IIT JEE physics (1978 to 2018)

IIT JEE PHYSICS
(1978–2018: 41 Years)
Topic-wise Complete Solutions

Combined Volume
Mechanics, Waves and Optics
Heat, Electromagnetism and Modern Physics

Jitender Singh
Shraddhesh Chaturvedi

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PsiPhiETC
2018

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ii

Published by PsiPhiETC
116, Nakshatra Colony, Balapur
Hyderbad 500005, Telangana, India.

IIT JEE Physics
Copyright c 2018 by Authors
All rights reserved.
No part of this publication may be reproduced or transmitted in any form or by any

means, electronic or mechanical, including photocopy, recording, or any information
storage and retrieval system, without permission in writing from the authors.
Request for permission to make copies of any part of the work should be mailed to:
116, Nakshatra Colony, Balapur, Hyderbad 500005, Telangana, India.
The authors have taken care in preparation of this book, but make no expressed or
implied warranty of any kind and assume no responsibility for errors or omissions.
No liability is assumed for incidental or consequential damages in connection with
or arising out of the use of the information contained herein.
Typeset in TEX.
Third Edition, 2018

2

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We dedicate this book to the hundreds of anonymous professors at IITs
who formulated the challenging problems for IIT-JEE. The book is a
showcase of their creation.

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v

Foreword
Physics starts with observing the nature. The systematic observation results in
simple rules which unlock the doors to the nature’s mystery. Having learned a
handful of simple rules, we can combine them logically to obtain more complicated
rules and gain an insight into the way this world works. The skill, to apply the
theoretical knowledge to solve any practical problem, comes with regular practice
of solving problems. The aim of the present collection of problems and solutions is
to develop this skill.
IIT JEE questions had been a challenge and a center of attraction for a big section
of students at intermediate and college level. Independent of their occurrence as
an evaluation tool, they have good potential to open up thinking threads in mind.
Jitender Singh and Shraddhesh Chaturvedi have used these questions to come up
with a teaching material that can benefit students. The explanations accompanying
the problems could bring conceptual clarity and develop the skills to approach any
unseen problem, step by step. These problems are arranged in a chapter sequence
that is used in my book Concepts of Physics. Thus a student using both the books
will find it as an additional asset.
Both Jitender Singh and Shraddhesh Chaturvedi have actually been my students
at IIT, Kanpur. Jitender Singh has been closely associated with me since long. It

gives me immense pleasure to see that my own students are furthering the cause of
Physics education. I wish them every success in this work and expect much more
contribution from them in future!
Dr. H C Verma
Professor of Physics
IIT Kanpur

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vii

Preface
This book provides a comprehensive collection of IIT JEE problems and their solutions. We have tried to keep our explanations simple so that any reader, with basic
knowledge of intermediate physics, can understand them on his/her own without
any external assistance. It can be, therefore, used for self-study.

To us, every problem in this book, is a valuable resource to unravel a deeper understanding of the underlying physical concepts. The time required to solve a problem
is immaterial as far as Physics is concerned. We believe that getting the right answer
is often not as important as the process followed to arrive at it. The emphasis in this
text remains on the correct understanding of the principles of Physics and on their
application to find the solution of the problems. If a student seriously attempts all
the problems in this book, he/she will naturally develop the ability to analyze and
solve complex problems in a simple and logical manner using a few, well-understood
principles.
For the convenience of the students, we have arranged the problems according to
the standard intermediate physics textbook. Some problems might be based on
the concepts explained in multiple chapters. These questions are placed in a later
chapter so that the student can try to solve them by using the concept(s) from
multiple chapters. This book can, thus, easily complement your favorite text book
as an advanced problem book.
The IIT JEE problems fall into one of the nine categories: (i) MCQ with single
correct answer (ii) MCQ with one or more correct answers (iii) Paragraph based
(iv) Assertion Reasoning based (v) Matrix matching (vi) True False type (vii) Fill
in the blanks (viii) Integer Type, and (ix) Subjective. Each chapter has sections
according to these categories. In each section, the questions are arranged in the
descending order of year of appearance in IIT JEE.
Detailed solutions are given for each problem. We advise you to solve each problem
yourself. You may cover the solution with a piece of paper to focus your attention to
the problem. If you can’t solve a problem, you can always look at the solution later.
However, trying it first will help you identify the critical points in the problems,
which in turn, will accelerate the learning process. Furthermore, it is advised that
even if you think that you know the answer to a problem, you should turn to its
solution and check it out, just to make sure you get all the critical points.
This book has a companion website, www.concepts-of-physics.com. The site will
host latest version of the errata list and other useful material. We would be glad
to hear from you for any suggestions on the improvement of the book. We have

tried our best to keep the errors to a minimum. However, they might still remain!
So, if you find any conceptual errors or typographical errors, howsoever small and
insignificant, please inform us so that it can be corrected in the later editions. We
believe, only a collaborative effort from the readers and the authors can make this
book absolutely error-free, so please contribute.
Many friends and colleagues have contributed greatly to the quality of this book.
First and foremost, we thank Dr. H. C. Verma, who was the inspiring force behind
this project. Our close friends and classmates from IIT Kanpur, Deepak Sharma,
Chandrashekhar Kumar and Akash Anand stood beside us throughout this work.
This work would not have been possible without the constant support of our wives
Reena and Nandini and children Akshaj, Viraj and Maitreyi.
Jitender Singh,
Shraddhesh Chaturvedi,

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Contents

I

Mechanics . . . . . . . . . . . . . . . . . . . . . . . .

1

22 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

1

Units and Measurements . . . . . . . . . . . . . . . . . . .

3

23 Laws of Thermodynamics . . . . . . . . . . . . . . . . . . 302

2

Rest and Motion: Kinematics . . . . . . . . . . . . . . .

9

24 Specific Heat Capacities of Gases . . . . . . . . . . . 309

3

Newton’s Laws of Motion . . . . . . . . . . . . . . . . . .

19

25 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

4

Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

5

Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . .

36

6

Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . .

41

7

Centre of Mass, Linear Momentum, Collision .

53

8

Rotational Mechanics . . . . . . . . . . . . . . . . . . . . . .

74

9

Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

V Electromagnetism . . . . . . . . . . . . . . . . . 345
26 Electric Field and Potential . . . . . . . . . . . . . . . . 347
27 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
28 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
29 Electric Current in Conductors . . . . . . . . . . . . . 390
10 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . 130

30 Thermal and Chemical Effects of Electric
Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414

11 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 148

31 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

12 Some Mechanical Properties of Matter . . . . . . . 165

32 Magnetic Field due to a Current . . . . . . . . . . . . 437

II Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

33 Permanent Magnets . . . . . . . . . . . . . . . . . . . . . . . 452

13 Wave Motion and Waves on a String . . . . . . . . 179

34 Electromagnetic Induction . . . . . . . . . . . . . . . . . 458

14 Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

35 Alternating Current . . . . . . . . . . . . . . . . . . . . . . . 483

15 Light Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

36 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . 490

III Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

VI Modern Physics . . . . . . . . . . . . . . . . . . . 491

16 Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . 233

37 Electric Current through Gases . . . . . . . . . . . . . 493

17 Optical Instruments . . . . . . . . . . . . . . . . . . . . . . . 273

38 Photoelectric Effect and Wave-Particle Duality 495

18 Dispersion and Spectra . . . . . . . . . . . . . . . . . . . . 275

39 Bohr’s Model and Physics of the Atom . . . . . . 505

19 Photometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

40 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521
41 Semiconductors and Semiconductor Devices . . 526

IV Thermodynamics . . . . . . . . . . . . . . . . . . 279

42 The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529

20 Heat and Temperature . . . . . . . . . . . . . . . . . . . . . 281

A Quick Reference Formulae . . . . . . . . . . . . . . . . . . 549

21 Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . 290
ix

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Part I

Mechanics

θ
L

m
v

M

1

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Chapter 1

Units and Measurements

One Option Correct

scale (m is counted beyond MSR). The value measured
by this calipers is

Q 1. There are two Vernier calipers both of which have
1 cm divided into 10 equal divisions on the main scale.
The Vernier scale of one of the calipers (C1 ) has 10
equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2 ) has
10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in
the figure. The measured values (in cm) by calipers C1
and C2 , respectively are
(2016)
2

3

X = MSR + x = MSR + mxm − vxv .

(1)

In calipers C1 , MSR1 = 2.8 cm, m1 = 7 and v1 = 7 and
in calipers C2 , MSR2 = 2.8 cm, m2 = 8 and v2 = 7.
Substitute these values in equation (1) to get
X1 = MSR1 + m1 xm1 − v1 xv1
= 2.8 + 7(0.1) − 7(0.09) = 2.87 cm.
X2 = MSR2 + m2 xm2 − v2 xv2

4

= 2.8 + 8(0.1) − 7(0.11) = 2.83 cm.

C1
0
2

5

10

3

The Vernier calipers are generally of type C1 having
m = v and least count LC = xm −xv . For these calipers,
equation (1) gives X = MSR + v × LC.
Ans. B

4

C2

0

(A) 2.85 and 2.82
(C) 2.87 and 2.86

5

10

Q 2. The diameter of a cylinder is measured using a
Vernier calipers with no zero error. It is found that
the zero of the Vernier scale lies between 5.10 cm and
5.15 cm of the main scale. The Vernier scale has 50
divisions equivalent to 2.45 cm. The 24th division of
the Vernier scale exactly coincides with one of the main
scale divisions. The diameter of the cylinder is (2013)
(A) 5.112 cm (B) 5.124 cm
(C) 5.136 cm (D) 5.148 cm

(B) 2.87 and 2.83
(D) 2.87 and 2.87

Sol. In both calipers C1 and C2 , 1 cm is divided into
10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to xm1 = xm2 =
1 cm/10 = 0.1 cm. In calipers C1 , 10 equal divisions
on the Vernier scale are equal to 9 main scale divisions.
Thus, 1 division on the Vernier scale of C1 is equal to
xv1 = 9xm1 /10 = 0.09 cm. In calipers C2 , 10 equal divisions on the Vernier scale are equal to 11 main scale
divisions. Thus, 1 division on the Vernier scale of C2 is
equal to xv2 = 11xm2 /10 = 0.11 cm.

MSR1

1 MSD = 5.15 cm − 5.10 cm = 0.05 cm,
2.45
cm = 0.049 cm.
1 VSD =
50

7xm1
x1

2

Sol. From the given data, a main scale division (MSD)
and a vernier scale division (VSD) are

3

4

C1
0

5
7xv1

X1
MSR2

LC = 1 MSD − 1 VSD = 0.001 cm.

8xm2
x2

2

The least count (LC) of the given Vernier calipers is

10

3

For the given measurement, main scale reading (MSR)
is 5.10 cm and the vernier scale reading (VSR) is 24.
Hence, diameter D of the cylinder is

4

C2
0
X2

5

10

D = MSR + VSR × LC

7xv2

= 5.10 + 24 × 0.001 = 5.124 cm.
Let main scale reading be MSR and v th division of the
Vernier scale coincides with mth division of the main

Ans. B
3

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4

Part I. Mechanics

Q 3. The density of a solid ball is to be determined in
an experiment. The diameter of the ball is measured
with a screw gauge, whose pitch is 0.5 mm and there
are 50 divisions on the circular scale. The reading on
the main scale is 2.5 mm and that on the circular scale
is 20 divisions. If the measured mass of the ball has
relative error of 2%, the percentage error in density is
(2011)

(A) 0.9 % (B) 2.4 % (C) 3.1 % (D) 4.2 %
Sol. The density of a ball of mass m and diameter D
is given by
ρ = 6m/ πD3 .

6

π

D3 dm − 3mD2 dD
D6

Sol. For the given Vernier calipers, a main scale division (MSD) is 1 MSD = 1 mm. Since 20 vernier
scale divisions (VSD) are equal to 16 MSD, we get
1 VSD = 16/20 = 0.8 mm. The least count (LC) is
given by
LC = 1 MSD − 1 VSD = 1.0 − 0.8 = 0.2 mm.
Ans. D

(1)

Differentiate equation (1) to get
dρ =

Q 4. A Vernier calipers has 1 mm marks on the main
scale. It has 20 equal divisions on the Vernier scale
which match with 16 main scale divisions. For this
Vernier calipers, the least count is
(2010)
(A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm

.

(2)

Q 5. In a screw gauge, the zero of main scale coincides
with fifth division of circular scale as shown in the figure (i). The circular scale of a screw gauge has 50 divisions and its pitch is 0.5 mm. The diameter of the ball

being measured in the figure (ii) is
(2006)

Divide equation (2) by (1) to get

0

dρ
dm
dD
=
−3
.
ρ
m
D

(3)

In error analysis, measured mass mmeasured and actual
mass mactual are related by

10
5
0

fig. (i)

0

30
25
20

fig. (ii)

(A) 1.2 mm (B) 1.25 mm (C) 2.20 mm (D) 2.25 mm
Sol. One division on main scale is p = 0.5 mm (pitch)
and one division on circular scale (Least Count) is

mactual = mmeasured ± ∆m,
where ∆m is a small positive number representing measurement error. Let ∆m and ∆D be the measurement
errors (both positive) in m and D. From equation (3),
dρ is maximum when dm = ∆m and dD = −∆D.
Thus, the error in ρ is
∆ρ
∆m
∆D
=
+3
.
ρ
m
D

(4)

The least count (LC) of a screw gauge, with pitch p
and total number of divisions on the circular scale N ,
is given by

LC = p/N = 0.5/50 = 0.01 mm.
From the figure (i), zero error corresponding to five circular scale divisions is
e = 5 × LC = 5 × 0.01 = 0.05 mm.
From the figure (ii), measurement by screw gauge corresponding to two main scale divisions and 25 circular
scale divisions is
Dmeasured = 2 × p + 25 × LC
= 2 × 0.5 + 25 × 0.01 = 1.25 mm.
The diameter of the ball is equal to its measured value
minus zero error of the screw gauge i.e.,

LC = p/N = 0.5/50 = 0.01 mm.
LC is error in measurement of the ball diameter D i.e.,
∆D = 0.01 mm. The measured diameter for the given
main scale reading (MSR) and circular scale reading (n)
is
D = MSR + (n) LC = 2.5 + (20)0.01 = 2.70 mm.
2
Given ∆m
m = 100 = 0.02. Substitute the values in equation (4) to get

∆ρ
0.01
= 0.02 + 3 ×
= 0.031 = 3.1%.
ρ
2.70
Ans. C

Dactual = Dmeasured − e = 1.25 − 0.05 = 1.20 mm.

Ans. A
Q 6. A student performs an experiment for determi2
l
nation of g = 4π
. The error in pendulum length
T2
l (≈ 1 m) is ∆l. To measure time period T , the student
takes total time of n oscillations with a stop watch of
least count ∆Tlc and (s)he commits a human error of
∆Th = 0.1 s. The amplitude of oscillation is A. The
measurement of g is most accurate for
(2006)
(A) ∆l = 5 mm, ∆Tlc = 0.2 s, n = 10, A = 5 mm
(B) ∆l = 5 mm, ∆Tlc = 0.2 s, n = 20, A = 5 mm
(C) ∆l = 5 mm, ∆Tlc = 0.1 s, n = 20, A = 1 mm
(D) ∆l = 1 mm, ∆Tlc = 0.1 s, n = 50, A = 1 mm

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Chapter 1. Units and Measurements

5

Sol. The time period of given pendulum is
T = 2π

One or More Option(s) Correct

l/g ≈ 2 s,

where we used l = 1 m and g = 9.8 m/s2 . Differentiate
and simplify the formula for g to get
∆g
∆l
∆T
=
+2
.
g
l
T

(1)

Let t be the total time of n oscillations. The error in
measurement of t is due to least count of stop watch
(∆Tlc ) and human error (∆Th ) i.e., ∆t = ∆Tlc + ∆Th .
Thus, measurement error in time period is
∆Tlc
∆Th
∆t
=
+
.
∆T =
n
n
n

Substitute ∆T in equation (1) to get
∆l
∆Tlc + ∆Th
∆g
=
+2
.
g
l
nT

(2)

In equation (2), substitute l = 1 m, T = 2 s, ∆Th =
0.1 s and values of ∆l, ∆Tlc and n from given options.
The measurement of g is most accurate for ∆l = 1 mm,
∆T = 0.1 s and n = 50. Relative error in this case is
∆g/g = 0.005.
Ans. (D)
Q 7. A wire has a mass m = 0.3 ± 0.003 g, radius
r = 0.5 ± 0.005 mm and length l = 6 ± 0.06 cm. The
maximum percentage error in the measurement of its
density is
(2004)
(A) 1 (B) 2 (C) 3 (D) 4

Q 9. In an experiment to determine the acceleration
due to gravity g, the formula used for the time period
of a periodic motion is T = 2π 7(R−r)
. The val5g

ues of R and r are measured to be (60 ± 1) mm and
(10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s,
0.57 s, 0.54 s and 0.59 s. The least count of the watch
used for the measurement of time period is 0.01 s.
Which of the following statement(s) is(are) true?
(2016)

(A)
(B)
(C)
(D)

m
m
= 2 .
V
πr l

er =

=
max

(2)

Sol. The volume V = (1.2 × 10−2 )3 = 1.728 × 10−6 m3
shall be recorded as 1.7 × 10−6 m3 upto two significant
figures.
Ans. A

5
1

where we have truncated at 2 significant figures. The
absolute error and the relative percentage error in the
measurement of T are given by
5
1

|Ti − T¯|
5
0.04 + 0 + 0.01 + 0.02 + 0.03
=
= 0.02.
5
∆T
0.02
× 100 = 3.57%.
eT = ¯ × 100 =
0.56
T

∆T =

Given equation for time period can be written as
g=

28π 2 R − r
.
5

T2

Differentiate to get the relative error in g as
∆g
∆(R − r)
∆T
∆R + ∆r
∆T
=
+2
=
+2
g
R−r
T
R−r
T
1+1
0.02
=
+2
= 0.11 = 11%.
60 − 10
0.56

Ans. D
Q 8. The edge of a cube is measured to be
1.2 × 10−2 m. Its volume should be recorded as (2003)
(A) 1.7 × 10−6 m3
(B) 1.73 × 10−6 m3

−6
3
(C) 1.70 × 10 m
(D) 1.728 × 10−6 m3

measurement of r is 10%.
measurement of T is 3.57%.
measurement of T is 2%.
determined value of g is 11%.

Ti
0.52 + 0.56 + 0.57 + 0.54 + 0.59
=
5
5
= 0.556 s ≈ 0.56 s,

From the given data, m = 0.3 g, ∆m = 0.003 g, r =
0.5 mm, ∆r = 0.005 mm, l = 6 cm, and ∆l = 0.06 cm.
Substitute the values in equation (2) to get
∆ρ
0.003 2 × 0.005 0.06
=
+
+
= 0.04 = 4%.
ρ
0.3
0.5
6

the
the
the
the

∆r
1
× 100 =
× 100 = 10%.
r
10

T¯ =

(1)

∆m
∆r
∆l
+2
+
.
m
r
l

in
in
in

in

The mean value of time period is given by

Differentiate equation (1) and divide it by ρ to get
∆ρ
ρ

error
error
error
error

Sol. The relative percentage error in the measurement
of r is given by

Sol. The density of a wire of mass m, length l, and
radius r is given by
ρ=

The
The
The
The

It is interesting to note that the time period T =
2π 7(R−r)
is same as the time period of a small spher5g
ical ball of mass m and radius r when it rolls without
slipping on a rough concave surface of large radius R.

We encourage you to derive it.
Ans. A, B, D

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6

Part I. Mechanics

Sol. In given Vernier callipers, each 1 cm is equally
divided into 8 main scale divisions (MSD). Thus,
1 MSD = 1/8 = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD)
i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD =
0.8 × 0.125 = 0.1 cm. The least count of the Vernier
callipers is given by
LC = 1 MSD − 1 VSD = 0.125 − 0.1 = 0.025 cm.
In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the
screw gauge is the distance travelled on the linear scale
when it makes one complete rotation. Since circular
scale moves by two divisions on the linear scale when it
makes one complete rotation, we get p = 2l. The least
count of the screw gauge is defined as ratio of the pitch
to the number of divisions on the circular scale (n) i.e.,
lc = p/n = 2l/100 = l/50.

Sol. From the given expression,
d = λ/(2 sin θ).
Differentiate to get

∆d = −(λ/2) csc θ cot θ ∆θ.
The absolute error in d is
|∆d| = (λ/2) csc θ cot θ |∆θ|.
The figure shows that csc θ cot θ (and hence |∆d|) decreases as θ increases from 0◦ to 90◦ .
csc θ cot θ

(2015)

(A) If the pitch of the screw gauge is twice the least
count of the Vernier callipers, the least count of
the screw gauge is 0.01 mm.
(B) If the pitch of the screw gauge is twice the least
count of the Vernier callipers, the least count of
the screw gauge is 0.005 mm.
(C) If the least count of the linear scale of the screw
gauge is twice the least count of the Vernier
callipers, the least count of the screw gauge is
0.01 mm.
(D) If the least count of the linear scale of the screw
gauge is twice the least count of the Vernier
callipers, the least count of the screw gauge is
0.005 mm.

θ in the range 0 to 90◦ . The wavelength λ is exactly
known and the error in θ is constant for all values of θ.
As θ increases from 0◦ ,
(2013)
(A) the absolute error in d remains constant.
(B) the absolute error in d increases.
(C) the fractional error in d remains constant.

(D) the fractional error in d decreases.

0◦

θ

90◦

The fractional error in d is
−(λ/2) csc θ cot θ∆θ
∆d
=
= − cot θ∆θ.
d
λ/(2 sin θ)

cot θ

Q 10. Consider a Vernier callipers in which each 1 cm
on the main scale is divided into 8 equal divisions and
a screw gauge with 100 divisions on its circular scale.
In the Vernier callipers, 5 divisions of the Vernier scale
coincide with 4 divisions on the main scale and in the
screw gauge, one complete rotation of the circular scale
moves it by two divisions on the linear scale. Then,

0◦

θ

90◦

The figure shows that cot θ (and hence ∆d/d) decreases as θ increases from 0◦ to 90◦ .
Ans. D

(1)
Integer Type

If p = 2 LC = 2(0.025) = 0.05 cm, then l = p/2 =
0.025 cm. Substitute l in equation (1) to get the least
count of the screw gauge
lc = 0.025/50 = 5 × 10−4 cm = 0.005 mm.
If l = 2 LC = 2(0.025) = 0.05 cm then equation (1)
gives

Q 12. The energy of a system as a function of time t is
given as E(t) = A2 exp(−αt), where α = 0.2 s−1 . The
measurement of A has an error of 1.25%. If the error in
the measurement of time is 1.50%, the percentage error
in the value of E(t) at t = 5 s is . . . . . . .
(2015)
Sol. Differentiate the expression E(t) = A2 e−αt to get
dE = 2Ae−αt dA − A2 αe−αt dt.

lc = 0.05/50 = 1 × 10−3 cm = 0.01 mm.

(1)

Divide equation (1) by E(t) and simplify to get

Ans. B, C
Q 11. Using the expression 2d sin θ = λ, one calculates
the value of d by measuring the corresponding angles

dA
dA
dt
dE
=2
− αdt = 2
− αt .
E
A
A
t

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(2)


Chapter 1. Units and Measurements

7

The error in measurement of a parameter x is generally
defined by

Since 10 vernier scale divisions (VSD) are equal to

9 MSD, we get

xactual = xmeasured ± ∆x,

1 VSD = 9/10 = 0.9 mm.

where ∆x is a small positive number representing measurement error. Let ∆A and ∆t be the measurement
errors (both positives) in A and t. From equation (2),
dE is maximum when dA = ∆A and dt = −∆t. Thus,
the percentage relative error in E(t) is given by
∆A
∆t
∆E
=2
+ αt
E
A
t
= 2(1.25%) + (0.2)(5)(1.5%) = 4%.

The least count (LC) is given by
LC = 1 MSD − 1 VSD = 1.0 − 0.9 = 0.1 mm.
Given, the main scale reading (MSR) is 10 and the
vernier scale reading (VSR) is one. The measured value
of the edge is given by
a = MSR × MSD + VSR × LC
= 10 × 1 + 1 × 0.1 = 10.1 mm.

We encourage you to derive equation (2) by taking logarithm on both sides of E(t) = A2 e−αt and then differentiating it.
Ans. 4

Q 13. To find the distance d over which a signal can
be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the
distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and
its frequency f . The engineer finds that d is proportional to S 1/n . The value of n is . . . . . . .
(2014)
Sol. From the given information,
d = k ρa S b f c ,

(1)

where k is some dimensionless proportionality constant
and a, b, c are unknown parameters. Substitute dimensions of physical quantities in equation (1) to get
[L1 ] = [ML−3 ]a [MT−3 ]b [T−1 ]c .

(2)

Equate the exponents of M and L in equation (2) to get
a + b = 0,

(3)

1 = −3a.

(4)

The measurement of a has three significant digits. The
volume of the cube is V = a3 = 1.03 cm3 and the density is
m/V = 2.736/1.03 = 2.6563 = 2.66 g/cm3 ,
(after rounding off to three significant digits).
Ans. 2.66 g/cm3

Q 15. The pitch of a screw gauge is 1 mm and there
are 100 divisions on the circular scale. While measuring
the diameter of a wire, the linear scale reads 1 mm and
47th division on the circular scale coincides with the
reference line. The length of the wire is 5.6 cm. Find the
curved surface area (in cm2 ) of the wire in appropriate
number of significant figures.
(2004)
Sol. The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm
(pitch). The number of divisions on the circular scale
is N = 100. Thus, one division on the circular scale is
LC = p/N = 1/100 = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR)
is 47. Thus, the diameter of the wire is
d = LSR + CSR × LC
= 1 + 47 × 0.01 = 1.47 mm = 0.147 cm.

Solve equations (3) and (4) to get b = 1/3.
Ans. 3

The curved surface area of the cylinder is

Descriptive

A = πdl = 3.14(0.147)(5.6) = 2.5848 cm2 .

Q 14. The edge of a cube is measured using a Vernier
calipers (9 divisions of the main scale are equal to 10
divisions of Vernier scale and 1 main scale division is
1 mm). The main scale division reading is 10 and first
division of Vernier scale was found to be coinciding with

the main scale. The mass of the cube is 2.736 g. Calculate the density in g/cm3 upto correct significant figures.
(2005)
Sol. From the given data, a main scale division (MSD)
is

Rounding off to two significant digits gives A =
2.6 cm2 .
Ans. 2.6 cm2
Q 16. N divisions on the main scale of a Vernier
calipers coincide with (N + 1) divisions on its Vernier
scale. If each division on the main scale is of a units,
determine the least count of instrument.
(2003)
Sol. Given, a main scale division (MSD) of the Vernier
calipers is

1 MSD = 1 mm.

1 MSD = a.

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8

Part I. Mechanics

Since (N + 1) vernier scale divisions (VSD) are equal to
N main scale divisions, we get

1 VSD =

N
Na
MSD =
.
N +1
N +1

The least count is given by
LC = 1 MSD − 1 VSD =

a
.
N +1
Ans.

a
N +1

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Chapter 2

Rest and Motion: Kinematics

The sound of impact travels upwards with a velocity v = 300 m/s. The time taken by the sound to travel
a distance L is t2 = L/v = 0.067 s. Thus, time interval

between dropping the stone and receiving the sound of
impact is

One Option Correct
Q 1. Consider an expanding sphere of instantaneous radius r whose total mass remains constant. The expansion is such that the instantaneous density ρ remains
uniform throughout the volume. The rate of fractional
is constant. The velocity v
change in density ρ1 dρ
dt
of any point on the surface of the expanding sphere is
proportional to
(2017)
(A) r (B) 1/r (C) r3 (D) r2/3

Differentiate to get

Sol. The velocity of any point, on the surface of the
expanding sphere of instantaneous radius r, is radially outwards. Its magnitude is given by v = dr/dt.
The density of the sphere of mass m is given by ρ =
m/( 43 πr3 ) which gives,
r=

3m
4πρ

2L/g + L/v.

T = t1 + t2 =

δT =

1
2

2
δL
δL
δL +
=
gL
v
L

1
2

2L L
+
g
v

,

which gives
δL
=
L

1/3

.

δT
1
2

2L
g

+

0.01

=

L
v

2(20)
10

1
2

+

≈ 0.01 = 1%.
20
300

Differentiate w.r.t. time t to get
dr
1
=−
dt
3

3m
4πρ

Note that m and

1 dρ
ρ dt

1/3

Ans. (A)

1 dρ
r 1 dρ
=−
.
ρ dt
3 ρ dt

are constants and

dρ
dt

Q 3. The given graph shows the variation of velocity
(v) with displacement (x). Which one of the following
graph correctly represents the variation of acceleration
(a) with displacement
(2005)

is negative.
Ans. (A)

Q 2. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well.
The error in his measurement of time is δT = 0.01 s and
he measures the depth of the well to be L = 20 m. Take
the acceleration due to gravity g = 10 m/s2 and the velocity of sound is 300 m/s. Then the fractional error in
the measurement, δL/L, is closest to
(2017)
(A) 1% (B) 5% (C) 3% (D) 0.2%

v
v0

O

(A)

Sol. The stone is dropped from rest and falls under
a constant acceleration g = 10 m/s2 before it hits the
bottom of the well at depth L = 20 m. The time taken
by the stone to reach the bottom of the well is given by
t1 = 2L/g = 2 s.

a

O

(C)

•

(B)
x

a

O

x

a

O

(D)
x

x0

x

a

O

x

v
L

Sol. Given graph is a straight line with equation

g

v = v0 −
9

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v0
x.
x0


10

Part I. Mechanics

Differentiate the velocity v w.r.t. time t to get the acceleration

resistance, its velocity v varies with height h above the

ground as
(2000)
v

(A)

dv
dv dx
dv
=
=v
dt
dx dt
dx
v0
v0
v2
v2
=−
v0 − x = − 0 + 02 x.
x0
x0
x0
x0

v

(B)

a=

O

Thus, a-x graph is a straight line with negative intercept
on a axis and a positive slope.
Ans. A
Q 4. A particle starts from rest. Its acceleration a (in
m/s2 ) versus time t (in s) is as shown in the figure. The
maximum speed of the particle will be
(2004)

O

h

v

(C)

O

h

v

(D)

O

h

h

Sol. The equation v 2 − u2 = 2gh gives velocity of the
ball at a height h during its downward journey as

a(m/s2 )
10

vdown = − 2g(d − h) ˆ.
O

t(s)

11

y
ˆ

(A) 110 m/s (B) 55 m/s (C) 550 m/s (D) 660 m/s

d

Sol. The acceleration a is decreasing but positive from
t = 0 to t = 11 s. Thus, speed is maximum at t = 11 s
and is given by the area under t-a graph i.e.,
vmax =

1
2

× 11 × 10 = 55 m/s.

We encourage you to solve this problem analytically.
t
10
5 2
Hint: a = 10 − 11
t, v = 0 a dt = 10t − 11
t , and v is
maximum at t = 11 s.
Ans. B
Q 5. A small block slides without friction down an inclined plane starting from rest. Let sn be the distance
sn
travelled from t = n − 1 to t = n. Then sn+1
is (2004)
2n−1
2n+1
2n−1
2n
(A) 2n (B) 2n−1 (C) 2n+1 (D) 2n+1
Sol. The downward acceleration of the block on an inclined plane with inclination angle θ is given by a =
g sin θ. Initial velocity of the block is u = 0. The distance travelled in the first n seconds is given by
s(n) = un + 21 g sin θ n2 =

1
2

Thus, the distances travelled in the n
seconds are

vup =

which is positive and becomes zero at h = d/2.
Ans. A
Q 7. In 1.0 s, a particle goes from point A to point B,
moving in a semicircle of radius 1.0 m (see figure). The
magnitude of the average velocity is
(1999)

1m
•

and the (n+1)

B

th

(A) 3.14 m/s (B) 2.0 m/s (C) 1.0 m/s (D) zero
(1)

1
2

sn+1 = s(n + 1) − s(n) = (2n + 1) g sin θ.
sn
sn+1

gd − 2gh ˆ,

A

sn = s(n) − s(n − 1) = (2n − 1) 12 g sin θ,

Divide equation (1) by (2) to get

x

This velocity remains negative till the ball strikes
the ground. After bouncing from the ground the ball
rises to a√height d/2. Thus the speed just after bouncing
is u = gd. The velocity of the ball during upward
motion at a height h is

g sin θ n2 .
th

d
2

=

(2)

2n−1
2n+1 .

Ans. C
Q 6. A ball is dropped vertically from a height d above

the ground. It hits the ground and bounces up vertically
to a height d/2. Neglecting subsequent motion and air

Sol. The average velocity is given by v = ∆r/∆t, where
∆r is the change in the displacement vector and ∆t
is time interval. The magnitude of the displacement
vector changes by |∆r| = |AB| = 2 m in time interval
∆t = 1 s. Thus,
|v| = |∆r|/∆t = 2/1 = 2 m/s.

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Ans. B


Chapter 2. Rest and Motion: Kinematics

11

Q 8. A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this
instant of time, the horizontal component of its velocity
is v. A bead Q of the same mass as P is ejected from
A at t = 0 along the horizontal string AB, with the
speed v. Friction between the bead and the string may
be neglected. Let tP and tQ be the respective times
taken by P and Q to reach the point B. Then, (1993)
Q

A

Let vb and vr be the velocities of the boat and the
river current w.r.t. the ground. The velocity of the boat
in still water is equal to the relative velocity of the boat
w.r.t. water i.e.,
vb/r = vb − vr ,
which gives
|vb/r |2 = |vb |2 + |vr |2 .

|vb | =
C

Sol. Let l be the length of the chord AB. The time
taken by the particle Q and the particle P to reach the
point B are
tQ = l/v, and tP = l/¯
vh ,
where v¯h is the average horizontal velocity of the particle P.
Q

B

N

N

P

Q 10. A river is flowing from west to east at a speed
of 5 m/min. A man on the south bank of the river,

capable of swimming at 10 m/min in still water, wants
to swim across the river in the shortest time. He should
(1983)
swim in a direction
(B) 30◦ east of north
(A) due north
(C) 30◦ west of north (D) 60◦ east of north
Sol. Let vm and vr be velocities of the man and the
river current w.r.t. the ground. The velocity of the man
in still water is equal to the relative velocity of the man
w.r.t. water i.e.,

•

•
mg

vm/r = vm − vr .

mg

C

The forces acting on the particle P are its weight
mg and normal reaction N . The weight is always in
downward vertical direction and hence cannot affect
horizontal velocity vh . However, N has a horizontal
component that accelerates particle P from A to C and
retards it from C to B. By symmetry, P will have same
velocity on points symmetrically located about C. At B,

velocity is again vh and at every point between A and
B it is greater than vh . Thus, the average horizontal
velocity of P is more than its starting value i.e., v¯h > v.
Hence, tP < tQ .
Ans. A

Q
R vr T
vm/r

α
P

vb/r

vb

d = 1km

Sol. The boat crosses the river by the shortest path if
it moves perpendicular to the river current.

θ

vm

N
E

Let vm/r and vm make angle α and θ with the

north direction. Let d be the width of the river. The
time taken by the man to cross the river is given by
width of the river
component of man velocity along north
d
=
|vm | cos θ
d
.
=
|vm/r | cos α

t=

Q 9. A boat, which has a speed of 5 km/h in still water,
crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water (in
(1988)
km/h ) is
√
(A) 1 (B) 3 (C) 4 (D) 41

vr

1
d
= 4 km/h.
=
15/60
t

Substitute |vb | and |vb/r | in equation (1) to get |vr | =
3 km/h.
Ans. B

(B) tP = tQ
length of arc ACB
P
(D) ttQ
= length
of chord AB

A

(1)

Given, speed of the boat in still water |vb/r | =
5 km/h and the speed of the boat relative to the ground

B

P

(A) tP < tQ
(C) tP > tQ

(∵ vb ⊥ vr ).

(1)

(∵ PQ = t|vm | cos θ = t|vm/r | cos α).

From equation (1), t is minimum when denominator
is maximum. Given, |vm/r | = 10 m/min, a constant.
Thus, t become minimum when cos α = 1 i.e., α = 0.
Hence, the man takes the shortest time when he swims
perpendicular to the river velocity i.e., towards north.
Note that man will reach the point T and not Q.
Ans. A

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12

Part I. Mechanics

Q 11. In the arrangement shown in the figure the end P
and Q of an unstretchable string move downwards with
uniform speed U . Pulleys A and B are fixed. Mass M
moves upwards with a speed
(1982)
(A) 2U cos θ (B) U/ cos θ (C) 2U/ cos θ (D) U cos θ

N
−vi

vf
−

45◦

vi

Sol. In the right angled triangle AOM, let side AO = x,
OM = y, and MA = r. Pythagoras theorem gives
x2 + y 2 = r 2 .

(1)
A

O

P

M

The average acceleration of the particle is given by
∆v
vf − vi
5n
ˆ − 5 eˆ
=
=
∆t
tf − ti
10
√
n
ˆ − eˆ
n

ˆ − eˆ
√
√
= 0.5 2
= 0.707
.
2
2

Thus, the magnitude of the average acceleration is
0.707 m/s2 and its direction is north-west.
Ans. C

Q

y dy/dt = r dr/dt.

(2)

In triangle AOM, y/r = cos θ. As the string is unstretchable, the rate of decrease of r (side AM) is equal
to the rate at which P is pulled down i.e.,
dr/dt = U.
Substitute in equation (2) to get the upward speed of
M as
V = dy/dt = (dr/dt)r/y = U r/y = U/ cos θ.

One or More Option(s) Correct
Q 13. A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0,
and comes to rest at t = 1 at the point x = 1. No other
information is available about its motion at intermediate times (0 < t < 1). If α denotes the instantaneous

acceleration of the particle, then,
(1993)
(A) α cannot remain positive for all t in the interval
0 ≤ t ≤ 1.
(B) |α| cannot exceed 2 at any point in its path.
(C) |α| must be ≥ 4 at some point or points in its path.
(D) α must change sign during the motion, but no other
assertion can be made with the information given.
Sol. The positions of the particle at t = 0 and at t = 1 s
are

Note that V increases when string is pulled down and
becomes infinite at θ = 90◦ .
Aliter: Let V be the upward velocity of the mass
M . The tangential velocity of any point on the string
should be a constant, U (because string is unstretchable). Consider a point of string which is in contact
with M . The component of V along the string direction MA is V cos θ. Hence, V cos θ = U .
Ans. B
Q 12. A particle is moving eastwards with a velocity
of 5 m/s. In 10 s the velocity changes to 5 m/s northwards. The average acceleration in this time is (1982)
(A) zero.
(B) √12 m/s2 towards north-east.
√1
2

E

vi

aavg =

B

When the ends P and Q of the string are pulled
down, r and y decrease but x remains constant. Differentiate equation (1) w.r.t. time to get

(C)

O

θ

θ

vf

x(0) = 0 m,

x(1) = 1 m.

The particle is at rest at t = 0 and comes to rest at
t = 1 s. Thus, initial and final velocities of the particle
are
v(0) = 0,

v(1) = 0.

(1)

The velocity v(1) of a particle having instantaneous acceleration α is given by

1

v(1) = v(0) +

α dt.

(2)

0

α

m/s2 towards north-west.
1

(D) 1/2 m/s2 towards north.
Sol. Let eˆ and n
ˆ be the unit vectors along east and
north directions, respectively. Initially, velocity of the
particle is vi = 5 ˆe m/s. The particle velocity after t =
10 s is vf = 5 n
ˆ m/s.

t(s)
1

The equations (1) and (2) give, 0 αdt = 0. Thus,
area under the α-t curve is zero. If α = 0 for all 0 < t <
1 then area becomes zero but this also makes x(1) = 0,
which is not true. If α > 0 for all 0 < t < 1 then

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Chapter 2. Rest and Motion: Kinematics

13
x

area under the curve is positive and if α < 0 for all
0 < t < 1 then area under the curve is negative. Thus,
α must change sign at some time in 0 < t < 1. Only
1
conclusions that can be made are 0 αdt = 0, α = 0,
and α must change sign (at least once) in 0 < t < 1.

•O2

O2 •

x
d2
α
α1
0
−α2

•O1

•

θ1
d1
tc

1

θ2

t

Lens

v(m/s)
2

tc> 21

tc< 21
0

O1

tc

1

•
I1

•
I2

Retina
•
I1,I2

t

If α change sign only once (say at t = tc ) and
α = α1 in 0 < t < tc and α = −α2 in tc < t <
1, where α1 and α2 are some positive constants, then
condition x(1) = 1 m is violated if α < 4 at all points.
We encourage you to explore area under v-t graph to
show that α ≥ 4 at some point(s). Hint: The area
under v-t graph, 21 × 1 × vmax = 1, gives vmax = 2 m/s.
Now, vmax = 2 = α1 tc = α2 (1 − tc ). If tc < 21 than
α1 > 4 otherwise α2 > 4.
Ans. A, C

The explanation has to do with how human
mind/eye perceive the motion. Take a simple situation
of two point objects O1 and O2 (see figure). The images I1 and I2 formed on the retina of eye are point images. As the object moves, the image on the retina also
moves. The distance moved by the image on the retina
is perceived as motion/velocity. Since eye is a lens with
distance of retina/screen w.r.t. lens almost fixed, the
angular displacement of the image is perceived as velocity. For same relative velocity of objects O1 and O2 ,
angular displacement of nearby object (θ1 in the figure)
is more than the angular displacement of the far-off object (θ2 ). Hence nearby objects appear to move faster.

Ans. B
True False Type

Assertion Reasoning Type
Q 14. Statement 1: For an observer looking out
through the window of a fast moving train, the nearby
objects appear to move in the opposite direction to the
train, while the distant objects appear to be stationary.

Q 15. A projectile fired from the ground follows a
parabolic path. The speed of the projectile is minimum at the top of its path.
(1984)
Sol. Let the projectile be fired with an initial velocity
v at an angle θ from the horizontal.
y

Statement 2: If the observer and the object are
moving at velocities V1 and V2 respectively with reference to a laboratory frame, the velocity of the object
with respect to the observer is V2 − V1 .
(2008)
(A) Statement 1 is true, statement 2 is true; statement
2 is a correct explanation for statement 1.
(B) Statement 1 is true, statement 2 is true; statement
2 is not a correct explanation for statement 1.
(C) Statement 1 is true, statement 2 is false.
(D) Statement 1 is false, statement 2 is true.
Sol. Both, statement 1 and statement 2 are true but
statement 2 is not a correct explanation of statement 1.
The relative velocity of the nearby as well as the far-off
objects is same.

P
•

v cos θ

v sin θ

v

θ
O v cos θ

x

The horizontal component of velocity, v cos θ, remains constant throughout the trajectory because there
is no acceleration in the horizontal direction. The vertical component of velocity starts decreasing from the
initial value v sin θ to zero at the topmost point P and
then again increases to v sin θ when projectile hits the
ground. Thus, the speed of the projectile at the top,
v cos θ, is minimum. We encourage you to use conservation of mechanical energy to arrive at the same conclusion.
Ans. T

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14

Part I. Mechanics

Q 16. Two balls of different masses are thrown vertically upwards with the same speed. They pass through
the point of projection in their downward motion with
the same speed. [Neglect air resistance.]
(1983)
Sol. Apply v 2 − u2 = 2as (or the conservation of mechanical energy) to show that speeds of the balls at the
projection point (in the downward journey) are equal
to their initial projection speeds.
Ans. T
Fill in the Blank Type
Q 17. The trajectory of a projectile in a vertical plane
is y = ax−bx2 , where a, b are constants, and x and y are
respectively the horizontal and vertical distances of the
projectile from the point of projection. The maximum
height attained is . . . . . . and the angle of projection
from the horizontal is . . . . . .
(1997)
Sol. Trajectory equation is y = ax − bx2 . The slope of
the trajectory is zero at maximum height i.e.,
dy
= a − 2bx = 0.
dx

(1)

Solve equation (1) to get x = a/(2b). Substitute it
in the given trajectory equation to get the maximum
height
a
a

−b
2b
2b

ymax = a

2

=

a2
.
4b

The slope of trajectory at the projection point is given
by
tan θ =

dy
dx

= a − 2b(0) = a.
x=0

Consider a small time interval ∆t at the beginning
of journey. In this time interval, all of them travel a
distance v∆t in the direction of person they are facing
i.e., person K moves a distance v∆t along the line KL,
person L moves a distance v∆t along the line LM and
so on. In the next time interval, K starts moving in

the direction of new position of L, L starts moving in
the direction of new position of M and so on. Through
this process, all persons finally meet at the centre of
the square O. The symmetry plays a key role in this
problem. Four persons always remain at the corners of
a square. The sides of this square decrease and rotate
continuously but its centre remains fixed at O.
Now, concentrate on the motion of K. The magnitude of its velocity is v and the direction of its velocity at any instant is along the side K L of the square
formed by the four persons at that instant (see figure).
Thus, its velocity vector always makes 45◦ angle with
the line joining current position of K and centre of the
square O. Hence, the
√ component of its velocity towards
O is v cos 45◦ = v/ 2 (a constant). The net displacement of K from the initial position K to√the final posi# »
tion O is vector KO with magnitude d/ 2. Thus, the
time taken by K to reach O is the ratio of ‘displacement’ and √‘component of velocity along displacement’
d/ 2
√ = d.
i.e., t = v/
v
2
Aliter: The velocity of K is v along KL. The velocity of L is zero along KL (because it is perpendicular
to the line KL). Thus, the separation KL decreases at
a rate v. Since this rate is constant, the time taken to
reduce the separation from d to zero is d/v.
Ans. d/v
Q 19. A particle moves in a circle of radius R. In half
the period of revolution its displacement is . . . . . . and
distance covered is . . . . . .
(1983)

Thus, the angle of projection is θ = tan−1 a.
2
Ans. a4b , tan−1 a
Q 18. Four persons K, L, M, N are initially at the four
corners of a square of side d. Each person now moves
with a uniform speed v in such a way that K always
moves directly towards L, L directly towards M, M directly towards N and N directly towards K. The four
persons will meet at a time . . . . . .
(1984)

Sol. Let the particle travels from the point P to the
point Q in half of the time period.

R
P•

•

O

•

Q

Sol. The trajectory followed by the four persons is very
interesting (see figure).
N

M

M

N

O

v

L

Integer Type

v
√
2

K

The distance travelled by the particle is half of the
circle perimeter i.e., πR. The displacement of the particle is a vector from the initial point to the final point
# »
i.e., PQ. The magnitude of the displacement vector is
# »
|PQ| = 2R.
Ans. 2R, πR

v

K

v
d

L

Q 20. A ball is projected from the ground at an angle
of 45◦ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon

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Chapter 2. Rest and Motion: Kinematics

15

hitting the ground for the first time, it loses half of
its kinetic energy. Immediately after the bounce, the
velocity of the ball makes an angle of 30◦ with the horizontal surface. The maximum height it reaches after
the bounce, in meters, is . . . . . . .
(2018)
Sol. The maximum height of the ball projected from
the ground with an initial speed u at an angle θ with
the horizontal is given by
H=

u2 sin2 θ
.
2g

(1)

y

Sol. Let us analyse the problem in the laboratory
frame. Let v0 be the speed of the rocket when balls
are thrown, say at time t = 0. In the laboratory
frame, the speed of a ball thrown from the left end is
vl = v0 +0.3 and that of a ball thrown from the right end
is vr = v0 −0.2. Since the rocket is accelerating towards
the right with an acceleration a = 2 m/s2 , its velocity
keeps on increasing and the left face of the rocket collides with a ball thrown from the left end before this
ball collides with the ball thrown from the right end.
This collision takes place at a time t when the distance
travelled by the rocket is equal to the distance travelled
by the ball thrown from the left i.e.,

u

H

v0 t + 12 (2)t2 = (v0 + 0.3)t.

u

θ

θ

O

P

H

x
Q

By conservation of energy, the kinetic energy of the
ball just before it hits the ground is equal to its initial
kinetic energy (K = mu2 /2). The ball loses half of its
kinetic energy upon hitting the ground. Thus, kinetic
energy of the ball just after hitting the ground is
K =

1 1
1
1
2
mu = K = · mu2 ,
2 2
2
2

(2)

2

which gives u = u2 /2. The velocity u makes an angle

θ = 30◦ with the horizontal. Apply equation (1) to get
the maximum height after hitting the ground

Solve to get t = 0.3 s. The displacement of left
ball relative to the left face of rocket at time t is
(v0 + 0.3)t − (v0 t + 12 (2)t2 ) which attains a maximum
value of 2.25 cm at time t = 0.15 s. Type of collision
(elastic/inelastic) between the rocket and the ball is not
given in the question. Since the relative distance of the
ball thrown from the left w.r.t. rocket’s left face is negligible in comparison to the chamber length, we assume
that the ball and the left face are located approximately
at the same distance from the right face. Thus, we need
to find the time at which a ball thrown from the right
end collides with the left face of the rocket. The right
ball collides with the left face when their displacements
are equal i.e.,

2

u2 sin2 θ
u sin2 θ
·
=
2g
2
2g
2
H sin2 θ
1 2gH sin θ
=

·
= ·
2
2g
2 sin θ
2 sin2 θ
(120) sin2 30◦
= 30 m.
=
2 sin2 45◦

H =

(using (1))

Q 21. A rocket is moving in a gravity free space with a
constant acceleration of 2 m/s2 along +x direction (see
figure). The length of a chamber inside the rocket is
4 m. A ball is thrown from the left end of the chamber
in +x direction with a speed of 0.3 m/s relative to the
rocket. At the same time, another ball is thrown in
−x direction with a speed of 0.2 m/s from its right end
relative to the rocket. The time in seconds when the
(2014)
two balls hit each other is . . . . . . .

0.2 m/s
4m

Solve to get t = 1.9 s ≈ 2 s.

Ans. 2

We encourage you to find the range OQ of the ball when
it hits the ground for the second time.
Ans. 30

0.3 m/s

v0 t + 12 (2)t2 = 4 + (v0 − 0.2)t.

(using (2))

Q 22. Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30◦ and 60◦
as shown in
with respect to the horizontal respectively
√
the figure. The speed of A is 100 3 m/s. At time
t = 0 s, an observer in A finds B at a distance of 500 m.
This observer sees B moving with a constant velocity
perpendicular to the line of motion of A. If at t = t0 , A
just escapes being hit by B, t0 in seconds is . . . . . . .
(2014)

A

30◦

B
60◦

a = 2 m/s2
x

Sol. Let VA and VB be the velocity vectors of airplane
A and B in a frame attached to the ground. The figure
shows VA , VB , and VB/A , the velocity of B relative to A.

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