1. The document discusses various concepts related to mechanics including dimensional analysis, units of measurement in SI system, Newton's laws of motion, kinematics equations, work, power, elasticity, and Bernoulli's principle. 2. Key points covered include defining dimension, dimensional formula, and dimensional equation with an example of velocity. SI base units for length, mass, time and other quantities are also listed. 3. Newton's three laws of motion and the concept of impulse are explained. Equations for displacement under uniform acceleration and work done by a couple are derived. Stress and strain concepts and Hooke's law are distinguished along with definitions of the three elastic moduli.

1. OPTC Edavanna Page 1
MODULE I
1. Differentiate between dimension, dimensional formula and dimensional equation
with a suitable example
Dimension of a physical quantity is powers to which the fundamental units of mass,
length and time must be raised to represent a derived unit of the quantity.
Dimensional formula is an expression which tells us which and how the physical
quantity depends on the fundamental units.
Dimensional equation is the equation obtained when a physical quantity is equated to its
Dimensional formula.
Eg: velocity:
Dimension is zero in mass (M), +1 in length (L) and -1 in time (T).
Dimensional formula is LT-1
Dimensional equation: V = [M°L1
T-1
]
2. Give the physical quantities and their units in which SI is based on ?
Quantity Unit
Length meter
Mass kg
Time S
Electric current ampere
Temperature K
Amount of substance mol
Luminous intensity cd
Plane angle rad
3. Write the advantages of SI system over other systems of unit?
 Universally accepted system
 Comprehensive
 Coherent system
 Requires no separate practical units
Orphanage Polytechnic College, Edavanna
REVISION PACKAGE
Engineering Physics-I ( Revision 2015)

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4. Write the limitations of dimensional analysis?
 Numerical constants, trigonometric ratios, angles etc appearing in a formula
cannot be deduced from dimensional method
 This fails if a physical quantity depends on more than 3 others quantities
5. Define a Newton. Write dimensional formula for force ?
Newton is the SI unit of force. One Newton is that force will cause a mass of 1 kg to
have an acceleration 1m/s2
.
Force = Mass X Acceleration.
Therefore dimensional formula for force is MLT-2
.
6. Write down the equation of motion of the body moving under gravity
V = u + at
S = ut + t2
V2
= U2
+ 2as
When a body is moving under gravity a = g
V = u + gt
S = ut + t2
V2
= U2
+ 2gs
7. Analyse the statement – Newton’s first law defines force and second law provides a
means to measure force
Newton’s first law states that a body continues to be in it state of rest or of uniform
motion along a straight line until and unless it is compelled by an external force. From the
statement , force can be defined as that which changes or tends to changes the state of
rest or that of uniform motion.
Second law tells what happens when a force is actually exerted on a body. When a force
acts on a body and there occurs a change in its momentum, then according to second law,
the rate of change of momentum is proportional to the applied force and in its direction

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8. When a body is thrown up, show that time of ascent is equal to the time of descent
Let a body be projected vertically up with a velocity u. Let time taken to reach the
maximum height (time of ascent) be t1.
At the highest point, velocity is zero using v = u+ at, we get
0 = u – gt1 or t1 = u/g --- (A)
Let h be the maximum height reached. V2
= U2
+ 2as
0 = U2
– 2gh
H = ------------- (1)
Let t2 be the time of ascent for downward travel, initial velocity is zero
S = ut + t2
h = 0 + t2
2
Substitute (1) in above equation
U2
/2g = t2
2
t2
2
= U2
/ g2
t2 = u/g ------ (B)
Comparing (A) and (B), t1 = t2
ie, time of ascent = time of descent
9. What is impulse? Derive equation for impulse.
Large force acting on a body for short time is called impulse (I).
Impulse, I = F.t
F = ma = m (v – u)/t
I =
= m (v – u) = mv - mu
Means impulse of a force is measured by the change it produces in momentum.

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10. State the law of conservation of momentum. Prove it in the case of collision of two
bodies moving in the same direction
Law of conservation of momentum states that when two or more bodies collide, the
sum of their momenta before impact is equal to the sum of momenta after
impact.
Momentum of m2 before Collision = m2u2
Momentum of m2 after Collision = m2v2
Changes of momentum in t seconds = m2v2 - m2u2
Rate of change of momentum m2 =
Change of momentum of first body in t seconds = m1v1=m1u1
Rate of change of momentum of the first body =
This rate of change of first body is the reaction. Since action and reaction are
equal and opposite.
= -
m1u1 + m2u2 = m1v1 + m2v2
i.e., total momentum before collision is equal to the total momentum after
collision.
11. State Newton’s laws of motion
First law: Everybody continues in its state of rest or of uniform motion along a straight
line unless it is compelled by an external force to change its state
Second law: The rate of change of momentum of a body is proportional to the applied
force and takes place in the direction in which the force acts.

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Third law: Every action has an equal and opposite reaction
12. Prove Newton’s second law of motion and derive expression for force
Consider a body of mass moving with a velocity u. When an unbalanced force F acts
on it for a time t, its velocity changes to v.
So, its initial momentum = mu
Final momentum = mv
Therefore changes in momentum = mv-mu
Rate of change of momentum =
According to the second law , the rate of change of momentum is proportional to force
F. Thus,
F ma F=k ma , k = 1
F = ma
13. A gun weighting 10 kg fires a bullet of 30g with a velocity of 330m/s. With what
velocity does the gun recoil?
Mass of gun, M=10 kg
Mass of bullet, m=30 kg
Velocity of bullet, v = 330 m/s
Recoil velocity of gun, V = = 9.90m/s
14. Derive the equation for the displacement of body during the nth
second of its motion
S1 = un + an2
S2 = u(n-1) + (n-1)2
Distance Sn covered in nth seconds, Sn = S1 – S2
= un + an2
– [ u(n-1) + (n-1)2
]
= u + an - a
= u + a(n -

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15. Explain the term ‘Elastic fatigue’.
The property of an elastic body due to which its behaviour becomes less elastic under the
action of repeated alternating deforming forces
16. A body covers 120m in the 4thsecond.If it travels 240m in 8s ,calculate its acceleration,
initial velocity and velocity at the end of 8th second
S = u + a(n -
120 = u + a(4 -
S = ut + at2
240 = 8u + x a x 82
The equations get simplified as 240 = 2u + 7a and 480 = 16u + 64a
Solving, we get a= -180 m/s2
, u = 750 m/s
By 8 seconds, the distance covered by the body be S=ut+ ½ at2
S = 750 x 8 + ½ (-180) x 82
= 240 m
Using v2 = u2 + 2as = 7502 – 2 x 180 x 240 = 476100
Therefore, v = 690 m/s
I.e. by the end of 8 seconds, the body will have a velocity 690 m/s
17. A body moving with uniform acceleration describes 10m in the 2nd second and 20m
in the 40th second of its motion. Calculate the distance moved by it in the 5th second
of its motion
S = u + a (n -
10 = u + a (2 -
20 = u +
20 = 2u+3a --------- (1)
20 = u + a (4 -
20 = u +

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40 = 2u+7a………….. (2)
Solving, a = 5m/s2
, u = 2.5 m/s
S5 = 2.5 + 5 (5 -1/2 ) = 25m
18. Find the initial velocity and acceleration of a particle travelling with a uniform
acceleration in a straight line, if it traverses 55m in the 8th second and 85m in the
13th second of its motion.
S = u + a (n -
55 = u + a (8 -
20 = u +
110 = 2u + 15a ------ (1)
85 = u + a (13 -
85 = u +
170 = 2u + 25a ------ (2)
Solving, a = 6 cm/s2
, u = 10 m/s
19. A stone is dropped into water from a bridge 44.1 m above the water level. Another
stone is thrown vertically downward one second later. Both stones reach water
surface simultaneously. Find the downward velocity of the second stone (g = 9.8
m/s2
)
Time taken for the first stone to fall freely from the bridge from 44.1 m height will be t,
S = ut + at2
44.1 = 0 + X 9.8 X t2
t = √ = 3 s

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20. A neutron having a mass of 1.67 x 10-27
kg and moving at 108 m/s collides with a
deuteron at rest and sticks to it . Calculate the speed of the combination (mass of
deutron=3.34*10-27kg)
m1u1 + m2u2 = m1v1 + m2v2
Here v1 = v2 = v , u2 = 0
Thus, m1u1+0=(m1+m2) v
Therefore v =
=
= 3.33x107m/s

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MODULE II
1. What is mean by concurrent forces?
Forces whose lines of action pass through a common point
2. Explain analytical method to find the resultant of two forces?
OC2
= OD2
+ CD2
R2
= (OA + AD)2
+ CD2
= OA2
+ 2.OA.AD + AD2
+ CD2
, But AD2
+ CD2
= AC2
R2
= OA2
+ 2.OA.AD + AC2
Since OA = P and AC = OB = Q
R2
= OA2
+ 2.P.AD + Q2
From right angled triangle ADC,
cos = AD/AC
AD = AC.cos = Q. cos
R2
= P2
+ 2.P.Q.cos + Q2
Resultant, R = √
tan =
From triangle ADC,
CD = Q.sin

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AD = Q.cos
tan =
=
3. What is mean by Lami’s theorem ?
If three forces acting on a body keep the body in equilibrium, each force is
proportional to the sine of the angle between the other two.
4. A body of mass 1 kg is suspended by two strings AC and BC of length 5 cm
and 12 cm respectively from two hooks separated by 13 cm. Calculate the
tension in the strings
Downward force,F = mg = 1 x 9.8 = 9.8 N
Applying Lami’s theorem, we get,

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T1 = 9.8 sinӨ
T2 = 9.8 cosӨ
But, sinӨ = 12/13 and cosӨ = 5/13
Substituting above, we get, T1 = 9.05 N and T2 = 3.77 N
5. Derive expression for work done by a couple.
FF – couple acting on a body
Moment of the couple C, C = F.AB
W = F.AA1 + F.BB1
Ө =
AA1 = AO. Ө
BB1 = BO. Ө
W = F. AO. Ө + F. BO. Ө
= F (AO + BO ). Ө
= F.AB. Ө
= C Ө
6. What is mean by power ?
Work done per second is called power. P = 2 NC

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MODULE III
1. Distinguish between stress and strain. Deduce the expression for young’s
modulus
Stress is the restoring force setup per unit area inside the body. It is measured by the
applied force per unit area.
Stress = F/A
Strain is the fractional deformation resulting from a stress. It is measured by the ratio
of change in dimension of a body to the original dimension.
Longitudinal strain =
young’s modulus is the ratio of longitudinal stress to the longitudinal strain
Y =
2. Write applications of Bernoulli’s principle
 Atomiser
 Air foil
3. Explain Hook’s law. Formulate the three elastic moduli.
Hooke’s law states that the strain produced in a body is directly proportional to the
stress provided the stress is not very large.
(1)Modulus of elasticity
= Constant.
This Constant is called Modulus of elasticity
Young’s modulus : It is ratio of the longitudinal stress to the longitudinal strain
Y = =
(2)Modulus of Rigidity or Rigidity modulus
Ratio of shearing stress to shearing strain

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Rigidity modulus, =
(3)Bulk Modulus
Bulk modulus comes into play when a body is subjected to a uniform normal
force distributed over the whole of its surface. If V is original volume and ⱴ the
change in volume, bulk strain is ⱴ/V.
K =
4. A cable is re placed by another of the same length and material but twice the
diameter. Analyze how it effects the elongation under a given load
Y = =
Since both cables have same length, load and material, but diameter differs,
the eg for elongation can be given as,
l1 = FL / r1
2
y)
l2 = FL / r2
2
y)
Since r2 = 2r, l1/l2 = (2r1)2
/r1
2
= 4
i.e. l1 = 4l2
Elongation becomes four times compared to the first cable
5. A brass wire of young’s modulus 90 GPa of length 3.14 and a diameter 0.4 mm is
stretched. Find the mass to be suspended to produce an elongation of 0.98 cm.
Y= 90 GPa = 90 X 109
Pa
L= 3.14 m
Diameter = 0.4 mm, Radius = 0.2 mm
L = 0.98 cm = 0.98 X 10-2
m.
Y =  F = = = [90 X 109
x 0.98 X 10-2
x (0.2x10-3)2
] / 3.14
F = mg = 35.298 N
Therefore mass to be suspended, m = 35.298/9.8 = 3.6 kg

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6. Explain principle of continuity?
 Relation representing conservation of mass
 For a steady, non-viscous, non-uniform flow, the product of the area of cross
section and the velocity of flow remains the same at every point in the tube
Volume of liquid flowing per second at P = a1V1
Volume of liquid flowing per second at Q = a2V2
Volume entering at P = volume leaving at Q
a1V1 = a2V2 = a constant
This expression is called equation of continuity.

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Module IV
1. Write examples of simple harmonic motion?
 Simple pendulum
 Loaded spring
 U tube
 Test tube floating in water
 Freely suspended magnet
2. Explain characteristics of simple harmonic motion
Period (T): time taken for one complete vibration
Frequency (f): number of vibrations executed per second. F =
Amplitude (a): maximum displacement of particle in simple harmonic motion
Phase: stage of vibration through which the particle is passing
3. Define wave length?
It is the linear distance between any two nearest particles of the medium which are in
the same state of vibration.
4. Derive equation for simple harmonic motion?
The projection of a uniform circular motion along a diameter of circle is called simple
harmonic motion.
y = asin t
Differentiating,
= - 2
y [y = asinwt ]
This is the differential equation for simple harmonic motion
5. Derive the relation v = f
T - time period
– distance travelled by the wave
Velocity , v = = (since f = 1/T )

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6. What is mean by resonance?
 A phenomenon occurring in vibratory bodies having the same natural
frequency
 At resonance the body vibrates with very large amplitude
7. What are applications of ultrasonic waves
 Sonar (echo sounding)
 Ultrasonic signalling
 Emulsification of immiscible liquids
 Metal testing
 Under water communication
 Metallurgical uses
 Medical and biological uses
 Ultrasonic cleaners
8. Explain second mode of vibration? What is first overtone?
 Two nodes and two anti-nodes
 Fundamental frequency, f =
 Frequency of vibration in this mode, f1 = ----- (1)
 Vibrating length, L = + +
L =
1=
Substitute value of 1 in (1), we get,
f1 = = 3 x = 3f
This frequency is called first overtone or third harmonic

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9. Calculate the velocity of sound using resonance column experiment.
 A tuning fork is sounded on upper end of tube
 Tube raised slowly until resonance occurs.
 At this point, frequency of tuning fork = frequency of air column
 Length of air column above water level is measured using a meter scale
( l1 + e ) = -------------- (1)
= 4 ( l1 + e )
e – end correction = 0.3d
 Velocity of sound in air, v = f = f x 4 ( l1 + e )
 End correction can be eliminated if the second resonant length l2 is measured
( l2 + e ) = -------------- (2)
(2) – (1)  (l2 – l1) = or = 2 (l2 – l1)
= f = 2 f (l2 – l1)