44 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Section - B : Multiple Choice Type Questions
1. Answer (1, 2, 3)
(i) Between inner and outer sphere E is absent hence potential is constant.
(ii) When we connect a body to earth, potential becomes zero.
(iii) Vinner sphere
= 0 = R
kQ
R
kq
2
++
q
R2R
Q
s
2
2. Answer (1, 2, 3)
(i) Charge will appear on outer surface of conductor.
+
+ +
(ii) At the surface of conductor surface charge density () )( curvature of radius
1
R
Surface charge density is not uniform for irregular shape
(iii) Electric field is zero inside conductor.
3. Answer (2, 4)
3 = 4y x
y
x
z
Electric field vector is perpendicular to the surface, there are infinite
planes which cut xy plane in line 3y = 4x. All vectors in form of
)ˆˆ3ˆ4( kcijCE are perpendicular.
4. Answer (1, 2, 3)
All motions straight line, circular and helical are possible. Examples are like these
Straight line If particle is at rest and electric field is uniform.
Circular Motion of negatively charged particle around positive infinite line charge.
Helical Motion of negatively charged particle around positive infinite line charge as shown in diagram
v1
v2
wireof axis along motion for eresponsibl
velocity tangential
2
1
⇒
⇒
v
v
45Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
5. Answer (1, 2, 3, 4)
Q1
Q2
E Point P
(0, 0, 0) ( , 0, 0)a
( /2, , 0)a a
y
If Ex at P is zero, then magnitude and sign of charges must be same.
Field at ⎟⎠
⎞⎜⎝
⎛0,0,
2
a
is zero while on point (2a, 0, 0) and (–2a, 0, 0) E field is along x axis while at point ⎟⎠
⎞⎜⎝
⎛a
a,0,
2 E
is along z axis.
6. Answer (2, 4)
z
vE
y
vE
x
vE zyx
;;
Ey and E
z are zero. So change in potential along y and z-axis are zero
∫∫ dxEvx
y
z
x
1 m
O2
2
0xEv + constant (non-linear)
Electric flux through the cubical volume 000 )0(1 EEE
Gauss theorem implies
0
0E
qE
00Eq
7. Answer (1, 2, 3)
At a point, two potentials are not possible.
8. Answer (1, 2, 4)
Induced charge per unit area of the outer surface, 21
4 b
Q
Electric field at the outer surface 2
00
1
4 b
Q
Electric field at inner surface = 2
04 a
Q
If inner or outer surface of the sphere is earthed, it comes to zero potential. If a charge is brought from infinity
to the sphere, no work should be done charge on outer surface of sphere is zero.
46 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
9. Answer (1, 3, 4)
As dielectric constant varies from left to right, strength of electric field will be difference at difference points on a
line to plates.
If we draw a gaussian surface, flux will not be zero so charge enclosed will also be non-zero.
10. Answer (2, 4)
Capacitance,
d
AC
0
If d increases, C decreases. Potential difference V is constant
Charge, Q = CV decreases
Electric field, d
VE decreases.
Energy stored in capacitor, 2
2
1CVU decreases. Energy flows back to the source of voltage and/or gets
dissipated as heat
Also, work done by the external agent goes to the source of voltage and/or gets dissipated as heat.
11. Answer (1, 2, 3)
+2 – –
O q1
– q1
q2
– q2 O
Option 1 in modified charge on outer surfaces of 1 and 3 plate will not be changed
O + q1 – q
2 + O = + Q [Conservation of charge on plates 1 and 3]
q1 – q
2 = Q ...(i)
V1 – V
3 = 0 =
1 2
0 0
q d q d
A A
...(iii)
By solving these two equations q1 = q
2 =
2
Q
12. Answer (3, 4)
Before removing cell,
Ceq
= 3
2C
Charge on each capacitor
3
2 C
After removing cell and reconnecting the two capacitors, charge on capacitor with capacitance C
02
CCC
Similarly charge on capacitor 2C = 0
So, all energy stored in capacitors is lost.
13. Answer (1, 4)
C1 = 4
0 (1.5) a = 6
0a
C2 = a
a
a
aa
aa
0
2
00 12)5.0(
)5.1()4(
)5.1(
5.14
C3 = a
a
aa
a
0
0
2
0 185.0
5.15.14
)5.1(
5.15.14
C3 > C
2 > C
1
47Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
14. Answer (1, 3)
15. Answer (1, 2, 3)
S is open S is closed
V 2V
+–
q CV =
V 2V
CV–CV +CV
+CV
–CV
Work done by V = CV2
Work done by 2V = 0
Heat dissipated
2
2 2
2
CVCV CV
2
2
CV
16. Answer (1, 2, 3)
Charge will flow only when the circuit is closed, i.e., all keys are closed.
Initial charges q1 = 100 C and q
2 = 80 C
If a charge q flows anticlockwise, new charges are 1
100 –q q and 2
40 –q q
KVL 100 – 40 –
01 2
q q q = 80 C
New charges are 20 C and – 40 C capacitor C2 does not lose energy finally
17. Answer (1, 2, 3)
By ohm’s law i
VR and if increase voltage across voltmeter, current divides and resistance of circuit gets
changed.
Voltmeter R1, Ammeter R
2
321
1RRR
RV
32
1
11
1
RRRr
rRRr
rRV
A
V
R3
r ))(( 3211
1
RRRrrR
rR
)(1 321
1
1
RRr
RR
R
⎟⎠
⎞⎜⎝
⎛
)( 321
321
1
RRr
RRRR
R
VV
48 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
18. Answer (2, 3)
For maximum current
5.22
m
n
nEI
2.5
nE
nE
m
mn
E
mn
mnE
5
90
5
2
For maximum current 0545
–2
m
m = 3 & n = 15
19. Answer (2, 3, 4)
If resistances of A and B is R, resistance of C is 2R.
Power consumption in C, R
VPC
2
2
Power consumption in B, R
V
R
VPB
4
)2/( 22
Power consumption in A, R
V
R
VPA
4
)2/( 22
20. Answer (1, 3)
∫ ∫ ⎟⎠
⎞⎜⎝
⎛
R L
A
dx
xL
LdR
0 0
0 )2ln(0
A
LR
i = current = )2ln(0L
VA
R
V
Current density, )2ln(0L
V
A
iJ
= constant
J = 6(sigma) E E = J
)(2ln)2ln(0
0
xL
V
L
V
xL
LE
⎟
⎠
⎞⎜⎝
⎛
at x = 0 ; L
VE
)2(ln
at x = L ; L
VE
)4(ln
21. Answer (1, 4)
Random motion of electron depends on temperature.
Average speed is non-zero for any moving object. In absence of current, net flow of charge is zero.
22. Answer (1, 4)
Points B and D are at same potential. Equivalent circuit diagram is as follows:
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
AB,DR/2
R/2
R/2O
R
R
C
Equivalent resistance between O and A,
15
7RR
eq
49Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
23. Answer (1, 3)
Given that after every one second, potential or charge falls by a factor of 0 × 8.
After 2 second, Q = 0.64Q0
U = (0.64)2U0 (∵ U Q2)
Loss by a factor 2 369
1– (0.64)625
Potential V = 0.64 × 100 V = 64 V
0 00.8
⎛ ⎞ ⎜ ⎟
⎝ ⎠
t
q Q Q e at t = 1 s
1
ln(5 / 4) second
24. Answer (1, 2, 3)
If K1 is closed, no current flows through this wire reading of ammeter does not change,
when both K1 and K
2 are closed, even then no current flows Ammeter reading remains same.
When all the keys are closed, the middle part of each wire is shorted Req.
decreases
reading of ammeter increases.
25. Answer (1, 3, 4)
A4
I = 0.3 A
1.5 V, r
1.50.3
4
r
r = 1 Cell is non-ideal (1) is correct
Now, with a 4 resistance, current through the cell
I 1.5 V, 1
A4
4
I
2
I
2
1.50.5 A
2 1
I
Current through ammeter is 0.25 A
Power dissipated in Ammeter = 0.252 × 4 = 0.25 W
in cell = 0.52 × 1 = 0.25 W
50 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
in 4 resistance = 0.252 × 4 = 0.25 W
One third power is dissipated in cell (3) is correct
Reading of voltmeter 1.5
4 1.2 V4
r
(4) is correct
26. Answer (2, 3, 4)
Accuracy is proportional to length and does not depend on material. Also, accuracy is independent of cross-
sectional area. Meter bridge works on Wheatstone bridge.
27. Answer (2, 3)
xxxxxx
0
V
B =
A
B
C
B = 0
q60º
B = 0
xxxxxxxxxxxxxxx
In regions (A) and (C) charged particle will go undeflected
In region (B)
Path will be helical
28. Answer (1, 4)
a = 1
2
2
12
m
T qB
mT
qB
b = 1
2
sin37º
3
sin53º 4
mv
R qB
mvR
qB
c = 1 1
2 2
cos37º 4
sin37º 3
P v T
P v T
3 4
1 14 3
abc
a bc
29. Answer (2, 3)
Magnetic field due to a moving charged particle is given by
3
0
4 r
rvqB
��
�
ir ˆ05.0�
; meter
v
�
should be such that rv
��
should point along positive y-axis.
51Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
30. Answer (2, 3)
If 21 andVV are parallel, then deviation produced by
21 and BB are opposite
2
22
1
11 sinsin
r
d
r
d ⇒
1V
2V
d1
d2
1 = –
2
sin1 = –sin
2
2
2
1
1
2
2
1
1
qB
mV
d
qB
mV
d
r
d
r
d ⇒
d1B
1 = d
2B
2
1
2
2
1
2
1
B
B
d
d
r
r
31. Answer (2, 3)
Force on the wire at the instant shown in figure,
iilBF ˆ;�
kBilriilBjr ˆˆˆ �
kBilr ˆ
external�
BilrkkBilr )ˆ()ˆ(Powerext
��
32. Answer (3, 4)
V
d
If charge particle is positive
For turning
d > x
d > 2
rr
qB
Vmd
2
d
x
30°
60°C
1
qB
Vm
qB
mV
2
VV 2
52 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
If it is negative
2
3
2
rrrx
For turning
d > x
2
3rd
d
x
rr/2
30°
qB
Vm
qB
mV 2
3
VV 3
2
32(a). Answer (1, 3, 4) IIT-JEE 2008
When radius r > l, particle will move to region III.
qB
mV > l V >
m
Blq
when V = m
Blq, the particle moves in the biggest semicircle possible in region II.
Time spent t = Bq
m in region II, provided particle returns to region I.
32(b). Answer (1, 3) JEE (Advanced)-2013
Mt
qB
Clearly θ = 30° = 6
C
F
x L=
ˆ2 j
�
v
ˆ2 3i
ˆ4ix = 03
100 50
6 36 10 10
M M MB
q Qq
B
�
must be in –z direction.
33. Answer (1, 4)
Take a cubical Gaussian surface and find flux through it
0∫
qsdE�
�
ixEE ˆ
0�
implies there is distribution of positive charge in the region which is possible
0∫ sdB�
�
(Gauss law for magnetism)
This is not valid for ixBB ˆ
0�
34. Answer (1, 2, 3)
qB
mKr
2
For He+2 and O+2 radius is same, neutron is neutral hence turning of neutron is not possible.
53Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
35. Answer (1, 4)
For equilibrium,
0 1 2 0 1 2 0 1 2,
2 2 2
i i i i i iF l Mg r
r Mg mg
⇒
�
Force on the moving charge is perpendicular either left or right.
36. Answer (1, 2, 3)
(Using Faraday's and Lenz's law)
37. Answer (1, 3, 4)
i = 10e–2t
–2
– 20
tdie
dt
At t = 0
i
A 5 V B2
At t
10
A 5 V B2 Ldi
dt
= 40 V
– 20 – 5 40 A B
V V
– –15 V
A BV V
At t 0, 0⎛ ⎞ ⎜ ⎟⎝ ⎠
dii
dt
VA – V
B = 5 V
U = 21
2Li = 0
38. Answer (1, 2, 3)
22
2
2
22
2
2
BbBaVV
BbVV
BaVV
AO
AD
DO
B C
A
b
a
OD
B
VOA
can never be zero.
2
)2( 2 dcaBVV CO and it is independent of ‘b’.
54 Electromagnetism Success Magnet-Solutions (Part-II)
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
39. Answer (1, 3, 4)
Lxir
dt
diLir
rA
A
i r
C
QEVV
EC
QVV
AC
AB
2
1
2
A
B
BE
C
C
C1
i
E
B
Lx
Q1 = CE
Q2 = 2CE
1
2
1
2 Q
Q
40. Answer (1, 2, 3)
Bulb 2 dies as soon as key is switched into 1, because there
LR1
B1 B
2
21
is no induced e.m.f. Initially current in L is 2R
E
, total heat
developed = energy stored in solenoid =
2
22
1
⎟⎟⎠
⎞⎜⎜⎝
⎛
R
EL
41. Answer (1, 3, 4)
Loop tries to maximise its area and time required is very small and final shape is circular and Fnet
is zero on loop,
center of mass is at rest.
Ai = ab
2
⎟⎠
⎞⎜⎝
⎛
ba
Af
R
Bab
ba
R
ABQ
⎥⎥⎦
⎤
⎢⎢⎣
⎡
–
)( 2
42. Answer (2, 4)
R
Ei
Edt
diL
max
i
tFor Circuit 2 Circuit 1
dt
di
L
E 33
13
12
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛
dt
di
dt
di
R
E2i
0i0
R
L
3
2
3 2
3
2
1
55Success Magnet-Solutions (Part-II) Electromagnetism
Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
43. Answer (1, 2, 3)
i3
i1
i2
+q
–qF
Bi3
i1
i2
–q
+q
BR
Bvl
at = 0t
V
44. Answer (1, 3)
22a
dt
dBae
22
2a
a
a
R
ei (anticlockwise)
Electric field lines are circular and PR is along the diameter of circular field lines. Hence electric field along
PR is zero. So, no current flows in PR.
45. Answer (2, 3, 4)
If resistor is present in a circuit, power will be always dissipated, capacitor stores electrical energy, inductor
stores magnetic energy.
46. Answer (2, 3, 4)
For purely resistive circuit, instantaneous power is P = V0I0sin2t this expression has never zero average value
for any interval of time.
47. Answer (2, 3, 4)
222
222
)(500
)(
CLR
CLR
VVV
VVVV
R L C
VR < 500
VL ~ V
C < 500
Hence only first option is not possible
48. Answer (1, 2, 3)
In steady state
25 Ai
4
2 100 V
Shorted
–3
2
14 10 625
2L
U
= 1250 × 10–3
= 125 J
Potential difference across L2 at (t )
0di
Ldt
� � �