Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018)

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A Complete Resource Book in

PHYSICS

for

JEE Main 2019
Sanjeev Kumar

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The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an
attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been
taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear
any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or
factual) that may have found their way into this book.
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ISBN 978-93-530-6216-3
eISBN 9789353063429

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Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
JEE Mains 2018 Paper

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

JEE Mains 2017 Paper

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvi

Chapter 1

Unit and Dimension. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1–1.56

Chapter 2

Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1–2.68

Chapter 3

Newton’s Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1–3.70

Chapter 4

Work, Energy, and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1–4.48

Chapter 5

Impulse and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1–5.64

Chapter 6

Rigid Body Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1–6.52

Chapter 7

Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1–7.36

Chapter 8

Properties of Solids and Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1–8.60

Chapter 9

Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1–9.102

Chapter 10

Heat and Thermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1–10.40

Chapter 11

Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1–11.20

Chapter 12

Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1–12.34

Chapter 13

Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13.1–13.108

Chapter 14

Current Electricity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1–14.56

Chapter 15

Magnetism and Magnetic Effect of Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1–15.66

Chapter 16

Electromagnetic Induction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1–16.58

Chapter 17

Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1–17.12

Chapter 18

Ray Optics and Wave Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1–18.86

Chapter 19

Modern Physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1–19.64

Chapter 20

Semiconductor and Communication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1–20.54

Mock Test Paper 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M1.1–M1.6
Mock Test Paper 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M2.1–M2.6
Mock Test Paper 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M3.1–M3.6
Mock Test Paper 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M4.1–M4.6
Mock Test Paper 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M5.1–M5.6

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Preface
About the Series
A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination.

There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen
the fundamental concepts and prepare students for various engineering entrance examinations. It provides class-tested
course material and numerical applications that will supplement any ready material available as student resource.
To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced
faculties.
About the Book
It gives me immense pleasure to present this book A Complete Resource Book in Physics For JEE Main 2019. This book
will help the students in building the analytical and quantitative skills necessary to face the examination with confidence.
This title is designed as per the latest JEE Main syllabus, spread across 20 chapters. It has been structured in an user
friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then
practice questions. The brain-map section in every chapter will help the students to revise the important formulae. The
chapter end exercises are structured in line with JEE questions; where ample number of questions on single choice correct
question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and
integer type questions are included for extensive practice. Previous 15 years’ questions of JEE Main and AIEEE are also
added in every chapter. Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and
numerical applications.
Series Features
•
•
•
•
•
•

Complete coverage of topics along with ample number of solved examples.
Includes various types of practice problems with complete solutions.
Chapter-wise Previous 15 years’ AIEEE/JEE Main questions.
Fully solved JEE Main 2017 and 2018 questions are included in the book.
5 Mock Tests based on JEE Main pattern.
5 Free Online Mock Tests as per recent JEE Main pattern.

I dedicate this book to my family for their immense support and love. Special thanks to my parents for their support
and encouragement and to my wife Pallawi and my sons Haardik and Saarthak for sustaining me throughout this project.
I would like to express my heartfelt gratitude to the Pearson team, without them I would not have been able to bring out
this book.
Any suggestions and comments from the readers would be highly appreciated. Please communicate to us if there are
any errors, misprints or other such concerns.
Sanjeev Kumar

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Physics Trend Analysis
(2007 to 2018)
S. No

Chapters

07

08

09

10

12

13

14

15

16

17

18

1

Unit, Dimensions and Vectors

–

2

1

1

–

2

1

1

2

1

2

–

1

2

Kinematics

2

–

1

3

1

2

2

–

1

1

–

–

1

3

Laws of Motion

1

1

0

0

1

1

1

2

2

1

1

1

1

4

Work, Power and Energy

1

2

0

1

1

–

1

1

2

2

1

2

1

5

Centre of Mass, Impulse
and Momentum

1

2

1

2

–

–

–

1

2

2

–

2

2

6

Rotation

3

1

1

2

–

2

–

1

2

–

2

–

2

7

Gravitation

–

1

1

–

1

1

1

–

1

1

1

1

1

8

Simple Harmonic Motion

3

–

1

–

2

2

1

2

1

–

1

1

1

9

Solids and Fluids

–

3

1

1

3

2

1

1

1

1

–

2

1

10

Waves

2

3

2

1

2

1

1

2

1

1

2

2

1

11

Heat and Thermodynamics

4

1

5

1

3

4

4

2

3

3

3

4

1

12

Optics

2

4

2

3

5

3

3

4

3

3

3

2

2

13

Current Electricity

1

4

1

3

2

–

2

8

1

2

1

3

3

14

Electrostatics

5

4

4

4

3

3

2

3

1

3

2

3

2

15

Magnetics

4

2

2

1

2

1

2

–

1

3

3

1

2

16

Electromagnetic Inductions
& AC

2

1

1

1

–

2

2

1

1

2

1

1

2

17

Modern Physics

9

4

6

6

4

4

6

1

5

4

7

5

6

Total No of Questions

40

35

30

30

30

30

30

30

30

30

30

30

30

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11

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JEE Mains 2018 Paper
1.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

(A) 2.5%
(B) 3.5%

(C) 4.5%
(D) 6%

2.All the graphs below are intended to represent the same
motion. One of them does it incorrectly. Pick it up.

(A)  18.3 kg
(C)  43.3 kg

(B)  27.3 kg
(D)  10.3 kg

4.A particle is moving in a circular path of radius a
k
under the action of an attractive potential U = − 2 .
2r
Its total energy is
k
k
(A)  − 2 (B) 
2a 2
4a
3 k
−

(C) 
Zero(D) 
2 a2

Velocity

(A) 

5.In a collinear collision, a particle with an initial speed
v0 strikes a stationary particle of the same mass. If the
final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is
v
2v0
(A)  0 (B) 
4
v
v0
(C)  0 (D) 
2
2

Time

Velocity

(B) 

Time

6.Seven identical circular planar discs, each of mass M
and radius R are welded symmetrically as shown. The
moment of inertia of the arrangement about the axis
normal to the plane and passing through the point P is

Velocity

(C) 

Time

P

Velocity
O

(D) 

Time

3.Two masses m1 = 5 kg and m2 = 10 kg, connected by an
inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of
horizontal surface is 0.15. The minimum weight m that
should be put on top of m2 to stop the motion is
m

m2

T

T
m1

m1g

A01_KUMAR_0283_01_SE_PREL.indd 8

(A) 

19
55
MR 2 (B) 
MR 2
2
2

(C) 

73
181
MR 2 (D) 
MR 2
2
2

7.From a uniform circular disc of radius R and mass 9M,
R
a small disc of radius
is removed as shown in the
3
figure. The moment of inertia of the remaining disc
about an axis perpendicular to the plane of the disc and
passing through centre of disc is

5/18/2018 5:13:32 PM


JEE Mains 2018 Paper    ix

12. A silver atom in a solid oscillates in simple harmonic
motion in some direction with a frequency of 1012/sec.
What is the force constant of the bonds connecting one
atom with the other? (Mole wt. of silver = 108 and
Avogadro’s number = 6.02 × 1023 gm mole–1)
(A)  6.4 N/m
(B)  7.1 N/m
(C)  2.2 N/m
(D)  5.5 N/m

2R
3
R

40
MR 2
(A) 4MR2(B) 
9
37
MR 2
(C) 10MR2(D) 
9
8.A particle is moving with a uniform speed in a circular
orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of
the particle is T, then
(A) T ∝ R3/2 for any n(B) 
T ∝ R n /2 +1
(C) T ∝ R(n+1)/2

(D) 
T ∝ Rn/2

9.A solid sphere of radius r made of a soft material of
bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the
surface of the liquid, covering entire cross section of
cylindrical container. When a mass m is placed on the
surface of the piston to compress the liquid, the frac dr 
tional decrement in the radius of the sphere,   , is
 r 
Ka
Ka
(A) 

(B) 
mg
3mg
(C) 

mg

3Ka

mg
(D) 
Ka

10. Two moles of an ideal monoatomic gas occupies a
volume V at 27°C. The gas expands adiabatically to a
volume 2V. Calculate (a) the final temperature of the

gas and (b) change in its internal energy.
(A)  (a) 189 K  (b) 2.7 kJ
(B)  (a) 195 K  (b) -2.7 kJ
(C)  (a) 189 K  (b) -2.7 kJ
(D)  (a) 195 K  (b) 2.7 kJ
11. The mass of a hydrogen molecule is 3.32 × 10-27 kg.
If 1023 hydrogen molecules strike, per second, a fixed
wall of area 2 cm2 at an angle of 45° to the normal, and
rebound elastically with a speed of 103 m/s, then the
pressure on the wall is nearly
(A) 2.35 × 103 N/m2
(B) 4.70 × 103 N/m2
2
2
(C) 2.35 × 10 N/m
(D) 4.70 × 102 N/m2

A01_KUMAR_0283_01_SE_PREL.indd 9

13. A granite rod of 60 cm length is clamped at its middle
point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental
frequency of the longitudinal vibrations?
(A)  5 kHz
(B)  2.5 kHz
(C)  10 kHz
(D)  7.5 kHz
14. Three concentric metal shells A, B and C of respective
radii a, b and c (a < b < c) have surface charge densities +σ, -σ and +σ respectively. The potential of shell
B is



σ  a2 − b2
σ  a2 − b2
+ c  (B) 
+ c
(A)  

∈0  a
∈0  b


(C) 



σ  b2 − c2
σ  b2 − c2
+ a  (D) 
+ a


∈0  b
∈
c
0 



15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material
5

of dielectric constant K = is inserted between the
3
plates, the magnitude of the induced charge will be
(A)  1.2 nC
(B)  0.3 nC
(C)  2.4 nC
(D)  0.9 nC
16. In an a.c. circuit, the instantaneous emf and current
are given by

e = 100 sin30t
π

i = 20 sin  30t − 

4

In one cycle of a.c., the average power consumed by
the circuit and the wattless current are, respectively:
1000
, 10

(A) 
50, 10(B) 
2
50
, 0 (D) 
(C) 
50, 0
2

17. Two batteries with emf 12 V and 13 V are connected
in parallel across a load resistor of 10 Ω. The internal
resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between
(A)  11.6 V and 11.7 V
(B)  11.5 V and 11.6 V
(C)  11.4 V and 11.5 V
(D)  11.7 V and 11.8 V

5/18/2018 5:13:34 PM


x    JEE Mains 2018 Paper
18. An electron, a proton and an alpha particle having the
same kinetic energy are moving in circular orbits of
radii re, rp, rα respectively in a uniform magnetic field
B. The relation between re, rp, rα is
(A) re > rp = rα
(B) re < rp = rα
(C) re < rp < rα
(D) re < rα < rp
19.

The dipole moment of a circular loop carrying a current I is m and the magnetic field at the centre of the
loop is B1. When the dipole moment is doubled by
keeping the current constant the magnetic field at the
B
centre of the loop is B2. The ratio 1 is
B2

3

(A) 
2(B) 
1
(C)  2 (D) 
2
20. For an RLC circuit driven with voltage of amplitude vm
1
the current exhibits resoand frequency ω0 =
LC
nance. The quality factor, Q is given by
ω L
ω0 R
(A)  0 (B) 
R
L
R
CR
(C) 
(D) 
(ω0C )
ω0
21. An EM wave from air enters a medium. The elec
 z 
tric fields are E1 = E01 xˆ cos  2π v  − t   in air and
 c 

E2 = E02 xˆ cos[ k ( 2 z − ct )] in medium, where the
wave number k and frequency v refer to their values in
air. The medium is non-magnetic. If ∈r1 and ∈r2 refer

to relative permittivities of air and medium respectively, which of the following options is correct?
(A) 
(C) 

∈r1
∈r2
∈r1

∈r2

= 4
=

1

4

(B) 
(D) 

∈r1
∈r2
∈r1
∈r2

=2
=

1
2

22. Unpolarized light of intensity I passes through an ideal
polarizer A. Another identical polarizer B is placed
behind A. The intensity of light beyond B is found to
1
be . Now another identical polarizer C is placed
2
between A and B. The intensity beyond B is now found
1
to be . The angle between polarizer A and C is
8

A01_KUMAR_0283_01_SE_PREL.indd 10

(A) 0°
(C) 45°

(B) 30°
(D) 60°

23. The angular width of the central maximum at a single
slit diffraction pattern is 60°. The width of the slit is
1 μm. The slit is illuminated by monochromatic plane
waves. If another slit of same width is made near it,
Young’s fringes can be observed on a screen placed at
a distance 50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance? (i.e.,
distance between the centres of each slit.)
(A) 25 μm(B) 
50 μm

(C) 75 μm(D) 
100 μm
24. An electron from various excited states of hydrogen
atom emit radiation to come to the ground state. Let
λn, λg be the de Broglie wavelength of the electron in
nth state and the ground state respectively. Let Λn be
the wavelength of the emitted photon in the transition
from the nth state to the ground state. For large n, (A, B
are constants)
B
(A)  Λ n ≈ A + 2
λn
(B)  Λ n ≈ A + Bλn
(C)  Λ 2n ≈ A + Bλn2
(D)  Λ 2n ≈ λ
25. If the series limit frequency of the Lyman series is vL,
then the series limit frequency of the Pfund series is.
(A) 25vL(B) 
16vL
(C) vL/16(D) 
vL/25
26. It is found that if a neutron suffers an elastic collinear
collision with deuterium at rest, fractional loss of its
energy is Pd; while for its similar collision with carbon
nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively
(A)  (0.89, 0.28)
(B)  (0.28, 0.89)
(C)  (0, 0)
(D)  (0, 1)
27. The reading of the ammeter for a silicon diode in the

given circuit is
200 Ω

3V

(A) 0
(C)  11.5 mA

(B) 15 mA
(D)  13.5 mA

5/18/2018 5:13:38 PM


JEE Mains 2018 Paper    xi
28. A telephonic communication service is working at
carrier frequency of 10 GHz. Only 10% of it is utilized
for transmission. How many telephonic channels can
be transmitted simultaneously if each channel requires
a bandwidth of 5 kHz?
(A) 2 × 103(B) 
2 × 104
5
(C) 2 × 10 (D) 
2 × 106


29. In a potentiometer experiment, it is found that no current
passes through the galvanometer when the terminal of
the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω,

a ­balance is found when the cell is connected across
40 cm of the wire. Find the internal resistance of the cell.
(A) 1 Ω(B) 
1.5 Ω
(C) 2 Ω(D) 
2.5 Ω
30. On interchanging the resistances, the balance point of
a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much
was the resistance on the left slot before interchanging
the resistances?
(A) 990 Ω

(B) 505 Ω
(C) 550 Ω(D) 
910 Ω

Answer keys
1. (C)2. 
(A)3. 
(B)4. 
(C)5. 
(B)6. 
(D)7. 
(A)8. 
(C)9. 
(C)10. 
(C)
11.  (A)12. 
(B)13. 

(A)14. 
(B)15. 
(A)16. 
(B)17. (B)18. 
(B)19. 
(C)20. 
(A)
21.  (C)22. 
(C)23. 
(A)24. 
(A)25. 
(D)26. 
(A)27. 
(C)28. 
(C)29. 
(B)30. 
(C)

Hints and Solutions
m m
=
v l3

1. Density= p=

∴

∆p ∆m
∆l
=

+3
p
m
l


⇒

∆p
∆m
∆l
× 100 =
× 100 + 3 × 100
p
m
l

4.U = −

k
2r 2


Force = F = −

dU
d  k  −k
= − − 2  = 3
dr
dr  2r  r

|F | =

k mv 2
k
=
⇒ mv 2 = 2
3
r
r
r


Hence, the correct option is (C).

=
KE


1 2 1 k
=
mv
2
2 r2

2.Incorrect option is (A) because it represents two values of
velocity at the same instant, which is impossible.

Total energy = KE + PE


Hence, the correct option is (A).



 = 1.5 + 3 × 1 = 4.5%



3.For stopping the motion:

T = m1 g(1)

T = μ(m + m2)g(2)

∴ m1 g = μ(m + m2)g

⇒m=

m1
− m2
µ

5
− 10 = 23.33 kg
0.15

∴ Minimum weight m = 27.3 kg

=


Hence, the correct option is (B).

A01_KUMAR_0283_01_SE_PREL.indd 11

=

k
k
−
=0
2r 2 2r 2


Hence, the correct option is (C).
5.Momentum conservation yields:

mv0 = mv1 + mv2
⇒ v0


= v1 + v2 (1)

1
1
1

mv22 + mv12 = 1.5 × mv02
2
2
2
⇒ v12 + v22 =


3 2
v0 (2)
2

5/18/2018 5:13:41 PM


xii    JEE Mains 2018 Paper

Solving the two equations, we get


⇒ T2 = 189 K

v1 − v2 = 2v0


 dQ = dU + dW


Hence, the correct option is (B).

∴ dU = − dW [∵ dQ = 0]

 MR 2
MR 2
2
6. I 0 =  2 + M ( 2 R)  × 6 + 2




=


By parallel axis theorem:


Hence, the correct option is (C).


∴ dU = -2.7 kJ

I p = I 0 + 7M (3R) 2


11. F = ∆p = 2mv cosθ × n
∆t

 MR 2

MR 2 + 7M (3R)2

∴Ip = 
+ M ( 2 R) 2  × 6 +
2
 2



Pressure = P =

181
=
MR 2
2




Hence, the correct option is (D).
7.Mass per unit area =

9M
π R2

9 MR 2  M  R 
 2R 
−   +M

2
2

3
 3 
  
2

= 4MR2

Hence, the correct option is (A).
k
8. F = k ⇒
= n
Rn
R
R

∴

m  2π r 
k
= n
R  T 
R

⇒


1
1
∝
∴T ∝ R

T 2 R n +1

n +1
2

9. K = ∆P
∆V
V

∴




 = 2.35 × 103 N/m 2

2 × 10 −4 × 2

12. Time period = T = 2π
2





Frequency = f =

Where, m =

m

k

1
1
=
T 2π

k
m

108 × 10 −3
kg
6.02 × 10 23

1
k × 6.02 × 10 23
2 × 3.14
108 × 10 −3
k × 6.02 × 10 23
1
2 × 3.14
108 × 10 −3


⇒ k = 7.1 N/m


Hence, the correct option is (C).

⇒

 =


⇒ 1012 =

2

2 × 3.32 × 10 −27 × 103 × 10 23




∴f =

mv 2

F 2mv cosθ × n
=
A
A


Hence, the correct option is (A).

π R2 9 M π R2

Mass of
=

×
=M
9
9
π R2

Moment of inertia of the remaining disc

I = I total − I hole =

− nR
−2 × 8.3
(T1 − T2) =
(300 − 189)
γ −1
0.6


Hence, the correct option is (B).
13. For fundamental frequency:
v
l
λ l
= ⇒
=
4 2
4f 2

∴f =

∆V ∆P mg
=
=
V
K
Ka

2v v
=
4l 2l


⇒ f =

dr 1 ∆V
mg
=
=
3Ka
r 3 V



 =


Hence, the correct option is (C).

1
2l

y
p

1
2 × 60 × 10 −2
5 kHz

9.27 × 1010
2.7 × 103

10. For adiabatic expansion


Hence, the correct option is (A).

T1V1γ −1


14. Potential of shell B is given by:

= T2V2γ −1
γ −1

V 

T2 = T1  1 
 V2 

A01_KUMAR_0283_01_SE_PREL.indd 12

5
−1
 1 3

= 300  
2


VB =

1 qA
1 qB
1 qC
+
+
4π ∈0 b 4π ∈0 b 4π ∈0 c

5/18/2018 5:13:47 PM


JEE Mains 2018 Paper    xiii

Where q A = σ 4π a 2 ; qB = σ 4π b 2 ; qC = σ 4π c 2

18. r =


Putting the values of qA, qB, and qC, we get

mv
qB
1 2
mv ⇒ v =
2

2k
m


σ  a2 − b2

VB =
+ c

∈0  b



KE = k =


Hence, the correct option is (B).


∴r =

m 2K
2mK
=

qB m
qB

1

15. Induced charge Qi = Q 1 − 
 K


re =

2me k
; rp =
eB

2m p k

1

∴ Qi = ( KC )V 1 − 

K




rα =

2 × 4m p k

2m p k




= KCV − CV = CV ( K − 1)
=

2eB

eB

=


 3 − 1



eB


Hence, the correct option is (B).
19. Dipole moment = M = iA

= 1.2 nC

Hence, the correct option is (A).

∴ M = iπ r 2 ⇒ M ∝ r 2

16. Wattless current = irms sin φ


B=

io



= 10 A

2

20

sin φ =

=

2

=



=

100

2

ì

20
2

ì

ì

1
2




1
à oi
B
2r
r

r1
=
r2

M1
M
r2 1

1 = 12 =
M2
M 2 r2 2

r1 B2

Also, =
=
r2 B1


Average power = Vrmsirms cos φ


1
2

1000


∴

1
2

B1
= 2
B2

Hence, the correct option is (C).

2

ω0 L
R


Hence, the correct option is (B).

20. Quality factor = Q =

17. Equivalent emf = E = E1r2 + E2 r1
r1 + r2


Hence, the correct option is (A).


∴E =

12 × 2 + 13 × 1 37
=
V
1+ 2
3


Equivalent resistance = r = r1r2
r1 + r2

∴r =

1× 2 2
= Ω
1+ 2 3


Current = i =


VR = iR =

37
3

2
+ 10
3

21. E1 = E01xˆ cos  2π v  z − t  

 c 



 2π c z  z    

or, E1 = E01xˆ cos 2π v   − t − t  
 λ  c  c    

= E01xˆ cos [ k ( z − ct )]


We know that, k = ε r
=

37
A
32

37
× 10 = 11.56 V
32


Hence, the correct option is (B).

A01_KUMAR_0283_01_SE_PREL.indd 13

2mα k
2eB

∴ re < rp = rα


5
90 × 10 −12 × 20

; rα =


∴

ε r1  k1 
= 
ε r2  k2 

2

   z c 
ˆ cos  2π
k vz− −tt

Also, EE21 = E02
01x
   c 2 

∴ k1 = k ; k2 = 2k

5/18/2018 5:13:53 PM


xiv    JEE Mains 2018 Paper

⇒

ε r1  K 2 1
=
=
4
ε r2  2 K 



1 = R  1 − 1 
2
2
Λ
n
n
2
 1


Hence, the correct option is (C).


∴

22.
I

I I
= cos2θ
2 2

I
2

A

B

1
1 1 
= R − 2 
Λn
1 n 


1
1 

⇒
= R 1 − 2 2 
Λn
 λn c 

⇒ Λn =

λn2c 2
2
(λn c 2 − 1) R

=

cos 2 θ = 1 ⇒ θ = 0


Let the angle between the pass axes of A and C be b

I
A



I
2

C

IC

I
I C = cos 2 β

2
I


I B =  cos 2 β  cos 2 β
2

I I
1

∴ = (cos 2 β ) 2 ⇒ (cos 2 β ) 2 =
8 2
4
1
1

∴ cos 2 β = ⇒ cos β =
2
2

∴ β = 45°

Hence, the correct option is (C).
23. 2θ = 60° ⇒ θ = 30°

Condition for diffraction minima:

B

IB



=

1
1 
1 −


R  λn2c 2 



=

1
1 
1 +

R  λn2c 2 
A+



1

1 
1 − 2 2  R
 λn c 
−1

B
λn2


Hence, the correct option is (A).


25. 1 = R  1 − 1 

2
2
λ
n
n
2
 1
v
1 1 
L = R  − 
C
1 ∞ 

⇒ vL = RC (1)

For Pfund
 1
1

vP = CR  2 − 
(
5
)
∞



∴ vP =

CR vL

=
25 25

d sin θ = λ ⇒ λ = 10 −6 sin 30° = 0.5 × 10 −6



Hence, the correct option is (D).


For double slit:

26. Velocity of neutron after the collision is:


β=

λD
d

0.5 × 10 −6 × 0.5
d

⇒ d = 25 µm


⇒ 10 −2 =


Hence, the correct option is (A).

V1 =

m1 − m2
u
( m1 + m2 )

1
1
m1v 2 − m1v12
2
2

Fractional loss in KE =
1
m1v 2
2

2
24. KE= k= 13.6 = P
2
n
2me

=

v 2 − v12
v2

k h
= ; where k is a constant
n λ
nh

∴ λn =
= nc; where c is a constant
k

=

( m1 + m2 ) 2 − ( m1 − m2 ) 2
( m1 + m2 ) 2

=

4 m1m2
( m1 + m2 ) 2


⇒P=

A01_KUMAR_0283_01_SE_PREL.indd 14



5/18/2018 5:13:59 PM


JEE Mains 2018 Paper    xv

For collision with deuterium,
=
m1 m=
; m2 2m

∴ Pd =

4 × m × 2m 8
= = 0.89
( m + 2m) 2 9


For collision with carbon,
=
m1 m=
; m2 12m

PC =

4 × m × 2m
48
=
= 0.28
2
( m + 12m)
169


40λ = E − ir = E −

E
r
( R + r)

r 

∴ 40λ = E 1 −

 R+ r
r 

⇒ 40λ = 52λ 1 −

R
+ r 


⇒

R
10
=
R + r 13

27. For a silicon diode the barrier potential is 0.7 volts


⇒

5
10
=
⇒ r = 1.5 Ω
5 + r 13

 3 − 0.7 
3

∴i = 
 × 10 = 11.5 mA.
 200 


Hence, the correct option is (B).


Hence, the correct option is (C).

30.


Hence, the correct option is (A).

28. 10% of 10 GHz = 10 × 109 ×

∴ Number of channels =

10
= 109 Hz

100

109
= 2 × 105
5 × 103


Hence, the correct option is (C).
29. Let the voltage gradient be λ

∴ Terminal voltage = 40λ

A01_KUMAR_0283_01_SE_PREL.indd 15

R1
R2
(1)
=
x (100 − x )

R
R1
(2)
2 =
x − 10 (110 − x )

Also, R1 + R2 = 1000 (3)

Solving these equations, we get

R2 = 450 Ω; R1 = 550 Ω

Hence, the correct option is (C).

5/18/2018 5:14:01 PM


JEE Mains 2017 Paper
1.An observer is moving with half the speed of light
towards a stationary microwave source, emitting
waves at frequency 10 GHz. What is the frequency of
the microwave measured by the observer?

(Speed of light, c = 3 × 108 m/s)
(A)  12.1 GHz
(C)  15.3 GHz

(B)  17.3 GHz
(D)  10.1 GHz

2.The following observations were taken for determining surface tension T of water by capillary method:
Diameter of capillary, D = 1.25 × 10-2 m; rise of water,
h = 1.45 × 10-2 m. Using g = 9.80 m/s2 and the simplirhg
fied relation, T =
× 103 N/m, the possible error in
2
surface tension is closest to
(A) 1.5%
(B) 2.4%
(C) 10%

(D) 0.15%
3.Some energy levels of a molecule are shown in the
­figure. The ratio of the wavelengths, r = λ1 /λ2 , is
given by

3
(B) 
1
2
3
3
(C) 
(D) 
2
2
7.A radioactive nucleus A with a half life T, decays into
a nucleus B. At t = 0, there is no nucleus B. At some
value of t, the ratio of the number of B to that of A is
0.3. Then, t is given by
log 1.3
(A)  t = T
log 2
(A) 

(B) t = T log(1.3)
T
(C)  t =
log(1.3)

−E

λ2
−4E
3

λ1
−2E

−3E

(A)  r =

2
3
r=
(B) 
4
3

(C)  r =

1
4
r=
(D) 
3
3

4.A body of mass, m =10-2 kg is moving in a medium
and experiences a frictional force, F = -kv2. Its initial

speed is, v0 = 10 m/s. If, after 10 seconds, its energy is
1 2
mv0 , then the value of k will be
8
(A) 10-3 kg/s
(B)  10-4 kg/m
-1
(C) 10 kg/m/s
(D)  10-3 kg/m
5.Cp and Cv are specific heats at constant pressure and
constant volume, respectively. It is observed that

A01_KUMAR_0283_01_SE_PREL.indd 16

Cp – Cv = a, for hydrogen gas; Cp – Cv = b, for nitrogen
gas. The correct relation between a and b is
(A) a = b(B) 
a = 14b
1
a= b
(C) a = 28b(D) 
14
6.The moment of inertia of a uniform cylinder of length
l and radius R about its perpendicular bisector is I.
What will be the ratio of l to R, such that the moment
of inertia is minimum?

T log 2
2 log 1.3
8.Which of the following statements is false?

(A) In a balanced wheatstone bridge if the cell and
the galvanometer are exchanged, the null point is
disturbed.
(B)  A rheostat can be used as a potential divider.
(C) 
Kirchhoff 
’s second law represents energy
conservation.
(D) 
Wheatstone bridge is the most sensitive when
all the four resistances are of the same order of
magnitude.
(D)  t =

9.A capacitance of 2 µF is required in an electrical circuit
across a potential difference of 1.0 kV. A large number
of 1 µF capacitors are available which can withstand a
potential difference of not more than 300 V. The minimum number of capacitors required to achieve this, is
(A) 16
(B) 24
(C) 32
(D) 2
10. In the given circuit diagram, when the current reaches
steady state in the circuit, the charge on the capacitor
of capacitance C will be

5/18/2018 5:14:04 PM


JEE Mains 2017 Paper    xvii

E

and 520 nm, is used to obtain interference fringes on
the screen. The least distance from the common central maximum to the point where the bright fringes due
to both the wavelengths coincide, is
(A)  7.8 mm
(B)  9.75 mm
(C)  15.6 mm
(D)  1.56 mm

r

r1
C

r2

(A)  CE

r1

( r2 + r )

r2
CE
(B) 
( r + r2 )

(C)  CE

r1

( r1 + r )

(D) 
CE

11.

2V

2V

1Ω
2V

2V

1Ω
2V

1Ω
2V

16. An electric dipole has a fixed dipole moment p,
which makes an angle θ with respect to X-axis. When
subjected to an electric field, E1 = Eiˆ, it experiences
a torque, T1 = τ kˆ. When subjected to another electric
field, E2 = 3E1 ˆj , it experiences a torque, T2 = −T1 .
Then, the angle θ is

(A) 45° (B) 
60°
(C) 90° (D) 
30°
17. A slender uniform rod of mass M and length l is pivoted at one end, so that it can rotate in a vertical plane
(see figure). There is negligible friction at the pivot.
The free end is held vertically above the pivot and then
released. The angular acceleration of the rod when it
makes an angle θ with the vertical, is


In the above circuit the current in each resistance is
(A)  0.25 A
(B)  0.5 A
(C)  0 A
(D)  1 A
12. In amplitude modulation, sinusoidal carrier frequency
used is denoted by ωc and the signal frequency is
denoted by ωm. The bandwidth (∆ωm) of the signal is
such that, ∆ωm << ωc. Which of the following frequencies is not contained in the modulated wave?
(A) ωc(B) 
ωm + ωc
(C) ωc - ωm(D) 
ωm
13. In a common emitter amplifier circuit using an n-p-n
transistor, the phase difference between the input and
the output voltages will be:
(A) 90° (B) 
135°
(C) 180° (D) 

45°
14. A copper ball of mass 100 gm is at a temperature
T. It is dropped in a copper calorimeter of mass 100
gm, filled with 170 gm of water at room temperature.
Subsequently, the temperature of the system is found
to be 75°C. T is given by

(Given: Room temperature = 30°C, specific heat of
copper = 0.1 cal/gm°C)
(A) 885°C (B) 
1250°C
(C) 825°C (D) 
800°C
15. In a Young’s double slit experiment, slits are separated
by 0.5 mm, and the screen is placed 150 cm away. A
beam of light consisting of two wavelengths 650 nm

A01_KUMAR_0283_01_SE_PREL.indd 17

Z

θ

X

2g
3g
sin θ (B) 
cos θ
3l

2l
2g
3g
sin θ
(C)  cos θ (D) 
3l
2l
18. An external pressure P is applied on a cube at 0°C,
so that it is equally compressed from all sides. K is
the bulk modulus of the material of the cube and α is
its coefficient of linear expansion. Suppose we want
to bring the cube to its original size by heating. The
temperature should be raised by
(A) 

P
3α
(B) 
αK
PK
P
(C) 3PKα (D) 
3α K
19. A diverging lens with magnitude of focal length 25 cm is
placed at a distance of 15 cm from a converging lens of
magnitude of focal length 20 cm. A beam of parallel light
falls on the diverging lens. The final image formed is
(A) 

5/18/2018 5:14:07 PM

xviii    JEE Mains 2017 Paper
(A) Virtual and at a distance of 40 cm from convergent lens.
(B) Real and at a distance of 40 cm from the divergent
lens.
(C) Real and at a distance of 6 cm from the convergent lens.
(D) Real and at a distance of 40 cm from convergent
lens.
20. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It
produces continuous as well as characteristic X-rays.
If λmin is the smallest possible wavelength of X-rays
in the spectrum, the variation of log λmin with log V is
correctly represented in

(A) log λ min
log V

10

Current
(amp)

Time

(A)  225 Wb
(C)  275 Wb

0.5 sec

(B)  250 Wb
(D)  200 Wb

23. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full
scale deflection. The value of the resistance to be put
in series with the galvanometer to convert it into a
voltmeter of range 0–10 V is
(A)  2.045 × 103 Ω
(B)  2.535 × 103Ω
3
(C)  4.005 × 10 Ω
(D)  1.985 × 103Ω
24. A time dependent force, F = 6t acts on a particle of
mass 1 kg. If the particle starts from rest, then the work
done by the force during the first 1 second will be
(A)  22 J
(B)  9 J
(C)  18 J
(D)  4.5 J

(B) log λ min
log V

(C) log λ min
log V

25. A magnetic needle of magnetic moment 6.7 × 10–2
Am2 and moment of inertia 7.5 × 10–6 kgm2 is performing simple harmonic oscillations in a magnetic field of
0.01T. Time taken for 10 complete oscillations is
(A)  8.89 s

(B)  6.98 s
(C)  8.76 s
(D)  6.65 s
26. The variation of acceleration due to gravity g with distance d from centre of the earth, is best represented by
(R = Earth’s radius)
g

(D) log λ min
log V

21. The temperature of an open room of volume 30 m3
increases from 17°C to 27°C due to the sunshine. The
atmospheric pressure in the room remains 1 × 105 Pa.
If ni and nf are the number of molecules in the room
before and after heating, then nf – ni will be
(A)  1.38 × 1023
(B)  2.5 × 1025
25
(C)  –2.5 × 10
(D)  –1.61 × 1023
22. In a coil of resistance 100 Ω, a current is induced by
changing the magnetic flux through it, as shown in the
­figure. The magnitude of change in flux through the coil is

A01_KUMAR_0283_01_SE_PREL.indd 18

(A) 

O

R

d

g

(B) 

O

R

d

5/18/2018 5:14:07 PM


JEE Mains 2017 Paper    xix
g

28. A particle A of mass m and initial velocity of v collides
m
with a particle B of mass which is at rest. The colli2
sion is head on, and elastic. The ratio of the de Broglie
wavelengths λA to λB after the collision is

(C) 

O

d

R

(A) 

λA
=2
λB

(B) 

λA 2
=
λB 3

(C) 

λA 1
=
λB 2

(D) 

λA 1
=
λB 3

g

(D) 
d

O

27. A body is thrown vertically upwards. Which one of
the following graphs correctly represent the velocity
vs time?

29. A particle is executing simple harmonic motion with
a time period T. At time t = 0, it is at its position of
­equilibrium. The kinetic energy-time graph of the particle will look like

v

KE

(A) 

(A) 

t

0

T

t

KE

(B) 

0

v

T
2

T

t

T
4

T
2

T t

T
2

T

KE

(B) 

(C) 

t

0

KE

(D) 
v

(C) 

0

T

t

30. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by
a factor of

t

1
(A)  (B) 
81
9

v

(D) 

(C) 

1
(D) 
9
81

t

A01_KUMAR_0283_01_SE_PREL.indd 19

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xx    JEE Mains 2017 Paper

Answer keys
1. (B)2. 
(A)3. 
(C)4. 
(B)5. 
(B)6. 
(D)7. 
(A)8. 
(A)9. 
(C)10. 

(B)
11.  (C)12. 
(D)13. 
(C)14. 
(A)15. 
(A)16. 
(B)17. (D)18. 
(D)19. 
(D)20. 
(D)
21.  (C)22. 
(B)23. 
(D)24. 
(D)25. 
(D)26. 
(C)27. 
(B)28. 
(A)29. 
(C)30. 
(D)

Hints and Solutions
v0

1. f ′ = f

c+u
c−u

= 10 ×

c + c/2
GHz = 17.3 GHz.
c − c/2


− k (10)

⇒


=

rhg
 Drhg 
3
× 103 = 
 × 10
2
 4 

∆T ∆D ∆h
=
+
+ (g is constant)
T
r
h

C p − Cυ =


0.01 0.01
+
1.25 1.45

⇒a=



= 0.0149

∴  %error =

 1 2
− v 
  v0

2 1

− k (10) = −10 −2  − 
 v0 v0 

⇒ k = 10-4 kg/m.

Hence, the correct option is (B).
5.For H2,

Hence, the correct option is (B).
2.T =

= 10 −2

R
( M H = 2)
2

R
(1) 
2


For N2,
∆T
× 100 = 1.49% 1.5%.
T


C p − Cυ =


Hence, the correct option is (A).
hc 4 E
E
3. =
−E = 
λ2
3

3


⇒b=
(1) 

R
( M N = 28)
28

R
(2) 
28


From Eqs. (1) and (2), we get

a
hc
= 14
= 2 E − E = E (2) 
b
λ1

a = 14b.

Dividing Eq. (1) by Eq. (2), we get

Hence, the correct option is (B).
λ1 1

= .
MR 2 Ml 2
λ2 3
+
6. I =
4
12

Hence, the correct option is (C).
2

π R ρl = M
4.F = ma
M

l=
dv

− kv 2 = m
π R2 ρ
dt

Now,

v
1 2 1 2
mv = mv0 ∴ v = 0
2
8
2

10

v0 / 2

0

v0


∴ − k ∫ dt = m

A01_KUMAR_0283_01_SE_PREL.indd 20

∫

dv
v2


I=

MR 2 M
M2
+
+ 2 4 2
4
12 π R ρ

I=

1 
M 2
M2
R + 2 2 × 4 
4 
3π ρ
R 

5/18/2018 5:14:13 PM


JEE Mains 2017 Paper    xxi
1/4 μF

2


dI = M  2 R − 4 M

2
2
dR 4 
3π ρ R5 


For I to be minimum,
dI
= 0
dR

2R =

R=


⇒

8 rows

4M 2
3π 2 ρ 2 R5
250 V

2l 2
3R

l
=
R

250 V

250 V

250 V

1000 V

3

.
2


∴ The minimum number of capacitors required = 4 × 8 = 32.

Hence, the correct option is (C).
10. In steady state no current will pass through capacitor.

Current through cell, i =

R

l

Axis

E
( r + r2 )


Potential difference across capacitor = Potential difference
across r2

V = ir2

=

Er2
( r + r2 )

∴  Charge on capacitor = CV
=


Hence, the correct option is (D).
7.

A 
→ B
at t = 0, N 0
0
at t = t , N


Hence, the correct option is (B).
11. 6 V

4V

2V

0V

6V

4V

2V

A
0V

N0 − N

N −N
0
= 0.3
N

N0 = 1.3N

N0 = 1.3 N0e-λt

⇒ λt = ln(1.3)

t=

CEr2
.
( r + r2 )

1
ln(1.3)
λ


Taking voltage of point A as = 0

Then voltage at other points can be written as shown in
figure

Hence voltage across all resistance is zero.

Hence current = 0

log(1.3)
t =T

log( 2)

12. Refer NCERT Page No. 526

Three frequencies are contained
ω + ωc , ωc − ωm and ωc
m


Hence, the correct option is (A).
8.Conceptual.

Hence, the correct option is (A).
9.To get capacitance of 2 µF, such that potential difference
across any one should not be more than 300 V, can be
arranged as follows:

13. Conceptual

Hence, the correct option is (C).

14. From principle of calorimetry,

Heat lost by copper sphere = 
Heat gained by calorimeter
+ Heat gained by water

⇒ 100 × 0.1 × (T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

A01_KUMAR_0283_01_SE_PREL.indd 21

5/18/2018 5:14:17 PM


xxii    JEE Mains 2017 Paper

⇒ T – 75 = 810

⇒ T = 885°C.

Hence, the correct option is (A).
15. Let n1th order maxima of wavelength 520 nm coincides with
n2th order maxima of wavelength 650 nm.

That is, n1 × 520 = n2 × 650
n
5

⇒ 1 =
n2 4

So, 5th order maxima of wavelength 520 nm coincides with
4th order maxima of wavelength 650 nm. Then, the least distance from central maxima is,

Y=

18. dV = –V(3α) dT(1) 

K =−







(2) 


From Eqs. (1) and (2), we get dT =

P
.
3Kα


Hence, the correct option is (D).
19.

L1

5λ D
d

= 5×


P
 dV
 V


L2

I2

I1

(520 × 10 −9 ) × 150 × 10 −2
5 × 10 −4

f = 25 cm


= 7.8 mm.

Hence, the correct option is (A).

f = 20 cm
15 cm

For ‘L1,’ image will be formed at a distance of 25 cm from
‘L1’ (left)

For ‘L2,’ u = –40 cm

16. τ1 = τ 2

⇒ PE sin θ = P × 3E sin(90 − θ )
⇒ tan θ = 3


⇒ θ = 60°.




f = +20 cm




 v = +40 cm (using lens formula)


⇒ Real image will be formed at a distance of 40 cm from L2,
i.e., converging lens (right of L2).

Y

Hence, the correct option is (D).
P

θ
O


Hence, the correct option is (B).
17.

M

θ

l
Mg

20.

X

hc
= eV
λmin

eV
1
=
λmin hc

e
 1 
= nV + n
n

λ
hc
 min 
e
− n(λmin ) = nV + n
hc

 e 
n(λmin ) = − nV − n  
 hc 


It is a straight line with -ve slope.
21. Using, PV = NRT (N = Number of moles), we get

105 × 30 = NiR × 290

and 105 × 30 = Nf R × 300

l sin θ
2

l
Ml 2

Mg sin θ =
α
2
3
3
g sin θ .
2l

Hence, the correct option is (D).


⇒ N f − Ni =

=

105 × 30  −10 
R  300 × 290 


=

−105
R × 290


α=

A01_KUMAR_0283_01_SE_PREL.indd 22

105 × 30  1
1 
−
R  300 290 

5/18/2018 5:14:21 PM


JEE Mains 2017 Paper    xxiii

⇒ For molecules,


For d ≥ R,

n f − ni = N 0 ( N f − N i )



g=

−6.023 × 10 23 × 105
8.314 × 290


=


Hence, the correct option is (C).
27. v = u – gt

= –249.8 × 1023


GM
.
d2

v

−2.5 × 10 25.


Hence, the correct option is (C).
22. ε = iR

t

dφ
= iR
dt

Hence, the correct option is (B).


∫ dφ = R ∫ idt
1
1

= 100 ×  × 10 × 
2
2

28.

m
A

m
2
B
Rest

v


= 250 Wb.

Hence, the correct option is (B).
R

23.

5 mA


( R + 15) ×

mv = mv1 +

15 Ω
G

5
= 10
1000


R + 15 = 2000

R = 1985 Ω.

Hence, the correct option is (D).
24. a = 6t

v = v1 +


v

B

v2

m
v2
2

v2
(1) 
2


Also, e =

v2 − v1
=1
v


v = v2 – v1(2) 

Adding Eqs. (1) and (2), we get
2 v = v2 +


t

v2 3
4v
= v2   v2 =
2 2
3


v=

4v

− v1
3


v1 =

v
4v
−v =
3
3


∫0 dv = 6 ∫0 tdt
6t 2
=
v = 3t 2

2

v = 3 m/s
W = ∆KE =


1 2
mv0 = 4.5 J.
2


=

PA mv
=
1

I
= 0.665 s
MB


For 10 oscillations, time = 10T = 6.65 s.

Hence, the correct option is (D).
26. For d ≤ R,
GM
d
R3

A01_KUMAR_0283_01_SE_PREL.indd 23

mv
3


PB =

m
m 4v
v2 = ×
2
2 3

PB =

2mv
3


Hence, the correct option is (D).

g=


v1


From Eq. (1)  ,

dV
= 6t
dt

25. T = 2π

A


P=

h

λ

mv
PA = λB = 3
PB λ A 2 mv
3

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xxiv    JEE Mains 2017 Paper
1
λ
B =
λA 2

∴

T/4 T/2

λA
= 2.
λB


Hence, the correct option is (C).


Hence, the correct option is (A).
29.

T

t=0
mean


v = vmax cos ω t
1 2
KE = mvmax
cos 2 ωt

2

A01_KUMAR_0283_01_SE_PREL.indd 24

30. Stress, σ =

F mg ρ LgA
=
=
= ρ Lg
A
A
A


Now when length increases,

σ′ =

mg ρ A′9 Lg
= 9 ρ Lg.
=
A′
A′


Hence, the correct option is (D).

5/18/2018 5:14:29 PM