos f, and 2 1 2 sin tan cos A A A f d = + f As intensity I µ A2 So, resultant intensity I = I1 + I2 + 2 1 2 cos I I f For constructive interference (maximumintensity) : Phase difference, f= 2np and path difference = nl where n = 0, 1, 2, 3, ... Þ Amax = A1 + A2 and Imax = I1 + I2 + 2 1 2 I I = 2 1 2 ( ) I I + For destructive interference (minimumintensity) : Phase difference, f= (2n + 1)p, and path difference = ( ) 2n 1 2 l - ; where n = 0, 1, 2, 3, ... Þ Amin = A1 – A2 and Imin = I1 + I2 – 2 1 2 I I = 2 1 2 ( – ) I I Results : (1) The ratio of maximum and minimum intensities in any interference wave form. 2 2 1 2 max 1 2 min 1 2 1 2 I I I A A I A A I I æ ö + æ ö + = = ç ÷ ç ÷ - è ø - è ø (2) Average intensity of interference in wave form : Iav = max min 2 I I + Put the value of Imax and Imin or Iav = I1 + I2 If A = A1 = A2 and I1 = I2 = I then Imax = 4I, Imin = 0 and Iav = 2I (3) Condition of maximum contrast in interference wave form A1 = A2 and I1 = I2 then Imax = 4I and Imin = 0 For perfect destructive interference we have a maximum contrast in interference wave form. ReflectionofWaves : A mechanical wave is reflected and refracted at a boundary separating two media according to the usual laws of reflection and refraction. When sound wave is reflected from a rigid boundary or denser medium, the wave suffers a phase reversal of p but the nature does not change i.e., on reflection the compression is reflected back as compression and rarefaction as rarefaction. When sound wave is reflected from an open boundary or rarer medium, there is no phase change but the nature of wave is changed i.e., on reflection, the compression is reflected back as rarefaction and rarefaction as compression. Keep in Memory (i) For a wave, v = f l (ii) The wave velocity of sound in air E Y RT v M r g = = = r r (iii) Particle velocity is given by dt dx vp = . It changes with time. The wave velocity is the velocity with which disturbances travel in the medium and is given by k vw w = . (iv) When a wave reflects from denser medium thephase change is p and when the wave reflects from rarer medium, the phase change is zero. (v) In a tuning fork, the waves produced in the prongs is transverse whereas in the stem is longitudinal. (vi) A medium in which the speed ofwave is independent ofthe frequency of the waves is called non-dispersive. For exampleair is anon-dispersivemedium for the soundwaves. (vii) Transverse waves can propagate in medium with shear modulus of elasticitye.g., solid whereas longitudinal waves need bulk modulus of elasticityhence can propagate in all media solid, liquid and gas. EnergyTransportedby aHormonicWaveAlong a String: Kinetic energyof a small element of length dx is 2 t y ) dx ( 2 1 dk ÷ ø ö ç è æ ¶ ¶ m = where m = mass per unit length 2 2 2 1 cos ( ) 2 dk A kx t dx é ù = m w - w ë û and potential energy stored 2 2 2 2 1 1 ( ) cos ( ) 2 2 y dU T dx A kx t dx x ¶ æ ö = = m w - w ç ÷ ¶ è ø Example1. The displacement y (in cm)produced by a simple harmonic wave is given by y = (10/p) sin (2000 pt – px /17). What will be the periodic time and maximum velocity of the particles in the medium? Solution : ÷ ø ö ç è æ p - p p = 17 x t 2000 sin 10 y or ÷ ø ö ç è æ - p p = 34 x t 1000 2 sin 10 y The standard equation of S.H.M. is, ú û ù ê ë é l - p = x T t 2 sin a y ;
- 380. 376 PHYSICS Bycomparison of the standard equation to above given equation, we get . sec 10 1000 1 T 3 - = = Particle velocity ÷ ø ö ç è æ p - p p ´ p = = 17 x t 2000 cos 2000 10 dt dy s / m 200 s / cm 000 , 20 dt dy max = = ÷ ø ö ç è æ Þ (as vmax = A Aw) Example2. A progressive wave of frequency 500 Hz is travelling with a velocity of 360 m/s. Howfar apart are two points 60o out of phase? Solution : We known that for a wave v = f l so l= v 360 f 500 = = 0.72 m Now as in a wave path difference is related to phase difference by the relation. Phase difference Df= 60o = (p/180) x 60 = (p/3) rad Path difference Dx= 2 l p (Df)= 0.72 2 3 p p = 0.12 m Example3. Determine the change in volume of 6 litres of alcohol if the pressure is decreased from 200 cm of Hg to 75 cm. [velocity of sound in alcohol is 1280 m/s, density of alcohol = 0.81 gm/cc, density of Hg = 13.6 gm/cc and g = 9.81 m/s2]. Solution : For propagation of sound in liquid v = ( ) B / r , i.e., B = v2r But by definition B= – V P V D D so –V P V D D = v2r, i.e., DV = 2 V( P) v -D r Here DP = H2rg – H1rg = (75 – 200) ´ 13.6 ´ 981 = –1.667 ´ 106 dyne/cm2 so DV = ( )( ) ( ) 3 6 2 5 6 10 1.667 10 0.81 1.280 10 ´ ´ ´ ´ = 0.75 cc Example4. (a) Speed of sound in air is 332 m/s at NTP. What will be the speed of sound in hydrogen at NTP if the density of hydrogen at NTP is (1/16) that of air? (b) Calculate the ratio of the speed of sound in neon to that in water vapour at any temperature. [Molecular weight of neon = 2.02 ´ 10–2 kg/mol and for water vapours = 1.8 ´ 10–2 kg/mol] Solution : The velocity of sound in air is given by v = E r = P g r = RT M g (a) In terms of density and pressure H air v v = air H H air P P r ´ r = air H r r [as Pair = PH] or vH = vair ´ air H r r = 332 ´ 16 1 = 1328 m/s (b) In terms of temperature and molecular weight Ne W v v = Ne W Ne W M M g ´ g [as TN = TW] Now as neon is monatomic (g = 5/3) while water vapours polyatomic (g = 4/3) so Ne W v v = ( ) ( ) 2 2 5 / 3 1.8 10 4 / 3 2.02 10 - - ´ ´ ´ ´ = 5 1.8 4 2.02 ´ = 1.05 BEATS When two wave trains slightly differing in frequencies travel along the same straight line in the same direction, then the resultant amplitude is alternately maximum and minimum at a point in the medium. This phenomenon of waxing and waning of sound is called beats. Let two sound waves of frequencies n1 and n2 are propagating simultaneouslyand in same direction. Then at x=0 y1 = A sin 2p n1t, and y2 = A sin 2p n2t, For simplicity we take amplitude of both waves to be same. By principle of superposition, the resultant displacement at any instant is y = y1 + y2 = 2A cos 2p nAt sin 2p navt where 2 n n n 2 1 av + = , 2 n n n 2 1 A - = Þ y = Abeat sin 2p navt ..................(i) It is clear from the above expression (i) that
- 381. 377 Waves (i) Abeat =2A cos 2p nAt, amplitude of resultant wave varies periodically as frequency ÷ ø ö ç è æ - = 2 n n n 2 1 A Aismaximum when 1 | cos | max = q A 2 | A | max beat = Ais minimum when 0 | cos | min = q 0 | A | min beat = (ii) Since lntensity is proportional to amplitude i.e., 2 beat A Ia For Imax cos 2p nAt = ± 1 For Imin i.e., 2pnAt=0,p,2p 2pnAt = p/2, 3p/2 i.e., t = 0, 1/2nA, 2/2nA t = 1/4nA, 3/4nA....... So time interval between two consecutive beat is ÷ ÷ ø ö ç ç è æ = - = D - A 1 n n n 2 1 t t t Number of beats per sec is given by 1 2 1 2 2( ) 1 2 2 beat A n n n n n n t - = = - = - D So beat frequency is equal to the difference of frequency of two interferring waves. To hear beats, the number of beats per second should not be more than 10. (due to hearing capabilities of human beings) Filing/Loading a Tuning Fork On filing the prongs of tuning fork, raises its frequency and on loading it decreases the frequency. (i) When a tuning fork of frequency n produces Dn beats per second with a standard tuning fork of frequency n0, then n D ± n = n 0 Ifthe beat frequencydecreases or reduces to zero or remains the same on filling the unknown fork, then n D - n = n 0 (ii) Ifthe beat frequencydecreases or reduces to zero or remains the same on loading the unknown fork with a little wax, then n D + n = n 0 If the beat frequency increases on loading, then n D - n = n 0 Example5. Tuning fork A has frequency 1% greater than that of standard fork B, while tuning fork C has frequency 2% smaller than that of B. When A an C are sounded together, the number of beats heard per second is 5. What is the frequency of each fork? Solution : Let the frequencies of forks be n1, n2 and n3 respectively. Then n1 = n2 (1 + 0.01) = 1.01 n2 and n3 =n2 (1 - 0.02) = 0.98 n2 Further, n1 – n3 = 5 Substituting the values, we get (1.01 n2 – 0.98 n2) = 5 n2 =166.7Hz Now n1 =1.01× 166.7=168.3Hz and n3 = 0.98 ×166.7= 163.3Hz. Example6. Two tuning forks A and B sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are 24 cm and 25 cm respectively. Calculate the frequencies of forks. Solution : Let the frequency of the first fork be f1 and that of second be f2. We then have, 1 v f 4 24 = ´ and 2 v f 4 25 = ´ We also see that f1 > f2 f1 – f2= 6 …(i) and 1 2 f 24 f 25 = …(ii) Solving eqns. (i) and (ii), we get f1 = 150Hz and f2 = 144 Hz DOPPLER EFFECT When a source of sound and an observer or both are in motion relative to each other there is an apparent change in frequency of sound as heard by the observer. This phenomenon is called the Doppler's effect . Apparent change in frequency : (a) When source is in motion and observer at rest (i) when source moving towards observer 1 0 S V V V æ ö n = n ç ÷ - è ø (ii) when source moving away from observer 1 0 S V V V æ ö ¢ n = n ç ÷ + è ø Here V = velocity of sound VS = velocity of source n0 = source frequency. (b) When source is at rest and observer in motion (i) when observer moving towards source 0 2 0 V V V + æ ö n = n ç ÷ è ø (ii) when observer moving away from source and V0 = velocity of observer. 0 2 0 V V V - æ ö ¢ n = n ç ÷ è ø (c) Whensource and observer both are in motion (i) If source and observer both move away from each other. 0 3 0 s V V V V æ ö - n = n ç ÷ - è ø
- 382. 378 PHYSICS Solution : Inthis case, we can assume that both the source and observer are moving towards each other with velocityv. If c be the velocity of signal, then n v c v c n ÷ ø ö ç è æ - + = ¢ or n ) v c ( ) v c ( ) v c ( n 2 ÷ ÷ ø ö ç ç è æ - - + = ¢ or 2 2 2 2 2 2 (c v ) c n n n; (as c v) c v 2c v c 2cv - ¢ ¢ = Þ = >> + - - n v 2 c c n ÷ ÷ ø ö ç ç è æ - = ¢ STATIONARY OR STANDING WAVES When two progressive waves having the same amplitude, velocity and time period but travelling in opposite directions superimpose, then stationary wave is produced. Let two waves ofsame amplitude and frequencytravel in opposite direction at same speed, then y1 = Asin (wt –kx) and y2 = Asin (wt + kx) By principle of superposition y = y1 + y2 = (2A cos kx) sin wt ...(i) s y A sin ωt = It is clear that amplitude ofstationary waveAs varywith position (a) As = 0, when cos kx = 0 i.e., kx = p/2, 3p/2............ i.e.,x =l/4, 3l/4...................[ask=2p/l] These points are called nodes and spacing between two nodes is l/2. (b) As is maximum, when cos kx is max i.e., kx= 0,p,2p,3pi.e.,x=0,ll/2,2l/2.... It is clear that antinode (where As is maximum) are also equally spaced with spacing l/2. (c) The distance between node and antinode is l/4 (see figure) Antinode Antinode segment 1 segment 2 segment 3 Node x l l /2 /4 o 2A Keep in Memory 1. When a string vibrates in one segment, the sound produced is called fundamental note. The string is said to vibrate in fundamental mode. 2. The fundamental note is called first harmonic, and is given by l 2 v 0 = n , where v = speed of wave. 3. If the fundamental frequencybe 0 n then 2 0 n , 3 0 n , 4 0 n ... are respectively called second third, fourth ... harmonics respectively. 4. If an instrument produces notes of frequencies 4 3 2 1 , , , n n n n .... where 1 2 3 4 ν ν ν ν < < < ....., then 2 n is (ii) If source and observer both move towards each other. 0 3 0 s V V V V æ ö + ¢ n = n ç ÷ - è ø ; Whenthewindblows inthedirection of sound, then in all above formulaeVisreplacedby(V+ W) whereWis thevelocityofwind. Ifthe wind blowsin theoppositedirectiontosoundthenVisreplaced by(V– W). Keep in Memory 1. The motion of the listener causes change in number of waves receivedbythe listener andthis producesan apparent change in frequency. 2. The motion of the source of sound causes change in wavelength of the sound waves, which produces apparent change in frequency. 3. If a star goes away from the earth with velocity v, then the frequencyof the light emitted from it changes from n ton'. n' = n (1–v/c), where c is the velocity of light and or v v c c Dn Dl = = n l where l D is called Doppler’s shift. If wavelength of the observed waves decreases then the object from which the waves are coming is moving towards the listener and vice versa. Example7. Two engines cross each other travelling in opposite direction at 72 km/hour. One engine sounds a whistle of frequency 1088 cps. What are the frequencies as heard by an observer on the other engine before and after crossing. Take the speed of sound as 340 m/s. Solution : The apparent frequency before crossing, ÷ ÷ ø ö ç ç è æ - + = ¢ s 0 v v v v n n HereVo = 72 km/hour = 20 m/s vs = 72 km/hour = 20 m/s ÷ ø ö ç è æ - + ´ = ¢ 20 340 20 340 1088 n Hz 1224 320 360 1088 = ´ = The apparent frequency after crossing 0 s v v n'' n v v æ ö - = ç ÷ + è ø ÷ ø ö ç è æ + - ´ = 20 340 20 340 1088 Hz 11 . 967 36 32 1088 = ´ = Example8. A rocket is going towards moon with a speed v. The astronaut in the rocket sends signals of frequency n towards the moon and receives them back on reflection from the moon. What will be the frequency of the signal received by the astronaut? (Take v << c)
- 383. 379 Waves called first overtone,3 n is called second overtone, 4 n is called third overtone ... so on. 5. Harmonics are the integral multiples of the fundamental frequency.If n0 be the fundamental frequency, then 0 nn is the frequencyof nth harmonic. 6. Overtones are the notes of frequency higher than the fundamental frequencyactuallyproduced bythe instrument. 7. In the strings all harmonics are produced. StationaryWaves in an Organ Pipe : In the open organ pipe all the harmonics are produced. In an open organ pipe, the fundamental frequency or first harmonic is 0 2 v n = l , where v is velocity of sound and l is the length of air column [see fig. (a)] (a) l (b) l 2 l = l , 1 2l = l 2 2l = l , 2 2l = l (c) l 2 3l = l , 3 L 3 = l Similarly the frequency of second harmonic or first overtone is [see fig (b)], 01 2 2 v n = l Similarly the frequency of third harmonic and second overtone is, [(see fig. (c)] 02 3 2 v n = l Similarly 03 04 4 5 , 2 2 v v n n = = l l .................. In the closed organ pipeonlythe odd harmonics are produced. In a closed organ pipe, the fundamental frequency (or first harmonic) is (see fig. a) 4 c v n = l (a) l (b) l (c) l 4 l = l 4 3l = l 4 5l = l Similarly the frequencyof third harmonic or first overtone (IInd harmonic absent) is (see fig. b) l 4 v 3 n 2 c = Similarly l l 4 v 7 n , 4 v 5 n 4 c 3 c = = ........ End Correction It is observed that the antinode actually occurs a little above the open end. A correction is applied for this which is known as end correction and is denoted bye. (i) For closed organ pipe : l is replaced by l+ e where e = 0.3D, D is the diameter of the tube. (ii) For open organ pipe: l is replaced by l + 2e where e=0.3D In resonance tube, the velocity of sound in air given by ( ) 2 1 2ν v l l = - where n = frequency of tuning fork, ll = 1st resonating length, l2 = 2nd resonating length. ResonanceTube : It is used to determine velocity of sound in air with the help of a tuning fork of known frequency. n n A e l 1 e l 2 Let l1 and l2 are lengths of first and second resonances then 1 e 4 l + = l and 2 3 e 4 l + = l 2 1 2 l Þ - = l l 2 1 2( ) Þ l = - l l Speed of sound in air is v = ul where u is the frequency 2 1 2 ( ) v = u - l l
- 384. 380 PHYSICS For vibratingstrings/ open organ pipe. Mode of vibration First or Fundamental First Fundamental tone 2l Second Second First overtone Second overtone 3n 2n Third Third Harmonic Tone Frequency Wavelength Shape for string N N N A A N N A v 2l n = 2 3 l 2 2 l For closed organ pipe. Mode of vibration First or Fundamental First Fundamental tone 4l Second Third First overtone Second overtone 5n 3n Third fifth Harmonic Tone Frequency Wavelength Shape for string v 4l n = 4 3 l 4 5 l Comparision of Progressive (or travelling) and Stationary (orstanding) Wave: SI. Progressive Wave Stationary Wave 1. The wave advances with a constant speed The wave does not advance but remains confined in a particular region. 2. The amplitude is the same for all the particles in the path of the wave The amplitude varies according to position, being zero at the nodes and maximum at the antinodes. 3. All particles within one wavelength have different phases. Phase of all particles between two adjacent nodes is the same. Particles in adjacent segments of length p/2 have opposite phase. 4. Energy is transmitted in the direction of propagation of the wave Energy is associated with the wave, but there is no transfer of energy across any section of the medium.
- 385. 381 Waves ComparativeStudyofInterference, BeatsandStationaryWave: Interference Superposition oftwo waves of nearly same amplitude or same amplitude having a constant phase difference (or no phase difference) travelling in the same direction and with the same wavelength show the phenomenon of interference. Maxima and minima are fixed at their locations. Beats Superposition of two waves of same amplitude having zero phase difference moving in the same direction having small difference in frequency (less than 10 Hz) show the phenomenon of beats. Maxima and minima vary periodically with time at every location. Stationary wave Superposition of two waves of same amplitude having a constant phase difference (or no phase difference) moving in opposite direction having same frequency give rise to stationary waves. Maxima and minima are fixed at their locations. CHARACTERISTICS OF SOUND 1. A musical sound consists of quick, regular and periodic succession of compressions and rarefactions without a sudden change in amplitude. 2. A noise, consists ofslow, irregular and a periodic succession of compressions and rarefactions, that may have sudden changes in amplitude. 3. (i) Pitch, (ii) loudnessand (iii) quality arethecharacteristics of musical sound. 4. Pitch depends on frequency, loudness depends on intensity and, quality depends on the number and intensity of overtones. 5. The ratio of the frequencies of the two notes is called the interval between them. For example interval between two notes offrequencies 512 Hz and 1024 Hz is 1 : 2 (or 1/2). 6. Two notes are said to be in unison if their frequencies are equal, i.e., if the interval between them is 1 : 1. Some other common intervals, found useful in producing musical sound are the following: Octave (1 : 2), majortone (8 : 9), minortone (9 : 10) and semitone (15 : 16) 7. Major diatonic scale : It consists of eight n