Volume 24
Managing Editor
Mahabir Singh

February 2016

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CONTENTS

Editor
Anil Ahlawat
(BE, MBA)

No. 2

Physics Musing Problem Set 31

8

JEE Main Practice Paper

10

Core Concept

22

JEE Workouts

26

JEE Accelerated Learning Series

31

Brain Map

46

Ace Your Way CBSE XII

54

AIPMT Practice Paper

63

Physics Musing Solution Set 30

73

Exam Prep 2016

75


You Ask We Answer

82

Live Physics

83

At a Glance 2015

84

Crossword

85

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Physics for you | FEBRUARY ‘16

7


P

PHYSICS

MUSING

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment
the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.
In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed
solutions of these problems will be published in next issue of Physics For You.
The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who
send atleast five correct solutions will be published in the next issue.
We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.


31
single oPtion correct tyPe
1. A particle starts from rest at A and moves with
uniform acceleration a m s–2 in a straight line. After
1/a seconds, a second particle starts from A and
moves with uniform velocity u in the same line and
same direction. If u > 2 m s–1 then during the entire
motion, the second particle remains ahead of first
particle for a duration
a
(a) 2 u(u − 2)
(b)
u(u − 2)
2
a
1
2
(c)
(d)
u(u − 2)
u(u − 2)
a
a
2. A block of mass 100 g moves with a speed
of 5 m s–1 at the highest point in a closed circular
tube of radius 10 cm kept in a vertical plane. The
cross-section of the tube is such that the block just fits
in it. The block makes several oscillations inside the
tube and finally stops at the lowest point. The work

done by the tube on the block during the process is
(a) 1.45 J
(b) – 1.45 J
(c) 0.2 J
(d) zero
3. Two identical balls A and B are released from the
positions shown in figure. They collide elastically
on horizontal portion MN. All surfaces are smooth.
The ratio of heights attained by A and B after
collision will be (neglect energy loss at M and N)
(a) 1 : 4
B
A
(b) 2 : 1
4h
(c) 4 : 1
h
45° M N 60°
(d) 2 : 5

4. A stone of mass m, tied to the end of a string, is
whirled around in a horizontal circle. (Neglect the
force due to gravity.) The length of the string is
reduced gradually, keeping the angular momentum
of the stone about the centre of the circle constant.
Then, the tension in the string is given by T = Arn,
where A is a constant, r is the instantaneous radius
of the circle, and n is
(a) 1
(b) – 1

(c) – 2
(d) – 3
8

Physics for you | february ‘16

5. Consider two hollow glass spheres, one containing
water and the other containing mercury. Each liquid
fills about one-tenth of the volume of the sphere. In
zero gravity environment
(a) water and mercury float freely inside the
spheres
(b) water forms a layer on the glass while mercury
floats
(c) mercury forms a layer on the glass while water
floats
(d) water and mercury both form a layer on the
glass.
subjective tyPe
6. A body starts from rest and moving with uniform
acceleration of 4 m s–2 covers half of its total path
during the last second of its motion. Find the time
taken and the total distance covered.
7. A cone of height h and base radius r is fixed base to
base on a hemisphere of equal radius. Find h so that
the centre of gravity of the composite solid lies on
the common base. (Assume same density for both
objects.)
8. A rod PQ of length l is pivoted at an end P and
freely rotated in a horizontal plane at an angular

speed w about a vertical axis passing through P. If
coefficient of linear expansion of material of rod is
a, find the percentage change in its angular velocity ,
if temperature of system is increased by DT.
9. Distance between the centres of two stars is 10a. The
masses of these stars are M and 16 M and their radii
a and 2a, respectively. A body of mass m is fired
straight from the surface of the large star towards
the smaller star. What should be its minimum initial
speed to reach the surface of the smaller star?
10. The first overtone of an open organ pipe beats with
the first overtone of a closed organ pipe with a beat
frequency of 2.2 Hz. The fundamental frequency of
the closed organ pipe is 110 Hz. Find the lengths of
the pipes. (Take, speed of sound in air = 330 m s–1)
nn



Motion in a Plane
PaPer-i (single oPtion correct tyPe)

1. From a certain height, two bodies are projected
horizontally with velocities 5 m s–1 and 15 m s–1.
They reach the ground in time t1 and t2 respectively.
Then
(b) 3t1 = t2
(a) t1 = t2
(c) t1= 3t2
(d) none of these

2. Two balls A and B are thrown with speeds u and
u/2 respectively. Both the balls cover the same
horizontal distance before returning to the plane of
projection. If the angle of projection of ball B is 15°
with the horizontal, then the angle of projection of
A is
(a) (1/2) sin–1 (1/8) (b) (1/4) sin–1 (1/8)
(c) (1/3) sin–1 (1/8) (d) sin–1 (1/8)
3. A projectile is fired at an angle 30° to the horizontal
such that the vertical component of its initial
velocity is 80 m s–1. Find approximate velocity of
the projectile at time T/4 where T is time of flight.
(a) 155 m s–1
(b) 130 m s–1
(c) 145 m s–1
(d) 180 m s–1
4. A particle is projected with a velocity u so that
the horizontal range is twice the greatest height
attained, then the greatest height attained is
(a) 2u/5g
(c) 2u2/5g

(b) 4u2/2g
(d) u2/2g

5. A bomb is dropped from a plane flying horizontally
with velocity 720 km h–1 at an altitude of 980 m.
The bomb will hit the ground after (Take g = 9.8 m s–2)
(a) 14.14 s
(b) 1.414 s

(c) 7.2 s
(d) 141.4 s
6. The trajectory equation of a particle is given as
y = 4x – 2x2 where x and y are the horizontal and

vertical displacements of the particle in m from
origin(point of projection). Find the maximum
distance of the projectile from x-axis.
(a) 1.5 m
(b) 3 m
(c) 1 m
(d) 2 m
7. A car is moving horizontally along a straight line
with a uniform velocity of 25 m s–1. A projectile is,
to be fired from this car in such a way that it will
return to it after it has moved 100 m. The speed of
the projectile with respect to car should be
(Take g = 9.8 m s–2)
(a) 19.6 m s–1
(b) 15.6 m s–1
–1
(c) 9.8 m s
(d) 24.6 m s–1
8. The ceiling of a hall is 40 m high. For maximum
horizontal distance, the angle at which the ball may
be thrown with a speed of 56 m s–1 without hitting
the ceiling of the hall is (Take g = 9.8 m s–2)
(a) 45°
(b) 60°
(c) 30°

(d) 25°
9. Two paper screens A and B are separated by 150 m.
A bullet pierces A and then B. The hole in B is 15 cm
below the hole in A. If the bullet is travelling
horizontally at the time of hitting A, then the
velocity of the bullet at A is (Take g = 10 m s–2)

(a) 500 3 m s −1
(b) 200 3 m s −1
(c) 100 3 m s −1
(d) 300 3 m s −1
10. A flag is mounted on a car moving due north with
velocity of 20 km h–1. Strong winds are blowing due
west with velocity of 20 km h–1. The flag will point
in direction
(a) east
(b) north-east
(c) south-east
(d) south-west

11. The graph shows position as a function of time
for two trains running on parallel tracks. Which
statement is true?

Contributed by : Shiv R Goel, Intelli Quest, 9878359179

10

Physics For you | FEBRUARY ‘16




(a) At time tB, both trains have the same velocity.
(b) Both trains have the same velocity at some time
after tB.
(c) Both trains have the same velocity at some time
before tB.
(d) Nowhere the trains have some velocity.
12. A wedge is placed on a smooth horizontal plain
and a rat runs on its sloping side. The velocity of
wedge is v = 4 m s–1 towards right. What should be
the velocity of rat with respect to wedge (u), so that
the rat appear to move in vertical direction to an
observer standing on ground?

(a) 2 m s–1

(b) 4 m s–1

(c) 8 m s–1

(d) 4 2 m s −1

13. A plank is moving on ground with a velocity v and a
block is moving on the plank with respect to it with
a velocity u as shown in figure. What is the velocity
of block with respect to ground?

(a) v–u towards right (b) v–u towards left
(c) u towards right (d) none of these

14. A man is crossing a river flowing with velocity
of 5 m s–1. He reaches a point directly across at a
distance of 60 m in 5 s. His velocity in still water
should be

(a) 12 m s–1
(c) 5 m s–1
12

(b) 13 m s–1
(d) 10 m s–1

Physics For you | FEBRUARY ‘16

15. Two particles P1 and P2 are moving with velocities
v1 and v2 respectively. Which of the statements
about their relative velocity vr is true?
(a) vr cannot be greater than v1 + v2
(b) vr cannot be greater than v1 – v2
(c) vr > (v1 + v2 )
(d) vr < (v1 – v2 )

16. A boat having a speed of 5 km h–1 in still water,
crosses a river of width 1 km along the shortest
possible path in 15 minutes. The speed of the river
in km h–1 is
(a) 1
(b) 3
(c) 4
(d) 41

17. A man can swim in still water with a speed of 2 m s–1. If
he wants to cross a river of water current of speed
3 m s–1 along shortest possible path, then in which
direction should he swim?
(a) At an angle 120° to the water current
(b) At an angle 150° to the water current
(c) At an angle 90° to the water current
(d) None of these.
18. A ship is travelling due east at 10 km h–1. What
must be the speed of a second ship heading 30° east
of north if it is always due north of the first ship?
(b) 25 km h–1
(a) 30 km h–1
(c) 15 km h–1
(d) 20 km h–1
19. Width of a river is 100 m. Water flows in the river
with a velocity of 0.5 m s–1. A boat starts travelling
in water from one bank. If the direction of boat with
respect to water makes an angle 60° with upstream,
find the velocity of boat with which it should travel
in water to reach the other bank by shortest route.
(a) 0.5 m s–1
(b) 1 m s–1
–1
(c) 2 m s
(d) 1.5 m s–1
20. A bird flies to and fro between two cars which move
with velocities v1 = 20 m s–1 and v2 = 30 m s–1. If
the speed of the bird is v3 = 10 m s–1 and the initial
distance of separation between them is d = 2 km,

find the total distance covered by the bird till the
cars meet.



(a) 2000 m
(c) 400 m

(b) 1000 m
(d) 200 m

21. A car is moving towards a wall with a fixed velocity
of 20 m s–1. When its distance from the wall is 100 m,
a bee starts to move towards the jeep with a constant
velocity 5 m s–1. The time taken by bee to reach the
jeep is
(a) 5 s
(b) 4 s
(c) 2 s
(d) none of these
22. A stone is allowed to fall from the top of a tower and
covers half of the height of tower in the last second
of the journey. The time taken by the stone to reach
the foot of the tower is
(b) 4 s
(a) (2 − 2 ) s
(c) (2 + 2 ) s
(d) (2 ± 2 ) s
  
  

23. If a, b, c are unit vectors such that a + b − c = 0,


then the angle between a and b is
π
π
(b)
(a)
3
6
2π
π
(c)
(d)
3
2
24. The sum of magnitudes of two forces acting at a
point is 16 N and magnitude of their resultant is
8 3 N. If the resultant is at 90° with the force of
smaller magnitude, then their magnitudes(in N)
are
(a) 3, 13
(b) 2, 14
(c) 5, 11
(d) 4, 12
25. The resultant of two forces acting at an angle of 150°
is 10 N and is perpendicular to one of the forces.
The other force is
(a) 20/ 3 N
(b) 10 3 N

(c) 20 N
(d) 20 3 N
PaPer-ii (one or More oPtions correct tyPe)

26. The velocity time graph of two bodies A and B is
given here. Choose correct statements.

(a) Acceleration of B > acceleration of A.
(b) Acceleration of A > acceleration of B.
14

Physics For you | FEBRUARY ‘16

(c) Both are starting from same point.
(d) A covers greater distance than B in the same time.
27. A man wishes to throw two darts one by one at the
target at T so that they arrive at T at the same time
as shown in figure. Mark the correct statements
about the two projections.

(a) Projectile that travels along trajectory A was
projected earlier.
(b) Projectile that travels along trajectory B was
projected earlier.
(c) Both were projected at same time.
(d) Darts must be projected such that qA + qB = 90°.
28. In the figure shown, two boats start simultaneously
with different speeds relative to water. Water flow
speed is same for both the boats. Mark the correct
statements. qA and qB are angles from y-axis at

which boats are heading at initial moment.

(a) If vA > vB then for reaching the other bank
simultaneously qA > qB.
(b) In option (a), drift of boat A greater than boat B.
(c) If vB > vA and qA > qB, boat B reaches other
bank earlier than boat A.
(d) If vB = vA and qA > qB, drift of A is greater.
29. A particle has an initial velocity of 4i + 4 j m s −1
and an acceleration of −0.4i m s–2, at what time
will its speed be 5 m s–1?
(a) 2.5 s
(b) 17.5 s
(c) 7 2 s
(d) 8.5 s
30. A boat is traveling due east at 12 m s–1. A flag on the
boat flaps at 53° north of west. Another flag on the
shore flaps due north.
(a) Speed of wind with respect to ground is 16 m s–1
(b) Speed of wind with respect to ground is 20 m s–1
(c) Speed of wind with respect to boat is 20 m s–1
(d) Speed of wind with respect to boat is 16 m s–1


laws oF Motion; work, energy and Power
PaPer-i (single oPtion correct tyPe)

31. A child is sliding down a slide in a playground with
a constant speed.


(a)

Statement-1 : His kinetic energy is constant.
Statement-2 : His mechanical energy is constant.
(a) Statement-1 is true, statement-2 is true
and statement-2 is correct explanation for
statement-1.
(b) Statement-1 is true, statement-2 is true and
statement-2 is not the correct explanation for
statement 1.
(c) Statement-1 is true, statement-2 is false.
(d) Statement-1 is false, statement-2 is true.
32. Statement - 1 : Force F1 required to just lift block A
of mass m in case (i) is more than that in case (ii).

(b)

(c)

Statement -2 : Less work has to be done in case (ii)
to lift the block from rest to rest by a distance h.
(d)

F2

F1
m A
(i)

h


B m
(ii)

(a) Statement-1 is true, statement-2 is true
and statement-2 is correct explanation for
statement-1.
(b) Statement-1 is true, statement-2 is true and
statement-2 is not the correct explanation for
statement 1.
(c) Statement-1 is true, statement-2 is false.
(d) Statement-1 is false, statement-2 is true.
33. A particle initially at rest is displaced from x = –10 m to
x = +10 m under the influence of force F as shown
in the figure. Now the kinetic energy vs position
graph of the particle is

34. The potential energy of an object is given by
U(x) = 3x2 – 2x3, where U is in joules and x is in
meters.
(a) x = 0 is stable and x = 1 is unstable.
(b) x = 0 is unstable and x = 1 is stable.
(c) x = 0 is stable and x = 1 is stable.
(d) x = 0 is unstable and x = 1 is unstable.
35. A boy blowing a whistle sends out air at one gram
per second with a speed of 200 m s–1. Find his lung
power.
(a) 20 W (b) 0.2 W (c) 2 W (d) 200 W
36. A stone tied to a string of length 2 m is whirled in
a vertical circle with the other end of the string at

the centre. At a certain instant of time, the stone is
at its lowest position and has a speed 10 m s–1. The
magnitude of the change in its velocity as it reaches
a position where the string is horizontal, is
(a) 60 m s–1
(b) 40 m s–1
(c) 80 m s–1
(d) 160 m s–1
37. A ball of mass m is hung on a thread. The thread
is held taut and horizontal, and the ball is released
as shown. At what angle between the thread and
vertical will the tension in thread be equal to weight
in magnitude?

Physics For you | FEBRUARY ‘16

15


(a) 30°

−1  2 
(b) cos  
3

−1  1 
(c) cos  
(d) never
3
38. A ball whose size is slightly smaller than width of the

tube of radius 2.5 m is projected from bottommost
point of a smooth tube fixed in a vertical plane with
velocity of 10 m s–1. If N1 and N2 are the normal
reactions exerted by inner side and outer side of the
tube on the ball

(a) N1> 0 for motion in ABC, N2 > 0 for motion in
CDA
(b) N1> 0 for motion in CDA, N2 > 0 for motion in
ABC
(c) N2> 0 for motion in ABC and part of CDA
(d) N1 is always zero.
39. A catapult on a level field tosses a 3 kg stone, a
horizontal distance of 100 m. At second time, 3 kg
stone tossed in an identical fashion breaks apart in
the air into 2 pieces, one with a mass of 1 kg and
one with a mass of 2 kg. Both of the pieces hit the
ground at the same time. If the 1 kg piece lands a
distance of 180 m away from the catapult, how far
away from the catapult does the 2 kg piece land?
Ignore air resistance.
(a) 20 m
(b) 100 m
(c) 180 m
(d) 60 m
40. A small particle of mass m is at rest on a horizontal
circular platform that is free to rotate about a
vertical axis through its center. The particle is
located at a radius r from the axis, as shown in the
figure below. The platform begins to rotate with

constant angular acceleration a. Because of friction
between the particle and the platform, the particle
remains at rest with respect to the platform. When
the platform has reached angular speed w, the
angle q between the static frictional force fs and the
inward radial direction is



fs

r

16

Physics For you | FEBRUARY ‘16

(a) q =

w 2r
g

a
(b) q = 2
w

 w2 
 a 
(c) q = tan −1   (d) q = tan −1  2 
w 

 a 
41. Which of the following statements is true for a
particle moving in a circle with a constant angular
acceleration?
(a) The magnitude of acceleration is constant.
(b) The acceleration vector is along the tangent to
the circle.
(c) The velocity vector points along tangent to the
circle.
(d) The velocity and acceleration vectors are always
perpendicular to each other.

42. A 50 kg boy runs at a speed of 10 m s–1 and jumps
onto a cart as shown in the figure. The cart is initially
at rest. If the speed of the cart with the boy on it is
2.50 m s–1, what is the mass of the cart?

(a) 150 kg
(c) 175 kg

(b) 210 kg
(d) 260 kg

43. In a one-dimensional collision, a particle of mass
2m collides with a particle of mass m at rest. If the
particles stick together after the collision, what
fraction of the initial kinetic energy is lost in the
collision?
(a) 1/4
(b) 1/3

(c) 1/2
(d) none of these
44. When the momentum of a body increases by 100%,
its KE increases by
(a) 400%
(b) 100%
(c) 300%
(d) none of these
45. A spaceship of speed v0 travelling along +y axis
suddenly shots out one fourth of its part with speed
2v0 along +x-axis. xy axes are fixed with respect to
ground. The velocity of the remaining part is
2
20
(b)
(a) v0
v
3
3 0
5
13
(c) 3 v0
(d)
v
3 0
46. A highly elastic ball moving at a speed of 3 m s–1
approaches a wall moving towards it with a speed
of 3 m s–1 (figure). After the collision, the speed of
the ball will be



mass of the plank is M, then the distance that the
man moves relative to ground is
(a) 3L/4
(b) L/4
(c) 4L/5
(d) L/3
–1

(a) 3 m s
(c) 9 m s–1

–1

(b) 6 m s
(d) zero

47. On a frictionless surface, a ball of mass M moving
at speed v collides elastically with another ball of
the same mass that is initially at rest. After the
collision, the first ball moves at an angle q to its
initial direction and has a speed v/2. The second
ball's speed after the collision is
3
v
(b)
v
(a)
4
2

v
3
v
(d) v + cos q
2
2
48. A particle of mass m is moving along the x-axis with
speed v when it collides with a particle of mass 2m
initially at rest. After the collision, the first particle
has come to rest, and the second particle has split
into two equal mass pieces that move at equal angle
q = 30° with the x-axis, as shown in the figure
below. Which of the following statements correctly
describes the speeds of the two pieces?
(c)

51. Block A, with a mass of 4 kg, is moving with a speed
of 2.0 m s–1 while block B, with a mass of 8 kg, is
moving in the opposite direction with a speed of
3 m s–1. The center of mass of the two block system
is moving with a velocity of
(a) 1.3 3 m s–1 in the same direction as A
(b) 1.3 3 m s–1 in the same direction as B
(c) 2.7 3 m s–1 in the same direction as A
(d) 1.0 3 m s–1 in the same direction as B.
PaPer-ii (one or More oPtions correct tyPe)

52. You lift a suitcase from the floor and keep it on a
table. The work done by you on the suitcase does
not depend on

(a) the path taken by the suitcase
(b) the time taken by you in doing so
(c) the weight of the suitcase
(d) your weight.
53. Figure (a) shows a frictionless roller coaster track.
Figures (i), (ii), (iii) show potential energy and total
energy for a car on roller coaster. Figures (i), (ii),
(iii) show drawing of actual roller coaster track
because gravitational potential energy is directly
proportional to height, it is also U vs x graph.

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(a) Each piece moves with speed v.
(b) One of the pieces moves with speed v, the other
moves with speed less than v.
(c) One of the pieces moves with speed v/2, the
other moves with speed greater than v/2.
(d) Each piece moves with speed greater than v/2.
49. Two objects of different mass and with same initial
speed, moving in a horizontal plane, collide head
on and move together at half their initial speed after
the collision. Ratio of their masses is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 5 : 1
50. A man of mass 3 M stands at one end of a plank of

length L which lies at rest on a frictionless surface.
The man walks to other end of the plank. If the

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including syllabus are also available at website-www.wbjeeb.nic.in/
www.wbjeeb.in.
The examinations will be conducted on 17.05.2016
Physics For you | FEBRUARY ‘16

17


unstretched spring of length L = 20 cm, and of
spring constant 16 N m–1. The mass of the spring
is negligible. At a certain moment the bodies are
given an initial speed of v0 = 0.36 m s–1, towards the
wall on the right. The body at the right collides with
the wall totally elastically.


U
Etotal
(i)
A

B

D
x

C

U
Etotal

(ii)
A

B

C

D
x

U

(a) There will be 2 collisions with the wall.
(b) After 1st collision, centre of mass comes to rest.
(c) After 2nd collision, centre of mass moves to left

with speed v0.
(d) After all collisions are over, the system oscillates
about the centre of mass.
solutions

1. (a) : Initial vertical velocity of both bodies is same
i.e, 0.
u2
u2 sin 2q
; for B, R =
g
8g
1
1
Comparing; sin 2q = 1/8 ⇒ q = sin −1
2
8
uy
2u y
3. (c) : T =
⇒ v yT /4 =
= 40 m s −1
g
2
⇒ v xT / 4 = ux = 80 3 m s −1

2. (a) : For A, R =
(iii)
A


B

C

Etotal
D
x

(a) In case (i), the car can negotiate the hill at C and
reach D.
(b) In case (ii), the motion is confined between
two turning points where the total energy and
potential energy curves intersect.
(c) In case (iii), motion of car will be confined in
first valley between A and B.
(d) The turning points are attained where Etotal = U.
54. An object of mass 3m, initially at rest on a frictionless
horizontal surface, explodes breaking into two
fragments of mass m and 2m respectively. Which
one of the following statements after the explosion
is true?
(a) Velocity of center of mass increases.
(b) Speed of smaller fragment will be twice that of
larger fragment.
(c) Fragments have equal magnitude of momentum
in ground frame but different magnitude of
momentum in center of mass frame.
(d) Kinetic energy of system increases.
55. Bodies of mass 0.5 kg, resting on a horizontal
frictionless tabletop, are connected with an

18

Physics For you | FEBRUARY ‘16

\
4. (c) :

vT /4 = v 2xT /4 + v 2yT /4 ≈ 145 m s −1
2u2 sin q cos q
u2 sin2 q
=2×
g
2g

tan q = 2 ⇒ sin q = 2 / 5

\H=

2u2
5g

1
5. (a) : 980 = 9.8t 2 ⇒ t = 10 2 = 14.14 s
2
dy
6. (d) :
= 4 − 4x = 0 ⇒ x = 1 ⇒ y = 2 m
dx
7. (a) : The projectile is fired vertically upwards with
respect to the car. Time taken to return = 4 s.

For maximum height, t = 2 s.
\ 0 = u – 9.8 × 2, u = 19.6 m s–1.
8. (c) : Usually maximum range corresponds to q = 45.
But in this case height is more than ceiling. So q <
45°.
u2y
28 1
= H ; u y = 40 × 19.6 ⇒ sinq = = ⇒ q = 30
2g
56 2


1
9. (a) : 0.15 = × 10t 2 , t = 0.1 3
2
150
= 500 3 m s −1
⇒ v A = vH =
0. 1 3



10. (d) : v = −20i, v = 20 j
w

18. (d) : Speed of both boats
towards east should be same
\ v2 sin 30° = 10
⇒ v2 = 20 km h–1
19. (b) :


c

Here we have to look for velocity of wind with
respect to car. So,



vw /c = vw − vc = −20i − 20 j

This is in south-west direction.
11. (c) : At some time before tB, slope of B will be
equal to slope of A. Acceleration of A is zero always
whereas that of B is not zero.
 

12. (c) : vr = vrw + vw

1 0.5
u
⇒ =
⇒ v = 1 m s −1
2 v
v
d
20. (c) : Time to meet the cars, t =
v1 + v2
Distance travelled by bird in this time
v d
10 × 2000

s = v3t = 3 =
= 400 m
v1 + v2 (20 + 30)
sin 30 =

21. (b) : t =

v
= cos q
u
4
v
⇒ u=
=
= 8 m s −1

cos q cos 60
 

13. (a) : vb = vbp + v p = (−u) + v = v − u towards right.
⇒

14. (b) : t =

d

2

v −u


2

⇒5=

60

2

2

v −5

⇒ v = 13 m s −1

15. (a) : vr lies between (v1 – v2) and (v1 + v2) depending
upon angle between v1 and v2.
16. (b) :

100 m

= 4 seconds
25 m s −1
1
22. (c) : h = gt 2
...(i)
2
h 1
= g (t − 1)2
2 2
...(ii)

From equations (i) and (ii)
t =2± 2 s
But t cannot be 2 − 2 seconds since it is less than
1 second which is not possible.
23. (d)
24. (b) : A + B = 16
...(i)
...(ii)
and B2 = R2 + A2
Given, R = 8 3 N
On solving, A = 2 N, B = 14 N
25. (c) : R = 10 N ⇒

t=
17. (b) :

d
v 2 − u2

⇒

15
1
=
⇒ u = 3 km h −1
2
2
60
5 −u


3
⇒ q = 60
2
Hence at an angle of 150° to the water current.
sinq =

R
= cos60°
B

10 1
= ⇒ B = 20 N
B 2
26. (b, d) : Slope of A is greater than slope of B.
\ aA > aB
1
S = at 2 ( u = 0) \ SA > SB
2
Nothing can be said about starting positions of
bodies from velocity-time graph.
⇒

Physics For you | FEBRUARY ‘16

19


27. (b) : Since maximum height attained by B is more,
so it will take more time to reach the target. Hence
it should be projected earlier so that both reach

simultaneously. For same range, sum of projection
angles is 90° if speed of projection is same, but here
speeds may be different.
28. (a, b, c, d) : For reaching the other bank
simultaneously, their velocities along y direction
should be same.
So, vA cos qA = vB cos qB , if vA > vB then cos qA < cos qB,
⇒ qA > qB. Hence (a) is correct.
Drift: x = (v sin q + u)t
For option (a): vA sin qA > vB sin qB, hence drift of
A is greater than B. So (b) is correct.
Same will be true for option (d), hence (d) is also
correct.
For option (c): vB cos qB > vA cos qA, so boat B
reaches earlier than A. Hence (c) is correct.
29. (a, b) : Since acceleration is in x direction only,
velocity in y-direction will not change.
52 = vx2 + vy2 = vx2 + 42 ⇒ vx = ± 3 m s–1

v − ux
vx = ux + axt ⇒ t = x
ax
3−4
−3 − 4
\ t1 =
= 2.5 s , t 2 =
= 17.5 s
−0.4
0. 4


32. (c) :

But work done from rest to rest is equal, because
change in K.E. is same.
33. (c)
34. (a) :

dU
= 6 x − 6 x 2 = 0 ⇒ 6 x(1 − x ) = 0 ⇒ x = 0,1
dx
d 2U

= 6 − 12 x
dx 2
d 2U
at x = 0,
> 0 ⇒ stable equilibrium
dx 2
d 2U
< 0 ⇒ unstable equilibrium
at x = 1,
dx 2
35. (a)
36. (d) : v = (10)2 − 2 × 10 × 2 = 60 m s–1
–1

∆v = (60) + 100 = 160 m s
2
30. (a, c) : Velocity of wind with respect to boat is 53° 37. (c) : mgl cos q = 1 mv 2 ⇒ 2mg cos q = mv
2

l
north of west.

2
1
mv
2
Actual velocity of wind is towards north,mgl
because
...(i)
cos q a= mv ⇒ 2mg cos q =
T
2
l
flag on shore flaps towards north. We have



2
mv
vw = vw /b + vb
T − mg cos q =
mg
l
2
mv
= 2 mg cos q ...[using (i)]
mg – mg cos q =
l
1

1

cos q = ⇒ q = cos −1  

3
3
38. (c)
vw
v
4
= tan 53° ⇒ w = ⇒ vw = 16 m s −1
vb
12 3

39. (d) :

vb
3
= cos 53° =
vw /b
5
⇒
31. (c)
20

5
5
vw /b = vb = × 12 = 20 m s −1
3
3


Physics For you | FEBRUARY ‘16

100 =

2 x + 180
⇒ x = 60 m
3


a
Ra
 a 
40. (d) : tan q = t =
= tan −1  2 
2
w 
ar Rw
41. (c) : Net acceleration = a2 R2 + (w2 R)2
As w increases, the net acceleration increases.

42. (a) : 50 × 10 = (50 + m)2.5 ⇒ m = 150 kg
43. (b)
44. (c)
45. (b)




46. (c) : v BW = v B − vW ; v BW = 3 − (−3) = 6 m s −1

Before collision : vBW = 6 m s–1
After collision : v′BW = 6 m s–1
vB′ = v′BW + vW = 6 + 3 = 9 m s–1
47. (c) : Kinetic energy is conserved,

nn 
solution oF January 2016 crossword

2

1
1 v  1
Mv 2 = M   + Mv ′2
2
2 2 2
v2 −

3
v2
= v ′2 ⇒ v ′ =
v
4
2

48. (d) : mv = mv′ cos q + mv′cos q
v
v
v′ =
⇒ v′ >
2 cos q

2
49. (b) : m1v – m2v = (m1 + m2)v/2
2m1 – 2m2 = m1 + m2
m1 = 3m2
m1 3
=
m2 1
50. (b) : 4x = L ⇒ x = L/4
51. (b)
52. (a, b, d)
53. (a, b, c, d)
54. (b, d)
55. (a, b, c) :
Winner January 2016
1.

Anu Sharma (Delhi)

2.

Shruti Gupta (Delhi)

3.

Rohan Kashyap (Haryana)

4.

Atriz Roy (WB)


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1.

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2.

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Physics For you | FEBRUARY ‘16

21


Capacitive Circuits
In this segment, we would learn and gradually master
solving capacitive circuit questions.
To begin with, let me start asking two basic questions:
1. What is meant by series combination?
2. What is a parallel combination?
Maximum book says that in a series combination the
charges on all capacitors are identical. But is it really
so? Always?
Let us consider a series combination of n capacitors
connected to a battery.
V1
+Q –Q
C1

V2

+Q –Q
C2

Vn

+Q

–Q
Cn

–Q
V

Considering, initially all capacitors are uncharged. There
was no potential difference across the combination
before closing the switch, but on connecting the battery
a potential difference equal to emf of the cell has to be
maintained across the combination which can come only
if electric field is set up between the plates of capacitor.
What the battery does is, it pulls out electrons from
one plate of capacitor and deposits it to the other plate
of the other extreme end's capacitor and thereafter by
induction and conduction charges are induced on other
plates and capacitors.
Consider any intermediate capacitors, say C1 and C2.
From left plate of C 1
isolated
charge –Q was pulled
–Q
by battery due to which

+Q appeared on it. By
+Q
C2
C1
induction –Q appears on
right plate of C1 but right plate of C1 and left plate of
C2 forms an isolated part of circuit. By saying isolated
we mean, they are not physically connected to others,
hence the net charge on them has to remain conserved

and hence since before charging the capacitors were
uncharged, the summation of charges on these plates
has to be zero i.e.
Sqjunction = 0 [Kirchhoff 's junction law]
Hence the –Q appearing on right plate of C1 comes
from the left face of C2 due to which +Q appears on
it (one gains, other loses by same amount).
But what if the capacitors were already charged before
connecting them to the battery?
Then the Sqjunction ≠ 0 and hence the charges on the
capacitors would not be equal even if they were in
series combination.
Just remember one line - If between any two terminals,
say A and B, there exists only one path which leads
from terminals A to B, the capacitors are said to be in
series but use series combination formula i.e.
1
1
1
1

=
+
+ ... +
Ceq C1 C2
Cn
only if none of the capacitors are initially charged to
find the charges on any capacitor.
V2
V1
Supposedly now, none
C2
C
of the capacitors were
1
initially charged, what is
+Q –Q
+Q –Q
the potential difference
across each capacitor?
Since Q = CV
= constant for both.
V
1
\ V∝
C
 C1 
 C2 
V and V2 = 
V
⇒ V1 = 


 C1 + C2 
 C1 + C2 
It is interesting to note that the potential difference
across capacitors in series combination is not dependent
upon the value of individual capacitances but rather
the ratio of capacitances. For example if both C1 and
C2 are doubled, the potential difference across both
remains unchanged.

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

22

physics for you | february ‘16


Now let me answer the 2 nd question raised at the
beginning.
We say two or more capacitors to be in parallel if their
ends are maintained at common potential. They need
not be placed geometrically parallel to each other.
For example,
C1
B
C3

C2

A


Notice that left plates of all the three capacitors are
connected to terminal A whereas their right plates to
terminal B through connecting conducting wires and
since all the points of a conductor are equipotential at
electrostatic condition, we conclude that a common
potential difference is maintained across each, hence
equivalent capacitance,
Ceq = C1 + C2 + C3

With these basic understanding being cleared, now let
us move to our next segment where we learn how to
solve complicated circuits involving multiple batteries
and capacitors.
Our approach will be slightly different from what is
followed by most books.
Just follow these basic steps:
1. If none of the point in the given circuit is grounded,
we can choose any one arbitrary point in the circuit
to be at zero potential. Depending on the choice
of zero potential point, the potential of all other
points will be dependent but the potential difference
between any two points would not change and for
any capacitor it is the potential difference which
is important and not the potential of each of its
terminals.
I prefer taking zero volt to be at the lower potential
terminal of the largest e.m.f 's cell connected. This
simplifies calculation.
2. Distribute the potential of all points (junctions) in

the circuit, either in terms of known or unknown
variables, by using the fact that (a) All points of conducting wire are at same potential.
(b) The potential difference across the cell is equal
to its emf (at electrostatic condition).
For example,
(x + 10)

10 V

(x)

(c) If the charge on a capacitor is not known, the
potential difference across it would not be known,
hence new variables would be required for it.
3. Count the number of variables taken, since we need
to frame exactly same number of equations by using
Kirchhoff 's junction law, i.e.
0; if all capacitors of the junction

 were initially uncharged.

∑ qjunction = initial charge(qi );if one or more
capacitors connected to junction were

initially charged.
4. Solve simultaneous equations obtained.
Now, let us apply whatever we learnt.
Q.1 : A capacitor of capacitance 5 mF is charged with an
initial charge of 50 mC and then connected to another
uncharged capacitor of 20 mF with a battery of emf

20 V as shown.
20 F

5 F
+50 C –50 C

20 V

Find the charges on both the capacitors after closing
the switch.
Soln.: After closing the switch, the potential of different
points has been shown as below:
5 F
(20)

(20)

20 F
(x)

20 V

(0)

(0)

∑ qx = −50 mC
(initial charge on isolated part marked)
⇒ 5(x – 20) + 20(x – 0) = –50
⇒ x=2V

\ Charge on 5 mF, q5 = 5(20 – 2) = 90 mC
Charge on 20 mF, q20 = 20(2 – 0) = 40 mC
Note that the charges are different on both.
physics for you | february ‘16

23


Q.2 : The shown network
is a small segment of
a large circuit and the
potential of the three
terminals are marked.
Find the potential
C2
of junction.
V2
Soln.: Here, since the
potential of certain points
are already marked, the
choice of zero potential
is not our own.
C2
Let the junction be at
x V then
V

V1
C1


C3

V3

V1
C1
(x)

From the diagram,
9e
= e − 40 = x
13
4e
⇒
= 40 ⇒ e = 130 V
13
\ Potential of points x and y are
x = 130 – 40 = 90 V
2
1
y = e = × 130 V
6
3
130 140
\ x − y = 90 −
=
V
3
3
Q.4 : Find the charge on 4 mF capacitor.


V3

2

S qx = 0

3 F

Q.3 : In the given circuit, after steady state, charge on
3 mF is found to be 120 mC then find
(i) emf of cell
(ii) potential difference between points x and y.
4 F

10 V

(x + 10)

4 F

(0)

(x)
4 F


Soln.: Since q3 = 120 mC on 3 mF
\ Potential difference across it
120

∆V3 =
= 40 V
3
This is also the potential difference across a parallel
combination of 2 mF, 3 mF and 4 mF. Hence,
(y)

4 F

2

2+4

()
(x)

(0)

(0)

20 V

9

9+4

physics for you | february ‘16

(20)


Now apply, Sqx = 0.
Remember that when we apply this we need to take the
sum of charges of all the plates of the capacitor which
can give/take charge from the considered junction.
\ 2(x – 20) + 4(x – 0) + 2(x + 10) = 0
⇒ x = 2.5 V
\ q4 = 4(x – 0) = 4 × 2.5 = 10 mC
Q.5 : Find the charge on 2 mF capacitor.
4 F

20 V

20 V

4 F
4 F

4 F



24

(20)

2 F

3 F

40 V


2 F

(x)
4 F

2 F

9 F

20 V

Sol.: Clearly 6 mF and 3 mF are in series, hence we
6×3
= 2 mF and distribute the
can replace them with
6+3
potential of all points as below:

(y)

2 F

4 F

6 F

⇒ C1(x – V1) + C2(x – V2) + C3(x – V3) = 0
C V + C2V2 + C3V3
⇒ x= 1 1

C1 + C2 + C3

2 F

2 F

10 V

C3

2 F
10 V


Soln.:
(20)

20 V

4 F (x)

(x – 20)

4 F

20 V

4 F

(0)

(0)

(x – 20)

2 F
(x – 10)

(x – 20)

10 V

S qx = 0
⇒ 4(x – 20) + 4(x – 0) + 4(x – 20) + 2(x – 10) = 0
90
⇒ x=
V
7
 90

\ q2 = 2(x – 10 – 0) = 2  − 10 
7

20 40
=2× =
mC
7
7
Q.6 : Find (i) charge flown through the switch,
(ii) work done by both the batteries after closing the
switch in the circuit below.


So alternatively, the charge could have been found
out imagining a single capacitor and single cell in the
loop.
\ q3 = q6 = Cloop eloop
3×6
=
× (20 − 10) = 20 mC
 3 + 6 
Hence, the higher potential terminal of both the
capacitors will have +20 mC while the other has
–20 mC charge.
40
3

3 F

+20 C –20 C

Soln.: Before closing the switch:
3 F

20 V
(0)
(20)

(x)

(x – 10)
10 V


6 F

(0)

Sqx = 0 ⇒ 3(x – 20) + 6(x – 10 – 0) = 0

40
V
3
40 

\ q3 = 3  20 −  = 20 mC

3
⇒

x=

 40

q6 = 6(x − 10 − 0) = 6  − 10  = 20 mC
 3

The result shows both the capacitors have identical
charge and that should have been since they are in
series.

10
3


10 V

6 F

(0)

After closing the switch:
3 F
(30)

(20)

20 V
(0)

+30 C –30 C

10 V

6 F

(0)

40
3

(30)

10 V


20 V

–20 C +20 C

20 V

3 F

(20)

+120 C –120 C
(20)
6 F

(0)

Clearly, potential difference across 3 mF and 6 mF are
10 V and 20 V respectively now.
\ q3 = 3 × 10 = 30 mC
q6 = 6 × 20 = 120 mC
The right plate of 6 mF capacitor initially had –20 mC
but now has –120 mC, hence 100 mC charge flows
through the wire connected to it and hence through
20 V battery also.
\ Work done by 20 V battery = qflown × emf
= 100 × 20 = 2000 mJ = 2 mJ
Similarly, charge flown through 10 V battery = 50 mC
\ Work done by 10 V battery = 50 × 10
= 500 mJ = 0.5 mJ

For charge flown through switch, consider the upper
junction.
3 C 50 C

20 V

100 C

q = 50 + 100 = 150 C

nn
physics for you | february ‘16

25


class-Xii

ParagraPh based questions
Paragraph-1
In a certain region of space, there exists a uniform
and constant electric field of magnitude E along the
positive y-axis of a co-ordinate system. A charged
particle of mass m and charge –q (q > 0) is projected
from the origin with speed 2v at an angle of 60° with
the positive x-axis in x–y plane. When the x-coordinate
of particle becomes 3mv 2 /qE , a uniform and constant
magnetic field of strength B is also switched on along
positive y-axis.
1. Velocity of the particle just before the magnetic

field is switched on is
ˆ
(a) vi

3v ˆ
(b) viˆ +
j
2

3v ˆ
3v ˆ
(c) viˆ −
(d) 2viˆ −
j
j
2
2
2. x-coordinate of the particle as a function of time
after the magnetic field is switched on is
3mv 2 mv  qB 
(a)
−
sin  t 
m 
qE
qB
(b)

3mv 2 mv
 qB 

sin  t 
+
m 
qE
qB

(c)

3mv 2 mv
 qB 
cos  t 
−

qE
qB
m 

(d)

3mv 2 mv
 qB 
cos  t 
+
m 
qE
qB

3. z-coordinate of the particle as a function of time
after the magnetic field is switched on is
mv 

 qB  
(a)
1 − cos  t  
qB 
m 
26

Physics for you | February ‘16

(b) −

mv 
 qB  
1 + cos  t  
qB 
m 

(c) −

mv 
 qB  
1 − cos  t  
qB 
m 

(d)

mv 
 qB  
1+ cos  t  

qB 
m 

Paragraph-2
There is a uniformly charged
ring having radius R. An
infinite line charge (charge per
unit length l) is placed along a
diameter of the ring (in gravity
free space). Total charge on the
ring Q = 4 2lR. An electron
of mass m is released from
rest on the axis of the ring at
a distance x = 3R from the
centre.
4. Magnitude of initial acceleration of the electron is
(a)



el  3 − 2 2 
(b) el  3 + 2 2 


πe0 mR  4 6 
πe0 mR  4 6 

(c)

el  3 + 2 2  (d) none of these

πe0 mR  4 3 

5. The distance from centre of ring on the axis where
the net force on the electron is zero, is
(a) 2R
(c) R

(b)

2R
(d) none of these


Physics for you | February ‘16

27


6. Potential difference between points A(x =
and B(x = R) i.e., (VA – VB) is
(a) −
(b)

l
πe0

l
πe0

3R)



1  ln 3 
 1 −

−
2 4 



1  ln 3 
 1 −

−
2 4 



1  ln 3 
 1 +

−
2 4 

(d) none of these
(c) −

l
πe0


Paragraph-3
Pulfrich refractometer is used to
measure the refractive index
of solids and liquid. It consists
of right angled prism A having
its two faces perfectly plane.
One of the face is horizontal
and the other is vertical as
shown in figure. The solid B whose refractive index
is to be determined is taken having two faces cut
perpendicular to one another. Light is incident in a
direction parallel to the horizontal surface so that the
light entering the prism A is at critical angle C. Finally, it
emerges from the prism at an angle i. Let the refractive
index of the solid be m and that of the prism A be m0
(which is known). Here m0 > m and by measuring i, m
can be determined.

Paragraph-4
In a photoelectric setup, a point source of light of power
3.2 × 10–3 W emits monoenergetic photons of energy 5.0
eV. The source is located at a distance of 0.8 m from the
center of a stationary metallic sphere of work function
3.0 eV and of radius 8.0 × 10 –3 m. The efficiency of
photoelectron emission is 1 for every 106 incident photons.
Assume that the sphere is isolated and initially neutral
and that photoelectrons are instantaneously swept away
after emission.
10. Calculate the number of photoelectrons emited per
second.

(a) 103
(b) 104
4
(c) 5 × 10
(d) 105
11. It is observed that photoelectron emission stops at
a certain time t after the light source is switched on.
It is due to the retarding potential developed in the
metallic sphere due to left over positive charges. The
stopping potential (V) can be represented as
(a) 2(KEmax/e)
(b) (KEmax/e)
(c) (KEmax/3e)
(d) (KEmax/2e)
12. Evaluate time t, mentioned in question 11.
(a) 1.85 min
(b) 2.36 min
(c) 2.75 min
(d) 0.78 min
Paragraph-5
In figure shown, the rod has a resistance R, the horizontal
rails have negligible friction. A battery of e.m.f e and
negligible internal resistance is connected between points
a and b. The rod is initially at rest.

7. Refractive index of the solid (m) in terms of m0 and i is
(a)

m 20 + sin2 i


(b) m0 + sin2i

(c)

m 20 − 2 sin2 i

(d)

m20 − sin2 i

8. If m0 = 2 and the ray just fails to emerge from the
prism, refractive index m of the solid will be
(a) 1.21 (b) 2
(c) 1
(d) 3 / 2
9. A ray of light is incident normally
on the horizontal face of the slab
and just fails to emerge from the
diagonal face of the prism. If prism
angle is 30°, refractive index of the
prism can be
(a) 2
(b) 3
(c) slightly greater than 2 (d) slightly less than 2
28

Physics for you | February ‘16

13. The velocity of the rod as a function of time t
(where t = mR/B2l2) is

e
e
(a)
(b)
(1− e −t /t )
(1+ e −t /t )
Bl
Bl
(c)

2 e
(1 + e −t /t )
3 Bl

(d)

e
(1 − e −t /t )
2 Bl

14. After some time, the rod will approach a terminal
speed. The expression for it, is
3 e
e
e
2e
(a)
(b)
(c)
(d)

2 Bl
2Bl
Bl
Bl


15. The net current through the circuit when the rod
attains its terminal speed is
2e
e
3e
(a)
(b)
(c)
(d) zero
R
R
2R
solutions

1. (a) : At first, particle will travel along parabolic path
−qE
OA. Let time taken from O to A is t 0 .a y =
m

4. (a) : Electric field at distance x,
Q
x
l
+

E=–
2
2 πe0 x 4 πe0 (R + x 2 )3/2
(considering right direction as positive)
=−
=

x 4 2 lR
l 1
+
2 πe0 x 4 πe0 (R2 + x 2 )3/2

l  1
2 2 xR 
− + 2

2 πe0  x (R + x 2 )3/2 

Initially x =

3R
l  1 2 2 3
\ E=
+
−

2 πe0 R  3
8 
l  −2 2 + 3 
l 3 − 2 2 

=

=


2 πe0 R  3 (2 2 )  2 πe0 R  2 6 
\ Acceleration, a =
x0 =

3mv 2
3mv
= (2v cos 60°)t 0 ⇒ t 0 =
qE
qE

qE 3mv
=0
m qE
Hence at point A, velocity will be purely along
x-axis and it will be 2vcos60° = v.
2. (b) : Now magnetic field is switched on along y-axis.
Now its path will be helical as shown below with
increasing pitch towards negative y-axis.
v y = u y + a y t 0 = 2v sin 60° −

(−e)(E )
el  3 − 2 2 
=−
m
πe0 mR  4 6 


5. (c) : Force on electron is zero at point where E = 0
⇒x=R
6. (b) : Potential difference between two points
DV = – E dx
Potential difference due to line charge between
x = R and x = 3 R
3R
ldx
l  ln 3 
VB − VA = − ∫
−
=


R
2 πe0 x πe0  4 
Potential difference due to ring between x = 3 R
and x = R
1  4 2 lR 4 2 lR  l  1 − 1 
VA − VB =
−

=

4 πe0  2 R
2R  πe0 
2
\


Net potential difference
l 
1 ln 3 
VA − VB =
1−
−
πe0 
4 
2

7. (d) : sin C =

mv
r=
qB
mv
x = x0 + r sin q = (2v cos 60°)t 0 +
sin ωt
qB
mv 2 mv  qB 
sin  t 
= 3
+
qE qB
m 
3. (c) : z = – (r – r cos q) = –

m
m
⇒ sin[90° − r ] =

m0
m0

m
m0
sin i
sin i
Also, m0 =
⇒ sin r =
sin r
m0

... (i)

⇒ cos r =

2

mv 
 qB  
1− cos  t  
qB 
 m 

[Q C + r = 90°]

... (ii)
2

 m   sini 

From (i) and (ii),   + 
 =1
 m0   m0 
⇒ m = m20 − sin2 i
Physics for you | February ‘16

29


8. (c) : If ray just fails to emerge, i = 90°
Given, m 0 = 2
\

m = ( 2 )2 − sin2 90° = 1

9. (c) : Here 30° > C
⇒ sin 30° > sin C
1 1
⇒
>
⇒ m0 > 2
2 m0
10. (d) : If P is the power of point source of light, the
P
intensity at a distance r is I =
4 πr 2
The energy intercepted per unit time by the metallic
sphere is
P
× πR2

E = Intensity × projected area of sphere =
2
4 πr

r

S

R

If E is the energy of the single photon and h the
efficiency of the photon to liberate an electron, the
number of ejected electrons is
n=h

PR2
4r 2 E

=

(10−6 )(3.2 × 10−3 )(8 × 10−3 )2
4 × (0.8)2 × (5 × 1.6 × 10−19 )

= 105 electron s–1
11. (b) : The emission of electrons from a metallic sphere
leaves it positively charged. As the potential of the
charged sphere begins to rise, it attracts emitted
electrons. The emission of electrons will stop when
the kinetic energy of the electrons is neutralised by
the retarding potential of the sphere. So, we have

 KE 
eV = KEmax or V =  max 
 e 
12. (a) : From Einstein’s photoelectric equation,
KEmax = hu – f = (5 – 3) = 2 eV
The potential of a charged sphere is
1 q
1  Ne 
=
V=
4 πe0 R 4 πe0  R 
1  Ne 
\
  =2
4 πe0  R 
30

Physics for you | February ‘16

4 πe0 2R

2 × 8 × 10 −3

= 1.11 × 107
9
−19
×
×
×
.

9
10
1
6
10
e
The photoelectric emission will stop when
1.11 × 107 electrons have been emitted.
The time taken by it to emit 1.11 × 107 electrons,
1.11 × 107
= 111s = 1.85 min
t=
105
13. (a) : The current due to the battery at any instant, I = e/R.
The magnetic force due to this current
eBl
FB = IBl =
R
This magnetic force will accelerate the rod from its
position of rest. The motional e.m.f. developed in
the rod is Blv.
Blv
The induced current, I induced =
R
The magnetic force due to the induced current
B2l 2v
Finduced =
R
From Fleming’s left hand rule, force FB is to the
right and Finduced is to the left.

Net force on the rod = FB – Finduced.
From Newton’s second law,
dv
dv eBl B 2 l 2 v
−
=m
FB − Finduced = m ,
dt
R
dt R
On separating variables and integrating speed from
0 to v and time from 0 to t, we have
v
dv
Bl t
dv
Bl
dt
=
=
dt , ∫0
(e − Bvl ) mR ∫0
e − Bvl mR
N=

=

22

 e − Bvl 

B2 l 2 e − Bvl − B l t
t,
ln 
=−
= e mR
mR
 e 
e
mR
e
(1 − e −t /t ) where t =
( Bl )2
Bl
14. (c) : The rod will attain a terminal velocity at t → ∞
i.e., when e–t/t = 0, the velocity is independent of time.
e
vT =
Bl
\

v=

15. (d) : The induced current Iinduced = Blv/R. When the
rod has attained terminal speed,
Bl  e 
I induced = ×   = e/R
R  Bl 
The current due to battery and the induced current
are of same magnitude, hence net current through
the circuit is zero.

