JEE VP AIR (2025-26) Physics

PRACTICE SHEET
[Electromagnetic Induction]

JEE MAIN
[Induced EMF] (1) Bulb A goes out bulb B gets brighter
1. A metallic ring connected to a rod oscillates freely (2) Bulb B goes out bulb A gets brighter
like a pendulum. If now a magnetic field is (3) Bulb A goes out bulb B gets dimmer
applied in horizontal direction so that the (4) Bulb B goes out bulb A gets dimmer
pendulum now swings through the field as shown
in the figure, the pendulum will 4. A uniform but time varying magnetic field is
present in a circular region of radius R. The
magnetic field is perpendicular and into the plane
of the loop and the magnitude of field is
increasing at a constant rate . There is a straight
conducting rod of length 2R placed as shown in
figure. The magnitude of induced emf across the
rod is
(1) Keep oscillating with the old time period
(2) Keep oscillating with a smaller time period
(3) Keep oscillating with a larger time period
(4) Keep oscillating with decreasing amplitude
and eventually come to rest soon.
R2
(1) R 2  (2)
2. A long horizontal metallic rod with length along 2
the east-west direction is falling under gravity. R2 R2
The potential difference between its two ends will (3) (4)
2 4
(assume negligible air resistance)
(1) Be zero
5. In the given figure, there are two concentric
(2) Be constant
cylindrical region in which time varying magnetic
(3) Increase with time
field is present. From the centre of radius R
(4) Decrease with time
magnetic field is perpendicular in to the plane
dB
3. In figure-(a) a solenoid produce a magnetic field varying as = 2k and in a region from R to 2R
dt
whose strength increases into the plane of the
magnetic field is perpendicular out of the plane
page. An induced emf is established in a
dB
conduction loop surrounding the solenoid, and varying as = 4k . Find the induced emf across
dt
this emf lights bulbs A and B. In figure-(b) point P
an arc AB of radius 3R. (k is positive constant)
and Q are shorted. After the short is inserted,
P Solenoid P
xxx x xx
xxxxx xxxxx
A xx xx x B A xx xx x B
xxxxx xxxx x
x x x xxx x x x xxx
x xx x xx

Q Q (1) 6R2k (2) 5R2k

Figure (a) Figure (b) (3) 7R2k (4) None of these
1
6. A wire loop enclosing a semicircle of radius R is 9. A circular loop of wire of radius a is bent about its
located on the boundary of a uniform magnetic diameter such that the angle between the plane of
field B. At the moment t = 0, the loop is set into two halves become 60°. The resistance per unit
rotation with constant angular acceleration α
length of the wire is r0. The magnetic field is
about an axis O. The clockwise emf direction is
varying as B = B0t along the (+) Z-axis. Find the
taken to be positive
total charge flown through the loop in first t
seconds.

The variation of emf as a function of time is

1 3 2
(1) BR 2 t (2) BR  tA
2 2
BR2 t B0 at
(3) 3BR2t (4) (1)
2 2r0
B0 at
(2)
7. An electron is moving in a circular orbit of radius 8r0
R with an angular acceleration α. At the centre of
B0 at
the orbit is kept a conducting loop of radius r, (r (3)
<< R). The e.m.f induced in the smaller loop due
r0
to the motion of the electron is 2B0 at
(4)
(1) zero, since charge on electron in constant r0
μ0er 2
(2) α
4R 10. In a very long solenoid of radius R, if the
μ0er 2 magnetic field changes at the rate of dB / dt. AB =
(3) α
4πR BC. The induced emf for the triangular circuit
(4) none of these ABC shown in figure is

8. The figure shows a square loop L of side 5 cm

which is connected to a network of resistances.
The whole setup is moving towards right with a
constant speed of 1 cm s–1. At some instant, a part
of L is in a uniform magnetic field of 1T,
perpendicular to the plane of the loop. If the
resistance of L is 1.7 Ω, the current in the loop at
that instant will be close to  dB 
(1) R 2  
 dt 
 dB 
(2) 4R2  
 dt 
1 2  dB 
2  dt 
(3) R
(1) 60 μA
(2) 115 μA  dB 
(4) 2R2  
(3) 150 μA  dt 
(4) 170 μA
2
11. There is a horizontal cylindrical uniform but time-
varying magnetic field increasing at a constant
rate dB / dt as shown in figure. A charged particle
having charge q and mass m is kept in
equilibrium, at the top of a spring of spring
constant K, in such a way that it is on the
horizontal line passing through the centre of the
magnetic field as shown in figure. The μ0 I d+ 
2πR  d 
(1) ln 
compression in the spring will be
μ0 I  2d + 
2πR  2d 
(2) ln 

μ0 I  d+ 
2πR  2d + 
(3) ln 

μ0 I  2d + 2 
2πR  2d + 
(4) ln 

1 qR 2 dB 
(1)  mg −  14. In the figure, a long thin wire carrying a varying
K 2 dt 
current i = i0 sinωt lies at a distance y above one
1 qR 2 dB  edge of a rectangular wire loop of length L and
(2)  mg + 
K dt  width W lying in the x-z plane. The maximum emf
induced in the loop.
1 2qR 2 dB 
(3) mg + 
K dt 
1 qR 2 dB 
(4)  mg + 
K 2 dt 

12. A flexible wire loop in the shape of a circle has a

radius that grows linearly with time. There is a
magnetic field perpendicular to the plane of the μ 0i0Wω  L2 
loop that has a magnitude inversely proportional (1) ln  2 + 1
4π Y 
to the distance from the centre of loop i.e.
μ 0i0Wω  L 2

1
B(r )  . How does the emf E vary with time? (2) ln  2 − 1
r 4π Y 
(1) E ∝ t2 μ 0i0Wω  L2 
(2) E ∝ t (3) ln  2 + 1
2π Y 
(3) E ∝ t
μ 0i0Wω  L 2

(4) E is constant (4) ln  2 − 1
2π Y 
13. A square loop of side length  is placed near an
15. A rectangular coil of single turn, having area A,
infinitely long straight current carrying wire. rotates in a uniform magnetic field B with an
Current in the wire is I. Distance of the nearer side angular velocity ω about an axis perpendicular to
of the loop from the wire is d. The wire and the the field. If initially the plane of coil is
loop are coplanar. Total resistance of the loop is perpendicular to the field, then the average
R. If the loop translates through distance 2d in a induced e.m.f. when it has rotated through 90° is
direction shown in the figure, with the loop ωBA ωBA
(1) (2)
always remaining coplanar with wire then the π 2π
charge flowing in it during the given displacement ωBA 2ωBA
will be : (3) (4)
4π π
3
16. A rod of length  is oscillating as a physical 20. At a place the value of horizontal component of
the earth’s magnetic field H is 3 × 10–5 Weber/m2.
pendulum about one of its end with small angular
A metallic rod AB of length 2 m placed in east-
amplitude α in a crossed magnetic field B. The
west direction, having the end A towards east,
maximum emf induced in the rod will be
falls vertically downward with a constant velocity
1 3
(1) Bα g 3
(2) Bα g 3 of 50 m/s. What is the value of induced potential
2 8 difference (in mV) between the end A and end B?
1
(3) Bα g 3
(4) Bα g 3
3 21. A uniform magnetic field B = 0.25T kˆ exists in a
circular region of radius R = 5 m. A loop of radius
17. Three identical conducting circular loops are R = 5m lying in x – y plane encloses the magnetic
placed in uniform magnetic fields. Inside each field at t = 0 and then pulled at uniform velocity
loop, there are two magnetic field regions,
v = 4m/s iˆ If the emf induced (in volts) in the
separated by dashed line that coincides with a
loop at t = 2 sec is-
diameter, as shown. Magnetic fields may either be
increasing (marked as INCR) or decreasing
(marked as DECR) in magnitude at the same
rates. If IA, IB and IC are the magnitudes of the
induced currents in the loops I, II and III
respectively then

22. A metal rod of resistance 20Ω is fixed along a

diameter of a conducting ring of radius 0.1 m and
(1) IA > IB = IC (2) IA = IC > IB lies on x-y plane. There is a magnetic field
(3) IA = IB = IC (4) IC > IA = IB B = (50 T) k . The ring rotates with an angular
velocity ω = 20 rad/sec about its axis. An external
18. A circular coil of wire consists of exactly 100 resistance of 10Ω is connected across the centre of
turns with a total resistance 0.20 . The area of the ring and rim. The current through external
the coil is 100 cm2. The coil is kept in a uniform resistance is equal to 1/n Amperes. Value of n is
magnetic field B as shown in figure. The equal to______.
magnitude field is increased at a constant rate of 2
T/s. Find the induced current in the coil (in Amp.) 23. A long circular tube of length 10 m and radius 0.3
m carries a current I along its curved surface as
shown. A wire-loop of resistance 0.005 ohm and
of radius 0.1 m is placed inside the tube with its
axis coinciding with the axis of the tube. The
current varies as I = I0 cos(300t) where I0 is
constant. If the magnetic moment of the loop is
19. In a coil of resistance 100 , a current is induced Nμ0I0 sin (300 t) then 'N' is (Take 10 = π )
by changing the magnetic flux through it as shown
in the figure. The magnitude of change in flux
(in mWb) through the coil is:

4
24. A uniform magnetic field B = 0.5 T exists in a
circular region of radius R = 5 m. A loop of radius
R = 5 m encloses the magnetic field at t = 0 and
then pulled at uniform speed v = 2 m/s in the
plane of the paper. Find the induced emf (in V) in
the loop at time t = 3 s (1) BLv (2) 2BLv
2 BvL
(3) 2BLv (4)


29. A conducting wire cd of length and mass m is

sliding without friction on conducting rails a-x
25. Two identical coils 1 and 2 lie closed to each and b-y as shown. The vertical rails are connected
other such that 50% of the magnetic flux to each other with a resistance R between a and b.
generated by one links with the other. A current of A uniform magnetic field B is applied
1 A in coil 1 generate in it flux of 10–5 Wb. If this perpendicular to the plane abcd such that cd
current is reversed in 0.1 sec., the average induced moves with a constant velocity of
emf in coil 2 is (in μV). R
a b
26. A conducting disc of conductivity σ has a radius a
and thickness t. If the magnetic field B is applied
c d
l
in a direction perpendicular to the plane of the
x y
dB
disc, changes with time at the rate of =α. mgR
dt (1)
Calculate the power dissipated (in mW) in the disc B
due to the induced current. [Take : α = 2 T/s, mgR
(2)
4  B2 2
a = 1 cm, t = 2 mm and σ = 108 −1 m−1  mgR
π  (3)
B3 3
27. A uniform magnetic field exists in a circular mgR
(4)
region of radius R. A loop of radius R encloses the B2
magnetic field at t = 0 and then pulled at uniform
speed v in the plane of the paper. The induced emf 30. A conducting ring of radius 2R rolls on a smooth
in the loop as a function of time is Bv nR2 − v2t 2 horizontal conducting surface with a velocity v as
shown in figure. A uniform horizontal magnetic
. Find n.
field B is perpendicular to the plane of the ring.
The potential of A with respect to O is:
B A

2R v

[Motional EMF]
28. A straight wire of length L is bent into a O
semicircle. It is moved in a uniform magnetic (1) 2 BvR
field with speed v with diameter perpendicular to 1
(2) BvR
the field. The induced emf between the ends of the 2
wire is (3) 8 BvR
(4) 4 BvR

5
31. A uniform conducting rod AC of length 4 is 34. Two identical conducting rings A & B of radius R
rotated about a point O in a uniform magnetic are in pure rolling over a horizontal conducting
field B directed into the paper. AO = and OC = plane with same speed of centre of mass v but in
3 . Then opposite direction. A constant horizontal magnetic
field B is exist in the space pointing inside the
B plane of paper. The potential difference between
O
A C the topmost points of the two rings is:

B 2
7 (1) Zero (2) 2BvR
(1) VA − VO = (2) VO − VC = B 2
2 2 (3) 4BvR (4) None of these
9
(3) VA − VC = 4B 2
(4) VC − VO = B 2
35. A uniform magnetic field 20 T exists on right side
2
of the boundary in a gravity free space as shown
32. A conducting rod AB of mass m and negligible in figure. The given circular arc of radius 2 cm
resistance and of length 0.5 m slide on a pair of made of conducting wire of total resistance 4Ω is
parallel metallic rails, which are connected by a rotated around point O at a constant angular speed
resistance R = 10  and capacitance C = 1 F as 2 rad per second. Power required to maintain the
shown in the figure. A magnetic field B = 1 T is constant angular velocity between time interval
π π
applied normal and into the plane of the circuit. At t = s to t = s is :
time t = 0, a force F = 0.4 N is applied on the rod 6 3
AB. The terminal speed acquired by rod will be:

(1) 64 μW (2) 32 μW
(3) 128 μW (4) 16 μW

36. The figure shows an apparatus suggested by

(1) 0.5 m/s (2) 4 m/s Faraday to generate electric current from a
(3) 8 m/s (4) 16 m/s flowing river. Two identical conducting plates of
length a and width b are placed parallel facing one
33. In the figure shown the section EDGF is fixed. A another on opposite sides of the river following
rod having resistance ‘R’ is moved with constant with velocity v at a distance d apart. Now both the
velocity in a uniform magnetic field B as shown in plates are connected by a load resistance R. Then
the figure. DE & FG are smooth and the current through the load R is:- (Consider
resistanceless. Initially capacitor is uncharged. vertical component of the magnetic field produced
The charge on the capacitor: by earth is Bv and the resistivity of river water is
)

Bvub Bvub
(1) (2)
(1) remains constant R ρd
R+
(2) increases exponentially with time ab
(3) increases linearly with time Bvud
(3) (4) None of the above
(4) oscillates with time ρd
R+
ab
6
37. A copper rod AB of length L, pivoted at one end 40. In the figure shown a conducting rod of length ,
A, rotates at constant angular velocity ω, at right
angles to a uniform magnetic field of induction B. resistance R & mass m can move vertically
The e.m.f developed between the mid point C of downward due to gravity. Other parts are kept
the rod and end B is fixed. Magnetic field (B) is constant. MN and PQ
are vertical, smooth, conducting rails. The
capacitance of the capacitor is C. The rod is
released from rest. Find the maximum current in
the circuit.

BωL2 BωL2
(1) (2)
4 2
3 Bω L2 3 Bω L2
(3) (4)
4 8

38. Figure shows a copper rod moving with velocity v

mglBC
parallel to a long straight wire carrying current = (1) imax =
m + B 2l 2C
100 A. Calculate the induced emf in the rod,
where v = 5 ms–1, a =1 cm, b =100 cm (2) imax =
mglBC
m − B 2l 2C
2mglBC
(3) imax =
m + 2 B 2l 2C
2mglBC
(4) imax =
m − B 2l 2C
(1) 0.23 mV (2) 0.46 mV
(3) 0.16 mV (4) 0.32 mV
41. A rectangular loop with a sliding connector of
39. AB is a resistanceless conducting rod which forms length l = 1.0 m is situated in a uniform magnetic
a diameter of a conducting ring of radius r field B = 2T perpendicular to the plane of loop.
rotating in a uniform magnetic field B as shown in
Resistance of connector is r=2. Two resistances
figure. The resistors R1 and R2 do not rotate. Then
the current through the resistor R1 is of 6 and 3 are connected as shown in Figure.
The external force (in N) required to keep the
connector moving with a constant velocity v = 2
ms–1 is:

B
6 v 3

Bωr 2 Bωr 2 42. A rod of length 10 cm made up of conducting and

(1) (2)
2R1 2R2 non-conducting material (shaded part is non-
conducting). The rod is rotated with constant
Bωr 2 Bωr 2
(3) ( R1 + R2 ) (4) 2 R + R angular velocity 10 rad/sec about point O, in
2R1R2 ( 1 2)
constant magnetic field of 2 tesla as shown in the
figure. The induced emf (in mV) between the
point A and B of rod will be-
7
 46. A long solenoid of length L, cross section A
having N1 turns has wound about its centre a small
coil of N2 turns as shown in fig. The mutual
inductance of two circuits is

43. A rod of mass m = 2 kg slide without friction

along two parallel rails at distance d =1.3 m from
each other (see figure). The rails are joined by a
resistor to a resistance R = 0.32 Ω and placed in a μ0 A( N1 / N2 )
vertical magnetic field of induction B = 0.4 T. The (1)
L
μ0 A( N1N2 )
rod is pushed with velocity v0 = 3.38 m/s. Find the
distance (in m) covered by the rod until it stops. (2)
L
(3) μ0 AN1N2 L

μ0 AN12 N2
(4)
L

47. In Fig, there is a conducting loop ABCDEF of

resistance λ per unit length placed near a long
44. Statement 1: A resistance R is connected between straight current-carrying wire. The dimensions are
the two ends of the parallel smooth conducting shown in the figure. The long wire lies in the
rails. A conducting rod lies on these fixed plane of the loop. The current in the long wire
horizontal rails and a uniform constant magnetic
varies as I = I0t
field B exists perpendicular to the plane of the
rails as shown in figure. If the rod is given a
velocity v and released as shown in figure, it will
stop after some time. The total work done by
magnetic field is negative

The mutual inductance of the pair is

μ0 a  2a + 
2π  
(1) In
Statement 2: If force acts opposite to direction of 
velocity its work done is negative μ0 a  2a − 
2π  
[Inductors, Self and Mutual Inductance] (2) In

45. When current in a coil change from 5A to 2A in
2μ0 a  a + 
0.1sec, average voltage of 50V is produced. The (3) In  
self-inductance of the coil is: π  
(1) 1.67 H (2) 6 H μ0 a  a + 
π  
(4) ln
(3) 3 H (4) 0.67 H 

8
48. Loop A of radius (r << R) moves towards loop B 52. A relatively long straight conductor and a
with a constant velocity V in such a way that their conducting rectangular loop lie in the same plane, as
planes are always parallel. What is the distance shown in figure. Taking h = 0.4 mm, w = 1.2 mm
between the two loops (x) when the induced emf and l = 2.7 mm, find their mutual inductance, in
in loop A is maximum pH (picohenry). (Take ln(2) = 0.7)
I
h

R
(1) R (2)
2 53. A long solenoid contains another coaxial solenoid
R  1  (whose radius R is half of its own). Their coils
(3) (4) R 1 − 
2  2 have the same number of turns per unit length and
initially both carry no current. At the same instant
49. Three identical large plates are fixed at separation current starts increasing linearly with time in both
of d from each other as shown. The area of each solenoids. At any moment the current flowing in
plate is A. Plate 1 is given charge Q0 while plates the inner coil is twice as large as that in the outer
2 and 3 are neutral and are connected to each one and their directions are the same. As a result
other through coil of inductances L and switch S. of the increasing currents a charged particle,
If resistance of all connected wires is neglected initially at rest between the solenoids, starts
the maximum current flow through coil after moving along a circular trajectory of radius r (see
closing switch is (C = 0A/d) (Neglect fringe r 2
effect) figure). The value of is_______.
R

Q0 [RL Circuit]
(1) 54. In the circuit shown in figure, L is an ideal
LC
inductor and E is an ideal cell. Switch is closed at
Q0
(2) t = 0.
2LC R
2Q0
(3) 5 R
LC R
E
Q0
(4) t=0 C L
2 LC

50. If in a coil rate of change of area is 5 m2 / millisecond (1) After a long time interval potential difference
and current become 1 amp from 2 amp in 2 × 10–3 across capacitor and inductor will be equal
sec. If magnitude of field is 1 tesla then self- (2) after a long time interval charge on capacitor
inductance of the coil is________ Henry. will be EC.
(3) After a long time interval current in the
51 A coil of wire of a certain radius has 600 turns inductor will be E/R.
and a self-inductance of 108 mH. The self (4) After a long time interval current through
inductance of a 2nd similar coil of 500 turns will battery will be equal to the current through it
be ________mH. initially.
9
55. In the circuit shown, capacitor is initially 58. In which of the following circuit the current
uncharged and the battery is ideal, if the switch is through the battery is maximum just after the
closed at t = 0. The ratio of current I through the switch S is closed

cell at t = 0 and at t =  will be

(1) (i) (2) (ii)

(3) (iii) (4) Both (ii) and (iii)

59. In the given circuit, let i1 be the current drawn from

(1) 1 : 3 (2) 1 : 2 battery at time t = 0 when switch S is closed and i2 be
i
(3) 2 : 1 (4) 3 : 1 steady current at t = , then the ratio 1 is:
i2

56. A circuit element is placed in a closed box. At

time t = 0, constant current generator supplying a
current of 1 amp, is connected across the box.
Potential difference across the box varies (1) 1.0 (2) 0.8
according to graph shown in figure. The element (3) 1.2 (4) 1.5
in the box is:
60. In Fig, there is a frame consisting of two square
V(volts)
loops having resistors and inductors as shown.
8 This frame is placed in a uniform but time varying
magnetic field in such a way that one of the loops
2
is placed in crossed magnetic field and the other is
t(s) placed in dot magnetic field. Both magnetic fields
3 are perpendicular to the planes of the loops. If the
(1) resistance of 2 magnetic field is given by B = (20 + 10t) Wb m–2
(2) battery of emf 6V in both regions ( = 20 cm, b = 10 cm, R = 10Ω,
(3) inductance of 2H L = 10 H)
(4) capacitance of 0.5F

57. The adjoining figure shows two bulbs B1 and B2

resistor R and an inductor L. When the switch S is
turned off
R S
B1
Which of the following is correct about the
L magnitude or the direction of induced current in
B2
the bigger loop
(1) Direction is Clockwise
1
(2) i = 1 − e−t  Amp
(1) Both B1 and B2 die out promptly 40
1
(2) Both B1 and B2 die out with some delay (3) i = 1 − e−t  Amp
20
(3) B1 dies out promptly but B2 with some delay
1
(4) i = 1 − e−t  Amp
(4) B2 dies out promptly but B1 with some delay 10  
10
61. In the given circuit Key is closed at t = 0. At what 64. Suppose the emf of the battery in the circuit
time the P.D across inductor is one fourth of emf shown varies with time t so the current is given by
of the cell. i(t) = 3 + 5t, where i(t) is in amperes & t is in
second (Take R = 4, L = 6H) an expression for
the battery emf as function of time. is a + bt Volt.
Then the value of a + b is

L L
(1) ln 2 (2) ln 4
R R
2L
(3) 0 (4)
R

62. The network shown in the figure is part of a

complete circuit. If at a certain instant, the current
I is 5A and it is decreasing at a rate of
103 Amp/ sec. then VB –VA equals (in volt)
1 + 5 mH
A B
1 15V

63. At a given instant the current in the circuit is

increasing with a rate a = 4 amp/s and current in
the circuit is 2 amp. The charge (in C) on the
capacitor at this given instant will be:

11
 JEE ADVANCED
[Induced EMF] (A) The emf induced in the left ring is zero
1. Two concentric coplanar circular loops made of (B) The emf induced in both the rings is non
wire, with resistance per unit length 10 Ω/m have zero.
diameters 0.2 m and 2 m. A time varying potential (C) The magnetic force acting on the right ring is
difference (4 + 2.5 t) volt is applied to the larger zero.
loop. Calculate the current in the smaller loop. (D) The magnetic force acting on both the rings
is non-zero.
2. In figure, the square loop of wire has sides of
5. Plane rectangular loop is placed in a magnetic
length 2.0 cm. A magnetic field is directed out of
field. The emf induced in the loop due to this field
the page; its magnitude is given by B = 4.0 t2y,
where B is in tesla, t in second, and y in metre. is i whose maximum value is im . The loop was
Determine the emf around the square at t = 2.5 s pulled out of the magnetic field at a variable
and give its direction. velocity. Assume the B is uniform and constant
i is plotted against time t as shown in the graph.
Which of the following are/is correct statement(s):

(A) im is independent of rate of removal of coil

from the field
3. A conducting circular loop is pulled with a (B) The total charge that passes through any
constant velocity v towards a region of uniform point of the loop in the process of complete
magnetic field of induction B as shown in figure removal of the loop does not depend on
then current induced in the loop is (d > 2r) velocity of removal.
(C) The total area under the curve (i vs t) is
independent of rate of removal of coil from
the field.
(D) The area under the curve is dependent on the
(A) clockwise while entering. rate of removal of the coil.
(B) anti clockwise while entering.
6. A conducting loop is placed in a uniform
(C) zero when completely inside.
magnetic field with its plane perpendicular to the
(D) clockwise while leaving.
field. An emf is induced in the loop if
(A) it is translated
4. In the diagram shown, a uniform constant (B) it is rotated about its axis
magnetic field is present perpendicular to the (C) it is rotated about axis through bisector
plane of the paper. Both the rings are identical and (D) it is expanded
have a constant resistance per unit length. The left
ring has been kept fixed at its position and the 7. Two metallic rings A and B, identical in shape
right ring is slid uniformly on the left ring towards and size but having different resistivities A and
the right side. Which of the following statements B, are kept on top of two identical solenoids as
is/are true? (Neglect the self-inductance and shown in the figure. When current I is switched on
mutual inductance of the coils.) in both the solenoids in identical manner, the rings
A and B jump to heights hA and hB, respectively,
with hA > hB. The possible relation(s) between
their resistivities and their masses mA and mB
is(are)

12
 (A) Change in magnetic flux is 100 Wb
(B) Rate of change of magnetic flux is
decreasing.
(C) Total heat produced in the resistor is 666.67 J.
(D) Maximum power during the flow of current
is 1000 W.

11. A nonconducting ring of uniform mass m, radius b

and uniform linear charge density ‘λ’ is suspended
(A) A > B and mA = mB as shown in figure in a gravity free space. There is
(B) A < B and mA = mB uniform coaxial magnetic field B0, pointing up in
(C) A > B and mA > mB a circular region of radius ‘a’ (< b). Now if this
(D) A < B and mA < mB field is switched off, then:-

8. A semicircle conducting ring of radius R is placed

in the xy plane, as shown in the figure. A uniform
magnetic field is set up along the x-axis. No net
emf, will be induced in the ring if

(A) There will be induced electric field on

periphery of ring, in anticlockwise sense
when seen from above
(B) Induced electric field imparts angular
(A) it moves along the x-axis momentum of magnitude λπa2b B0
(B) it moves along the y-axis
(C) it moves along the z-axis (C) Final angular velocity of ring will be more if
(D) it remains stationary time taken to switch off the field (B0) is small
(D) Final angular velocity will always be
9. Figure shown below is made of a conductor independent of time taken to switch off the
located in a magnetic field along the inward field (B0).
normal to the plane of the figure. The magnetic
field starts diminishing. Then the induced current 12. An equilateral triangle ABC of side a is placed in
the magnetic field with side AC and its centre
coinciding with the centre of the magnetic field.
The magnetic field varies with time as B = kt. The
emf induced across side AB is

(A) at point P is clockwise

(B) at point Q is anticlockwise
(C) at point Q is clockwise
(D) at point R is zero
3 2
10. A variable current flows through a 10 Ω resistor (A) ak
4
coil kept in changing magnetic field for 2 seconds.
(B) Zero
3 2
(C) ak
8
3 2
(D) ak
8

13
13. Consider a perfectly conducting uniform disc of 14. List-I shows the cylindrical region of radius r
where a downward magnetic field B exists,
mass m and radius 'a' hinged in vertical plane
where B is increasing at the rate of dB/dt. A rod
from its centre and free to rotate with respect to PQ is placed in different situation as shown.
Match the List-I with the correct statement in List-
hinge. A resistance R is connected between centre
II regarding the induced emf in rod
of the disc and periphery by using two sliding List–I List–II
I P Induced emf
contacts C1 & C2. A long non conducting massless in rod PQ is
1 2 dB
string is wrapped around the disc, whose another r 
2 dt
end is attached with a block of mass m. There II Q Induced emf
in rod PQ is
exist a uniform horizontal magnetic field B. less than
1 2 dB
Whole arrangement is shown in the figure. r 
2 dt
III R End P is
positive with
respect to
point Q

IV S End Q is
positive with
respect to
point P

T None of these
Given system is released from rest at t = 0. I II III IV
(A) P, S P, R Q, S P, Q
Assume friction between string and disc is
(B) Q, R P, S P, R Q, S
sufficient so that there is no slipping between (C) P, R Q, S P, S P, S
(D) Q, S P, S P, S P, R
them. Let at any time t, velocity of block is v,

angular velocity of disc is ω and current in Paragraph Questions (15 to 19)

Passing a current through two conductors and measuring
resistance is i. the force between them provides an absolute
determination of the current itself. The “Current
(A) From the energy equation Balance” designed by Lord Kelvin in 1882 exploits this
dv d  2 method. It consists of six identical single turn coils
mgv = mv +I +i R C1...C6 of radius a, connected in series. As shown in
dt dt
figure, the fixed coils C1, C3, C4, and C6 are on two
(B) The work done by the magnetic field is zero horizontal planes separated by a small distance 2h. The
coils C2 and C5 are carried on balance arms of length d,
but it converts some part of the mechanical and they are, in equilibrium, equidistant from both
energy into heat. planes. The current I flows through the various coils in
such a direction that the magnetic force on C2 is upwards
(C) The velocity v, of the block as a function of while that on C5 is downwards. A mass m at a distance x
from the fulcrum O is required to restore the balance to

time is v =

(
1 − e−t ) the equilibrium position described above when the
current flows through the circuit. Let M be the mass of
the balance (except for m and the hanging parts), G its
g −t
(D) The acceleration of the block is e centre of mass and l the distance OG .
3
14
 19. Choose correct options from following :-
(A) Electrical energy stored in capacitor is
maximum when rod is at its lower extreme
position
(B) Flectrical energy stored in capacitor is
maximum when rod is at its mean position
(C) Current in rod is maximum at mean position
of rod
(D) If magnetic field is switched off then mean
position of rod will change
15. Compute the force F on C2 due to the magnetic
interaction with C1. For simplicity assume that the
20. A uniformly charged non conducting disc of
force per unit length is the one corresponding to
radius 2 m, mass 1 kg and charge Q = 4 C is lying
two long, straight wires carrying parallel currents.
over smooth horizontal plane. The space carries a
μ0 I 2 μ0 I 2a uniform vertical magnetic field of 1 Tesla find the
(A) (B)
2πh h angular velocity of disc (in rad/sec) just after
2μ0 I 2 a μ0 I 2 reversal in direction of magnetic field.
(C) (D)
h ah
21. A triangular wire frame (each side = 2m) is placed
16. The current I is measured when the balance is in in a region of time variant magnetic field having
equilibrium. Give the value of I in terms of the dB / dt = 3 T / s . The magnetic field is
physical parameters of the system. The
perpendicular to the plane of the triangle. The
dimensions of the apparatus are such that we can
base of the triangle AB has a resistance 1Ω while
neglect the mutual effects of the coils on the left
and on the right. the other two sides have resistance 2Ω each. The
1/2 1/2 magnitude of potential difference between the
 mghx   mghx  points A and B will be _______V.
(A)   (B)  
 4μ 0 ad   8μ 0 ad 
 mghx 
1/2
 mgax 
1/2 22. A metal rod OA of mass m and length r is kept
(C)   (D)   rotating with a constant angular speed ω in a
 2μ 0 ad   4μ 0 hd  vertical plane about a horizontal axis at the end O.
The free end A is arranged to slide without friction
17. The balance equilibrium is stable against along a fixed conducting circular ring in the same
deviations producing small changes δz in the plane as that of rotation. A uniform and constant
height of C2 and –δz in C5. Compute the magnetic induction B is applied perpendicular
maximum value δz-max so that the balance still and into the plane of rotation as shown in figure.
returns towards the equilibrium position when it is An inductor L and an external resistance R are
released. (Consider that the coils centres remain connected through a switch S between the point O
approximately aligned) and a point C on the ring to form an electrical
circuit. Neglect the resistance of the ring and the
M h2 M h2
(A) (B) rod. Initially, the switch is open.
2mxd mxd
M h M h
(C) (D)
mxd mx

18. Find time period of oscillation of rod :-

m B 2 2C
(A) 2π (B) 2π
k K
m + B 2 2C B 2 2C + m
(C) π (D) 2π
k 2K
15
 (a) What is the induced emf across the terminals
of the switch?
(b) (i) Obtain an expression for the current as a
function of time after switch S is closed.
(ii) Obtain the time dependence of the torque
required to maintain the constant angular
speed, given that the rod OA was along the
26. A magnetic field B = ( B0 y / a ) k is into the plane
positive X-axis at t = 0.
of paper in the +z direction. B0 and a are positive
23. Two straight conducting rails form a right angle constants. A square loop EFGH of side a, mass m
where their ends are joined. A conducting bar and resistance R, in x-y plane, starts falling under
contact with the rails starts at vertex at the time t = the influence of gravity. Note the directions of x
0 and moves symmetrically with a constant and y axes in the figure. Find
velocity of 5.2 m/s to the right as shown in figure.
A 0.35 T magnetic field points out of the page.
Calculate:
(i) The flux through the triangle by the rails &
bar at t = 3.0 s.
(ii) The emf around the triangle at that time.
(iii) In what manner does the emf around the (a) the induced current in the loop and indicate
triangle vary with time. its direction,
(b) the total Lorentz force acting on the loop and
indicate its direction,
(c) an expression for the speed of the loop, v(t)
and its terminal value.

[Motional EMF]
27. PQ is an infinite current-carrying conductor. AB
24. A long straight wire is arranged along the and CD are smooth conducting rods on which a
symmetry axis of a radial coil of rectangular conductor EF moves with constant velocity V as
cross-section, whose dimensions are given in the shown in figure. The force needed to maintain
figure. The number of turns on the coil is N, and constant speed of EF is
relative permeability of the surrounding medium
is unity. Find the amplitude of the emf induced in
this coil, if the current i = im cos ωt flows along
the straight wire.

2
1  0 IV  b  
In   
VR  2
(A)
 a 
2
  IV  b   3
(B)  0 In   
 2  a   VR
2
25. A variable magnetic field creates a constant emf E   IV  b  V
(C)  0 In   
in a conductor ABCDA. The resistances of portion  2  a  R
ABC, CDA and AMC are R1, R2 and R3 2
respectively. What current will be shown by meter V  0 IV  b  
In   
R  
(D)
M? The magnetic field is concentrated near the  a 
axis of the circular conductor.
16
28. The loop shown moves with a velocity v in a 31. The arrangement shown which is confined in a
uniform magnetic field of magnitude B, directed vertical plane has two rails inclined at angle θ
into the paper. The potential difference between with horizontal. A horizontal rod of length 
points P and Q is e. Then, moves on the rails with constant speed v, in the
region with transverse field B. Choose the correct
alternative(s).
The rod starts moving at time t = 0

1
(A) e = BLv
2
(B) e = BLv
(C) P is positive with respect to Q
(D) Q is positive with respect to P

29. A thin conducting rod of length l is moved such 2Bvl sinθ

(A) At t = 0, current in the circuit is
that it's end B moves along the X-axis while end r
A moves along the Y-axis. A uniform magnetic Blv sin θ
(B) At t = ∞, current in the circuit is
field B = B0 kˆ exist in the region. At some instant, r
(C) At any time t (except at t = 0) point A is at
velocity of end B is v and the rod makes an angle higher potential than point B
of  = 60° with the X-axis as shown in the figure. (D) At any time t (except at t = 0) point D is at
Then, at this instant lower potential than C

32. A uniform magnitude field exists in a region given

by B = 3iˆ + 4 ˆj + 5kˆ . A rod of length 5 m along
y-axis moves with a constant speed of 1 m/s along
x-axis. Then the induced emf in the rod will be
(A) 0, if rod is insulating
2v (B) 25 V, if rod is conducting
(A) angular speed of rod AB is  = (C) 25 V, if rod is insulating
3l
(D) 15 V, if rod is conducting
3v
(B) angular speed of rod AB is  =
2l 33. A disc of radius r is rolling without sliding on a
Blv horizontal surface with a velocity of centre of
(C) e.m.f. induced in rod AB is mass v and angular velocity ω in a uniform
3 magnetic field B which is perpendicular to the
Blv plane of the disc as shown in figure. O is the
(D) e.m.f. induced in rod AB is
2 3 centre of the disc and P, Q, R and S are the four
points on the disc. Which of the following
statements is true
30. A conducting rod of length l is hinged at point O.
It is free to rotate in a vertical plane. There exists
a uniform magnetic field B in horizontal
direction. The rod is released from the position
shown in figure. Potential difference between the
(A) Due to translation, induced emf across PS =
two ends of the rod is proportional to
Bvr
(B) Due to rotation, induced emf across QS = 0
(C) Due to translation, induced emf across RO = 0
(D) Due to rotation, induced emf across
Br 2
(A) l3/2
(B) l 2 OQ =
2
(C) sin (D) (sin ())1/2
17
Paragraph Queston (34 to 36)
A pair of parallel horizontals conducting rails of
negligible resistance shorted at one end is fixed on a
table. The distance between the rails is L. A conducting
massless rod of resistance R can slide on the rails
without friction. The rod is tied to a massless string
which passes over a pulley fixed to the edge of the table.
A mass m, tied to the other end of the string, hangs 37. The acceleration of connector as a function of x
vertically. A constant magnetic field B exists Kx Kx
(A) (B)
perpendicular to the table. The system is released from m m − B 2 L2C
rest. (Figure) Kx
(C) (D) None
m + B 2 L2C

38. The rod will execute (after projecting it at t = 0)

(A) SHM
(B) harmonic motion but not SHM
(C) The rod will come to rest at certain position
then afterward it will not move
(D) None of these
34. The acceleration of the mass m moving in the
downward direction is 39. The maximum compression in the spring
m − B2 L2C m + B2 L2C
2 2
B Lv
(A) g (B) (A) V0 (B) V0
mR K K
 B 2 L2 v   B 2 L2v  m
(C)  g −  (D)  g +  (C) V0 (D) None of therse
 mR   mR  K

40. Consider the situation shown in figure. The wires

35. The terminal velocity acquired by the rod is
P1Q1 and P2Q2 are made to slide on the rails with
(A) g (B) gR the same speed 5 cm/s. Find the electric current in
mgR the 19Ω resistor if :
mgR
(C) (D) (a) both the wires move towards right
BL B2 L2 (b) if P1Q1 moves towards left but P2Q2 moves
towards right.
36. The acceleration of mass m when the velocity of
the rod is half the terminal velocity is
g
(A) g (B)
2
g g
(C) (D)
3 4
41. A uniform magnetic field B fills a cylindrical
Paragraph Question (37 to 39)
volumes of radius R. A metal rod CD of length l is
The conducting connector ab mass m and length L can placed inside the cylinder along
freely slide on a horizontal long conducting parallel rails
connected by capacitor C at one end (as shown in
figure). A non-conducting light spring (spring const. K)
connected to the connector ab and it is in a relaxed state.
The whole system is placed in uniform magnetic field of
strength B directed into the plane of rails (as shown in
a chord of the circular cross-section as shown in
figure) Now at time t = 0, connector is suddenly given a
the figure. If the magnitude of magnetic field
velocity V0 in rightward direction. If resistance of circuit
increases in the direction of field at a constant rate
is negligible then dB/dt, find the magnitude and direction of the
EMF induced in the rod.
18
42. Two parallel vertical metallic rails AB & CD are 45. In the figure, a conducting rod of length l = 1
separated by 1 m. They are connected at the two meter and mass m = 1 kg moves with initial
ends by resistance R1 & R2 as shown in the figure. velocity u = 5 m/s on a fixed horizontal frame
A horizontally metallic bar L of mass 0.2 kg slides containing inductor L = 2 H and resistance R = 1
without friction, vertically down the rails under Ω. PQ and MN are smooth, conducting wires.
the action of gravity. There is a uniform There is a uniform magnetic field of strength B =
horizontal magnetic field of 0.6 T perpendicular to 1 T. Initially there is no current in the inductor.
the plane of the rails, it is observed that when the Find the total charge in coulomb, flown through
terminal velocity is attained, the power dissipated the inductor by the time velocity of rod becomes
in R1 & R2 are 0.76 W & 1.2 W respectively. Find vf = 1 m/s and the rod has travelled a distance x =
the terminal velocity of bar L & value R1 and R2. 3 meter.

[Inductors, Self and Mutual Inductance]

46. Two concentric and coplanar coils have radii a
43. A square loop ABCD of side ℓ is moving in xy
and b (>> a) as shown in fig. Resistance of the
plane with velocity v = βtjˆ . There exists a inner coil is R. Current in the outer is increased
nonuniform magnetic field from 0 to i, then the total change circulating the
( )
B = −B0 1 + αy kˆ ( y  0), where B0 and α are
2 inner coil is

positive constants. Initially, the upper wire of the

loop is at y = 0. Find the induced voltage across
the resistance R as a function of time. Neglect the
magnetic force due to induced current.

μ0ia2 π μ 0iab
(A) (B)
2Rb 2R
μ iab πb2 μ0ib
(C) 0 (D)
2a R 2πR
44. In the figure shown ‘PQRS’ is a fixed
resistanceless conducting frame in a uniform and 47. Two different coils have self-inductance L1 = 8
constant magnetic field of strength B. A rod ‘EF’ mH, L2 = 2mH. The current in one coil is
of mass ‘m’, length ‘l’ and resistance R can increased at a constant rate. The current in the
smoothly move on this frame. A capacitor charged second coil is also increased at the same rate. At a
to a potential difference ‘V0’ initially is connected certain instant of time, the power given to the two
as shown in the figure. Find the velocity of the rod coils is the same. At that time the current, the
as function of time ‘t’ if it is released at t = 0 from induced voltage and the energy stored in the first
rest. coil are i1, V1 and W1 respectively. Corresponding
values for the second coil at the same instant are
i2, V2 and W2 respectively. Then
i 1 i
(A) 1 = (B) 1 = 48
i2 4 i2
W2 V2 1
(C) =4 (D) =
W1 V1 4

19
48. A capacitor with charge Q on it is connected to an
inductor L as shown in diagram at t = 0. When the
switch is flipped from position 1 to 2, the current
in the circuit is observed to be at half of its
maximum value. Then

50. Value of I1, is

C
(A) 2V0
L
(A) Charge on capacitor at that time was Q/2
6C
Q 3 (B) V0
(B) Charge on capacitor at that time was L
2
3C
Q 2 (C) V0
(C) Total energy lost is L
4C
2C
Q2 (D) V0
(D) Total energy lost is L
8C
51. Value of I2, is
49. Two capacitors of capacitances 2C and C are
2C 5 C
connected in series with an inductor of inductance (A) 2V0 (B) V0
L 2 L
L. Initially capacitors have charge such that VB –
VA = 4V0 and Vc – VD = V0. Initial current in the 7C 3 C
(C) V0 (D) V0
circuit is zero. Find: L 2 L

52. Minimum value of t0, is

2LC
(A) 2π
3
π 2LC
(B)
(A) Maximum current that will flow in the circuit 2 3
6C 2
is (C) π 2LC
L 3
(B) Potential difference across each capacitor at 2LC
that instant is 3V0 (D) π
3
(C) Maximum current that will flow in the circuit
3C 53. In figure a 120-turn coil of radius 1.8 cm and
is
L resistance 5.3 Ω is placed outside a solenoid. The
(D) Potential difference across each capacitor at current in the solenoid is 1.5 A and it reduces to
that instant is 2V0 zero at a steady rate in 25 ms. What current
appears in the coil? The number of turns per unit
Paragraph questions (50 to 52) length of the solenoid is 220 turns/cm and its
For the circuit shown in figure, capacitors C, 2C are diameter D = 3.2 cm.
charged to potential V0 and 2V0 respectively. Switch S1
is closed at t = 0. At the instant t = t0, when current in
inductor is maximum and is equal to I1, switch S2 is also
closed. There after maximum current through inductor
becomes I2. Based on the given information, answer the
following questions.

20
54. The current in a coil of self-induction 2.0 henry is [RL Circuit]
increasing according to i = 2 sin t2 ampere. Find 58. In fig, i1 = 10e –2t
A, i2 = 4 A and VC = 3e–2t V
the amount of energy spent during the period
when the current changes from 0 to 2 ampere.

55. A coil with 1500 turns, a radius of 5.0 cm and a

resistance of 12 Ω surrounds a solenoids with 240
turns/cm and a radius of 4 cm as shown in figure.
The current in the solenoid changes at a constant
rate from 0 to 20 A in 0.10 s. Calculate the
magnitude of the induced current (in Amp, upto 1
decimal place) in the 1500 turn coil (π2 = 10, The current iL is
neglect self inductance of the coil). (A) [2 + 2e–2t] Amp (B) [4 + 2e–2t] Amp
(C) [1 + 2e–2t] Amp (D) [2 – 2e–2t] Amp

R
59. A solenoid of inductance L and resistance is
2
connected in parallel to resistance R. An ideal
battery of emf E is connected across the parallel
combination as shown in figure, switch S is kept
56. A thin wire ring of radius a and resistance r is closed for long time and it is opened at time t = 0.
located inside a long solenoid so that their axes The current through solenoid immediately after
coincide. The length of the solenoid is equal to l, switch opened is I and total heat generated in the
its cross-sectional radius, to b. At a certain solenoid after switch opened is H. Then choose
moment the solenoid was connected to a source of
the correct option(s):
a constant voltage V. The total resistance of the
circuit is equal to R. Assuming the inductance of
the ring to be negligible, find the maximum value
of the radial force acting per unit length of the
ring.

57. A conducting frame ABCD is kept fixed in a

E 2E
vertical plane. A conducting rod EF of mass m (A) I = (B) I =
can slide smoothly on it remaining horizontal R R
always. The resistance of the loop is negligible 2 LE 2 LE 2
and inductance is constant having value L. The (C) H = (D) H = 2 2
3 R2 R
rod is left from rest and allowed to fall under
gravity and inductor has no initial current. A
60. In the given circuit, the switch is closed at t = 0.
uniform magnetic field of magnitude B is present
throughout the loop pointing inwards. Determine. Choose the correct answer

(A) Current in the inductor when the circuit

reaches the steady state is 4 A
(B) The net change in flux in the inductor is 1.5
(a) Position of the rod as a function of time
Wb
assuming initial position of the rod to be x =
0 and vertically downward as the positive X- (C) The current through the inductor at steady
axis. state is 1.5 A
(b) Maximum current in the circuit (D) The charge stored in the capacitor in steady
(c) Maximum velocity of the rod. state is 1.2 mC
21
61. Figure shows a circuit with two resistors and an 64. Key is in position 2 for time t. Thereafter, it is in
ideal inductor. position 1. Resistances of the bulb and inductance
of inductor are marked in the figure choose the
correct alternative

(A) Bulb 2 dies as soon as key is switched into

(A) The current in R1 is zero just after closing the
position 1
switch S.
(B) The current in R1 is maximum just after (B) Time in which brightness of bulb 1 becomes
closing the switch S. half its maximum brightness does not depend
(C) The current in R2 is zero just after closing of on t.
the switch S. (C) If t = , total heat produced in bulb 1 is
(D) The currents in the resistors are maximum of L2
their values a long time after closing the 2R22
switch S.
(D) Ratio of maximum power consumption of
62. L, C and R represent inductance, capacitance and bulbs depends on time
resistance respectively. Which of the following
have dimensions of frequency? 65. A source of constant voltage V is connected to a
L 1 resistance R and two ideal inductors L1 and L2
(A) (B)
C LC through a switch S as shown. There is no mutual
R 1 inductance between the two inductors. The switch
(C) (D) S is initially open. At t = 0, the switch is closed
L RC
and current begins to flow. Which of the
63. Keys K1 and K2 are simultaneously closed at t = 0. following options is/are correct?
At any time t current through K1 is i and current in
inductor is increasing at the rate x. Current in the
resistor is zero. Choose the correct alternative

(A) After a long time, the current through L1 with

V L2
be
R L1 + L2
(B) After a long time, the current through L2 will
(A) ε – ir = Lx V L1
be
Lx R L1 + L2
(B) ε + ir =
2 (C) The ratio of the currents through L1 and L2 is
(C) Q2 = 2Q1 fixed at all times (t > 0)
Q (D) At t = 0, the current through the resistance R
(D) 1 = Lx
C V
is
R

22
66. Consider a so-called Maxwell’s bridge shown in List–I List–II
figure below, which is used for measuring the I P Voltage across
inductance L and the ohmic resistance R of a inductor can be
greater than E at t =
inductor. To that end, the other parameter are 0.
adjusted so that the voltage reading will be zero.
Assuming that such a state has been achieved
II Q Voltage across
inductor would be
less than E at t = 0.

III R After long time,

energy stored in
(A) L = R1R2C (B) L = 2R1R2C inductor is zero.
R1R2 2R1R2
(C) R = (D) R =
RC RC
IV S After long time,
energy stored in
67. In the circuit shown, L1 = 2 H, L2 = 3 H, R1 = 4 kΩ inductor is non-
zero
and R2 = 2 kΩ . In steady state, match List-I to
value in List-II
T Voltage across
inductor increases
as time progress.
I II III IV
(A) P, R Q, R Q, S P, R
(B) Q Q, R P, R Q, S
(C) R Q, R Q, R P, R
List–I List–II (D) P, S Q R P, S, T
I The current through P 16
inductor L1 in mA. Paragraph Questions (69 to 71 )
II The voltage across Q 10 In figure shown, the rod PQ has mass m, length l and
inductor L2 in volt. resistance R, the horizontal rails have negligible friction.
III The energy stored in R 4 A battery of emf E and negligible internal resistance is
inductor L1 in μJ connected between point a and b. The rod is released
IV The energy stored in S 0 from rest in a uniform magnetic field B (into the plane of
inductor L2 in μJ paper)
T 24
I II III IV
(A) P Q R S
(B) R S P T
(C) R S T Q
(D) Q P T R 69. The velocity of the rod as function of time is
 mR 
68. In List–I some circuit are given. In all the circuits  Here  is  = 2 2 
 B l 
except in (A) switch S remains closed for long
time and then it is opened at t = 0 while for (A), (A)
E
Bl
(
1 − e −t /  ) (B)
E
Bl
(
1 + e −t /  )
the situation is reversed. List–II tells something
about the circuit quantities. Match the entries of (C)
3 E
2 Bl
(
1 − e −t /  )
(D)
E
2Bl
(
1 − e −t /  )
List–I with the entries of List–II.
23
70. After some time, the rod will approach a terminal 73. The circuit shown in figure is in the steady state
speed. Find an expression for it. with switch S1 closed and S2 to open. At t = 0, S1
3E E is opened and S2 is closed. The first instant t,
(A) (B)
2 Bl 2 Bl when the energy in the inductor becomes one-
E 2E π
(C) (D) third of that in the capacitor C2 is millisecond.
Bl Bl α
Find the value of α.
71. The current when the rod attains its terminal speed
is
2E
(A) (B) zero
R
3E E
(C) (D)
2R 2R

72. In the figure shown switch S1 remains connected

and switch S2 remains open for a long time. Now 74. In the LR circuit shown, what is the variation of
S2 is also closed. Assuming  = 10 V and L = 1 H. the current I as a function of time? The switch is
Find the magnitude of rate of change of current closed at time t = 0 sec.
(in A/s) in inductor just after the switch S2 is
closed.

24
 ANSWER KEY

JEE MAIN

1. (4) 23. (6) 45. (1)

2. (3) 24. (8) 46. (2)
3. (1) 25. (50) 47. (1)
4. (4) 26. (4) 48. (2)
5. (2) 27. (4) 49. (4)
6. (1) 28. (4) 50. (10)
7. (2) 29. (2) 51. (75)
8. (4) 30. (1) 52. (756)
9. (2) 31. (3) 53. (2)
10. (1) 32. (4) 54. (4)
11. (4) 33. (2) 55. (4)
12. (4) 34. (3) 56. (4)
13. (4) 35. (4) 57. (2)
14. (1) 36. (3) 58. (2)
15. (4) 37. (4) 59. (2)
16. (2) 38. (2) 60. (2)
17. (2) 39. (1) 61. (1)
18. (5) 40. (1) 62. (15)
19. (250) 41. (2) 63. (6)
20. (3) 42. (51) 64. (62)
21. (6) 43. (8)
22. (3) 44. (4)

25
 JEE ADVANCED

1. (1.25 A) ER1 46. (A)

25.
2. (6. 80 μV, clockwise.) R1R2 + R2 R3 + R3 R1 47. (A, C, D)
48. (B, D)
B0 av
3. (B, C, D) 26. (a) i = in 49. (A, B)
R
4. (A, D) 50. (B)
anticlockwise direction,
5. (B, C) 51. (B)
v is velocity at time t,
52. (B)
6. (C, D) (b) Fnet = B02 a2v / R , 53. 30 mA
mgR  
7. (B, D) −
B02 a 2t 54. 4J
(c) v = 2 2 1 − e mR  55. (3.8)
8. (A, B, C, D) B0 a  
  0 a 2V 2
9. (A, C, D) 56.
4rBlb2
10. (A, B, D) 27. (A)
g
11. (A, B, D) 28. (A, C) 57. (a) x= [1 − cosωt ]
ω2
12. (A) 29. (A, D) 2mg
(b) I max =
13. (A, B, C) 30. (B, D) B
g
14. (B) 31. (A, B) (c) vmax =
ω
15. (B) 32. (A, B)
58. (B)
16. (A) 33. (A, B, C, D) 59. (A, C)
17. (B) 34. (C) 60. (A, B)
18. (D) 61. (B, C, D)
35. (D)
62. (B, C, D)
19. (B) 36. (B) 63. (A, C, D)
20. (4.00) 37. (C) 64. (A, B, C)
21. (0.4) 65. (A, B, C)
38. (A)
66. (A, C)
1 39. (B)
22. (a) E = Bωr 2 67. (B)
2 40. (a) 0.1 mA, 68. (C)
(b) (i) (b) zero 69. (A)
Bωr 1 − e− Rt / L 
2 70. (C)
I= , 1 dB l2 71. (B)
2R 41. R2 −
2 dt 4 72. (10.00)
(ii)
42. V = 1 m/s 73. (300)
Rt
V −L
( )
mgr ωB2 r 4 R1 = 0.47 Ω 74. e
τ= cosωt + 1 − e− Rt / L R
2 4R R2 = 0.30 Ω
 α  β 2t 5 
23. (i) 85.22 Tm ; 2 43. ε = − B0 Iβ  t + 
 4 
(ii) 56.8 V;
  B2 2 1  
− + t
44. v=
B CV0 
1 − e  mR RC  
(iii) linearly 2 2 
m+ B C  
 
− B2
( )
2

μ0 hωim N b x − m vf − u
24. ln 45. Q= R = 1C
2π a B

26
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