71_JEE_Main_Physics_2002_2020_Chapterwise_Solved_Papers@StudyAffinity (1).pdf

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Contents
1. Physical World, Units and Measurements

P-1 – P-13
Topic 1 : Unit of Physical Quantities
Topic 2 : Dimensions of Physical Quantities
Topic 3 : Errors in Measurements
2. Motion in a Straight Line

P-14 – P-25
Topic 1 : Distance, Displacement Uniform Motion
Topic 2 : Non-uniform Motion
Topic 3 : Relative Velocity
Topic 4 : Motion Under Gravity
3. Motion in a Plane

P-26 – P-35
Topic 1 : Vectors
Topic 2 : Motion in a Plane with Constant Acceleration
Topic 3 : Projectile Motion
Topic 4 : Relative Velocity in Two Dimensions Uniform Circular Motion
4. Laws of Motion

P-36 – P-53
Topic 1 : Ist, IInd IIIrd Laws of Motion
Topic 2 : Motion of Connected Bodies, Pulley Equilibrium of Forces
Topic 3 : Friction
Topic 4 : Circular Motion, Banking of Road
5. Work, Energy and Power

P-54 – P-75
Topic 1 : Work
Topic 2 : Energy
Topic 3 : Power
Topic 4 : Collisions
6. System of Particles and Rotational Motion

P-76 – P-112
Topic 1 : Centre of Mass, Centre of Gravity Principle of Moments
Topic 2 : Angular Displacement, Velocity and Acceleration
Topic 3 : Torque, Couple and Angular Momentum
Topic 4 : Moment of Inertia and Rotational K.E.
Topic 5 : Rolling Motion

7. Gravitation

P-113 – P-130
Topic 1 : Kepler’s Laws of Planetary Motion
Topic 2 : Newton’s Universal Law of Gravitation
Topic 3 : Acceleration due to Gravity
Topic 4 : Gravitational Field and Potential Energy
Topic 5 : Motion of Satellites, Escape Speed and Orbital Velocity
8. Mechanical Properties of Solids

P-131 – P-138
Topic 1 : Hooke’s Law Young’s Modulus
Topic 2 : Bulk and Rigidity Modulus and Work Done in Stretching a Wire
9. Mechanical Properties of Fluids

P-139 – P-154
Topic 1 : Pressure, Density, Pascal’s Law and Archimedes’ Principle
Topic 2 : Fluid Flow, Reynold’s Number and Bernoulli’s Principle
Topic 3 : Viscosity and Terminal Velocity
Topic 4 : Surface Tension, Surface Energy and Capillarity
10. Termal Properties of Matter

P-155 – P-168
Topic 1 : Termometer Termal Expansion
Topic 2 : Calorimetry and Heat Transfer
Topic 3 : Newton’s Law of Cooling
11. Termodynamics

P-169 – P-185
Topic 1 : First Law of Termodynamics
Topic 2 : Specifc Heat Capacity and Termodynamical Processes
Topic 3 : Carnot Engine, Refrigerators and Second Law of Termodynamics
12. Kinetic Teory

P-186 – P-198
Topic 1 : Kinetic Teory of an Ideal Gas and Gas Laws
Topic 2 : Speed of Gas, Pressure and Kinetic Energy
Topic 3 : Degree of Freedom, Specifc Heat Capacity, and Mean Free Path
13. Oscillations

P-199 – P-218
Topic 1 : Displacement, Phase, Velocity and Acceleration in S.H.M.
Topic 2 : Energy in Simple Harmonic Motion
Topic 3 : Time Period, Frequency, Simple Pendulum and Spring Pendulum
Topic 4 : Damped, Forced Oscillations and Resonance
14. Waves

P-219 – P-234
Topic 1 : Basic of Mechanical Waves, Progressive and Stationary Waves
Topic 2 : Vibration of String and Organ Pipe

Topic 3 : Beats, Interference and Superposition of Waves
Topic 4 : Musical Sound and Doppler’s Efect
15. Electric Charges and Fields

P-235 – P-253
Topic 1 : Electric Charges and Coulomb’s Law
Topic 2 : Electric Field and Electric Field Lines
Topic 3 : Electric Dipole, Electric Flux and Gauss’s Law
16. Electrostatic Potential and Capacitance

P-254 – P-280
Topic 1 : Electrostatic Potential and Equipotential Surfaces
Topic 2 : Electric Potential Energy and Work Done in Carrying a Charge
Topic 3 : Capacitors, Grouping of Capacitor and Energy Stored in a Capacitor
17. Current Electricity

P-281 – P-311
Topic 1 : Electric Current, Drif of Electrons, Ohm’s Law, Resistance and Resistivity
Topic 2 : Combination of Resistances
Topic 3 : Kirchhof’s Laws, Cells, Termo e.m.f. Electrolysis
Topic 4 : Heating Efect of Current
Topic 5 : Wheatstone Bridge and Diferent Measuring Instruments
18. Moving Charges and Magnetism

P-312 – P-339
Topic 1 : Motion of Charged Particle in Magnetic Field
Topic 2 : Magnetic Field Lines, Biot-Savart’s Law and Ampere’s Circuital Law
Topic 3 : Force and Torque on Current Carrying Conductor
Topic 4 : Galvanometer and its Conversion into Ammeter and Voltmeter
19. Magnetism and Matter

P-340 – P-347
Topic 1 : Magnetism, Gauss’s Law, Magnetic Moment, Properties of Magnet
Topic 2 : Te Earth Magnetism, Magnetic Materials and their Properties
Topic 3 : Magnetic Equipment
20. Electromagnetic Induction

P-348 – P-360
Topic 1 : Magnetic Flux, Faraday’s and Lenz’s Law
Topic 2 : Motional and Static EMI and Application of EMI
21. Alternating Current

P-361 – P-376
Topic 1 : Alternating Current, Voltage and Power
Topic 2 : AC Circuit, LCR Circuit, Quality and Power Factor
Topic 3 : Transformers and LC Oscillations

22. Electromagnetic Waves

P-377 – P-388
Topic 1 : Electromagnetic Waves, Conduction and Displacement Current
Topic 2 : Electromagnetic Spectrum
23. Ray Optics and Optical Instruments

P-389 – P-414
Topic 1 : Plane Mirror, Spherical Mirror and Refection of Light
Topic 2 : Refraction of Light at Plane Surface and Total Internal Refection
Topic 3 : Refraction at Curved Surface Lenses and Power of Lens
Topic 4 : Prism and Dispersion of Light
Topic 5 : Optical Instruments
24. Wave Optics

P-415 – P-432
Topic 1 : Wavefront, Interference of Light, Coherent and Incoherent Sources
Topic 2 : Young’s Double Slit Experiment
Topic 3 : Difraction, Polarisation of Light and Resolving Power
25. Dual Nature of Radiation and Matter

P-433 – P-448
Topic 1 : Matter Waves, Cathode and Positive Rays
Topic 2 : Photon, Photoelectric Efect X-rays and Davisson-Germer Experiment
26. Atoms

P-449 – P-460
Topic 1 : Atomic Structure and Rutherford’s Nuclear Model
Topic 2 : Bohr’s Model and the Spectra of the Hydrogen Atom
27. Nuclei

P-461 – P-472
Topic 1 : Composition and Size of the Nuclei
Topic 2 : Mass-Energy Equivalence and Nuclear Reactions
Topic 3 : Radioactivity
28. Semiconductor Electronics : Materials, Devices and Simple Circuits

P-473 – P-493
Topic 1 : Solids, Semiconductors and P-N Junction Diode
Topic 2 : Junction Transistor
Topic 3 : Digital Electronics and Logic Gates
29. Communication Systems

P-494 – P-500
Topic 1 : Communication Systems
Mock Test 1 with Solutions

MT-1 – MT-8
Mock Test 2 with Solutions

MT-9 – MT-16

Opening/ Closing Rank for TOP NITs List of NITs in India
Opening/ Closing Rank for Top NITS
College Name OR/CR CSE ECE ME EE/EEE
NIT Trichy ORF 2060 5325 4154 5708
CR 5317 8011 12970 10353
NIT Rourkela OR 2253 8571 11662 4084
CR 9420 12009 20304 19168
NIT Surathkal OR 960 3378 6315 3456
CR 3181 5608 11788 6801
NIT Warangal OR 978 2919 4340 5270
CR 2341 2919 10209 8152
NIT Calicut OR 2201 8023 10629 9703
CR 10222 14769 20480 18966
NIT Kurukshetra OR 2268 8320 11195 9454
CR 6170 12067 18115 16273
NIT Durgapur OR 5611 12509 14511 13595
CR 12095 16098 22753 19325
MNIT Allahabad OR 1449 3600 5884 5879
CR 4051 7128 11145 8790
NIT Silchar OR 8699 17899 21851 32579
CR 23882 40841 49215 56958
MNIT Jaipur OR 1148 3881 9277 4119
CR 3831 7868 11426 9179
List of NITs in India
After IITs, NITs form the second layer of topmost engineering Institutes in India
Rank Of (Amongst NITs) Name State NIRF Score NIRF Ranking
1 NIT Trichy Tamil Nadu 61.62 10
2 NIT Rourkela Odisha 57.75 16
3 NIT Karnataka Karnataka 55.25 21
4 NIT Warangal Telangana 53.21 26
5 NIT Calicut Kerala 52.69 28
6 V-NIT Maharashtra 51.27 31
7 NIT Kurukshetra Haryana 47.58 41
7 MN-NIT Uttar Pradesh 47.49 42
8 NIT Durgapur West Bengal 46.47 46
9 NIT Silchar Assam 45.61 51
10 M-NIT Rajasthan 45.20 53
11 SV-NIT Gujarat 41.88 58
12 NIT Hamirpur Himachal Pradesh 41.48 60
13 MA-NIT Madhya Pradesh 40.98 62
14 NITIE Maharashtra 40.48 66
15 NIT Meghalaya Meghalaya 40.32 67
16 NIT Agartala Tripura 39.53 70
17 NIT Raipur Chattisgarh 39.09 74
18 NIT Goa Goa 37.06 87

QUESTION: Which Colleges other than IITs accept JEE Ad-
vanced scores?
ANSWER:
1. Institute of Science (IISc), Bangalore
2. Indian Institute of Petroleum and Energy (IIPE), Visakhapatnam
3. Indian Institutes of Science Education and Research (IISER),
Bhopal
Mohali
Kolkata
Pune
Thiruvananthapuram
8. Indian Institute of Space Science and Technology (IIST), Thiru-
vananthapuram
9. Rajiv Gandhi Institute of Petroleum Technology (RGIPT), Rae
Bareli
QUESTION: If I am absent in one of the papers (Paper 1, Paper 2),
will my result be declared?
ANSWER: NO. You will be considered absent in JEE
(Advanced)-2018 and the result will not be prepared/declared. It is
compulsory to appear in both the papers for result preparation.
QUESTION: Do I have to choose my question paper language at the
time of JEE (Advanced)-2018 registration?
ANSWER: NO. There is no need to indicate question paper language
at the time of JEE (Advanced)-2018 registration. Candidates will
have the option to choose their preferred language (English or Hindi),
as the default language for viewing the questions, at the start of the
Computer Based Test (CBT) examination of JEE (Advanced)-2018.
QUESTION: Can I change the language (from English to Hindi
and vice versa) of viewing the questions during the CBT of JEE
(Advanced)-2018?
ANSWER: Questions will be displayed on the screen of the Candidate
in the chosen default language (English or Hindi). Further, the
candidate can also switch/toggle between English or Hindi languages,
as the viewing language of any question, anytime during the entire
period of the examination. The candidate will also be having the
option of changing default question viewing language anytime during
the examination.
QUESTION: Will I be given rough sheets for my calculations during
the CBT of JEE (Advanced)-2018?
ANSWER: Yes, you will be given “Scribble Pad” (containing
blank sheets, for rough work) at the start of every paper of JEE
(Advanced)-2018. You can do all your calculations inside this
“Scribble Pad”. Candidates MUST submit their signed Scribble Pads
at the end of each paper of the examination, given to them at the start
of the paper.
QUESTION: During examination can I change my answers?
ANSWER: Candidate will have the option to change previously
saved answer of any question, anytime during the entire duration of
the test.
QUESTION: How can I change a previously saved answer during the
CBT of JEE (Advanced)-2018?
ANSWER: To change the answer of a question that has already been
answered and saved, first select the corresponding question from
the Question Palette, then click on “Clear Response” to clear the
previously entered answer and subsequently follow the procedure for
answering that type of question.
QUESTION: Will I be given a printout/hard copy of the questions
papers along with my responses to questions in Paper-I and Paper-II
after the completion of the respective papers?
ANSWER: No.
QUESTION: How will I be getting a copy of the questions papers
and my responses to questions in Paper-I and Paper-II?
ANSWER: The responses of all the candidates who have appeared
for both Paper 1 and Paper 2, recorded during the exam, along with
the questions of each paper, will be electronically mailed to their
registered email ids, by Friday, May 25, 2018, 10:00 IST.
QUESTION: Suppose two candidates have same JEE
(Advanced)-2018 aggregate marks. Will the two candidates be given
the same rank?
ANSWER: If the aggregate marks scored by two or more candidates
are the same, then the following tie-break policy will be used for
awarding ranks: Step 1: Candidates having higher positive marks
will be awarded higher rank. If the tie breaking criterion at Step 1
fails to break the tie, then the following criterion at Step 2 will be
followed. Step 2: Higher rank will be assigned to the candidate who
has obtained higher marks in Mathematics. If this does not break the
tie, higher rank will be assigned to the candidate who has obtained
higher marks in Physics. If there is a tie even after this, candidates will
be assigned the same rank.
QUESTION: I have read in newspapers that for the academic year
2018-2019, supernumerary seats for female candidates would be there
in IITs. Does this mean that the non-females will get reduced number
of seats in IITs in 2018?
ANSWER:Adecision has been taken at the level of the IIT Council to,
inter alia, improve the gender balance in the undergraduate programs
at the IITs from the current (approximately) 8% to 14% in 2018-19
by creating supernumerary seats specifically for female candidates,
without any reduction in the number of seats that was made available
to non-female candidates in the previous academic year (i.e. academic
year 2017-2018).
FAQs - Frequently Asked Questions

Physical World, Units and Measurements P-1
TOPIC 1 Unit of PhysicalQuantities
1. The density of a material in SI unit is 128 kg m–3
. In
certain units in which the unit of length is 25 cm and
the unit of mass is 50 g, the numerical value of density
of the material is: [10 Jan. 2019 I]
(a) 40 (b) 16 (c) 640 (d) 410
2. AmetalsamplecarryingacurrentalongX-axiswithdensityJxis
subjectedtoamagneticfieldBz(alongz-axis).Theelectricfield
Ey developedalongY-axisisdirectlyproportionaltoJx aswell
as Bz. The constant of proportionalityhas SI unit
[Online April 25, 2013]
(a)
2
m
A
(b)
3
m
As
(c)
2
m
As
(d) 3
As
m
TOPIC 2
Dimensions of Physical
Quantities
3. The quantities
0 0
1
,
E
x y
B
= =
m e
and
1
z
CR
= are
defined where C-capacitance, R-Resistance, l-length,
E-Electric field, B-magnetic field and 0 0
, ,
e m - free
space permittivity and permeability respectively. Then :
[Sep. 05, 2020 (II)]
(a) x, y and z have the same dimension.
(b) Only x and z have the same dimension.
(c) Only x and y have the same dimension.
(d) Only y and z have the same dimension.
4. Dimensional formula for thermal conductivity is (here K
denotes the temperature : [Sep. 04, 2020 (I)]
(a) MLT–2 K (b) MLT–2 K–2
(c) MLT–3 K (d) MLT–3 K–1
5. A quantityx is given by(IFv2/WL4) in terms of moment of
inertia I, force F, velocity v, work W and Length L. The
dimensional formula for x is same as that of :
[Sep. 04, 2020 (II)]
(a) planck’s constant (b) force constant
(c) energy density (d) coefficient of viscosity
6. Amount of solar energy received on the earth's surface
per unit area per unit time is defined a solar constant.
Dimension of solar constant is : [Sep. 03, 2020 (II)]
(a) ML2T–2 (b) ML0T–3
(c) M2L0T–1 (d) MLT–2
7. If speed V, area A and force F are chosen as fundamental
units, then the dimension ofYoung's modulus will be :
[Sep. 02, 2020 (I)]
(a) FA2V–1 (b) FA2V–3
(c) FA2V–2 (d) FA–1V0
8. Ifmomentum (P), area (A) and time (T) aretaken to bethe
fundamental quantities then the dimensional formula for
energy is : [Sep. 02, 2020 (II)]
(a) [P2AT–2] (b) [PA–1T–2]
(c) 1/2 1
[PA T ]
-
(d) 1/2 1
[P AT ]
-
9. Which of the following combinations has the dimension
of electrical resistance (Î0
is the permittivity of vacuum
and m0
is the permeability of vacuum)?
[12 April 2019 I]
(a)
0
0
m
e (b)
0
0
m
e (c)
0
0
e
m (d)
0
0
e
m
10. In the formula X = 5YZ2
, X and Z have dimensions of
capacitance and magnetic field, respectively. What are
the dimensions of Y in SI units ? [10 April 2019 II]
(a) [M–3
L–2
T8
A4
] (b) [M–1
L–2
T4
A2
]
(c) [M–2
L0
T–4
A–2
] (d) [M–2
L–2
T6
A3
]
11. In SI units, the dimensions of
0
0
Î
m
is: [8 April 2019 I]
(a) A–1
TML3
(b) AT2
M–1
L–1
(c) AT–3
ML3/2
(d) A2
T3
M–1
L–2
12. Let l, r, c and v represent inductance, resistance,
capacitance and voltage, respectively. The dimension of
l
rcv
in SI units will be : [12 Jan. 2019 II]
(a) [LA – 2] (b) [A–1]
(c) [LTA] (d) [LT2]
Physical World, Units
and Measurements
1
Join- https://t.me/studyaffinity

P-2 Physics
13. The force of interaction between two atoms is given by
2
exp
x
F
kT
æ ö
= ab -
ç ÷
ç ÷
a
è ø
; where x is the distance, k is the
Boltzmann constant and T is temperature and a and b are
two constants. The dimensions of b is: [11 Jan. 2019 I]
(a) M0L2T–4 (b) M2LT–4
(c) MLT–2 (d) M2L2T–2
14. Ifspeed (V), acceleration (A) and force (F) are considered
as fundamental units, the dimension ofYoung’s modulus
will be : [11 Jan. 2019 II]
(a) V–2A2F–2 (b) V–2A2F2
(c) V–4A–2F (d) V–4A2F
15. A quantity f is given by f =
5
hc
G
where c is speed of
light, G universal gravitationalconstant andh isthePlanck’s
constant. Dimension of f is that of : [9 Jan. 2019 I]
(a) area (b) energy
(c) momentum (d) volume
16. Expression for time in terms ofG (universal gravitational
constant), h (Planck's constant) and c (speed of light) is
proportional to: [9 Jan. 2019 II]
(a)
5
hc
G
(b)
3
c
Gh
(c) 5
Gh
c
(d) 3
Gh
c
17. The dimensions of stopping potential V0
in photoelectric
effect in units of Planck’s constant ‘h’, speed of light ‘c’
and Gravitational constant ‘G’and ampere A is:
[8 Jan. 2019 I]
(a) hl/3
G2/3
cl/3
A –1
(b) h2/3
c5/3
G1/3
A –1
(c) h–2/3
e–1/3
G4/3
A–1
(d) h2
G3/2
C1/3
A–1
18. The dimensions of
2
0
2
B
m
, where B is magnetic field and m0
is the magnetic permeability of vacuum, is:
[8 Jan. 2019 II]
(a) MLT–2
(b) ML2
T–1
(c) ML2
T–2
(d) ML–1
T–2
19. Thecharacteristic distance at which quantum gravitational
effects aresignificant, thePlanck length, can bedetermined
from a suitable combination of the fundamental physical
constants G, h and c. Which of the following correctly
gives the Planck length? [Online April 15, 2018]
(a) G2hc (b)
1
2
3
Gh
c
æ ö
ç ÷
è ø
(c)
1
2
2
G h c
(d) Gh2c3
20. Time (T), velocity (C) and angular momentum (h) are
chosen as fundamental quantities instead of mass, length
and time. In terms ofthese, the dimensions of mass would
be : [Online April 8, 2017]
(a) [ M ]=[ T–1 C–2 h ] (b) [ M ]=[ T–1 C2 h ]
(c) [ M ]=[ T–1 C–2 h–1 ] (d) [ M ]=[ T C–2 h ]
21. A, B, C and D are four different physical quantities having
different dimensions. None of them is dimensionless. But
we know that the equation AD = C ln (BD) holds true.
Thenwhich ofthecombinationisnotameaningfulquantity?
(a)
2
C AD
BD C
- (b) A2 –B2C2
(c)
A
C
B
- (d)
(A C)
D
-
22. In the following 'I' refers to current and other symbols
have their usual meaning, Choose the option that
corresponds to the dimensions of electrical conductivity:
[OnlineApril 9, 2016]
(a) M–1L–3T3I (b) M–1L–3T3I2
(c) M–1L3T3I (d) ML–3T–3I2
23. If electronic charge e, electron mass m, speed of light in
vacuum cand Planck’sconstant h are taken asfundamental
quantities, the permeabilityof vacuum m0 can beexpressed
in units of : [Online April 11, 2015]
(a) 2
h
me
æ ö
ç ÷
è ø
(b) 2
hc
me
æ ö
ç ÷
è ø
(c) 2
h
ce
æ ö
ç ÷
è ø
(d)
2
2
mc
he
æ ö
ç ÷
ç ÷
è ø
24. Ifthe capacitance of a nanocapacitor is measured in terms
of a unit ‘u’ made by combining the electric charge ‘e’,
Bohr radius ‘a0’, Planck’s constant ‘h’and speed of light
‘c’then: [Online April 10, 2015]
(a)
2
0
e h
u
a
= (b) 2
0
hc
u
e a
=
(c)
2
0
e c
u
ha
= (d)
2
0
e a
u
hc
=
25. From the following combinations of physical constants
(expressed through their usual symbols) the only
combination, that would have the same value in different
systems of units, is: [Online April 12, 2014]
(a) 2
o
ch
2pe
(b)
2
2
o e
e
2 Gm
pe
(me = mass of electron)
(c)
o o
2 2
G
c he
m e
(d)
o o
2
2 h
G
ce
p m e

26. In terms ofresistanceRand time T, the dimensions ofratio
m
e
ofthe permeabilitym and permittivitye is:
(a) [RT–2](b) [R2T–1] (c) [R2] (d) [R2T2]
27. Let [ 0
Î ] denotethe dimensional formula of the permittivity
of vacuum. If M = mass, L = length, T = time and A =
electric current, then: [2013]
(a) 0
Î = [M–1 L–3 T
T2 A] (b) 0
Î = [M–1 L–3 T
T4 A2]
(c) 0
Î = [M1 L2 T
T1 A2] (d) 0
Î = [M1 L2 T
T1 A]
28. If the time period t of the oscillation of a drop of liquid of
densityd, radiusr, vibrating under surface tension sisgiven
by the formula 2 / 2
= b c a
t r s d . It is observed that the
time period is directly proportional to
d
s
. The value of b
should therefore be : [Online April 23, 2013]
(a)
3
4
(b) 3
(c)
3
2
(d)
2
3
29. The dimensions of angular momentum, latent heat and
capacitance are, respectively. [Online April 22, 2013]
(a) 2 1 2 2 2 1 2 2
ML T A , L T , M L T
- - -
(b) 2 2 2 2 1 2 4 2
ML T , L T , M L T A
- - -
(c) 2 1 2 2 2 2
ML T , L T , ML TA
- -
(d) 2 1 2 2 1 2 4 2
ML T , L T , M L T A
- - - -
30. Given that K = energy, V = velocity, T = time. If they are
chosen as the fundamental units, then what is dimensional
formula for surface tension? [Online May 7, 2012]
(a) [KV–2T–2] (b) [K2V2T–2]
(c) [K2V–2T–2] (d) [KV2T2]
31. The dimensions of magnetic field in M, L, T and C
(coulomb) is given as [2008]
(a) [MLT–1 C–1] (b) [MT2 C–2]
(c) [MT–1 C–1] (d) [MT–2 C–1]
32. Which of the following units denotes the dimension
2
2
Q
ML
, where Q denotes the electric charge? [2006]
(a) Wb/m2 (b) Henry (H)
(c) H/m2 (d) Weber (Wb)
33. Out of the following pair, which one does NOT have
identical dimensions ? [2005]
(a) Impulse and momentum
(b) Angular momentum and planck’s constant
(c) Work and torque
(d) Moment of inertia and moment of a force
34. Which one of the following represents the correct
dimensions of the coefficient of viscosity? [2004]
(a)
1 1
ML T
- -
é ù
ë û (b)
1
MLT-
é ù
ë û
(c)
1 2
ML T
- -
é ù
ë û (d)
2 2
ML T
- -
é ù
ë û
35. Dimensions of
o o
1
m e
, where symbols have their usual
meaning, are [2003]
(a) ]
T
L
[ 1
-
(b) ]
T
L
[ 2
2
-
(c) ]
T
L
[ 2
2 -
(d) ]
LT
[ 1
-
36. The physical quantities not having same dimensions are
(a) torque and work [2003]
(b) momentum and planck’s constant
(c) stress and young’s modulus
(d) speed and 2
/
1
o
o )
( -
e
m
37. Identify the pair whose dimensions are equal [2002]
(a) torque and work (b) stress and energy
(c) force and stress (d) force and work
TOPIC 3 Errors in Measurements
38. A screw gauge has 50 divisions on its circular scale. The
circular scale is 4 units ahead of the pitch scale marking,
prior to use. Upon one complete rotation of the circular
scale, a displacement of 0.5 mm is noticed on the pitch
scale. The nature of zero error involved, and the least
count of the screw gauge, are respectively :
[Sep. 06, 2020 (I)]
(a) Negative, 2 mm (b) Positive, 10 mm
(c) Positive, 0.1 mm (d) Positive, 0.1 mm
39. The density of a solid metal sphere is determined by
measuring its massand its diameter. Themaximum error in
the density of the sphere is
100
æ ö
ç ÷
è ø
x
%. Ifthe relative errors
in measuring themass and the diameter are6.0% and 1.5%
respectively, the value of x is ______.
[NA Sep. 06, 2020 (I)]
40. A student measuring the diameter of a pencil of circular
cross-section with the help of a vernier scale records the
following four readings 5.50 mm, 5.55 mm, 5.45 mm,
5.65 mm, The average of these four reading is 5.5375 mm
and the standard deviation of the data is 0.07395 mm. The
average diameter of the pencil should therefore be re-
corded as : [Sep. 06, 2020 (II)]
(a) (5.5375±0.0739)mm (b) (5.5375±0.0740)mm
(c) (5.538±0.074)mm (d) (5.54±0.07)mm
41. A physical quantity z depends on four observables a, b, c
and d, as z =
2
2 3
3
a b
cd
. The percentages of error in the mea-
surement ofa, b, c andd are 2%, 1.5%, 4% and 2.5% respec-
tively. The percentage of error in z is :
[Sep. 05, 2020 (I)]

P-4 Physics
(a) 12.25% (b) 16.5%
(c) 13.5% (d) 14.5%
42. Using screw gauge of pitch 0.1 cm and 50 divisions on its
circular scale, the thickness of an object is measured. It
should correctly be recorded as : [Sep. 03, 2020 (I)]
(a) 2.121cm (b) 2.124cm
(c) 2.125cm (d) 2.123cm
43. The least count of the main scale of a vernier callipers is
1 mm. Its vernier scale is divided into 10 divisions and
coincidewith 9 divisions ofthe main scale. When jaws are
touching each other, the 7th division of vernier scale
coincides with a division of main scale and the zero of
vernier scale is lying right side of the zero of main scale.
When this vernier is used to measure length of a cylinder
the zero of the vernier scale betwen 3.1 cm and 3.2 cm and
4th VSD coincides with a main scale division. The length
of the cylinder is : (VSD is vernier scale division)
[Sep. 02, 2020 (I)]
(a) 3.2cm (b) 3.21cm
(c) 3.07cm (d) 2.99cm
44. Ifthe screwon ascrew-gaugeisgiven six rotations, itmoves
by 3 mm on the main scale. If there are 50 divisions on the
circular scale the least count of the screw gauge is:
[9 Jan. 2020 I]
(a) 0.001cm (b) 0.02mm
(c) 0.01cm (d) 0.001mm
45. For the four sets of three measured physical quantities as
given below. Which of the following options is correct?
[9 Jan. 2020 II]
(A) A1
= 24.36, B1
= 0.0724, C1
=256.2
(B) A2
= 24.44, B2
= 16.082, C2
=240.2
(C) A3
= 25.2, B3
= 19.2812, C3
=236.183
(D) A4
= 25, B4
= 236.191, C4
=19.5
(a) A4
+B4
+C4
A1
+B1
+C1
A3
+B3
+C3
A2
+B2
+C2
(b) A1
+B1
+C1
=A2
+B2
+C2
=A3
+B3
+C3
=A4
+B4
+C4
(c) A4
+B4
+C4
A1
+B1
+C1
=A2
+B2
+C2
=A3
+B3
+C3
(d) A1
+B1
+C1
A3
+B3
+C3
A2
+B2
+C2
A4
+B4
+C4
46. A simple pendulum is being used to determine the value
ofgravitational acceleration gat a certain place. The length
of the pendulum is 25.0 cm and a stop watch with 1 s
resolution measures the time taken for 40 oscillations to
be 50 s. The accuracy in g is: [8 Jan. 2020 II]
(a) 5.40% (b) 3.40%
(c) 4.40% (d) 2.40%
47. In the density measurement of a cube, the mass and edge
length are measured as(10.00 ±0.10) kg and(0.10 ±0.01)
m, respectively. The error in the measurement of density
is: [9 April 2019 I]
(a) 0.01kg/m3
(b) 0.10kg/m3
(c) 0.013kg/m3
(d) 0.07kg/m3
48. The area of a square is 5.29 cm2
. The area of 7 such
squares taking into account the significant figures is:
[9 April 2019 II]
(a) 37cm2
(b) 37.030cm2
(c) 37.03cm2
(d) 37.0cm2
49. In a simple pendulum experiment for determination of
acceleration due to gravity (g), time taken for 20
oscillations is measured by using a watch of 1 second
least count. The mean valueof timetaken comes out to be
30 s. The length of pendulum is measured by using a
meter scale of least count 1 mm and the value obtained is
55.0 cm. The percentage error in the determination of gis
close to : [8 April 2019 II]
(a) 0.7% (b) 0.2% (c) 3.5% (d) 6.8%
50. The least count ofthe main scale ofa screwgauge is 1 mm.
The minimum number of divisions on its circular scale
required tomeasure 5 µm diameter of a wire is:
[12 Jan. 2019 I]
(a) 50 (b) 200 (c) 100 (d) 500
51. The diameter and height of a cylinder are measured by
a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm,
respectively. What will be the value of its volume in
appropriate significant figures? [10 Jan. 2019 II]
(a) 4264 ± 81 cm3
(b) 4264.4 ± 81.0 cm3
(c) 4260 ±80cm3
(d) 4300 ±80cm3
52. The pitch and the number of divisions, on the circular
scale for a given screw gauge are 0.5 mm and 100
respectively. When the screw gauge is fully tightened
without any object, the zero of its circular scale lies 3
division below the mean line.
The readings of the main scale and the circular scale, for
a thin sheet, are 5.5 mm and 48 respectively, the
thickness of the sheet is: [9 Jan. 2019 II]
(a) 5.755mm (b) 5.950mm
(c) 5.725mm (d) 5.740mm
53. The density of a material in the shape of a cube is
determined by measuring three sides of the cube and its
mass. If the relative errors in measuring the mass and
length are respectively 1.5% and 1%, the maximum error
in determining the density is: [2018]
(a) 2.5% (b) 3.5% (c) 4.5% (d) 6%
54. Thepercentage errors in quantities P, Q, Rand S are0.5%,
1%, 3% and 1.5% respectively in the measurement of a
physical quantity
3 2
P Q
A
RS
= .
Themaximum percentageerror in the value ofAwill be
(a) 8.5% (b) 6.0%
(c) 7.5% (d) 6.5%
55. The relative uncertaintyin the period of a satellite orbiting
around the earth is 10–2. If the relative uncertainty in the
radius ofthe orbit is negligible, the relative uncertaintyin
the mass of the earth is [Online April 16, 2018]
(a) 3×10–2 (b) 10–2
(c) 2 × 10–2 (d) 6 × 10–2

56. The relative error in the determination of the surface area
ofa sphereis a. Then therelative error in the determination
of its volume is [Online April 15, 2018]
(a)
2
3
a (b)
2
3
a (c)
3
2
a (d) a
57. In a screw gauge, 5 complete rotations of the screw cause
ittomove a lineardistanceof0.25cm. Thereare 100circular
scale divisions. The thickness of a wire measured by this
screw gauge gives a reading of 4 main scale divisions and
30 circular scaledivisions.Assumingnegligible zeroerror,
the thickness of the wire is: [OnlineApril 15, 2018]
(a) 0.0430cm (b) 0.3150cm
(c) 0.4300cm (d) 0.2150cm
58. The following observations were taken for determining
surface tensiton T of water by capillary method :
Diameter of capilary, D = 1.25 × 10–2
m
rise of water, h = 1.45 × 10–2
m
Using g = 9.80 m/s2
and the simplified relation
3
10
2
rhg
T = ´ N/m, the possible error in surface tension
is closest to : [2017]
(a) 2. 4 % (b) 10% (c) 0.15% (d) 1.5%
59. A physical quantity P is described by the relation
P = a1/2
b2
c3
d –4
If the relative errors in the measurement of a, b, c and d
respectively, are 2%, 1%, 3% and 5%, then the relative
error in P will be : [OnlineApril 9, 2017]
(a) 8% (b) 12% (c) 32% (d) 25%
60. A screw gauge with a pitch of 0.5 mm and a circular scale
with 50 divisions is used to measure the thickness of a
thin sheet ofAluminium. Before starting the measurement,
it is found that wen the two jaws of the screw gauge are
brought in contact, the 45th division coincides with the
main scale line and the zero of the main scale is barely
visible. What is the thickness of the sheet if the main scale
reading is 0.5 mm and the 25th division coincides with the
main scale line? [2016]
(a) 0.70 mm (b) 0.50 mm
(c) 0.75mm (d) 0.80mm
61. A student measures the time period of 100 oscillations of
a simple pendulum four times. The data set is 90 s, 91 s, 95
s, and 92 s. If the minimum division in the measuring clock
is 1 s, then the reported mean time should be: [2016]
(a) 92 ± 1.8 s (b) 92 ± 3s
(c) 92 ± 1.5 s (d) 92 ± 5.0 s
62. The period of oscillation of a simple pendulum is
T =
L
2
g
p . Measured value ofLis20.0 cmknown to1mm
accuracyand time for 100 oscillations of the pendulum is
found to be 90 s using a wrist watch of 1s resolution. The
accuracyin the determination of g is :
[2015]
(a) 1% (b) 5% (c) 2% (d) 3%
63. Diameter of asteel ball is measuredusing aVernier callipers
which has divisions of 0.1 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 as: [Online April 10, 2015]
S.No. MS(cm) VS divisions
1. 0.5 8
2. 0.5 4
3. 0.5 6
Ifthe zeroerror is– 0.03 cm, then mean corrected diameter
is:
(a) 0.52 cm (b) 0.59 cm
(c) 0.56cm (d) 0.53cm
64. The current voltage relation of a diode is given by
( )
1000V T
I e 1
= - mA, where the applied voltage V is in
voltsand the temperatureT is in degree kelvin. If a student
makes an error measuring 0.01
± V while measuring the
current of 5 mA at 300 K, what will be the error in the
value of current in mA? [2014]
(a) 0.2mA (b) 0.02mA (c) 0.5mA (d) 0.05mA
65. A student measured the length ofa rod and wrote it as 3.50
cm. Which instrument did he use to measure it?
[2014]
(a) A meter scale.
(b) A vernier calliper where the 10 divisions in vernier
scale matches with 9 division in main scale and main
scale has 10 divisions in 1 cm.
(c) A screw gauge having 100 divisions in the circular
scale and pitch as 1 mm.
(d) A screwgauge having50divisions in thecircular scale
and pitch as 1 mm.
66. Match List - I(Event) with List-II(Order ofthe timeinterval
for happening of the event) and select the correct option
from the options given below the lists:
(1) Rotation
period of earth
(i) 105
s
(2) Revolution
period of earth
(ii) 10
7
s
(3) Period of light
wave
(iii) 10
–15
s
(4) Period of
sound wave
(iv) 10–3
s
List - I List - II
(a) (1)-(i),(2)-(ii), (3)-(iii),(4)-(iv)
(b) (1)-(ii),(2)-(i),(3)-(iv),(4)-(iii)
(c) (1)-(i),(2)-(ii), (3)-(iv),(4)-(iii)
(d) (1)-(ii),(2)-(i), (3)-(iii),(4)-(iv)

P-6 Physics
67. In theexperiment ofcalibration ofvoltmeter, a standard cell
ofe.m.f. 1.1voltisbalanced against 440cmofpotentialwire.
The potential difference across the ends of resistance is
found to balance against 220 cm of the wire. The
corresponding reading ofvoltmeter is 0.5 volt. Theerror in
the reading of volmeter will be: [Online April 12, 2014]
(a) – 0. 15 volt (b) 0.15 volt
(c) 0.5 volt (d) – 0.05 volt
68. An experiment is performed to obtain the value of
acceleration due togravityg byusing a simple pendulum of
length L. In this experiment time for 100 oscillations is
measured byusing a watch of 1 second least count and the
value is 90.0 seconds. The length L is measured byusing a
meter scaleofleastcount1mm andthevalueis 20.0cm.The
error in the determination ofg wouldbe:
(a) 1.7% (b) 2.7% (c) 4.4% (d) 2.27%
69. Resistance of a given wire is obtained by measuring the
current flowingin it andthevoltagedifferenceapplied across
it. Ifthepercentage errorsin themeasurement ofthecurrent
and the voltage difference are 3% each, then error in the
value of resistance of the wire is [2012]
(a) 6% (b) zero (c) 1% (d) 3%
70. A spectrometer gives the following reading when used to
measure the angle of a prism.
Main scale reading : 58.5 degree
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5
degree. Total divisions on theVernier scale is30 and match
with 29 divisions of the main scale. The angle of the prism
from the above data is [2012]
(a) 58.59 degree (b) 58.77 degree
(c) 58.65 degree (d) 59 degree
71. N divisions on the main scale of a vernier calliper coincide
with (N+1) divisions ofthevernier scale. Ifeach division of
main scale is ‘a’units, then theleast count of theinstrument
is [Online May 19, 2012]
(a) a (b)
a
N
(c)
1
N
a
N
´
+
(d)
1
a
N +
72. A student measured the diameter of a wire using a screw
gauge with the least count 0.001 cm and listed the
measurements. The measured value should be recorded
as [Online May 12, 2012]
(a) 5.3200cm (b) 5.3cm
(c) 5.32cm (d) 5.320cm
73. A screw gauge gives the following reading when used to
measurethediameter ofa wire.
Main scalereading :0mm
Circular scalereading : 52 divisions
Given that 1mm on main scalecorresponds to100 divisions
of the circular scale. The diameter of wire from the above
data is [2011]
(a) 0.052cm (b) 0.026cm
(c) 0.005cm (d) 0.52cm
74. The respective number of significant figures for the
numbers23.023, 0.0003 and 2.1 × 10–3 are [2010]
(a) 5,1, 2 (b) 5,1, 5
(c) 5,5, 2 (d) 4,4, 2
75. In an experiment the angles are required to be measured
using an instrument, 29 divisions ofthe main scale exactly
coincide with the 30 divisions of the vernier scale. If the
smallest division of the main scale is half- a degree
(= 0.5°), then the least count ofthe instrument is: [2009]
(a) halfminute (b) one degree
(c) halfdegree (d) oneminute
76. A body of mass m = 3.513 kg is moving along the x-axis
with a speed of5.00ms–1. The magnitude ofits momentum
is recorded as [2008]
(a) 17.6kgms–1 (b) 17.565kgms–1
(c) 17.56kg ms–1 (d) 17.57kg ms–1
77. Twofull 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 of3 mm and thenumber of circular scale
divisions in line with the main scale as 35. The diameter of
the wire is [2008]
(a) 3.32mm (b) 3.73mm
(c) 3.67mm (d) 3.38mm

1. (a) Density of material in SI unit,
= 3
128kg
m
Density of material in new system
=
( )( )
( ) ( )
3 3
128 50g 20
25cm 4
= ( )
128
20 40units
64
=
2. (b) According to question
y x Z
E J B
µ
Constant of proportionality
3
y
Z x x
E C m
K
B J J As
= = =
[As
E
C
B
= (speed of light) and
I
J
Area
= ]
3. (a) We know that
Speed of light,
0 0
1
c x
= =
m e
Also,
E
c y
B
= =
Time constant, Rc t
t = =
Speed
l l
z
Rc t
= = =
Thus, x, y, z will have the same dimension of speed.
4. (d) From formula,
dQ dT
kA
dt dx
=
dQ
dt
k
dT
A
dx
æ ö
ç ÷
è ø
Þ =
æ ö
ç ÷
è ø
2 3
3 1
2 1
[ML T ]
[ ] [MLT K ]
[L ][KL ]
k
-
- -
-
= =
5. (c) Dimension of Force F = M1L1T–2
Dimension of velocity V = L1T–1
Dimension of work = M1L2T–2
Dimension of length = L
Moment of inertia = ML2
2
4
IFv
x
WL
=
1 2 1 1 2 1 2 2
1 2 2 4
(M L )(M L T )(L T )
(M L T )(L )
- -
-
=
1 2 2
1 1 2
3
M L T
M L T
L
- -
- -
= = = Energy density
6. (b) Solar constant
Energy
Time Area
=
Dimension of Energy, E = ML2T–2
Dimension of Time = T
Dimension of Area = L2
Dimension of Solar constant
1 2 2
1 0 3
2
M L T
M L T .
TL
-
-
= =
7. (d) Young's modulus,
stress
strain
Y =
–1 0
0
F
FA V
A
D
Þ = =
l
l
Y
8. (c) Energy, a b c
E A T P
µ
or, a b c
E kA T P
= ...(i)
where k is a dimensionless constant and a, b and c arethe
exponents.
Dimension of momentum, 1 1 1
P M LT -
=
Dimension of area, A = L2
Dimension of time, T = T1
Putting these value in equation (i), we get
1 2 2 2
c a c b c
M L T M L T
- + -
=
by comparison
c = 1
2a + c = 2
b – c = –2
c = 1, a = 1/2, b = –1
1/ 2 1 1
E A T P
-
=
9. (a)
2
0 0
0
0 0 0 0 0
1
c c
æ ö
m m
= = m =
ç ÷
e e m m e
è ø
m0
c ® MLT–2
A–2
× LT–1
ML2
T–3
A–2
Dimensions of resistance
10. (a) X= 5YZ2
2
X
Y
Z
Þ µ ...(i)

P-8 Physics
2 2 2
2 2
[ ]
Capacitance =
V [ ]
Q Q A T
X
W ML T -
= = =
X = [M–1
L–2
T4
A2
]
F
Z B
IL
= = [QF= ILB]
Z = [MT–2
A–1
]
1 2 4 2
2 1 2
[ ]
[ ]
M L T A
Y
MT A
- -
- -
=
Y = [M–3
L–2
T8
A4
] (Using (i))
11. (d)
2
0 0 0
0 0 0 0 0
é ù
é ù
e e e
= = ê ú
ê ú
m m e m e
ê ú ê ú
ë û ë û
= e0
C[LT
T–1
]×[e0
]
0 0
1
C
é ù
=
ê ú
m e
ê ú
ë û
Q
2
2
0
4
q
F
r
=
pe
Q
2
2 1 3 4
0 2 2
[ ]
[ ] [ ]
[ ] [ ]
AT
A M L T
MLT L
- -
-
Þ e = =
´
1 2 1 3 4
0
0
[ ] [ ]
LT A M L T
- - -
é ù
e
= ´
ê ú
m
ê ú
ë û
1 2 3 2
[ ]
M L T A
- -
=
12. (b) As we know,
[ ] [ ] [ ]
T and cv AT
r
é ù
= =
ê ú
ë û
l
–1
T
A
rcv AT
é ù é ù
é ù
= =ë û
ê ú ê ú
ë û ë û
l
13. (b) Force of interaction between two atoms,
2
x
kT
F e
æ ö
-
ç ÷
ç ÷
a
è ø
= ab
Since exponential terms are dimensionless

2
kT
x
a
é ù
ê ú
ê ú
ë û
= M0L0T0
[ ]
2
0 0 0
2 2
L
M L T
ML T-
Þ =
a
Þ [a] = M–1T2
[F] = [a] [b]
MLT–2 = M–1T2[b]
Þ [b] = M2LT–4
14. (d) Let [Y] = [V]a [F]b [A]c
[ML–1T–2] = [LT–1]a [MLT–2]b [LT–2]c
[ML–1T–2] = [MbLa+b+c T–a–2b–2c]
Comparing power both sideof similar terms we get,
b = 1, a + b + c = – 1, –a –2b –2c = – 2
solving above equations we get:
a = – 4, b= 1, c = 2
so [Y] = [V–4FA2] = [V–4A2F]
15. (b) Dimension of [h] = [ML2
T–1
]
[C] = [LT–1
]
[G] = [M–1
L3
T–2
]
Hence dimension of
2 1 5 5
5
1 3 2
=
ML T L T
hC
G M L T
- -
- -
é ù é ù
é ù ×
ë û ë û
ê ú
é ù
ê ú
ë û ë û
= [ML2
T–2
] = energy
16. (c) Let t µ Gx hy Cz
Dimensions of G = [M–1L3T–2],
h = [ML2T–1] and C = [LT–1]
[T] = [M–1L3T–2]x[ML2T–1]y[LT–1]z
[M0L0T1] = [M–x+y L3x+2y+z T–2x–y–z]
By comparing the powers of M, L, T both the sides
– x + y= 0 Þ x = y
3x +2y + z = 0Þ5x + z =0 .....(i)
–2x – y –z = 1 Þ 3x + z = –1 .....(ii)
Solving eqns. (i) and (ii),
1 5
x y , z
2 2
= = = - 5
Gh
t
C
µ
17. (None)
Stopping potential 0
( ) x y Z r
V h I G C
µ
Here,h = Planck’s constant 2 1
ML T -
é ù
= ë û
I = current = [A]
G = Gravitational constant = [M–1
L3
T–2
]
and c = speed of light = [LT–1
]
V0
= potential= [ML2
T–3
A–1
]
[ML2
T–3
A–1
]=[ML2
T–1
]x
[A]y
[M–1
L3
T–2
]z
[LT–1
]r
Mx – z
; L2x+3z+r
; T–x–2z–r
; Ay
Comparing dimension of M, L, T, A, we get
y= –1, x = 0, z = – 1 , r = 5
0 –1 –1 5
0
V h I G C
µ
18. (d) The quantity
2
0
B
2m
is the energydensity of magnetic
field.
2
3
0
Energy Force displacement
2 (displacement)
B
Volume
é ù ´
Þ = =
ê ú
m
ê ú
ë û
2 –2
–1 –2
3
ML T
ML T
L
é ù
= =
ê ú
ê ú
ë û

19. (b) Planklength isaunitoflength,lp=1.616229×10–35m
3
p
hG
l
c
=
20. (a) Let mass, related as M µ Tx
Cy
hz
M1
L0
T0
= (T')x
(L1
T–1
)y
(M1
L2
T–1
)z
M1
L0
T0
= Mz
Ly + 2z
+ Tx–y–z
z= 1
y+ 2z = 0 x – y– z = 0
y = –2 x+ 2– 1 = 0
x=–1
M = [T–1
C–2
h1
]
21. (d) Dimension of A¹ dimension of(C)
Hence A – C is not possible.
22. (b) We know that resistivity
RA
l
θ
Conductivity =
1
resistivity RA

l
I
VA

l
(Q V=RI)
2 2
2
[L][I]
[ML T
[L ]
[I][T]
,

é ù
ê ú´
ê ú
ê ú
ë û
W W
V
q it

Q
1 3 3 2 1 3 3 2
[M L T ][I ] [M L T I ]
, , , ,

23. (c) Let µ0 related with e, m, c and h as follows.
m0 = kea
mb
cc
hd
[MLT–2
A–2
] = [AT]a
[M]b
[LT–1
]c
[ML2
T–1
]d
= [Mb + d
Lc + 2d
Ta – c – d
Aa
]
On comparing both sides we get
a = – 2 ...(i)
b + d = 1 ...(ii)
c + 2d = 1 ...(iii)
a – c – d = –2 ...(iv)
By equation (i), (ii), (iii) (iv) we get,
a = – 2, b = 0, c = – 1, d = 1
0 2
[ ]
é ù
m = ê ú
ë û
h
ce
24. (d) Let unit ‘u’ related with e, a0
, h and c as follows.
[u] = [e]a [a0
]b [h]c [C]d
Using dimensional method,
[M–1L–2T+4A+2] = [A1T1]a[L]b[ML2T–1]c[LT–1]d
[M–1L–2T+4A+2] = [Mc Lb+2c+d Ta–c–d Aa]
a = 2, b = 1, c = – 1, d = – 1
u =
2
0
e a
hc
25. (b) The dimensional formulae of
0 0 1 1
e M L T A
é ù
=
ë û
1 3 4 2
0 M L T A
-
é ù
e =
ë û
1 3 2
G M L T
- -
é ù
=
ë û and
1 0 0
e
m M L T
é ù
=
ë û
Now,
2
2
0 e
e
2 Gm
pe
=
2
0 0 1 1
2
1 3 4 2 1 3 2 1 0 0
M L T A
2 M L T A M L T M L T
- - - -
é ù
ë û
é ù é ù é ù
p
ë û ë û ë û
=
2 2
1 1 2 3 3 4 2 2
T A
2 M L T A
- - + - + -
é ù
ë û
é ù
p
ë û
=
2 2
0 0 2 2
T A
2 M L T A
é ù
ë û
é ù
p
ë û
=
1
2p
Q
1
2p
is dimensionless thus the combination
2
2
0 e
e
2 Gm
pe
would have the same value in different systems of units.
26. (c) Dimensions of m = [MLT–2A–2]
Dimensions of Î = [M–1L–3T4A2]
Dimensions of R = [ML2T–3A–2]

Dimensionsof
Dimensionsof
m
Î
=
2 2
1 3 4 2
[MLT A ]
[M L T A ]
- -
- -
= [M2L4T–6A–4 ] = [R2]
27. (b) As we know, 1 2
2
0
1 q q
F
4 R
=
pe
Þ
1 2
0 2
q q
4 FR
e =
p
Hence,
2 2
0 2 2 2
C [AT]
N.m [MLT ][L ]
-
e = =
= [M–1 L–3 T4 A2]
28. (c)
29. (d) Angular momentum = m × v× r = ML2 T–1
Latent heat L =
2 2
Q ML T
m M
-
= = L2T–2
Capacitance C = 1 2 4 2
Charge
M L T A
P.d.
- -
=
30. (a) Surface tension,
2
2
. .
= =
l
l l l
F F T
T
T
(As, F.l = K (energy);
2
2
2
-
=
l
T
V )
Therefore, surface tension = [KV–2T–2]
31. (c) MagnitudeofLorentzformula F= qvB sin q
2
1 1
1
[ ]
F MLT
B MT C
qv C LT
-
- -
-
= = =
´

P-10 Physics
32. (b) Mutual inductance =
BA
I I
f
=
1 1 2
2 2
1
[ ]
[Henry]
[ ]
MT Q L
ML Q
QT
- -
-
-
= =
33. (d) Moment of Inertia, I = MR2
[I] = [ML2]
Moment of force, t
r
= r F
´
r
r
t
r
= 2 2 2
[ ][ ] [ ]
L MLT ML T
- -
=
34. (a) According to, Stokes law,
F = 6phrv
6
F
r v
Þ h =
p
–2
–1
[ ]
[ ][ ]
MLT
L LT
h = 1 1
[ ]
ML T
- -
Þ h =
35. (c) As we know, the velocity of light in free space is
given by
c =
1
o o
m e
2 2 2
1
0 0
1
e Z T
= =
m e
1
o o
m e = C2[m/s]2
= [LT–1]2
= [M0L2T–2]
36. (b) Momentum, = mv = [MLT–1]
Planck’s constant,
E
h
v
=
2 –2
–1
[ ]
[ ]
ML T
T
= = [ML2T–1]
37. (a) Work cos
W F s Fs
= × = q
r r
Q A B
×
r r
= AB cos q
2 2 2
[ ][ ] [ ]
MLT L ML T
- -
= = ;
Torque, r F
t = ´
r
r r
sin
rF
Þ t = q
Q A B
´
r r
= AB sin q
2 2 2
[ ] [ ] [ ]
L MLT ML T
- -
= =
38. (b) Given:No.ofdivisiononcircularscaleofscrewgauge=50
Pitch = 0.5mm
Least count of screw gauge
Pitch
No. of division on circular scale
=
5
0.5
mm 1 10 m = 10 m
50
= = ´ m
And nature of zero error is positive.
39. (1050)
Density, 3
4
3 2
M M
V D
r = =
æ ö
pç ÷
è ø
3
6
MD-
Þ r =
p
% 3 6 3 1.5 10.5%
m D
m D
æ ö
Dr D D
æ ö
= + = + ´ =
ç ÷
ç ÷ è ø
è r ø
1050
% % %
100 100
x
æ ö
Dr æ ö
= = ç ÷
ç ÷ è ø
è r ø
1050.00
x
=
40. (d) Averagediameter, dav = 5.5375 mm
Deviation of data, Dd= 0.07395 mm
As the measured data are upto two digits after decimal,
therefore answer should be in two digits after decimal.
(5.54 0.07) mm
d
= ±
41. (d) Given :
2 2/3
3
a b
Z
cd
=
Percentage error in Z,
=
Z
Z
D 2 2 1 3
3 2
a b c d
a b c d
D D D D
= + + +
2 1
2 2 1.5 4 3 2.5 14.5%.
3 2
= ´ + ´ + ´ + ´ =
42. (a) Thickness = M.S. Reading + Circular Scale Reading
(L.C.)
Here LC
Pitch 0.1
0.002
Circular scale division 50
= = = cm per
division
So, correct measurement is measurement of integral
multipleofL.C.
43. (c) L.C. ofvernier callipers= 1 MSD –1 VSD
9
1 1 0.1
10
æ ö
= - ´ =
ç ÷
è ø
mm=0.01cm
Here 7th division of vernier scale coincides with a division
of main scale and the zero of vernier scale is lying right
side of the zero of main scale.
Zeroerror =7×0.1 =0.7mm= 0.07cm.
Length of the cylinder = measured value – zero error
= (3.1 + 4 × 0.01) – 0.07 = 3.07cm.
44. (d) When screw on a screw-gauge is given six rotations,
it moves by 3mm on the main scale

3
Pitch 0.5mm
6
= =
Least count L.C.
Pitch 0.5mm
50
CSD
= =
1
mm 0.01 0.001cm
100
mm
= = =
45. (None)
D1
= A1
+ B1
+ C1
= 24.36 + 0.0724 + 256.2 = 280.6
D2
= A2
+ B2
+ C2
= 24.44 + 16.082 + 240.2 = 280.7
D3
= A3
+ B3
+ C3
= 25.2 + 19.2812 + 236.183= 280.7

D4
= A4
+ B4
+ C4
= 25 + 236.191 + 19.5 = 281
None of the option matches.
46. (c) Given, Length ofsimple pendulum, l = 25.0 cm
Time of 40 oscillation, T = 50s
Time period of pendulum
2
T
g
= p
l
2
2 4
T
g
p
Þ =
l 2
2
4
g
T
p
Þ =
l
Þ Fractional error in g =
2
g l T
g l T
D D D
= +
0.1 1
2 0.044
25.0 50
g
g
D æ ö æ ö
Þ = + =
ç ÷ ç ÷
è ø è ø
Percentageerrorin 100 4.4%
g
g
g
D
= ´ =
47. (Bonus) 3
3
M M
Ml
V l
-
d = = =
3
M l
M l
Dd D D
= +
d
0.10 0.01
3
10.00 0.10
æ ö
= + ç ÷
è ø = 0.31kg/m3
,
48. (d) A = 7× 5.29 = 37.03 cm2
The result should have three significant figures, so
A = 37.0 cm2
49. (d) We have
2
T
g
= p
l
or
2
2
4
g
T
= p
l
100 100 2 100
g R T
g Q T
D D D
´ = ´ + ´
0.1 1
100 2 100
55 30
æ ö
= ´ + ´
ç ÷
è ø
=0.18+6.67=6.8%
50. (b) Least count of main scale of screw gauge = 1mm
Least count of screw gauge
Pitch
Number of division on circular scale
=
3
6 10
5 10
N
-
-
´ =
Þ N= 200
51. (c)
52. (c) Least count of screw gauge,
LC=
Pitch
No. of division
= 0.5 × 10–3 = 0.5 × 10–2 mm+veerror = 3×0.5 ×10–2 mm
=1.5×10–2 mm= 0.015mm
Reading = MSR+ CSR– (+ve error)
= 5.5 mm + (48 × 0.5 × 10–2)–0.015
=5.5+0.24–0.015=5.725mm
53. (c) = 1.5 % + 3 (1%) = 4.5%
54. (d) Maximum percentage error in A
3(% error in P) 2(% error in Q)
= +
1
(% error in R) 1(% error in S)
2
+ +
1
3 0.5 2 1 3 1 1.5
2
= ´ + ´ + ´ + ´
=1.5+ 2+1.5+ 1.5=6.5%
55. (c) From Kepler's law, time period of a satellite,
3
r
T 2
Gm
= p
2
2 3
4
T r
GM
p
=
Relative uncertainty in the mass of the earth
2
M T
2 2 10
M T
-
D D
= = ´ (Q 4p G constant and
relative uncertaintyin radius
r
r
D
negligible)
56. (c) Relativeerror in Surface area,
s r
2
s r
D D
= ´ = a and
relative error in volume,
v r
3
v r
D D
= ´
Relative error in volume w.r.t. relative error in area,
v 3
v 2
D
= a
57. (d) Least count =
Value of 1 part on main scale
Number of parts on vernier scale
=
0.25
cm
5×100
= 5 × 10–4 cm
Reading = 4 × 0.05 cm + 30 × 5 × 10–4 cm
= (0.2 + 0.0150) cm = 0.2150 cm (Thickness of wire)
58. (d) Surface tension, 3
10
2
= ´
rhg
T
Relative error in surface tension,
0
D D D
= + +
T r h
T r h
(Q g, 2 103 are constant)
Percentage error
–2 –2
–2 –2
10 0.01 10 0.01
100 100
1.25 10 1.45 10
æ ö
D ´ ´
´ = +
ç ÷
´ ´
è ø
T
T
= (0.8 + 0.689)
= (1.489)= 1.489%@1.5%
59. (c) Given, P = a1/2
b2
c2
d–4
,
Maximumrelativeerror,
P 1 a b c d
2 3 4
P 2 a b c d
D D D D D
= + + +
1
2 2 1 3 3 4 5
2
= ´ + ´ + ´ + ´ =32%
60. (d) L.C.
0.5
50
= =0.01mm
Zeroerror = 5× 0.01 = 0.05mm(Negative)
Reading= (0.5+25× 0.01)+0.05= 0.80mm

P-12 Physics
61. (c)
1 2 3 4
| T | | T | | T | | T |
T
4
D + D + D + D
D =
2 1 3 0
1.5
4
+ + +
= =
As the resolution of measuring clock is 1.5 therefore
the mean time should be 92 ± 1.5
62. (d) As, g = 2
2
4
L
T
p
So, 100 100 2 100
g L T
g L T
D D D
´ = ´ + ´
=
0.1 1
100 2 100
20 90
´ + ´ ´ = 2.72 ; 3%
63. (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
64. (a) The current voltage relation of diode is
1000 /
( 1)
= -
V T
I e mA (given)
When, 1000 /
5 , 6
V T
I mA e mA
= =
Also, 1000 / 1000
( )
= ´
V T
dI e
T
Error = ± 0.01 (By exponential function)
=
1000
(6 ) (0.01)
300
´ ´
mA =0.2mA
65. (b) Measured length of rod = 3.50 cm
For Vernier Scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
66. (a) Rotation period of earth is about 24 hrs ; 105 s
Revolution period of earth is about 365 days ; 107 s
Speed of light wave C = 3 × 108 m/s
Wavelength of visible light of spectrum
l = 4000 – 7800 Å
C = f l
1
and T
f
æ ö
=
ç ÷
è ø
Therefore period of light wave is 10–15 s (approx)
67. (d) In a voltmeter
V l
µ
V = kl
Now, it is given E = 1.1 volt for l1 = 440 cm
and V = 0.5 volt for l2 = 220 cm
Let the error in reading of voltmeter be DV then,
1.1 = 400K and(0.5 – DV)=220K.
Þ
1.1 0.5 V
440 220
- D
=
V 0.05 volt
D = -
68. (b) According to the question.
t = (90 ± 1) or,
1
90
t
t
D
=
l = (20 ± 0.1) or,
0.1
20
l
l
D
=
% ?
g
g
D
=
As we know,
2
l
t
g
= p Þ 2
4 l
g
t
2
p
=
or, 2
D D D
æ ö
= ± +
ç ÷
è ø
g l t
g l t
=
0.1 1
2
20 90
æ ö
+ ´
ç ÷
è ø
= 0.027
% 2.7%
g
g
D
=
69. (a) According to ohm’s law, V = IR
R=
V
I
Percentage error =
2
Absolute error
10
Measurement
´
where, 100
V
V
D
´ = 100
I
I
D
´ = 3%
then, 100
R
R
D
´ =
2 2
10 10
V I
V I
D D
´ + ´
= 3% + 3% = 6%
70. (c) Q Reading of Vernier = Main scalereading
+ Vernier scale reading × least count.
Main scale reading = 58.5
Vernier scale reading = 09 division
least count of Vernier = 0.5°/30
Thus, R = 58.5° + 9 ×
0.5
30
°
R=58.65°
71. (d) No. of divisions on main scale = N
No. of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN = b (N + 1) Þ b =
1
aN
N +
Least count is a – b = a –
1
aN
N +
=
1
1
N N
a
N
+ -
é ù
ê ú
+
ë û
=
1
a
N +
72. (d) The least count (L.C.) ofa screwguage is the smallest
length which can be measured accurately with it.
As least count is 0.001 cm =
1
1000
cm
Hence measured value should be recorded upto 3 decimal
placesi.e., 5.320 cm

73. (a) Least count, L.C. =
1
100
mm
Diameter of wire = MSR + CSR × L.C.
Q 1 mm = 0.1 cm
= 0+
1
100
×52=0.52mm=0.052cm
74. (a) Number of significant figures in 23.023 = 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 ×10–3 = 2
So, the radiation belongs to X-rays part of the spectrum.
75. (d) 30 DivisionsofV.S. coincidewith 29 divisionsofM.S.
1V.S.D =
29
30
MSD
L.C.=1 MSD–1VSD
= 1 MSD
29
3 0
- MSD
=
1
MSD
30
=
1
0.5
30
´ ° = 1 minute.
76. (a) Momentum, p = m × v
Given, mass of a body = 3.513 kg speed of body
= (3.513) × (5.00) = 17.565 kg m/s
= 17.6 (Rounding offto get three significant figures)
77. (d) Least count of screw gauge = 0.01 mm
Q
0.5
50
mm
Reading = [M.S.R. + C.S.R. × L.C.] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm

14 Physics
TOPIC 1
Distance, Displacement
Uniform Motion
1. A particle is moving with speed v b x
= along positive
x-axis.Calculatethespeedoftheparticleat timet = t(assume
that the particle is at origin at t = 0). [12Apr. 2019 II]
(a)
2
4
b t
(b)
2
2
b t
(c) 2
b t (d)
2
2
b t
2. All the graphs below are intended to represent the same
motion. One of them does it incorrectly. Pick it up.
[2018]
(a) position
velocity
(b) time
distance
(c) time
position
(d) time
velocity
3. A car covers the first half of the distance between two
places at 40 km/h and other half at 60 km/h. The average
speed of the car is [Online May 7, 2012]
(a) 40 km/h (b) 45 km/h
(c) 48km/h (d) 60km/h
4. The velocity ofa particle is v = v0 + gt + ft2. If its position
is x = 0 at t = 0, then its displacement after unit time (t =
1) is [2007]
(a) v0 + g /2 + f (b) v0 + 2g + 3f
(c) v0 + g /2 + f/3 (d) v0 + g + f
5. A particle located at x = 0 at time t = 0, starts moving
along with the positive x-direction with a velocity 'v' that
varies as v = x
a . The displacement of the particle
varieswith timeas [2006]
(a) t2 (b) t (c) t1/2 (d) t3
TOPIC 2 Non-uniform Motion
6. The velocity (v) and time (t) graph of a body in a straight
line motion is shown in the figure. The point S is at 4.333
seconds. The total distance covered by the body in 6 s is:
[05 Sep. 2020 (II)]
4
2
0
–2
1 2 3 4 5 6
t (in s)
A B
S D
C
v (m/s)
(a)
37
3
m (b) 12m (c) 11m (d)
49
4
m
7. The speed verses time graph for a particle is shown in the
figure. The distance travelled (in m) bythe particle during
the time interval t = 0 to t = 5 s will be __________.
[NA 4 Sep. 2020 (II)]
1 2 3 4 5
2
4
6
8
10
u
(ms )
–1
time
( )
s
8. The distance x covered by a particle in one dimensional
motion varies with time t as x2
= at2
+ 2bt + c. If the
acceleration of the particle depends on x as x–
n, where n
is an integer, the value of n is ______. [NA9 Jan 2020 I]
9. A bullet of mass 20g has an initial speed of 1 ms–1
, just
before it starts penetrating a mud wall of thickness 20 cm.
Ifthe wall offers a mean resistance of2.5×10–2
N, the speed
of the bullet after emerging from the other side of the wall
is close to : [10Apr. 2019 II]
(a) 0.1 ms–1
(b) 0.7 ms–1
(c) 0.3 ms–1
(d) 0.4 ms–1
Motion in a Straight
Line
2

Motion in a Straight Line P-15
10. The position of a particle as a function of time t, is given
by
x(t) = at + bt2
– ct3
where, a, b and c are constants. When the particle attains
zero acceleration, then its velocity will be:
[9Apr. 2019II]
(a)
2
4
+
b
a
c
(b)
2
3
b
a
c
+
(c)
2
b
a
c
+ (d)
2
2
b
a
c
+
11. A particle starts from origin O from rest and moves with a
uniform acceleration along the positive x-axis. Identifyall
figures that correctly represents the motion qualitatively
(a = acceleration, v = velocity, x = displacement, t = time)
[8Apr. 2019 II]
(A) (B)
(C) (D)
(a) (B), (C) (b) (A)
(c) (A), (B), (C) (d) (A), (B), (D)
12. A particle starts from the origin at time t = 0 and moves
along the positive x-axis. The graph of velocity with
respect to time is shown in figure. What is the position
of the particle at time t = 5s? [10 Jan. 2019 II]
3
2
1
0
1
1 2 3 4 5 6 7 8 9 10
v
(m/s)
(a) 10 m (b) 6 m
(c) 3m (d) 9m
13. In a car race on straight road, car A takes a time t less
than car B at the finish and passes finishing point with
a speed 'v' more than of car B. Both the cars start from
rest and travel with constant acceleration a1 and a2
respectively. Then 'v' is equal to: [9 Jan. 2019 II]
(a) 1 2
1 2
2a a
t
a a
+
(b) 1 2
2a a t
(c) 1 2
a a t (d) 1 2
a a
t
2
+
14. An automobile, travelling at 40 km/h, can be stopped at a
distance of40m byapplying brakes. Ifthe sameautomobile
is travelling at 80 km/h, theminimum stopping distance, in
metres, is (assume no skidding) [Online April 15, 2018]
(a) 75m (b) 160m (c) 100m (d) 150m
15. The velocity-time graphs ofa car and a scooter are shown
in the figure. (i) the difference between the distance
travelled bythe car and the scooter in 15 s and (ii) the time
at which the car will catch up with the scooter are,
respectively [OnlineApril 15, 2018]
(a) 337.5m and 25s
5
0
15
30
A B
G
E
F
C
D
45
10 15
Time in (s) ®
Velocity
(ms
)
–1
®
20 25
O
Car
Scooter
(b) 225.5mand 10s
(c) 112.5mand 22.5s
(d) 11.2.5mand15s
16. Aman in a car at location Qon a straight highwayismoving
with speed v. He decides to reach a point P in a field at a
distance d from highway (point M) as shown in the figure.
Speed of the car in the field is half to that on the highway.
What should be the distance RM, so that the time taken to
reachPisminimum? [OnlineApril 15, 2018]
d
P
Q
R M
(a)
d
3
(b)
d
2
(c)
d
2
(d) d
17. Which graph corresponds to an object moving with a
constant negative acceleration and a positive velocity ?
[OnlineApril 8,2017]
(a)
Velocity
Time
(b)
Velocity
Time
(c)
Velocity
Distance
(d)
Velocity
Distance
18. The distance travelled by a body moving along a line in
time t is proportional to t3.
The acceleration-time (a, t) graph for the motion of the
body will be [Online May 12, 2012]

P-16 Physics
(a)
a
t
(b)
a
t
(c)
a
t
(d)
a
t
19. Thegraph of an object’s motion (along the x-axis) is shown
in the figure. The instantaneous velocity of the object at
points A and B are vA and vB respectively. Then
[Online May 7, 2012]
5
10
15
A
B
Dx = 4 m
10 20
x(m)
t(s)
0
Dt = 8
(a) vA = vB = 0.5 m/s (b) vA = 0.5 m/s vB
(c) vA = 0.5 m/s vB (d) vA = vB = 2 m/s
20. An object, moving with a speed of 6.25 m/s, is decelerated
at a rate given by
2.5
= -
dv
v
dt
where v is the instantaneous speed. The time
taken by the object, to come to rest, would be: [2011]
(a) 2 s (b) 4 s (c) 8 s (d) 1 s
21. A body is at rest at x = 0. At t = 0, it starts moving in the
positive x-direction with a constant acceleration. At the
same instant another body passes through x = 0 moving
in the positive x-direction with a constant speed. The
position of the first body is given by x1(t) after time ‘t’;
and that of the second body by x2(t) after the same time
interval. Which of the following graphs correctly
describes (x1 – x2) as a function of time ‘t’? [2008]
(a)
( – )
x x
1 2
t
O
(b)
( – )
x x
1 2
t
O
(c)
( – )
x x
1 2
t
O (d)
( – )
x x
1 2
t
O
22. Acar, starting from rest, accelerates at the rate f through a
distance S, then continues at constant speed for time t
and then decelerates at the rate
2
f
to come to rest. If the
total distance traversed is 15 S , then [2005]
(a) S = 2
1
6
ft (b) S = f t
(c) S =
2
1
4
ft (d) S =
2
1
72
ft
23. A particle is moving eastwards with a velocity of 5 ms–1.
In 10 seconds the velocity changes to 5 ms–1 northwards.
The average acceleration in this time is [2005]
(a) 2
ms
2
1 -
towards north
(b) 2
ms
2
1 - towards north - east
(c) 2
ms
2
1 -
towards north - west
(d) zero
24. The relation between time t and distance x is t = ax2 + bx
where a and b are constants. The acceleration is [2005]
(a) 2bv3 (b) –2abv2 (c) 2av2 (d) –2av3
25. An automobile travelling with a speed of 60 km/h, can
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
[2004]
(a) 60m (b) 40m (c) 20m (d) 80m
26. Acar, moving with a speed of 50 km/hr, can be stopped by
brakes after at least 6 m. If the same car is moving at a
speed of 100 km/hr, the minimum stopping distance is
[2003]
(a) 12m (b) 18m (c) 24m (d) 6m
27. If a body looses half of its velocity on penetrating 3 cm in
a wooden block, then how much will it penetrate more
before coming to rest? [2002]
(a) 1cm (b) 2cm (c) 3cm (d) 4cm.
28. Speeds of two identical cars are u and 4u at the specific
instant. The ratio of the respective distances in which the
two cars are stopped from that instant is [2002]
(a) 1 : 1 (b) 1 : 4 (c) 1 : 8 (d) 1 : 16
TOPIC 3 Relative Velocity
29. Train A and train B are running on parallel tracks in the
opposite directions with speeds of 36 km/hour and 72
km/hour, respectively. A person is walking in train A in
the direction opposite to its motion with a speed of 1.8

km/hour. Speed (in ms–1) of this person as observed from
train B will be close to : (take the distance between the
tracks as negligible) [2 Sep. 2020 (I)]
(a) 29.5 ms–1 (b) 28.5 ms–1
(c) 31.5 ms–1q (d) 30.5 ms–1
30. A passenger train of length 60 m travels at a speed of 80
km/hr. Another freight train of length 120 m travels at a
speed of 30 km/h. The ratio of times taken by the
passenger train to completelycross the freight train when:
(i) they are moving in same direction, and (ii) in the
opposite directions is: [12 Jan. 2019 II]
(a)
11
5
(b)
5
2
(c)
3
2
(d)
25
11
31. A person standing on an open ground hears the sound of
a jet aeroplane, coming from north at an angle 60º with
ground level. But he finds the aeroplane right vertically
above his position. If v is the speed of sound, speed of the
plane is: [12 Jan. 2019 II]
(a)
3
2
v (b)
2
3
v
(c) v (d)
2
v
32. A car is standing 200 m behind a bus, which is alsoat rest.
The two start moving at the same instant but with differ-
ent forward accelerations. The bus has acceleration 2 m/s2
and the car has acceleration 4 m/s2
. The car will catch up
with the bus after a time of :
(a) 110s (b) 120s
(c) 10 2s (d) 15 s
33. A person climbs up a stalled escalator in 60 s. If standing
on the same but escalator running with constant velocity
he takes 40 s. How much time is taken by the person to
walk up the moving escalator? [Online April 12, 2014]
(a) 37 s (b) 27 s (c) 24 s (d) 45 s
34. A goods train accelerating uniformlyon a straight railway
track, approaches an electric pole standing on the side of
track. Its engine passes the pole with velocity u and the
guard’s room passes with velocityv. Themiddlewagon of
the train passes the pole with a velocity.
[Online May 19, 2012]
(a)
2
u v
+
(b)
2 2
1
2
u v
+
(c) uv (d)
2 2
2
u v
æ ö
+
ç ÷
è ø
TOPIC 4 Motion Under Gravity
35. A helicopter rises from rest on the ground vertically up-
wards with a constant acceleration g. A food packet is
dropped from the helicopter when it is at a height h. The
time taken by the packet to reach the ground is close to
[g is the accelertion due to gravity] : [5 Sep. 2020 (I)]
(a)
2
3
h
t
g
æ ö
= ç ÷
è ø
(b) 1.8
h
t
g
=
(c) 3.4
h
t
g
æ ö
= ç ÷
è ø
(d)
2
3
h
t
g
=
36. A Tennis ball is released from a height h and after freely
falling on a wooden floor it rebounds and reaches height
2
h
. The velocityversus height of the ball during its motion
may be represented graphically by :
(graph are drawn schematicallyand on not to scale)
[4 Sep. 2020 (I)]
(a)
h/2
h
h v
( )
v
(b)
h/2
h
h v
( )
v
(c)
h/2
h
h v
( )
v
(d)
h/2
h
h v
( )
v
37. A ball is dropped from the top of a 100 m high tower on a
planet. In the last
1
2
s before hitting the ground, it covers a
distance of19 m.Acceleration due to gravity(in ms–2
) near
the surface on that planet is _______.
[NA 8 Jan. 2020 II]
38. A body is thrown vertically upwards. Which one of the
following graphs correctlyrepresent the velocity vs time?
[2017]
(a) (b)
(c) (d)
39. Two stones are thrown up simultaneously from the edge
of a cliff 240 m high with initial speed of 10 m/s and 40
m/s respectively. Which of the following graph best
represents the time variation of relative position of the
second stone with respect to the first ?

P-18 Physics
(Assume stones do not rebound after hitting the ground
and neglect air resistance, take g = 10 m/ s2) [2015]
(The figures are schematic and not drawn to scale)
(a) (y – y ) m
2 1
240
8 12
t(s)
(b) (y – y ) m
2 1
240
8 12
t(s)
(c)
(y – y ) m
2 1
240
8 12
t(s)
t®
(d)
(y – y ) m
2 1
240
12
t(s)
40. From a tower of height H, a particle is thrown vertically
upwards with a speed u. The time taken bythe particle, to
hit the ground, is n times that taken by it to reach the
highest point of its path. The relation between H, u and n
is: [2014]
(a) 2gH = n2u2 (b) gH = (n – 2)2 u2d
(c) 2gH = nu2 (n – 2) (d) gH = (n – 2)u2
41. Consider a rubber ball freelyfallingfrom a height h =4.9m
onto a horizontal elastic plate. Assume that the duration
of collision is negligible and the collision with the plate is
totallyelastic.
Then the velocity as a function of time and the height as
a function of time will be : [2009]
(a) t
+v1
v
O
–v1
y
h
t
(b)
v
+v1
O
–v1
t1 2t1 4t1
t t
y
h
t
(c) t
t1 2t1
O
y
h
t
(d)
v1
v
O t
t
y
h
42. Aparachutistafterbailingoutfalls50mwithoutfriction.When
parachuteopens,itdeceleratesat2m/s2 .Hereachestheground
with aspeedof3m/s.Atwhatheight, didhebailout? [2005]
(a) 182 m (b) 91 m
(c) 111m (d) 293m
43. A ballis releasedfrom thetop ofa tower of heighth meters.
It takes Tseconds to reach the ground. What is the position
of the ball at
3
T
second [2004]
(a)
8
9
h
meters from the ground
(b)
7
9
h
(c)
9
h
(d)
17
18
h
44. From a building two balls A and B are thrown such that A is
thrown upwards and B downwards (both vertically). If vA
and vB are their respective velocities on reaching the
ground, then [2002]
(a) vB vA
(b) vA = vB
(c) vA vB
(d) their velocities depend on their masses.

1. (b) Given, v = b x
or
1/2
dx
b x
dt
=
or
1/2
0 0
x t
x dx bdt
-
=
ò ò
or
1/2
1/ 2
x
= 6t or x =
2 2
4
b t
Differentiating w. r. t. time, we get
2
2
4
dx b t
dt
´
= (t = t)
or v =
2
2
b t
2. (b) Graphs in option (c) position-time and option (a)
velocity-position are corresponding to velocity-time graph
option (d) and its distance-time graph is as given below.
Hence distance-time graph option (b) is incorrect.
time
distance
3. (c) Average speed =
Total distance travelled x
Total time taken T
=
=
x
x x
2 40 2 60
+
´ ´
= 48 km/h
4. (c) We know that,
dx
v
dt
=
Þ dx = v dt
Integrating,
0 0
x t
dx v dt
=
ò ò
or 2
0
0
( )
t
x v gt ft dt
= + +
ò
2 3
0
0
2 3
t
gt ft
v t
é ù
= + +
ê ú
ê ú
ë û
or,
2 3
0
2 3
gt ft
x v t
= + +
At t = 1, 0
2 3
g f
x v
= + + .
5. (a) v x
= a ,
Þ
dx
x
dt
= a Þ
dx
dt
x
= a
Integrating both sides,
0 0
x t
dx
dt
x
= a
ò ò ; 0
0
2
[ ]
1
x
t
x
t
é ù
= a
ê ú
ë û
2 x t
Þ = a
2
2
4
x t
a
Þ =
6. (a) A B
S D
C
t
(in s)
v(m/s)
4
2
0
–2
1 2 3 4 5 6
O
1 13
4
3 3
OS = + =
1 5
2
3 3
SD = - =
Distance covered by the body = area of v-t graph
= ar (OABS) + ar (SCD)
1 13 1 5
1 4 2
2 3 2 3
æ ö
= + ´ + ´ ´
ç ÷
è ø
32 5 37
m
3 3 3
= + =
7. (20)
u
t
A
B
5
8
O
Distance travelled = Area of speed-time graph
1
5 8 20 m
2
= ´ ´ =
8. (3) Distance X varies with time t as x2
= at2
+ 2bt + c
2 2 2
dx
x at b
dt
Þ = +
( )
dx dx at b
x at b
dt dt x
+
Þ = + Þ =
2
2
2
d x dx
x a
dt
dt
æ ö
Þ + =
ç ÷
è ø
2 2
2
2
dx at b
a a
d x dt x
x x
dt
+
æ ö æ ö
- -
ç ÷ ç ÷
è ø è ø
Þ = =
( )
2 2
3 3
ax at b ac b
x x
2
- + -
= =
Þ a µx–3
Hence, n = 3

P-20 Physics
9. (b) From the third equation of motion
v2
– u2
= 2aS
But,
F
a
m
=
2 2
2
F
v u S
m
æ ö
= - ç ÷
è ø
2
2 2
3
2.5 10 20
(1) (2)
100
20 10
v
-
-
é ù
´
Þ = - ê ú
´
ê ú
ë û
Þ v2
=
1
1–
2
1
m/s 0.7m/s
2
v
Þ = =
10. (b) x = at + bt2
– ct3
Velocity,
2 3
( )
dx d
v at bt ct
dt dt
= = + +
= a + 2bt – 3ct2
Acceleration,
2
( 2 3 )
dv d
a bt ct
dt dt
= + -
or 0 = 2b – 3c × 2t
3
b
t
c
æ ö
= ç ÷
è ø
and v =
2
2 3
3 3
b b
a b c
c c
æ ö æ ö
+ -
ç ÷ ç ÷
è ø è ø
2
3
b
a
c
æ ö
= +
ç ÷
è ø
11. (d) For constant acceleration, there is straight line
parallel tot-axis on a t
®
- .
Inclined straight line on v t
®
- , and parabola on x t
®
- .
12. (d) Position of the particle,
S = area under graph (time t = 0 to5s)
1
2 2 2 2 3 1 9m
2
= ´ ´ + ´ + ´ =
13. (c) Let time taken byAto reach finishing point is t0
Time taken by B to reach finishing point = t0 + t
u = 0 v = a t
A 1 0
x
v = a + t)
B 2 0
(t
vA – vB = v
Þ v = a1 t0 – a2 (t0 + t) = (a1 – a2)t0–a2t ...(i)
2 2
B A 1 0 2 0
1 1
x x a t a (t t)
2 2
= = = +
( )
1 2 0
0
a t a t t
Þ = +
( )
1 2 0 2
a – a t a t
Þ =
Þ to = 2
1 2
a t
a – a
Putting this value of t0 in equation (i)
( ) 2
l 2 2
1 2
a t
v a – a –a t
a – a
=
( )
1 2 2 2
a a a t –a t
= + = 1 2 2 2
a a t a t –a t
+
or, 1 2
v a a t
=
14. (b) Accordingtoquestion,u1=40km/h,v1=0ands1 =40m
using v2 – u2 = 2as; 02 – 402 = 2a × 40 ...(i)
Again, 02 – 802 = 2as ...(ii)
From eqn. (i) and(ii)
Stopping distance, s = 160 m
15. (c) Using equation, a =
–
v u
t
and S = ut +
2
1
2
at
Distance travelled by car in 15 sec =
1 (45)
2 15
(15)2
=
675
2
m
Distancetravelled byscooter in 15 seconds = 30 × 15 = 450
(Q distance = speed × time)
Difference between distance travelled by car and scooter
in 15sec, 450–337.5= 112.5m
Let car catches scooter in time t;
675
45( –15) 30
2
t t
+ =
337.5 + 45t – 675 = 30t Þ 15t = 337.5
Þ t = 22.5 sec
16. (a) Let the car turn of the highway at a distance 'x' from
the point M. So, RM = x
And if speed of car in field is v, then time taken bythe car
to cover the distance QR = QM – x on the highway,
1
2
QM x
t
v
-
= .....(i)
Time taken to travel the distance 'RP' in the field
2 2
2
d x
t
v
+
= ..... (ii)
Total time elapsed to move the car from Q to P
2 2
1 2
2
QM x d x
t t t
v v
- +
= + = +
For 't' tobeminimum 0
dt
dx
=
d
P
Q
R M
2 2
1 1
0
2
x
v d x
é ù
- + =
ê ú
ê ú
+
ë û
or
2 3
2 1
d d
x = =
-

17. (c) According to question, object is moving with
constant negative acceleration i.e., a = – constant (C)
vdv
C
dx
= -
vdv = – Cdx
2
v
Cx k
2
= - +
2
v k
x
2C C
= - +
Hence, graph (3) represents correctly.
18. (b) Distance along a line i.e., displacement (s)
= t3 (Q 3
s t
µ given)
By double differentiation of displacement, we get
acceleration.
3
2
3
ds dt
V t
dt dt
= = = and
2
3
6
dv d t
a t
dt dt
= = =
a = 6t or a t
µ
Hence graph (b) is correct.
19. (a) Instantaneous velocity
D
=
D
x
v
t
From graph,
4
8
D
= =
D
A
A
A
x m
v
t s
= 0.5 m/s
and vB
8
16
D
= =
D
B
B
x m
t s
= 0.5 m/s
i.e., vA = vB = 0.5 m/s
20. (a) Given, 2.5
= -
dv
v
dt
Þ
dv
v
= – 2.5 dt
Integrating,
0 ½
6.25 0
2.5
-
= -
ò ò
t
v dv dt
Þ [ ]
0
½
0
6.25
2.5
(½)
+
é ù
= -
ê ú
ê ú
ë û
t
v
t
Þ – 2(6.25)½ = – 2.5t
Þ – 2 × 2.5 = –2.5t
Þ t = 25
21. (b) For the body starting from rest, distance travelled
(x1) is given by
x1 = 0 +
1
2
at2
Þ
2
1
1
2
=
x at
x x
1 2
–
v/a
t
Forthebodymovingwith constantspeed
x2 = vt
2
1 2
1
2
x x at vt
- = -
at t = 0, x1 – x2 = 0
This equation is of parabola.
For t
v
a
; the slope is negative
For t =
v
a
; the slope is zero
For t
v
a
; the slope is positive
These characteristics are represented by graph (b).
22. (d) Let car starts from A from rest and moves up to point
B with acceleration f.
Distance, AB = S =
2
1
1
2
ft
Distance, BC = (ft1)t
Distance, CD =
2
2
1
( )
2 2( /2)
ft
u
a f
= = ft1
2 = 2S
t
1
t 1
t
2
f 2
/
f
15 S
A B C D
Total distance, AD = AB + BC + CD = 15S
AD = S + BC + 2S
Þ 1 2 15
S f t t S S
+ + =
Þ 1 12
f t t S
= .............(i)
2
1
1
2
f t S
= ............ (ii)
Dividing (i) by (ii), we get 1
t =
6
t
Þ
2 2
1
2 6 72
t f t
S f
æ ö
= =
ç ÷
è ø
23. (c)
N
S
E
W
1
v
- 1
v
2
v
)
v
(
v
v 1
2 -
+
=
D
°
90
Initial velocity, 1
ˆ
5 ,
=
uu
r
v i

P-22 Physics
Final velocity, 2
ˆ
5 ,
=
uu
r
v j
Change in velocity 2 1
( )
v v v
D = -
uu
r u
r u
r
= 2 2
1 2 1 2
2 cos90
v v v v
+ +
= 0
5
5 2
2
+
+ = 5 2m/s
[As | 1
v | = | 2
v | = 5 m/s]
Avg. acceleration =
v
t
D
uu
r
2
s
/
m
2
1
10
2
5
=
=
1
5
5
tan -
=
-
=
q
which means q is in the second quadrant.
(towards north-west)
24. (d) Given, t = ax2 + bx;
Diff. with respect totime (t)
2
( ) ( ) .2
d d dx dx
t a x b a x
dt dt dt dt
= + = + b.v.
Þ 1 = 2axv + bv = v(2ax + b)(v = velocity)
2ax + b =
1
v
.
Again differentiating, we get
2
1
2 0
dx dv
a
dt dt
v
+ = -
Þ a =
dv
dt
= –2av3
dx
v
dt
æ ö
=
ç ÷
è ø
Q
25. (d) In first case speed,
5 50
60 m/s m/s
18 3
u = ´ =
d=20m,
Let retardation be a then
(0)2 – u2 = –2ad
or u2 = 2ad …(i)
In second case speed, u¢ =
5
120
18
´
=
100
m/s
3
and (0)2 – u¢2 = –2ad¢
or u¢2 = 2ad¢ …(ii)
(ii) divided by (i) gives,
'
4 ' 4 20 80m
d
d
d
= Þ = ´ =
26. (c) Fir first case : Initial velocity,
5
50 m / s,
18
0,s 6m,
u
v a a
= ´
= = =
Using, 2 2
2
v u as
- =
2
2 5
0 50 2 6
18
a
æ ö
Þ - ´ = ´ ´
ç ÷
è ø
2
5
50 2 6
18
a
æ ö
Þ - ´ = ´ ´
ç ÷
è ø
a =
250 250
–
324 2 6
´
´ ´
» = –16 ms–2.
Case-2 : Initial velocity, u = 100 km/hr
=
5
100
18
´ m/sec
v = 0, s = s, a = a
As v2 – u2 = 2as
Þ
2
2 5
0 100 2
18
æ ö
- ´ =
ç ÷
è ø
as
Þ
2
5
100
18
æ ö
- ´
ç ÷
è ø
= 2 × (–16)× 5
s =
500 500
324 32
´
´
= 24m
27. (a) In first case
u1 = u ; v1 =
2
u
, s1 = 3 cm, a1 = ?
Using, 2 2
1 1 1 1
2
- =
v u a s ...(i)
2
2
2
æ ö
-
ç ÷
è ø
u
u = 2 × a × 3
Þ a =
2
–
8
u
In second case:Assuming the same retardation
u2 = u /2 ; v2 = 0 ; s2 = ?;
2
2
8
-
=
u
a
2 2
2 2 2 2
2
- = ´
v u a s ...(ii)

2 2
2
–
0 2
4 8
æ ö
- = ´
ç ÷
è ø
u u
s
Þ s2 = 1 cm
28. (d) For first car
u1 = u, v1 = 0, a1 = – a, s1 = s1
As 2 2
1 1 1 1
2
- =
v u a s
Þ –u2 = –2as1
Þ u2 = 2as1
Þ s1 =
2
2
u
a
...(i)
For second car
u2 = 4u, v1 = 0, a2 = – a, s2 = s2
2 2
2 2 2 2
2
- =
v u a s
Þ –(4u)2 = 2(–a)s2

Þ 16 u2 = 2as2
Þ s2 =
2
8u
a
...(ii)
Dividing (i) and (ii),
2
1
2
2 2 8
= ×
s u a
s a u
=
1
16
29. (a) According to question, train A and B are running on
parallel tracks in the opposite direction.
36 km/h
A
1.8 km/h
36 km/h = 10 m/s
A
V =
72 km/h
B
72 km/h = –20 m/s
B
V = -
VMA = –1.8 km/h = –0.5 m/s
Vman, B = Vman, A + VA, B
= Vman, A + VA – VB = –0.5 + 10 – (–20)
= – 0.5 + 30 = 29.5 m/s.
30. (a)
31. (d)
R (Observer)
v
vP P
60
o
Q
Distance, PQ = vp × t (Distance = speed × time)
Distance, QR =V.t
PQ
cos60
QR
° =
p
p
v t
1 v
v
2 V.t 2
´
= Þ =
32. (c)
Car Bus
200 m
4 m/sec2
2 m/sec2
Given, uC
= uB
= 0, aC
= 4 m/s2
, aB
= 2 m/s2
hence relative acceleration, aCB
= 2 m/sec2
Now, we know, 2
1
s ut at
2
= +
2
1
200 2t u 0
2
= ´ =
Q
Hence, the car will catch up with the bus after time
t 10 2 second
=
33. (c) Person’s speed walking only is
1 escalator
60 second
Standing the escalator without walking the speed is
1 escalator
40 second
Walking with the escalator going, the speed add.
So, the person’s speed is
1 1 15
60 40 120
+ =
escalator
second
So, the time to go up the escalator
120
t
5
= = 24 second.
34 (d) Let 'S' be the distance between two ends 'a' be the
constant acceleration
As we know v2 – u2 = 2aS
or, aS =
2 2
2
-
v u
Let v be velocity at mid point.
Therefore,
2 2
2
2
- =
c
S
v u a
2 2
= +
c
v u aS
2 2
2 2
2
-
= +
c
v u
v u
vc =
2 2
2
u v
+
35. (c) For upward motion of helicopter,
2 2 2
2 0 2 2
v u gh v gh v gh
= + Þ = + Þ =
Now, packet will start moving under gravity.
Let 't' be the time taken by the food packet to reach the
ground.
2
1
2
s ut at
= +
2 2
1 1
2 2 0
2 2
h gh t gt gt gh t h
Þ - = - Þ - - =
or,
2 2 4
2
2
2
g
gh gh h
t
g
± + ´ ´
=
´
or,
2 2
(1 2) (1 2)
gh h
t t
g g
= + Þ = +
or, 3.4
h
t
g
=
36. (c) For uniformlyaccelerated/ deaccelerated motion :
2 2
2
= ±
v u gh
As equation is quadratic, so, v-h graph will be a parabola

P-24 Physics
1
3
2
v
2
h
d
at = 0,
t h = d
collision
takes
place
increases downwards
V
velocity changes its direction
V decreases upwards
1 2:
®
2 ®
2 3:
®
Initially velocity is downwards (–ve) and then after
collision it reverses its direction with lesser magnitude, i.e.
velocity is upwards (+ve).
Note that time t = 0 corresponds to the point on the graph
where h = d.
Next time collision takes place at 3.
37. (08.00) Let the ball takes time t to reach the ground
Using,
2
1
2
S ut gt
= +
2
1
0
2
S t gt
Þ = ´ +
Þ 200 = gt2
[ 2 100 ]
S m
=
Q
200
t
g
Þ = …(i)
In last
1
2
s, body travels a distance of 19 m, so in
1
–
2
t
æ ö
ç ÷
è ø
distance travelled = 81
Now,
2
1 1
– 81
2 2
g t
æ ö
=
ç ÷
è ø
2
1
– 81 2
2
g t
æ ö
= ´
ç ÷
è ø
1 81 2
–
2
t
g
´
æ ö
Þ =
ç ÷
è ø
1 1
( 200 – 81 2)
2 g
= ´ using (i)
2(10 2 – 9 2)
g
Þ =
2 2
g
Þ =
g = 8 m/s2
38. (a) For a body thrown vertically upwards acceleration
remains constant (a = – g) and velocity at anytime t is
given by V = u – gt
During rise velocity decreases linearly and during fall
velocity increases linearly and direction is opposite to
each other.
Hence graph (a) correctly depicts velocity versus time.
39. (b) y1 = 10t – 5t2 ; y2 = 40t – 5t2
for y1 = – 240m, t = 8s
y2 – y1 = 30t for t 8s.
for t 8s,
y2 – y1 = 240 – 40t –
1
2
gt2
40. (c) Speed on reaching ground
u
H
v = 2
2
+
u gh
Now, v = u + at
Þ 2
2
+ = - +
u gh u gt
Time taken to reach highest point is
u
t
g
= ,
Þ
2
2
+ +
= =
u u gH nu
t
g g
(from question)
Þ 2gH = n(n –2)u2
41. (b) For downward motion v = –gt
The velocity of the rubber ball increases in downward
direction and we get a straight line between v and t with a
negative slope.
Also applying 0
-
y y =
2
1
2
+
ut at
We get
2
1
2
y h gt
- = - 2
1
2
y h gt
Þ = -
The graph between y and t is a parabola with y = h at t = 0.
As time increases y decreases.
For upward motion.
The ball suffer elastic collision with the horizontal elastic
plate therefore the direction of velocityis reversed and the
magnituderemains the same.
Here v = u – gt where u is the velocity just after collision.
As t increases, v decreases. We get a straight line between v
and t with negative slope.
Also
2
1
2
= -
y ut gt
All these characteristics are represented by graph (b).
42. (d) Initial velocity of parachute
after bailing out,
u = 2gh
u = 50
8
.
9
2 ´
´ = 5
14
The velocity at ground,
s
/
m
2
a -
=
s
/
m
3
v
m
50
2
v = 3m/s
S =
2 2
2 2
-
´
v u
=
4
980
32
-
» 243 m
Initiallyhe has fallen 50 m.
Total height from where
he bailed out = 243 + 50 = 293 m

43. (a) We have 2
1
2
s ut gt
= + ,
Þ h = 0 × T +
1
2
gT2
Þ h =
2
1
2
gT
Vertical distance moved in time
3
T
is
2 2
1 1
' '
2 3 2 9 9
T gT h
h g h
æ ö
= Þ = ´ =
ç ÷
è ø
Position of ball from ground
9
h
h
= -
8
9
h
=
44. (b) Ball A is thrown upwards with
h
B
A
u
u
velocity u
from the building. During its
downward journeywhen it comes
back to the point of throw, its
speed is equal to the speed of
throw
(u). So, for the journey of both
the balls from point A to B.
We can apply v2 – u2 = 2gh.
As u, g, h are same for both the
balls, vA = vB

P-26 Physics
TOPIC 1 Vectors
1. A force $ $
( 2 3 )
F i j k
®
= + +
$ N acts at a point $ $
(4 3 )
i j k
+ -
$ m.
Then the magnitude oftorqueabout the point $ $
( 2 )
i j k
+ +
$ m
will be x N-m. The value of x is ______.
[NA Sep. 05, 2020 (I)]
2. The sum of two forces P
r
and Q
r
is R
r
such that | |
R
r
=
| |
P
r
. The angle q (in degrees) that the resultant of 2 P
r
and Q
r
will make with Q
r
is _______.
[NA7 Jan. 2020 II]
3. Let 1
A
uur
= 3, 2
A
uuu
r
= 5 and 1 2
A A
+
uur uuu
r
= 5. The value of
( ) ( )
1 2 1 2
2A 3A 3A 2A
+ · -
uur uuu
r uur uuu
r
is : [8 April 2020 II]
(a) – 106.5 (b) – 99.5
(c) –112.5 (d) –118.5
4. In the cube of side ‘a’ shown in the figure, the vector
from the central point of the face ABOD to the central
point of the face BEFO will be: [10 Jan. 2019 I]
(a) ( )
1 ˆ ˆ
a
2
k i
- (b) ( )
1 ˆ
ˆ
a
2
i k
-
(c) ( )
1 ˆ ˆ
a
2
j i
- (d) ( )
1 ˆ
ˆ
a
2
j k
-
5. Two forces Pand Q, of magnitude 2F and 3F, respectively,
are at an angle q with each other. If the force Q is
doubled, then their resultant also gets doubled. Then, the
angle q is: [10 Jan. 2019 II]
(a) 120° (b) 60°
(c) 90° (d) 30°
6. Two vectors A
ur
and B
u
r
have equal magnitudes. The
magnitudeof ( )
A B
+
ur u
r
is‘n’timesthemagnitudeof ( )
A B .
-
ur u
r
The angle between A
ur
and B
u
r
is: [10 Jan. 2019 II]
(a)
2
1
2
n 1
cos
n 1
- é ù
-
ê ú
+
ë û
(b) 1 n 1
cos
n 1
- -
é ù
ê ú
+
ë û
(c)
2
1
2
n 1
sin
n 1
- é ù
-
ê ú
+
ë û
(d) 1 n 1
sin
n 1
- -
é ù
ê ú
+
ë û
7. Let ˆ ˆ
A (i j)
= +
r
and ˆ ˆ
B (i j)
= -
r
. The magnitude of a
coplanar vector C
r
such that A.C B.C A.B
= =
r r r r
r r
is given
by [Online April 16, 2018]
(a)
5
9
(b)
10
9
(c)
20
9
(d)
9
12
8. A vector A
ur
is rotated by a small angle Dq radian (Dq 1)
to get a new vector B
u
r
. In that case B A
-
ur
u
r
is :
(a) A
ur
Dq (b) B A
Dq -
ur
u
r
(c) A
2
1
2
æ ö
Dq
-
ç ÷
ç ÷
è ø
ur
(d) 0
9. If ,
A B B A
´ = ´
r r
r r
then theanglebetweenAand Bis[2004]
(a)
2
p
(b)
3
p
(c) p (d)
4
p
Motion in a Plane
3

Motion in a Plane P-27
TOPIC 2
Motion in a Plane with
Constant Acceleration
10. Aballoon ismovingupin air verticallyaboveapointAonthe
ground. When it isat a height h1, a girl standingat a distance
d (point B) from A (see figure) sees it at an angle 45º with
respect to the vertical. When the balloon climbs up a further
height h2, it is seen at an angle 60º with respect tothevertical
ifthegirlmovesfurtherbyadistance2.464d(pointC).Thenthe
height h2 is (given tan 30º = 0.5774): [Sep. 05, 2020 (I)]
A B C
45° 60°
h1
h2
d 2.464d
(a) 1.464d (b) 0.732d
(c) 0.464d (d) d
11. Starting from the origin at time t = 0, with initial velocity
ˆ
5 j ms–1, a particle moves in the x–y plane with a constant
acceleration of ˆ ˆ
(10 4 )
i j
+ ms–2. At time t, its coordiantes
are (20 m, y0 m). The values of t and y0 are, respectively :
[Sep. 04, 2020 (I)]
(a) 2 s and 18 m (b) 4 s and 52 m
(c) 2 s and 24 m (d) 5 s and 25 m
12. The position vector of a particle changes with time
according to the relation $
2 2
(t) 15t (4 20t ) .
r i j
= + -
r
$
What is the magnitude of the acceleration at t = 1?
[9April 2019 II]
(a) 40 (b) 25 (c) 100 (d) 50
13. A particle moves from the point ( )
ˆ ˆ
2.0 4.0 m
i j
+ , at t = 0,
with an initial velocity ( ) 1
ˆ ˆ
5.0 4.0 ms
i j -
+ . It is acted upon
bya constant force which produces a constant acceleration
( ) 2
ˆ ˆ
4.0 4.0 ms
i j -
+ . What is the distance of the particle
from the origin at time 2s? [11 Jan. 2019 II]
(a) 15m (b) 20 2m
(c) 5m (d) 10 2m
14. A particle is moving with a velocityv
r = K (y ˆ
i + x ĵ ),
where K is a constant. The general equation for its path
is: [9 Jan. 2019 I]
(a) y = x2
+ constant (b) y2
= x + constant
(c) y2
= x2
+ constant (d) xy = constant
15. A particle starts from the origin at t = 0 with an initial
velocity of ˆ
3.0i m/s and moves in the x-y plane with a
constant acceleration ˆ ˆ
(6.0 4.0 )
i j
+ m/s2
. The x-
coordinate of the particle at the instant when its y-
coordinate is 32 m is D meters. The value of D is:
[9 Jan. 2020II]
(a) 32 (b) 50 (c) 60 (d) 40
16. A particle is moving along the x-axis with its coordinate
with time ‘t’given byx(t) = 10 + 8t – 3t2
.Another particleis
moving along they-axiswith its coordinate asa function of
time given by y(t) = 5 – 8t3
. At t = 1 s, the speed of the
second particle asmeasured in the frameofthe first particle
is given as v . Then v (in m/s) is____ [NA 8 Jan. 2020 I]
17. A particle moves such that its position vector r
r
(t) = cos
wt ˆ
i + sin wt ĵ where w is a constant and t is time. Then
which of the following statements is truefor the velocity v
r
(t) and acceleration a
r
(t) of the particle: [8 Jan. 2020 II]
(a) v
r
is perpendicular to r
r
and a
r
is directed awayfrom
the origin
(b) v
r
and a
r
both are perpendicular to r
r
(c) v
r
and a
r
both are parallel to r
r
(d) v
r
is perpendicular to r
r
and a
r
is directed towards
the origin
18. A particle is moving with velocity ˆ ˆ
( )
k yi xj
n = +
r
, where k
is a constant. The general equation for its path is [2010]
(a) y = x2 + constant (b) y2 = x + constant
(c) xy = constant (d) y2 = x2 + constant
19. A particle has an initial velocity of ˆ ˆ
3 4
+
i j and an
acceleration of ˆ ˆ
0.4 0.3
+
i j . Its speed after 10 s is : [2009]
(a) 7 2 units (b) 7 units
(c) 8.5 units (d) 10 units
20. The co-ordinates of a moving particle at any time ‘t’are
given by 3
x t
= a and 3
y t
= b . The speed of the particle
at time ‘t’ is given by [2003]
(a) 2 2
3t a + b (b) 2 2 2
3t a + b
(c) 2 2 2
t a + b (d) 2
2
b
+
a
TOPIC 3 Projectile Motion
21. A particle of mass m is projected with a speed u from the
ground at an angle q =
3
p
w.r.t. horizontal (x-axis). When
it has reached its maximum height, it collides completely
inelastically with another particle of the same mass and
velocity ˆ.
ui Thehorizontaldistancecoveredbythecombined
mass before reaching the ground is: [9 Jan. 2020 II]

P-28 Physics
(a)
2
3 3
8
u
g
(b)
2
3 2
4
u
g
(c)
2
5
8
u
g
(d)
2
2 2
u
g
22. The trajectory of a projectile near the surface of the earth
is given as y = 2x – 9x2
. If it were launched at an angle q0
with speed v0
then (g = 10 ms–2
): [12 April 2019 I]
(a) q0
= sin–1
1
5
and v0
=
5
3
ms–1
(b) q0
= cos–1
2
5
æ ö
ç ÷
è ø and v0
=
3
5
ms–1
(c) q0
= cos–1
1
5
æ ö
ç ÷
è ø and v0
=
9
3
ms–1
(d) q0
= sin–1
2
5
æ ö
ç ÷
è ø and v0
=
3
5
ms–1
23. A shell is fired from a fixed artillery gun with an initial
speed u such that it hits the target on the ground at a
distance R from it. If t1
and t2
are the values of the time
taken by it to hit the target in two possible ways, the
product t1
t2
is : [12 April 2019 I]
(a) R/4g (b) R/g (c) R/2g (d) 2R/g
24. Two particles are projected from the same point with the
same speed u such that they have the same range R, but
different maximum heights, h1
and h2
. Which of the
following is correct ? [12 April 2019 II]
(a) R2
= 4 h1
h2
(b) R2
=16 h1
h2
(c) R2
= 2 h1
h2
(d) R2
= h1
h2
25. A plane is inclined at an angle a = 30o
with respect to the
horizontal. A particle is projected with a speed u = 2 ms–1
,
from the base ofthe plane, as shown in figure. Thedistance
from the base, at which theparticle hits the planeis close to
: (Takeg=10 ms–2
) [10 April 2019 II]
(a) 20cm (b) 18cm (c) 26cm (d) 14cm
26. A body is projected at t = 0 with a velocity 10 ms–1 at an
angle of 60° with the horizontal. The radius of curvature
of its trajectory at t = 1s is R. Neglecting air resistance
and taking acceleration due to gravity g = 10 ms–2, the
value of R is: [11 Jan. 2019 I]
(a) 10.3 m (b) 2.8 m
(c) 2.5m (d) 5.1m
27. Two guns A and B can fire bullets at speeds 1 km/s and
2 km/s respectively. From a point on a horizontal
ground, they are fired in all possible directions. The
ratio of maximum areas covered by the bullets fired by
the two guns, on the ground is: [10 Jan. 2019 I]
(a) 1:16 (b) 1: 2 (c) 1: 4 (d) 1: 8
28. The initial speed ofa bullet fired from a rifle is630 m/s. The
rifleis fired at thecentre ofa target 700m awayat thesame
level as the target. How far above the centre of the target ?
(a) 1.0m (b) 4.2m (c) 6.1m (d) 9.8m
29. The position ofa projectile launched from the origin at t =
0 is given by ( )
ˆ ˆ
40 50 m
r i j
= +
r
at t = 2s. If the projectile
was launched at an angle q from the horizontal, then q is
(take g = 10 ms–2) [Online April 9, 2014]
(a)
1 2
tan
3
-
(b)
1 3
tan
2
-
(c)
1 7
tan
4
-
(d)
1 4
tan
5
-
30. A projectile is given an initial velocity of ˆ ˆ
( 2 )
i j
+ m/s,
where ˆ
i is along the ground and ĵ is along the vertical.
If g = 10 m/s2 , the equation of its trajectory is : [2013]
(a) 2
5
y x x
= - (b) 2
2 5
y x x
= -
(c) 2
4 2 5
y x x
= - (d) 2
4 2 25
y x x
= -
31. The maximum range of a bullet fired from a toy pistol
mountedon a car at rest is R0= 40 m. What will betheacute
angleofinclination ofthe pistol for maximum range when
the car is moving in the direction of firing with uniform
velocityv= 20m/s, on a horizontal surface ? (g = 10 m/s2)
(a) 30° (b) 60° (c) 75° (d) 45°
32. A ball projected from ground at an angle of 45° just clears
a wall in front. Ifpoint of projection is 4 m from the foot of
wall and ball strikes the ground at a distance of6 m on the
other side of the wall, the height of the wall is :
(a) 4.4m (b) 2.4m (c) 3.6m (d) 1.6m
33. A boycan throw a stone up toa maximum height of 10 m.
The maximum horizontal distance that the boy can throw
the same stone up to will be [2012]
(a) 20 2 m (b) 10 m
(c) 10 2 m (d) 20m
34. A water fountain on the ground sprinkles water all around
it. If the speed of water coming out of the fountain is v, the
total area around the fountain that gets wet is: [2011]
(a)
4
2
p
v
g
(b)
4
2
2
p v
g
(c)
2
2
p
v
g
(d)
2
p
v
g

35. A projectile can have the same range ‘R’ for two angles
of projection. If ‘T1’ and ‘T2’ to be time of flights in the
two cases, then the product of the two time of flights is
directly proportional to. [2004]
(a) R (b)
1
R
(c) 2
1
R
(d) R2
36. A ball is thrown from a point with a speed 0
' '
v at an
elevation angle of q. From the same point and at the same
instant, a person starts running with a constant speed
0
' '
2
v
tocatch the ball. Will the person beabletocatch the ball? If
yes, what should be the angle of projection q? [2004]
(a) No (b) Yes,30°
(c) Yes,60° (d) Yes,45°
37. A boy playing on the roof of a 10 m high building throws
a ball with a speed of 10m/s at an angle of 30º with the
horizontal. How far fromthethrowing point will the ball be
at the height of 10 m from the ground ? [2003]
[ 2 1 3
10m/s , sin30 , cos30
2 2
o o
g = = = ]
(a) 5.20m (b) 4.33m (c) 2.60m (d) 8.66m
TOPIC 4
Relative Velocity in Two
Dimensions Uniform
Circular Motion
38. A clock has a continuously moving second's hand of 0.1
m length. The average acceleration of the tip of the hand
(in units of ms–2) is of the order of: [Sep. 06, 2020 (I)]
(a) 10–3 (b) 10–4
(c) 10–2 (d) 10–1
39. When a carsit at rest, its driver sees raindrops falling on
it vertically. When driving the car with speed v, he sees
that raindrops are coming at an angle 60º from the hori-
zontal. On furter increasing the speed of the car to (1 +
b)v, this angle changes to 45º. The value of b is close to:
[Sep. 06, 2020 (II)]
(a) 0.50 (b) 0.41
(c) 0.37 (d) 0.73
40. The stream of a river is flowing with a speed of 2 km/h.
A swimmer can swim at a speed of 4 km/h. What should
be the direction of the swimmer with respect to the flow
of the river to cross the river straight? [9 April 2019 I]
(a) 90° (b) 150°
(c) 120° (d) 60°
41. Ship A is sailing towards north-east with velocity km/hr
where points east and , north. Ship Bis at a distance of 80
km east and 150 km north of ShipAand is sailing towards
west at 10 km/hr.Awill be at minimum distance from Bin:
[8April 2019 I]
(a) 4.2 hrs. (b) 2.6 hrs.
(c) 3.2 hrs. (d) 2.2 hrs.
42. Two particles A, B are moving on two concentric circles
of radii R1 and R2 with equal angular speed w. At t = 0,
their positions and direction of motion are shown in the
figure : [12 Jan. 2019 II]
B
A
Y
X
R1
R2
The relative velocity A B
® ®
-
v v and t =
2
p
w
is given by:
(a) w(R1 + R2) ˆ
i (b) –w(R1 + R2) ˆ
i
(c) w(R2 – R1) ˆ
i (d) w(R1 – R2) ˆ
i
43. A particle is moving along a circular path with a constant
speed of 10 ms–1. What is the magnitude of the change in
velocityof the particle, when it moves through an angleof
60° around the centre of the circle?
(a) 10 3m/s (b) zero
(c) 10 2m/s (d) 10m/s
44. Ifa bodymoving in circular path maintains constant speed
of10 ms–1, then which ofthe following correctlydescribes
relation between acceleration and radius?
(a)
r
a
(b)
r
a
(c)
r
a
(d)
r
a

P-30 Physics
1. (195)
Given : ˆ
ˆ ˆ
( 2 3 )
F i j k
= + +
r
N
And, ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
[(4 3 ) ( 2 )] 3 2
r i j k i j k i j k
= + - - + + = + -
r
Torque, ˆ ˆ
ˆ ˆ ˆ ˆ
(3 2 ) ( 2 3 )
r F i j k i j k
t = ´ = + - ´ + +
r
r
ˆ
ˆ ˆ
ˆ
ˆ ˆ
3 1 2 7 11 5
1 2 3
i j k
i j k
t = - = - +
Magnitude of torque, | | 195.
t =
r
2. (90) Given,
R P P Q P
= Þ + =
r
r r r r
q
a
2P Q
+
2P
Q
P2
+ Q2
+ 2PQ. cosq = P2
Þ Q + 2P cosq = 0
cos –
2
Q
P
Þ q = ..(i)
2 sin
tan ( 2P cos 0)
2 cos
P
Q
Q P
q
a = = ¥ q + =
+ q
Q
Þ a=90°
3. (d) Using,
R2
= A1
2
+ A2
2
+ 2A1
A2
cos q
52
= 32
+ 52
+ 2 × 3 × 5 cos q
or cos q = – 0.3
1 2 1 2
2 3 . 3 2
A A A A
® ® ® ®
æ ö æ ö
+ -
ç ÷ ç ÷
è ø è ø
= 2A1
× 3A1
+ (3A2
) (3A1
) cos q – (2A1
)(2A2
) cos q – 3A2
× 2A2
= 6A1
2
+ 9A1
A2
cos q – 4A1
A2
cos q – 6A2
2
= 6A1
2
6A2
2
+ 5A-
1
A2
cos q
= 6 × 32
– 6 × 52
+ 5 × 3 × 5 (– 0.3)
= – 118.5
4. (c) From figure,
G
a a
ˆ ˆ
r i k
2 2
= +
r
H
a a
ˆ ˆ
r j k
2 2
= +
r
( )
H G
a a a a a
ˆ ˆ ˆ ˆ ˆ ˆ
r – r j k – i k j– i
2 2 2 2 2
æ ö æ ö
= + + =
ç ÷ ç ÷
è ø è ø
r r
5. (a) Using, R2 = P2 + Q2 + 2PQcosq
4 F2 + 9F2 + 12F2 cos q = R2
When forces Q is doubled,
4 F2 + 36F2 + 24F2 cos q = 4R2
4 F2 + 36F2 + 24F2 cos q
= 4 (13F2 +12F2cosq)= 52 F2 + 48 F2 cosq
cos q = –
2
2
12F 1
–
24F 2
=
o
120
Þ q =
6. (a) Let magnitude of two vectors A
r
and B
r
= a
2 2 2
| A B| a a 2a cos and
+ = + + q
r r
( )
2 2 2
| A – B| a a – 2a cos 180 –
= + ° q
é ù
ë û
r r
= 2 2 2
a a –2a cos
+ q
and accroding to question,
| A B| n | A –B|
+ =
r r
r r
or,
2 2 2
2
2 2 2
a a 2a cos
n
a a – 2a cos
+ + q
=
+ q
2
a
Þ
( )
2
1 1 2cos
a
+ + q
( )
2
n
1 1– 2cos
+ q
( )
( )
2
1 cos
n
1–cos
+ q
Þ =
q
using componendo and dividendo theorem, we get
2
–1
2
n –1
cos
n 1
æ ö
q= ç ÷
+
è ø
7. (a) If ˆ ˆ
C ai bj
= +
r
then A.C A.B
=
r r r r
a + b = 1 ..... (i)
B.C A.B
=
r r
r r
2a – b = 1 ..... (ii)
Solving equation (i) and (ii) we get
1 2
a , b
3 3
= =
Magnitude of coplanar vector,
1 4 5
C
9 9 9
= + =
r
8. (a) Arc length = radius × angle
So, | – | | |
= D q
u
r u
r ur
B A A
B
A
B
A –
q
9. (c) 0
A B B A
´ - ´ =
r r
r r
0
A B A B
Þ ´ + ´ =
r r
r r
0
A B
´ =
r r
Anglebetween them is 0, p, or 2 p
from the given options, p
=
q

10. (d) From figure/ trigonometry,
1
tan 45
h
d
= ° 1
h d
=
h2
h1
45° 30°
A d B 2.464d C
And, 1 2
tan30
2.464
h h
d d
+
= °
+
1 2
( ) 3 3.46
h h d
Þ + ´ =
1 2 2
3.46 3.46
( )
3 3
d d
h h d h
Þ + = Þ + =
2
h d
=
11. (a) Given : ˆ
5 m/s
u j
=
r
Acceleration, ˆ ˆ
10 4
a i j
= +
r
and
final coordinate (20, y0) in time t.
2
1
2
x x x
S u t a t
= + [ 0]
x
u =
Q
2
1
20 0 10 2 s
2
t t
Þ = + ´ ´ Þ =
2
1
2
y y y
S u t a t
= ´ +
2
0
1
5 2 4 2 18 m
2
y = ´ + ´ ´ =
12. (d) 2 2
ˆ ˆ
15 (4 20 )
r t i t j
®
= + -
ˆ ˆ
30 40
d r
v ti tj
dt
®
®
= = -
Acceleration, ˆ ˆ
30 40
d v
a i j
dt
®
®
= = -
2 2 2
30 40 50m/s
a
= + =
13. (b) As 2
1
S ut at
2
= +
r r
r
1
ˆ ˆ ˆ ˆ
S (5i 4j)2 (4i 4 j)4
2
= + + +
r
ˆ ˆ ˆ ˆ
10i 8j 8i 8 j
= + + +
f i
ˆ ˆ
r r 18i 16j
- = +
r r
f i
[ass change in position r r ]
= = -
r r r
r
ˆ ˆ
r 20i 20j
= +
r
r
| r | 20 2
=
r
14. (c) From given equation,
( )
ˆ ˆ
V K yi xj
= +
r
dx dy
ky and kx
dt dt
= =
Now
dy
x dy
dt
dx y dx
dt
= =
,Þ ydy = xdx
Integrating both side
y2
= x2
+ c
15. (c) Using
2
1
2
S ut at
= +
2
1
(along Axis)
2
y y
y u t a t y
= +
2
1
32 0 (4)
2
t t
Þ = ´ +
2
1
4 32
2
t
Þ ´ ´ =
Þ t = 4 s
2
1
2
x x x
S u t a t
= + (Along x Axis)
2
1
3 4 6 4 60
2
x
Þ = ´ + ´ ´ =
16. (580)
For pariticle ‘A’ For particle ‘B’
XA
= –3t2
+ 8t + 10 YB
= 5 – 8t3
ˆ
(8 – 6 )
A
V t i
=
r 2 ˆ
–24
B
V t j
=
r
ˆ
–6
A
a i
=
r ˆ
48
B
a tj
= -
r
At t = 1 sec
ˆ ˆ ˆ
(8 – 6 ) 2 and –24
A B
V t i i v j
= = =
r r
/
ˆ ˆ
– –2 – 24
A B
B A
V v v i j
= + =
r r r
Speed of B w.r.t. A, v 2 2
2 24
= +
4 576 580
= + =
v = 580 (m/s)
17. (d) Given, Position vector,
ˆ ˆ
cos sin
r ti t j
= w + w
r
Velocity, ˆ ˆ
(–sin cos )
dr
v ti t j
dt
= = w w + w
r
r
Acceleration,
2 ˆ ˆ
(cos sin )
d v
a ti t j
dt
= = - w w + w
r
r
2
a r
= -w
r r
a

r
is antiparallel to r
r
Also . 0
v r =
r r
v r
^
r r
Thus, the particle is performing uniform circular motion.

P-32 Physics
18. (d) v = k(yi + xj)
v = kyi + kxj
dx
dt
= ky,
dy
dt
= kx
dy
dx
=
dy dt
dt dx
´
dy
dx
=
kx
ky
ydy = xdx ...(i)
Integrating equation (i)
ydy
ò = x dx
×
ò
y2 = x2 + c
19. (a) Given ˆ ˆ
3 4 ,
= +
r
u i j ˆ ˆ
0.4 0.3
= +
r
a i j ,t=10s
From 1st equatoin of motion.
a =
–
v u
t
v = at tu
Þ v = ( ) ( )
ˆ ˆ ˆ ˆ
0.4 0.3 10 3 4
i j i j
+ ´ + +
Þ ˆ ˆ ˆ ˆ
4 3 3 4
i j j j
+ + +
Þ v = ˆ ˆ
7 7
i j
+
Þ v
r
= 2 2
7 7
+ = 7 2 unit.
20. (b) Coordinates of moving particle at time ‘t’ are
x = at3 and y = bt3
2
3
x
dx
v t
dt
= = a and
2
3
y
dy
v t
dt
= = b
2 2 2 4 2 4
9 9
x y
v v v t t
= + = a + b
2 2 2
3t
= a + b
21. (a) Using principal of conservation of linear momentum
for horizontal motion, we have
2mvx
= mu + mu cos 60°
3
4
x
u
v =
For vertical motion
2
1
0
2
h gT
= +
2h
T
g
Þ =
Let R is the horizontal distance travelled by the body.
2
1
(0)( ) (For horizontalmotion)
2
x
R v T T
= +
R = vx
T
3 2
4
u h
g
= ´
Þ R
2
3 3
8
u
g
=
22. (c) Given, y = 2x – 9x2
On comparing with,
2
2 2
tan
2 cos
gx
y x
u
= q -
q
,
We have,
tan q = 2 or cos q =
1
5
and 2 2
9
2 cos
g
u
=
q
or 2 2
10
9
2 (1/ 5)
u
=
u = 5/3 m/s
23. (d) R will be same for q and 90° – q.
Time of flights:
t1
=
2 sin
u
g
q
and
t2
=
2 sin(90 ) 2 cos
u u
g g
° -q q
=
Now, t1
t2
=
2 sin 2 cos
u u
g g
æ ö æ ö
q q
ç ÷ ç ÷
è ø è ø
=
2
2 sin 2 2
u R
g g g
æ ö
q
=
ç ÷
è ø
24. (b) For same range, the angle of projections are :
q and 90° – q. So,
h1
=
2 2
sin
2
u
g
q
and
h2
=
2 2 2 2
sin (90 ) cos
2 2
u u
g g
° -q q
=
Also, R =
2
sin 2
u
g
q
h1
h2
=
2 2
sin
2
u
g
q
×
2 2
cos
2
u
g
q
=
2 2 2
2
(2sin cos )
16
u u
g
q q
=
2
16
R
or R2
= 16 h1
h2
25. (a) On an inclined plane, time of flight (T) is given by
2 sin
cos
u
T
g
q
=
a
Substituting the values, we get
(2)(2sin15 ) 4sin15
cos30 10cos30
T
g
° °
= =
° °
Distance, S
2
1
(2cos15 ) sin30 ( )
2
T g T
= ° - °
y 2 m/s
q =15°
a = 30°
gcos30
g
x
gsin30

2
2
4 sin15 1 16sin 15
(2cos15 ) 10sin30
10 10cos30 2 100cos 30
° °
æ ö
= ° - ´ °
ç ÷
è ø
° °
16 3 16
0.1952m 20cm
60
-
= ; ;
26. (b)
g
5
v
q
q
gcos
g
60
o
10 m/s
(10 5 3)
-
Horizontal component of velocity
vx = 10cos 60° = 5 m/s
vertical component of velocity
vy = 10cos 30° = 5 3m/s
After t = 1 sec.
Horizontal componentof velocityvx = 5 m/s
Vertical component of velocity
vy = ( )
| 5 3 –10 | m / s 10–5 3
=
Centripetal, acceleration an =
2
v
R
Þ
2 2
x y
n
v v 25 100 75–100 3
R
a 10cos
+ + +
= =
q
...(i)
From figure (using (i))
10–5 3
tan 2– 3 15
5
q= = Þq= °
( )
100 2– 3
R 2.8m
10cos15
= =
27. (a) As we know, range R
2
u sin 2
g
q
=
and, area A = p R2
Aµ R2
or,Aµ u4
4
4
1 1
4
2 2
A u 1 1
A 2 16
u
é ù
= = =
ê ú
ë û
28. (c) Let ‘t’ be the time taken bythe bullet to hit the target.
700m= 630ms–1 t
Þ t = 1
700m 10
sec
9
630ms-
=
For vertical motion,
Here, u = 0
h =
2
1
2
gt
=
2
1 10
10
2 9
æ ö
´ ´ ç ÷
è ø
=
500
m
81
=6.1m
Therefore, the rifle must be aimed 6.1 m above the centre
of the target to hit the target.
29. (c) From question,
Horizontal velocity(initial),
40
20m/s
2
= =
x
u
Vertical velocity (initial), 50 = uy t +
1
2
gt2
Þ uy × 2 +
1
2
(–10) ×4
or, 50 = 2uy – 20
or, uy =
70
35m /s
2
=

35 7
tan
20 4
q = = =
y
x
u
u
Þ Angle q = tan–1 7
4
30. (b) From equation, ˆ ˆ
2
v i j
= +
r
Þ x = t …(i)
2
1
2 (10 )
2
y t t
= - …(ii)
From (i) and(ii), y = 2x – 5x2
31. (b)
32. (b)
45°
A
P
wall
6 m
4 m
O
As ball is projected at an angle 45° to the horizontal
therefore Range = 4H
or 10 = 4HÞ
10
H 2.5m
4
= =
(QRange= 4m + 6 m =10m)
Maximum height, H=
2 2
u sin
2g
q
2
2 2
H 2g 2.5 2 10
u 100
sin 1
2
´ ´ ´
= = =
q æ ö
ç ÷
è ø
or, 1
u 100 10 ms-
= =
Height of wall PA
=
2
2 2
g(OA)
1
OA tan
2 u cos
q -
q
1 10 16
4 2.4m
1 1
2
10 10
2 2
´
= - ´ =
´ ´ ´

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