Post on 10-May-2023
Q. No. 1 One metallic sphere A is given positive charge where as another identical metallic
sphere B of exactly same mass as of A is given equal amount of negative charge. Then
Option 1 Mass of A and mass of B still remain equal
Option 2 Mass of A increases
Option 3 Mass of B decreases
Option 4 Mass of B increases
Correct Answer 4
Explanation Body a acquire +ve charge by loss of electrons and body B acquire -ve charge by gain of
electrons. Electrons have mass as well as in body B a mass will increase.
Q. No. 2 A body can be negatively charged by
Option 1 Giving excess of electrons to it
Option 2 Removing some electrons from it
Option 3 Giving some protons to it.
Option 4 Removing same neutrons from it
Correct Answer 1
Explanation Conceptual
Q. No. 3 Three charges +Q, +Q and -Q are placed at the corners of an equilateral triangle. The
ratio of the force on a positive charge to the force on the negative charge will be equal
to :
Option 1 1
Option 2 2
Option 3 3
Option 4 1
3
Correct Answer 4
Explanation
F is resultant of F and F → → →→ → →→ → →→ → →
AC BC
F = F +F +2 F F cos →→→→
θθθθ2 22 AC BC AC BC
F = F +F +2 F cos 120→→→→
2 2 2 0
F =F→→→→
F = F = F→ → →→ → →→ → →→ → →
AC BC AB
F = F +F +2 F F cos 60→→→→
2 2 02 CB AB AB CB
= 3 F
F1
=3
F
→→→→
→→→→
1
2
Q. No. 4 Two equal charges are separated by a distance d. A third charge placed on
perpendicular bisector at x distance will experience maximum coulomb force when
Option 1 dx =
2
Option 2 dx =
2
Option 3 dx =
2 2
Option 4 dx =
2 3
Correct Answer 3
Explanation
(((( ))))1 qx
E = .4
a + xπ∈π∈π∈π∈
P 3/22 20
(((( ))))1 Qqx
F = .4
a + xπ∈π∈π∈π∈ 3/2
2 20
dFF = maximum, when = 0
dx
ax =
2
da =
2
dx =
2 2
Q. No. 5 Point charges +4q, -q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a
respectively, then
Option 1 Only -q is in stable equilibrium
Option 2 None of the charges are in equilibrium
Option 3 All the charges are in unstable equilibrium
Option 4 All the charges are in stable equilibrium
Correct Answer 3
Explanation
Net force on –q
FAO = FOB
1 4q 1 4q. = .
4 4a aπ∈ π∈π∈ π∈π∈ π∈π∈ π∈
2 2
2 20 0
F = F→ →→ →→ →→ →
OA OB
-q is in unstable
Equilibrium
Force on charges at O and B are also same
System is in equilibrium (unstable)∴∴∴∴
Q. No. 6 Two point charges +9c and +e are at 16 cm away from each other. Where should
another charge q be placed between them so that the system remains in equilibrium
Option 1 24 cm from +9c
Option 2 12 cm from +9e
Option 3 24 cm from +e
Option 4 12 cm from +e
Correct Answer 2
Explanation
Charge q will be in equilibrium
If force due to +qe and +e will same
(((( ))))
qe.q q.ek. = k.
x 16 - x2 2
3 1=
x 16 - x
2 2
3 1= 48 - 3x = 3 4x = 48 x = 12cm
x 16 - x⇒ ⇒ ⇒⇒ ⇒ ⇒⇒ ⇒ ⇒⇒ ⇒ ⇒
12cm from +qe
Q. No. 7 Three charges q, 2q and 4q are connected by two strings of equal lengths as shown in
figure. What is the ratio of tensions in the strings AB and BC ?
Option 1 1:2
Option 2 2:1
Option 3 1:3
Option 4 3:1
Correct Answer 3
Explanation
Net force on A
2q 4q 3qF = k. +k. = k.
a 4a a
2 2 2
A 2 2 2
(((( ))))F = T Tensionin string ABA AB
Net force on C
8q 4qF = k. +k.
a 4a
2 2
C 2 2
9qT = F = k.
a
2
BC C 2
T 1=
T 3
AB
BC
Q. No. 8 Two particles, each of mass 10 g and having charge of 1 C are in equilibrium on aµµµµ
Horizontal table at a distance of 50cm. The minimum coefficient of friction between
the particles and the table is
Option 1 0.18
Option 2 0.54
Option 3 0.36
Option 4 0.72
Correct Answer 3
Explanation As particles are in equilibrium, therefore electrostatic force balance frictional force.
mg = Fµµµµ e
qmg = k.
dµµµµ
2
2
9 10 1 10 1 10=
25 10 10 10 10
× × × × ×× × × × ×× × × × ×× × × × ×µµµµ
× × × ×× × × ×× × × ×× × × ×
9 -6 -6
-2 -3
= 0.36µµµµ
Q. No. 9 A thin insulator rod is placed between two unlike point charges q1 and -q2. How will
the forces acting on the charges change.
Option 1 Total force acting on q1 will increase
Option 2 Total force acting on charge q1 will decrease
Option 3 Total force acting on q1 will remain unaffected
Option 4 None of these
Correct Answer 2
Explanation FF = , when insulator rod placed b / w two unlike point charges, total force acting on
K′′′′
either charges will decrease.
Q. No. 10 Two identical small balls lying on a horizontal plane are connected by a weightless
spring. One ball is fixed at point O and other is free. The balls are charged identically as
a result of which spring length increases two folds. What is ratio of frequency of
harmonic vibrations of the system after and before being charged?
Option 1 1
2
Option 2 2
Option 3 1
3
Option 4 2
Correct Answer 4
Explanation Initially
k
m=
1
2ρρρρ
ππππ0
When charge provided to balls
(((( ))))
1 q 1 qk = . k .
4 4 4l2l⇒⇒⇒⇒ ====
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2 2
2 30 0
ℓℓℓℓ
F =k x∆∆∆∆R 1
(((( ))))(((( ))))
1 qk + x - . = k x
4 + x∆ ∆∆ ∆∆ ∆∆ ∆
π∈π∈π∈π∈ ∆∆∆∆
2
120
ℓℓℓℓ
ℓℓℓℓ
(((( ))))(((( ))))
k.4k + x - = k x
2 + x∆ ∆∆ ∆∆ ∆∆ ∆
∆∆∆∆
3
12
ℓℓℓℓℓℓℓℓ
ℓℓℓℓ
4 xk + x - 1+ =k x
x4
∆∆∆∆ ∆ ∆∆ ∆∆ ∆∆ ∆
-23
12
ℓℓℓℓℓℓℓℓ
ℓℓℓℓ
xk + x - 1- =k x ∆∆∆∆
∆ ∆∆ ∆∆ ∆∆ ∆
1ℓ ℓℓ ℓℓ ℓℓ ℓℓℓℓℓ
K1= 2k
k= = 2
k
ρρρρ
ρρρρ1
0
Q. No. 11 Two point charges placed at a distance of 20cm in air repel each other with a certain
force. When a dielectric constant K is introduced between these point charges, force
of interaction becomes half of it’s previous value. Then K is approximately
Option 1 2
Option 2 4
Option 3 2
Option 4 1
Correct Answer 2
Explanation
(((( ))))
q qkF = .
4 20π∈π∈π∈π∈1 2
20
q qk FF = . =
4 2d- t + t k
′′′′π∈π∈π∈π∈
1 2
20
(((( ))))
q q1.
4 d- t + t kF=
q q1F.
4 20
π∈π∈π∈π∈ ′′′′
π∈π∈π∈π∈
1 22
0
1 22
0
20 1=
220 - 8 + 8 k
2
20 1=
12 + 8 k 2
12+ 8 k = 20 2
8 k = 20 2 -12
20 2 -12k =
8
k = 2 k = 4⇒⇒⇒⇒
Q. No. 12 Two identical blocks are kept on a frictionless horizontal table connected by a spring of
stiffness K and of original length . A total charge Q is distributed on the block such thatℓℓℓℓ
maximum elongation of the spring at equilibrium is equal to x. Value of Q is
Option 1 (((( ))))2 4 k + xπεπεπεπε0ℓ ℓℓ ℓℓ ℓℓ ℓ
Option 2 (((( )))) (((( ))))2 + x 4 k + xπεπεπεπε0ℓ ℓℓ ℓℓ ℓℓ ℓ
Option 3 (((( )))) (((( ))))2 + x 4 kπεπεπεπε0ℓ ℓℓ ℓℓ ℓℓ ℓ
Option 4 (((( )))) (((( ))))+ x 4 k xπεπεπεπε0ℓℓℓℓ
Correct Answer 2
Explanation
For maximum elongation
Charge distributed to both balls
q = =q2
θθθθ1 2
Restoring force = Electrostatic force
(((( ))))
1 1kx .
4 4 + x=
θθθθ××××
π∈π∈π∈π∈
2
20 ℓℓℓℓ
(((( ))))x= 16 k xθ π∈ +θ π∈ +θ π∈ +θ π∈ +2
0 ℓℓℓℓ
(((( ))))2 x 4 kx=θ + π∈θ + π∈θ + π∈θ + π∈0ℓℓℓℓ
Q. No. 13 Two point charges are kept separated by 4cm of air and 6cm of a dielectric relative
permittivity 4. The equivalent dielectric separation between them so that their
Coulombian interaction remains same is
Option 1 10cm
Option 2 8cm
Option 3 5cm
Option 4 16cm
Correct Answer 2
Explanation
Equivalent ⇒⇒⇒⇒
To equivalent dielectric separation
Fair = Fdielectric
q q q q
k. kk
= .r .x
1 2 1 2
2 2
r = k x
r = 4cm, k = 4
x = 2cm
d = 6cm + 2cm = 8cm
Q. No. 14 An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are
placed along X-axis at x = 1cm, x = 2cm, x = 4cm, x = 8cm …… and so on. In the setup if
the consecutive charges have opposite sign, then the electric field in Newton/Coulomb
1
at x = 0 is = 9 10 N-m / c4
××××
πεπεπεπε
9 2 2
0
Option 1 12 10×××× 4
Option 2 24 10×××× 4
Option 3 36 10×××× 4
Option 4 48 10×××× 4
Correct Answer 3
Explanation
(((( )))) (((( )))) (((( ))))1 5 10 5 10 5 10
E = - + - - - - - - -4
1 10 2 10 4 10
× × ×× × ×× × ×× × ×
π∈π∈π∈π∈ × × ×× × ×× × ×× × ×
-9 -9 -9
2 2 2-2 -2 -20
(((( )))) (((( )))) (((( ))))
1 5 10 1 1 1E = 1- + - - - - - - - -
4 10 2 4 8
×××× ××××
π∈π∈π∈π∈
-9
-4 2 2 20
(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))
1 1 1 1 1E = 195 10 1+ - - - - - - - - - + + + - - - - - - -
4 16 2 8 32
××××
4
2 2 2 2 2
(((( )))) (((( ))))
1 45 10 1 1E = 45 10 - 1+ + - - - - -
1 4 4 161-16
×××× × +× +× +× +
44
2 2
E = 48 10 -12 10 = 36 10 N / C× × ×× × ×× × ×× × ×4 4 4
Q. No. 15 Two identical point charges are placed at a separation of d. P is a point on the line
joining the charges, at a distance x from any one charge. The field at P is E, E is plotted
against x for values of x from close to zero to slightly less than d. Which of the
following represents the resulting curve
Option 1
Option 2
Option 3
Option 4
Correct Answer 4
Explanation
When point P is near charge at pt. A
Electric field is infinite and towards centre of line joining, as pt. P moves towards mid
point magnitude of electric field decreases and at mid point E. field will be zero. When
point ‘P’ is right gide of mid point direction of electric field reverses.
Q. No. 16 voltElectric field intensity at a point at a distance 60cm from charge is 2 then
metre
charge will be
Option 1 8 10 C×××× -11
Option 2 8 10 C×××× 11
Option 3 4 10 C×××× 11
Option 4 4 10 C×××× -11
Correct Answer 1
Explanation = = 60 cm =E 2 Vm , 0 md 60 1××××-1 -2
E =q
k.d2
2 3600 10= = = 800 10 = 8 10 C
9 10
E.dq
k
× ×× ×× ×× ×× ×× ×× ×× ×
××××
-4-1
23 -11
9
Q. No. 17 Electric field strengths due to a point charge of 5 C at a distance of 80cm from theµµµµ
charge is
Option 1 8 10 N / C×××× 4
Option 2 7 10 N / C×××× 4
Option 3 5 10 N / C×××× 4
Option 4 4 10 N / C×××× 4
Correct Answer 2
Explanation E =
qk.
d2
(((( ))))9 10 5 10 9 5 10
6400 1080 10
= =× × × × ×× × × × ×× × × × ×× × × × ×
××××××××
9 -6 3
2 -4-ℓℓℓℓ
0.007 10 7 /= = 10 N C× ×× ×× ×× ×7 4
Q. No. 18 A positive point charge 50 C is located in the plane xy at a point with radius vectorµµµµ
r = 2 i + 3 j . Evaluate electric field vector E at a point whose radius vector is r = 8 i - 5 j→ ∧ ∧ → → ∧ ∧→ ∧ ∧ → → ∧ ∧→ ∧ ∧ → → ∧ ∧→ ∧ ∧ → → ∧ ∧
0
where r0 and r are expressed in meters.
Option 1 1.8 i -2.6 j kNC
∧ ∧∧ ∧∧ ∧∧ ∧
-1
Option 2 1.4 i -2.6 j kNC
∧ ∧∧ ∧∧ ∧∧ ∧
-1
Option 3 2.7 i - 3.6 j kNC
∧ ∧∧ ∧∧ ∧∧ ∧
-1
Option 4 2.8 i -1.5 j kNC
∧ ∧∧ ∧∧ ∧∧ ∧
-1
Correct Answer 3
Explanation 1 qE = . r - r
4r - r
→ → →→ → →→ → →→ → →
→ →→ →→ →→ →
π∈π∈π∈π∈ 0
30
0
r - r = 8 i - 5 j - 2 i + 3 j = 6 i - 8 j→ → ∧ ∧ ∧ ∧ ∧ ∧→ → ∧ ∧ ∧ ∧ ∧ ∧→ → ∧ ∧ ∧ ∧ ∧ ∧→ → ∧ ∧ ∧ ∧ ∧ ∧
0
r - r = 6 + 8 = 10m→ →→ →→ →→ →
2 20
(((( ))))
9 10 50 10E = 6 i - 8 j
10
→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧ × × ×× × ×× × ×× × ×××××
9 6
3
E = 2.7 i -3.6 j kNC→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
-1
Q. No. 19 A cube of side b has a charge q at each of its vertices. The electric field due this charge
distribution at the center of this cube will be
Option 1 q
b2
Option 2 q
2b2
Option 3 32q
b2
Option 4 Zero
Correct Answer 4
Explanation In a regular body having equal charges at all vertices net electric field at centre is Zero.
Q. No. 20 A regular polygon has n sides each of length l. Each corner of the polygon is at a
distance r from the centre. Identical charges each equal to q are placed at (n -1)
corners of the polygon. What is the electric field at the centre of the polygon.
Option 1 n q
4 rπεπεπεπε 20
Option 2 n q
4πεπεπεπε 20 ℓℓℓℓ
Option 3 1 q
4 rπεπεπεπε 20
Option 4 1 q
4πεπεπεπε 20 ℓℓℓℓ
Correct Answer 3
Explanation 1 qE = .
4 rπ∈π∈π∈π∈ 20
Q. No. 21 Two very long line charges of uniform charge density and - are placed along sameλ λλ λλ λλ λ
line with the separation between the nearest ends being 2a as shown in figure. The
electric field intensity at the point O is
Option 1
2 a
λλλλ
πεπεπεπε0
Option 2
4 a
λλλλ
πεπεπεπε0
Option 3
3 a
λλλλ
πεπεπεπε0
Option 4
8 a
λλλλ
πεπεπεπε0
Correct Answer 1
Explanation
E = E = ; Net - electric field at 04 a
→ →→ →→ →→ → λλλλ
π ∈π ∈π ∈π ∈1 2
0
= 2 =4 a 2 a
λ λλ λλ λλ λ××××
π ∈ π ∈π ∈ π ∈π ∈ π ∈π ∈ π ∈0 0
Q. No. 22 Two point charges (+Q) and (-2Q) are fixed on the X-axis at positions a and 2a from
origin respectively. At what positions on the axis, the resultant electric field is zero
Option 1 Only x = 2 a
Option 2 Only x = - 2 a
Option 3 Both x = 2 a±±±±
Option 4 3 ax = only
2
Correct Answer 2
Explanation
Let at point P near change Q at a distance x net electric field is zero.
(((( ))))
Q 2Qk. = k.
x a+ x2 2
1 2=
x a+ x
22
a+ x = 2 x
(((( ))))a = x 2 -1
ax =
2 -1
(((( ))))x = 2 + 1 a
Position of pt. P from origin.
(((( ))))= a - 2 + 1 a
= - 2 a
Q. No. 23 In the figure distance of the point from A, where the electric field is zero is
Option 1 20cm
Option 2 10cm
Option 3 33cm
Option 4 None of these
Correct Answer 3
Explanation
Let at point P from A, net Electric field is zero
(((( ))))
10 10 20 10k. = k.
x 80 - x
× ×× ×× ×× ×-6 -6
2 2
1 2=
x 80 - x
22
80 - x = 2 x
(((( ))))80 = 2 + 1 x x = 33.14 cm = 33cm⇒⇒⇒⇒
Q. No. 24 In the diagram shown electric field intensity will be zero at position
Option 1 Between -q and +2q charges
Option 2 Towards +2q on the line drawn
Option 3 Away from the line towards 2q
Option 4 Away for the line towards -q
Correct Answer 4
Explanation In case of unlike charges, point where net electric field is zero lies outside the line
joining b/w the changes and near the charge having smaller magnitude.
Q. No. 25 A dust particle of radius 5 10 m lies in an electric field of 6.28 10 V /m. The× ×× ×× ×× ×-7 5
surrounding medium is air whose coefficient of viscosity is 1.6 10 N- s /m . If the×××× -5 2
particle moves with a horizontal uniform velocity of 0.02 m/s, the number of electrons
on it is
Option 1 10
Option 2 20
Option 3 30
Option 4 40
Correct Answer 3
Explanation Particle moves with uniform velocity in an electric field when force due to electric field
balance with drag force due to air.
F = 6 rπη νπη νπη νπη νe
qE = 6 rπη νπη νπη νπη ν
n = 6 rπη νπη νπη νπη νe (((( ))))q = ne
n6 r
=e
πη νπη νπη νπη ν
6 3.14 1.6 10 5 10 0.02=
1.6 10n
× × × × × ×× × × × × ×× × × × × ×× × × × × ×
××××
-5 7
-19
n = 30
Q. No. 26 The displacement of a charge Q in the electric field E = e i+ e j+ e k is r = ai+bj. The1 2 3
work done is
Option 1 (((( ))))Q ae +be1 2
Option 2 (((( )))) (((( ))))Q ae + be
2 2
1 2
Option 3 (((( ))))Q e + e a +b2 21 2
Option 4 (((( ))))Q e + e a+b2 21 2
Correct Answer 1
Explanation E = e i + e j+ e k→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
1 2 3
d = a i +b j→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
VE =
d
V = E . d = e a+ e b→ →→ →→ →→ →
1 2
(((( ))))W = Q.V = Q ae +be1 2
Q. No. 27 A charged particle of mass 0.003 mg is held stationary in space by placing it in a
downward direction of electric field of 6 10 N/ C. Then the magnitude of the charge is×××× 4
Option 1 5 10 C×××× -4
Option 2 5 10 C×××× -10
Option 3 -18 10 C×××× -6
Option 4 -5 10 C×××× -9
Correct Answer 2
Explanation Charge particle is stationary if
mg + qE = 0
qE = -mg
-mg 0.003 10 9.8q = =
E 6 10
− × ×− × ×− × ×− × ×
××××
-3
4
q = -4.9 10 C×××× -10
Magnitude of q= 4.9 10 C in 5 10 C× ×× ×× ×× ×-10 -10
Q. No. 28 Cathode rays travelling from east to west enter into region of electric field directed
towards north to south in the plane of paper. The deflection of cathode rays is towards
Option 1 East
Option 2 South
Option 3 West
Option 4 North
Correct Answer 4
Explanation Cathode rays deflect
Towards North
Q. No. 29 An electron and a proton are kept in a uniform electric field. The ratio of their
acceleration will be
Option 1 Unity
Option 2 Zero
Option 3 m
m
p
e
Option 4 m
m
e
p
Correct Answer 3
Explanation a =
qE
m
Both have same charge
But mass of proton is greater than electron
mp > me
a1
m∝∝∝∝
a=
a
m
m
e
p
p
e
Q. No. 30 A drop of 10-6
kg water carries 10-6
C charge. What electric field should be applied to
balance its weight (assume g = 10 m/s2)
Option 1 10V
m upward
Option 2 10V
m downward
Option 3 0.1V
m downward
Option 4 0.1V
m upward
Correct Answer 1
Explanation qE = mg
(((( ))))mg 10 10
E = = =10 V /m upwardq 10
××××-6
-6
Q. No. 31 The acceleration of an electron in an electric field of magnitude 50V/CM, if e/m value
of the electron is 1.76 10 C / kg,is×××× 11
Option 1 8.8 10 m / sec×××× 14 2
Option 2 6.2 10 m / sec×××× 13 2
Option 3 5.4 10 m / sec×××× 2 2
Option 4 Zero
Correct Answer 1
Explanation 50E = 50 V / cm = = 50 10 V / m
10×××× 2
-2
e=1.76 10 C /kg
m×××× 11
eE = ma
eEa = =1.76 10 5 10
m× × ×× × ×× × ×× × ×11 3
= 88 10 = 8.8 10 m / s× ×× ×× ×× ×13 14 2
Q. No. 32 An electron enters an electric field with its velocity in the direction of the electric lines
of force. Then
Option 1 The path of the electron will be a circle
Option 2 The path of the electron will be a parabola
Option 3 The velocity of the electron will decrease
Option 4 The velocity of the electron will increase
Correct Answer 3
Explanation Direction of electric field is +ve to -ve electron will experience force of repulsion in the
direction of electric field. Hence velocity of the electron will decrease.
Q. No. 33 A particle of mass m and charge q is placed at rest in a uniform electric field E and then
released. The kinetic energy attained by the particle after moving a distance y is
Option 1 qEy2
Option 2 qE2y
Option 3 qEy
Option 4 q2Ey
Correct Answer 3
Explanation u = 0
qEa =
m
2qEV -u = 2as V = y
m⇒⇒⇒⇒2 2 2
1K.E = mv qEy K.E = qEy
2==== ⇒⇒⇒⇒2
Q. No. 34 If an electron has an initial velocity in a direction different from that of an uniform
electric field, the path of the electron is
Option 1 A straight line
Option 2 A circle
Option 3 An ellipse
Option 4 A parabola
Correct Answer 4
Explanation
x x
= t =t
νννν ⇒⇒⇒⇒νννν
eEa =
m
1S = at
2
2
1 eE xy = .
2 m νννν
2
2
y x∝∝∝∝ 2
Path of electron will be Parabola.
Q. No. 35 A pendulum bob of mass 80mg and carrying a charger of 2 10 C is at rest in a×××× -8
horizontal uniform electric field of 20,000 V/m. The tension in the thread of the
pendulum is
Option 1 2.2 10 N×××× -4
Option 2 4.4 10 N×××× -4
Option 3 8.8 10 N×××× -4
Option 4 17.6 10 N×××× -4
Correct Answer 3
Explanation
m= 80 10 kg×××× -6
g = 9.8 m/s2
E = 20,000 V/m
T = Resultant of mg and qE when pendulum bob is at rest
(((( )))) (((( ))))T = mg + qE2 2
(((( )))) (((( ))))T = 80 10 10 + 4 10× × ×× × ×× × ×× × ×2 2
-6 -4
= 64 10 +16 10× ×× ×× ×× ×-8 -8
= 80 10 = 4 5 10× ×× ×× ×× ×-14 -4
= 8.8 10 N×××× -14
Q. No. 36 An electron falls through a distance of 8cm in a uniform electric field of 105 N/C. The
time taken by the electron in field will be
Option 1 3 10 s×××× -6
Option 2 3 10 s×××× -7
Option 3 3 10 s×××× -8
Option 4 3 10 s×××× -9
Correct Answer 4
Explanation E = 105 N/C
S = 8cm = 8 10 m×××× -2
u = 0
eEa =
m
1S =ut + at
2
2
1 eES = t
2 m
2
1 1.6 10 108 10 = t
2 9.1 10
× ×× ×× ×× ×× ×× ×× ×× ×
××××
-19 5-2 2
-31
8 9.1 10t =
8 10
× ×× ×× ×× ×
××××
-332
-15
t = 9 10××××2 -18
t = 3 10 sec×××× -9
Q. No. 37 A particle of specific charge (q/m) enters into uniform electric field E along the centre
qEline, with velocity v = 2q . After how much time it will collide with one of the plates
md
(figure)
Option 1 Not possible
Option 2 d
2V
Option 3 md
qE
Option 4 2md
qE
Correct Answer 3
Explanation qEv 2q
md====
dS = ,S =
2x yℓℓℓℓ
qEa =
m
uy = 0
V - V =2aS2 2y y y
V = 2aSy y
qE dV = 2
m 2× ×× ×× ×× ×y
qEV = d
my
Vy = uy + at
Vy = at
qEV m
t = =qEa
m
y
mdt =
qE
Q. No. 38 An electron moving with the speed 5 10 m / s is shooted parallel to the electric×××× 6
field of intensity 1 10 N/ C. Field is responsible for the retardation of motion of×××× 3
electron. Now evaluate the distance travelled by the electron before coming to rest
(((( ))))for an instant mass of e = 9 10 kg. charge = 1.6 10 C× ×× ×× ×× ×-31 -19
Option 1 7m
Option 2 0.7mm
Option 3 7cm
Option 4 0.7cm
Correct Answer 3
Explanation u = 5 10 m / s×××× 6
E = 1 10 N / C×××× 3
V = 0
eEa =
m
V2- u
2 = -2as
u2 = 2as
u u m 25 10 9.1 10S = = =
2a 2eE 2 1.6 10 1 10
× × ×× × ×× × ×× × ×
× × × ×× × × ×× × × ×× × × ×
2 2 12 -31
-19 3
S = 71.09 10×××× -3
= 7.109 10×××× -2
S = 7cm
Q. No. 39 A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a
string of length L. Time period of pendulum is T0. When parallel plates are charged, the
Ttime period changes to T. The ratio is equal to
T0
Option 1
qEg +
m
g
1
2
Option 2
g
qEg +
m
3
2
Option 3
g
qEg +
m
1
2
Option 4 None of these
Correct Answer 3
Explanation
qE
a =m
g = g + a′′′′
qEg = g +
m′′′′
LT = 2
gππππ0
T = 2g
ππππ′′′′
ℓℓℓℓ
T = 2qE
g +m
ππππ
ℓℓℓℓ
T g=
qETg +
m0
T g=
qETg +
m
1
2
0
Q. No. 40 A wire is bent in the form of a regular hexagon of side a and a total charge Q is
distributed uniformly over it. One side of the hexagon is removed. The electric field
due to the remaining sides at the centre of the hexagon is
Option 1 Q
12 3 aπεπεπεπε 20
Option 2 Q
16 3 aπεπεπεπε 20
Option 3 Q
8 2 aπεπεπεπε 20
Option 4 Q
8 2 aεεεε 20
Correct Answer 1
Explanation
d = dx6
θθθθθθθθ
dE = dEcosθθθθR
(((( )))) (((( ))))
d OP y d= =
AP4 AP6 4 a y + x
θ θ θθ θ θθ θ θθ θ θ××××
π∈π∈π∈π∈× π∈× π∈× π∈× π∈
2 3
0 2 2 20
E = dEcos θθθθ∫∫∫∫
( )
/
/
y dx=
6 4 ay + x
θθθθ
× ∈× ∈× ∈× ∈
a 2
3/22 20
-a 2
π ∫
x = y tan ,dx = y sec dθ θ θθ θ θθ θ θθ θ θ2
( )dx y sec d
= =y secy + x
θ θθ θθ θθ θ
θθθθ
2
3/2 3 22 2∫ ∫
1 1= cos d = sin
y yθ θ θθ θ θθ θ θθ θ θ
2 2∫
E =6 4 y a + 4y
θθθθ
× π∈× π∈× π∈× π∈ 2 20
E =3
6 4 a2
θθθθ
× π∈× π∈× π∈× π∈ 20
E =12 3 a
θθθθ
π ∈π ∈π ∈π ∈ 20
Q. No. 41 In the figure shown, if the linear charge density is , then the net electric field at Oλλλλ
will be
Option 1 Zero
Option 2 k
R
λλλλ
Option 3 2k
R
λλλλ
Option 4 2k
R
λλλλ
Correct Answer 1
Explanation
Segment 1
E = - i4 R
→ ∧→ ∧→ ∧→ ∧ λλλλ
π∈π∈π∈π∈ x
0
E = - j4 R
→ ∧→ ∧→ ∧→ ∧ λλλλ
π∈π∈π∈π∈ y
0
E = E + E = - i - j ... (i)4 R
→ → → ∧ ∧→ → → ∧ ∧→ → → ∧ ∧→ → → ∧ ∧ λλλλ
π∈π∈π∈π∈ 1 x y
0
Segment 2
E = j- i ...(ii)4 R
→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧ λλλλ
π∈π∈π∈π∈ 2
0
Segment 3
E = i2 R
→ ∧→ ∧→ ∧→ ∧λλλλ
π∈π∈π∈π∈3
0
E = E +E +E→ → → →→ → → →→ → → →→ → → →
1 2 3
= - i - y + j- i + i = 04 R 4 R 2 R
∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧ λ λ λλ λ λλ λ λλ λ λ
π∈ π∈ π∈π∈ π∈ π∈π∈ π∈ π∈π∈ π∈ π∈ 0 0 0
Q. No. 42 10 C charge is uniformly distributed over a thin ring radius 1m. A particle (mass = 0.9 gram,µµµµ
charge -1 C) is placed on the axis of ring. It is displaced towards centre of ring, thenµµµµ
time period of oscillations of particle
Option 1 0.6sec
Option 2 0.2sec
Option 3 0.3sec
Option 4 0.4sec
Correct Answer 1
Explanation
Q = 10 10 C×××× -6
q=1 10 C×××× -6
m= 0.9 10 kg×××× -3
R = 1m
4 mRT = 2
π∈π∈π∈π∈ππππ
30
(((( ))))0.9 10 1T = 2 = 2 10
9 10 10 10 10
× ×× ×× ×× ×π ππ ππ ππ π
× × × ×× × × ×× × × ×× × × ×
3-3-2
9 -6 -6
2T = = 0.6 sec
10
ππππ
Q. No. 43 A thin conducting ring of radius r has an electric charge +Q if a point charge q is placed
at the centre of the ring, then tension of the wire of ring will be
Option 1 Qq
8 rπ∈π∈π∈π∈ 20
Option 2 Qq
4 rπ∈π∈π∈π∈ 20
Option 3 Qq
8 rπ ∈π ∈π ∈π ∈2 20
Option 4 Qq
4 rπ ∈π ∈π ∈π ∈2 20
Correct Answer 3
Explanation
d k.d .q
2T sin =2 r
θ θθ θθ θθ θ2
k. qdlT d =
2 r
θθθθθθθθ
ππππ 3
dl k. qdlT =
r 2 r
θθθθ
ππππ 3
k qT =
2 r
θθθθ
ππππ 2
qT =
8 r
θθθθ
π ∈π ∈π ∈π ∈2 20
Q. No. 44 Four equal charges each q are held fixed at position (0, R), (0, -R), (R, R) and (R, -R)
respectively of a(x, y) co-ordinate system. The work done moving a charge Q from
point A(R, 0) to origin (0, 0) is
Option 1 Zero
Option 2 qQ 2 -1
4 2 Rπεπεπεπε0
Option 3 2qQ 2
Rπεπεπεπε0
Option 4 qQ 2 +1
4 2Rπεπεπεπε0
Correct Answer 1
Explanation
Potential at point A
k.2q k.2qV = +
R R 2A
(((( ))))k.qV = 2+ 2
RA
Potential at point B
(((( ))))k.qV = 2+ 2
RB
A change moves from B to Aθθθθ
[[[[ ]]]]V - V
W = = 0 as V = Vθθθθ
B ABA A B
Q. No. 45 Consider a rhombus ABCD, with angle at B is 1200. A charge +Q placed at corner A
produces field E and potential V at corner D. If we now added charges -2Q and +Q at
corners B and C respectively, the magnitude of field and potential at D will become,
respectively
Option 1 E and 0
Option 2 0 and V
Option 3 VE 2 and
2
Option 4 E V and
2 2
Correct Answer 1
Explanation
V = V + V + VD A B C
k.Q k.2Q k.Q
= - +a a a
VD = 0
E = E =E→ →→ →→ →→ →
A C
-1E = E +E + 2E
2××××2 2 2
AC
EAC = E
EB = 2E
Net Electric field at D = EB - EAC = E
Q. No. 46 Two charges of 4 C each are placed at the corners A and B of an equilateral triangle ofµµµµ
1 N-mside length 0.2 m in air. The electric potential at C is = 9 10
4 C
×××× πεπεπεπε
29
20
Option 1 9 10 V×××× 4
Option 2 18 10 V×××× 4
Option 3 36 10 V×××× 4
Option 4 72 10 V×××× 4
Correct Answer 3
Explanation
VC = VCA + VCB
4 10 4 10
= k. +k.2 10 2 10
× ×× ×× ×× ×
× ×× ×× ×× ×
-6 -6
-1 -1
= 2 9 10 2 10× × × ×× × × ×× × × ×× × × ×9 -5
= 36 10 V×××× 4
Q. No. 47 Electric charges of +10 C + 5 C,-3 C and + 8 C are placed at the corners of a square ofµ µ µ µµ µ µ µµ µ µ µµ µ µ µ
side 2m. The potential at the centre of the square is
Option 1 1.8V
Option 2 1.8 10 V×××× 6
Option 3 1.8 10 V×××× 5
Option 4 1.8 10 V×××× 4
Correct Answer 3
Explanation
OA = OB = OC = OD = 1m
Vcentre = VOA + VOB + VOC + VOD
0
1 10 10 5 10 3 10 8 10V = + - +
4 1 1 1 1
× × × ×× × × ×× × × ×× × × ×
π∈π∈π∈π∈
-6 -6 -6 -6
centre
[[[[ ]]]]= 9 10 10 15+ 5× ×× ×× ×× ×9 -6
= 180 10 = 1.8 10 V× ×× ×× ×× ×3 5
Q. No. 48 10Equal charg es + 10 are placed at each of the four corners of a square of side of 8cm.
3×××× -9
The potential at the intersection of the diagonals is
Option 1 150 2 volt
Option 2 1500 2 volt
Option 3 900 2 volt
Option 4 900 volt
Option 5
Option 6
Correct Answer 2
Explanation
10
q = q = q = q = q = 10 C3
×××× -9A B C D
8d = OA = OB = OC = OD = 10 m
2×××× -2
1 qV = 4 .
4 d××××
π∈π∈π∈π∈Catte
0
104 9 10 10
3=8
102
× × × ×× × × ×× × × ×× × × ×
××××
9 -9
-2
= 1.5 2 10 = 1500 2 volt×××× 3
Q. No. 49 In a regular polygon of n sides, each corner is at a distance r from the centre, identical
charges are placed at (n - 1) corners. At the centre, the intensity is E and the potential
is V, the ratio V/E has magnitude.
Option 1 rn
Option 2 r(n - 1)
Option 3 (((( ))))n-1
r
Option 4 (((( ))))r n-1
n
Correct Answer 2
Explanation qE = k.
r2
(((( ))))
qk.
E r=qV
k. n-1r
2
(((( ))))q
V =k. n-1r
(((( ))))E 1
=V n-1 r
(((( ))))V
= n-1 rE
Q. No. 50 Two electric charges 12 C and - 6 C are placed 20 cm apart in air. There will be a pointµ µµ µµ µµ µ
P one the line joining these charges and outside the region between them, at which
the electric potential is zero. The distance of P from -6 C charge isµµµµ
Option 1 0.10 m
Option 2 0.15m
Option 3 0.20m
Option 4 0.25m
Correct Answer 3
Explanation
Potential due to change at A + Potential due to change at B = 0
k.12 10 k.6 10- = 0
20 + x x
× ×× ×× ×× ×-6 -6
20 + x = 2x x = 20cm⇒⇒⇒⇒
x = 0.2m
Q. No. 51 Two unlike charges of magnitude q are separated by a distance 2d. The potential at a
point midway between them is
Option 1 Zero
Option 2 1
4πεπεπεπε0
Option 3 1 q.
4 dπεπεπεπε0
Option 4 1 2q.
4 dπεπεπεπε 20
Correct Answer 1
Explanation
Potential at point P = Potential at Point P due to charge at A + Potential at point P due
to charge at B
q qV = -k. +k. = 0
a a
Q. No. 52 Three charges 2q, -q, -q are located at the vertices of an equilateral triangle. At the
centre of the triangle
Option 1 The field is zero but potential is non-zero
Option 2 The field is non-zero but potential zero
Option 3 Both field and potential are zero
Option 4 Both field and potential are non-zero.
Correct Answer 2
Explanation
V = V + V + Vcentre OA OB OC
2q q q
=k. -k. -k. = 0a a a
Vcentre = 0
qE = E = k.
aB C 2
q
E = E +E + 2E E cos120 = k.a
2 2 0BC B C B C 2
2q
E = k.a
A 2
E = E + Ecentre A BC
(((( ))))3q
E = k. +vea
centre 2
Q. No. 53 Two points are at distances a and b(a > b) from a long string of charge per unit length
. The potential difference between the points is proportional toλλλλ
Option 1 b
a
Option 2 b
a
2
2
Option 3 b
a
Option 4 bIn
a
Correct Answer 4
Explanation
1 dq 1 dx
dV = . = .4 r 4 x + y
2 20 0
λλλλ
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
1 2E = .
4 r0
λλλλ
π∈π∈π∈π∈
V - V = -E dr
b
b a r
a
∫∫∫∫
1 2= . dr
4 r∫b
0a
λλλλ
π∈π∈π∈π∈
bV - V = ln
2 ab a
0
λλλλ
π∈π∈π∈π∈
bV - V ln
ab a
∝∝∝∝
Q. No. 54 An arc of radius r carries charge. The linear density of charge is and the arc λλλλ
subtends and angle at the centre. What is electric potential at the centre3
ππππ
Option 1
4 0
λλλλ
εεεε
Option 2
8 0
λλλλ
εεεε
Option 3
12 0
λλλλ
εεεε
Option 4
16 0
λλλλ
εεεε
Correct Answer 3
Explanation
V = K -λ αλ αλ αλ α
=3
λλλλαααα
V =K3
ππππλλλλ
1V = .
4 30
λπλπλπλπ
π∈π∈π∈π∈
V =12 0
λλλλ
∈∈∈∈
Q. No. 55 A hemisphere of radius R is charged uniformly with surface density of charge . Whatσσσσ
will be the potential at centre ?
Option 1 R
2 0
σσσσ
∈∈∈∈
Option 2
4 0
σσσσ
∈∈∈∈
Option 3
2 0
σσσσ
∈∈∈∈
Option 4 4 R
3 0
σσσσ
∈∈∈∈
Correct Answer 1
Explanation
Potential at Centre
QV = k
R
1 2 RV = .
4 R
2
0
π σπ σπ σπ σ
π∈π∈π∈π∈
RV =
2 0
σσσσ
∈∈∈∈
Q. No. 56 A solid sphere of radius R is charged uniformly through out the volume. At what
1distance from its surface is the electric potential of the potential a the centre ?
4
Option 1 8R
3
Option 2 R
3
Option 3 5R
3
Option 4 2R
3
Correct Answer 3
Explanation
1 Q
V = .4 R
surface0π∈π∈π∈π∈
3V = V
4centre urfaces
3 QV = k.
2 Rcentre
1Let Potential at point P = V
4centre××××
Q 1 3 kQk. =
R 4 2 R× ×× ×× ×× ×
1 3 8R= r =
r 8R 3⇒⇒⇒⇒
Distance of point P from surface of sphere
5r -R = R
3
Q. No. 57 In a hollow spherical shell potential (V) changes with respect to distance (r) from
centre O, as shown in graph
Option 1
Option 2
Option 3
Option 4
Correct Answer 2
Explanation
In case of hollow sphere
Vcentre = Vsurface
and outside the surface
QV = k.
r
1i.e., V
r∝∝∝∝
Q. No. 58 n small drops of same size are charged to V volt each. If they coalesce to form a single
large drop, then its potential will be
Option 1 Vn
Option 2 V n-1
Option 3
Vn
1
3
Option 4
Vn
2
3
Correct Answer 4
Explanation Let initially all drops having radius r and charge q.
After they coalesce, form new drop having radius R and charge Q.
Q = nq
New volume = n volume of small drop.××××
4 4R = n r
3 3
3 3π × ππ × ππ × ππ × π
R = nr R = n . r
1
3 3 3⇒⇒⇒⇒
Q nqV = k. = k.
Rn . r
1
3
′′′′
qV = n . k.
r
2
3′′′′
V = n .V
2
3′′′′
Q. No. 59 The charge Q and -2Q are placed at some distance. The locus of points in the plane of
the charges where the potential is zero will be
Option 1 straight line
Option 2 circle
Option 3 a parabola
Option 4 An ellipse
Correct Answer 2
Explanation Conceptual
Q. No. 60 Two copper spheres of same radii one hollow and other solid are charged to the same
potential then
Option 1 both will hold same charge
Option 2 solid will hold more charge
Option 3 hollow will hold more charge
Option 4 hollow can not be charged
Correct Answer 1
Explanation QV = k.
r
If both spheres (hollow and solid) having same radii and potential. Charge on both will
also same.
Q. No. 61 R and S are two non-identical metal spheres, placed near each other. R is positively
charged while S is negatively charged with the same quantity of charge. Then
Option 1 The charge on each sphere will be uniformly distributed over its surface.
Option 2 The potential on the part of the surface R which faces S is less than the potential on
the other part.
Option 3 Each sphere has the same potential throughout its volume but this potential on one is
same to that on the other sphere.
Option 4 The electrostatic force on the bigger sphere is same as that on the smaller sphere.
Correct Answer 4
Explanation Both spheres are opposite charged so they exerts electrostatic force (equal) on each
other.
Q. No. 62 The electric potential V at any point P(x, y, z) (all in metres) in space is given V = 4x2
volts. The electric field (in V/m) at the point (1m, 0, 2m) is :
Option 1 -8 i
∧∧∧∧
Option 2 8 i
∧∧∧∧
Option 3 -16 i
∧∧∧∧
Option 4 8 5 i
∧∧∧∧
Correct Answer 1
Explanation V = 4x V, x = 1 i + 2k
2→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
-dVE =
dx
(((( ))))-dE = 4x = -8x
dx
2
(((( ))))E = -8 i N / C1,0,2
→ ∧→ ∧→ ∧→ ∧
Q. No. 63 Equipotential surfaces are shown in figure. Then the electric field strength will be
Option 1 100 Vm
-1 along X-axis
Option 2 100 Vm-1
along Y-axis
Option 3 200 Vm-1
at an angle 1200 with X-axis
Option 4 50 Vm-1
at an angle 1200 with X-axis
Correct Answer 3
Explanation d = 10 sin 300 = 5cm
(((( ))))20 -10V 1000E = = = = 200V /m
d 55 10-2
∆∆∆∆
××××
Angle 1200 with x-axis.
Q. No. 64 A unit charge is taken from one point to another over an equipotential surface. Work
done in this process will be
Option 1 Zero
Option 2 Positive
Option 3 Negative
Option 4 Optimum
Correct Answer 1
Explanation Equipotential surface, potential at every point is same. i.e., potential difference = 0
(((( ))))V = 0∆∆∆∆
(((( ))))W = V q∆∆∆∆
W = 0
Q. No. 65 At a certain distance from a point charge the electric field is 500 V/m and potential is
3000V. What is this distance
Option 1 6m
Option 2 12m
Option 3 36m
Option 4 144m
Correct Answer 1
Explanation E = 500 V/m
V = 3000 V
Qk.
V 3000r= = r r = = 6mQE 500
k.r
2
⇒⇒⇒⇒
Q. No. 66 Electric potential at any point is V = - 5x + 3y + 15 z, then the magnitude of the electric
field is
Option 1 3 2
Option 2 4 2
Option 3 5 2
Option 4 7
Correct Answer 4
Explanation (((( )))) (((( )))) (((( ))))dV
E = - = - -5x + 3y + 15 zdr x y z
∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂
= 5-3- 15
(((( )))) (((( )))) (((( ))))E = 5 + -3 + 5 = 722 2
→→→→
Q. No. 67 In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a
potential difference 2400 V between the plates. To keep a drop of half the radius
stationary the potential difference had to be made 600 V. What is the charge on the
second drop
Option 1 Q
4
Option 2 Q
2
Option 3 Q
Option 4 3Q
2
Correct Answer 2
Explanation V 4QE = mg Q = R g
d 3
3 ⇒⇒⇒⇒ π ρπ ρπ ρπ ρ
Q R VR R 600Q =
RV Q R V 2400
2
3
331 1 2
2 2 1
∝∝∝∝ ⇒⇒⇒⇒ × ×× ×× ×× ×
Q Q= 2 Q =
Q 2
12
2
⇒⇒⇒⇒
Q. No. 68 100The intensity of electric field in a region of space is represented by E = V / m.
x2
The Potential difference between the points x = 10cm and x = 20cm will be
Option 1 15V
Option 2 10V
Option 3 5V
Option 4 1V
Correct Answer 3
Explanation 100E = V / m
x2
dV = Edr
100 100dV = dx dx
x x⇒∫ ∫ ∫
20 20 20
2 2
10 10 10
1V = -100
x
20
10
1-2V = -100
20
V = 5V
Q. No. 69 Two points A and B lying on Y-axis at distances 12.3 cm and 12.5 cm from the origin.
The potentials at these points are 56 V and 54.8 V respectively, then the component
of force on a charge of 4 C placed at A along Y - axis will beµµµµ
Option 1 0.12N
Option 2 48 10 N-3××××
Option 3 24 10 N-4××××
Option 4 96 10 N-2××××
Correct Answer 3
Explanation
VA = 56V, VB = 54.8V
VA - VB = 50 - 54.8
= 1.2V
V 1.2E = =
d 0.2 10 - 2
∆∆∆∆
××××
E = 600 N/C
F = qE
= 4 10 600-6× ×× ×× ×× ×
F = 24 10 N-4××××
Q. No. 70 An electric field of 100 Vm-1
exists along x-axis. The potential difference between a
point A(-1m, 0) and B (+3m, 0) is
Option 1 200V
Option 2 -200V
Option 3 400V
Option 4 -400V
Correct Answer 3
Explanation E = 100 Vm-1
t = 4m
V = E.d
V = 400V
Q. No. 71 An electric field E = 50 i + 75 j N / C exists in a certain region of space. Presuming the
→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
potential at the origin to be zero, the potential at point P (1m, 2m) will be
Option 1 100V
Option 2 -100V
Option 3 200V
Option 4 -200V
Correct Answer 4
Explanation E = 50 i + 75 j N / C → ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
d = - i -2 j m → ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
V = E . d→ →→ →→ →→ →
= -50 – 150
= -200V
Q. No. 72 In a uniform electric field, the potential is 10V at the origin of coordinates, and 8 V at
each of the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1, 1) will
be :
Option 1 0
Option 2 4V
Option 3 8V
Option 4 10V
Correct Answer 2
Explanation
-dV
E =dx
E = 2 i + 2 j+ 2k→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
dr = dx i + dy j+ dz k→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
(((( ))))
(((( ))))
V = E .dr
1,1
0,0,0
→ →→ →→ →→ →
∆∆∆∆ ∫∫∫∫
(((( ))))
(((( ))))
= - 2dx + 2dy + 2dz
1,1,1
0,0,0
∫∫∫∫
= (2 + 2 + 2) = -6V
Vf - Vi = 10
Vf + 6 = 10
Vf = 4V
Q. No. 73 The variation of potential with distance R from a fixed point is as shown below. The
electric field at R = 5m is
Option 1 2.5 volt/m
Option 2 -2.5 volt/m
Option 3 2volt / m
5
Option 4 2- volt / m
5
Correct Answer 1
Explanation
VE = -
R
∆
∆at R = 5m
5m5m
5m
5 - 0= -
4 - 6
Eat R = 5m = 2.5 V/m
Q. No. 74 The figure gives the electric potential V as a function of distance through five regions
on x-axis. Which of the following is true for the electric field E in these regions
Option 1 E1 > E2 > E3 > E4 > E5
Option 2 E1 = E3 = E5 and E2 < E4
Option 3 E2 = E4 = E5 and E1 < E3
Option 4 E1 < E2 < E3 < E4 < E5
Correct Answer 2
Explanation In region 1, 3, 5 V = constant.→
Therefore E1 = E3 = E5 = 0
In region 2 Slope is + ve→
dVE = -
dr2
In region 4 Slope is - ve→
dVE = -
dr4
E4 > E2
Q. No. 75 KElectric potential in an electric field is given as V = , (K being constant), if position
r
vector r 2 i 3 j 6 k , then electric field will be→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
= + += + += + += + +
Option 1 K2 i 3 j 6 k
243
∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +
Option 2 K2 i 3 j 6 k
343
∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +
Option 3 K3 i 2 j 6 k
243
∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +
Option 4 K6 i 2 j 3k
343
∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +
Correct Answer 2
Explanation dVE = -
dr
kV =
r
kE = r
r
3
→ →→ →→ →→ →
→→→→
(((( )))) (((( )))) (((( ))))
kE = . 2 i + 3 j+ 6k
2 + 3 + 63
2 2 2
→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
kE = 2 i + 3 j+ 6 k
343
→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧→ ∧ ∧ ∧
Q. No. 76 An electric field of strength 50 Vm-1
exists along the negative direction of Y-axis.
If 1 C of positive charge is shifted from a point A(1m, -1m) to B(1m, 3m), the work done byµµµµ
agent is
Option 1 0
Option 2 -0.2 mJ
Option 3 +0.2 mJ
Option 4 +0.8 mJ
Correct Answer 3
Explanation W = qV
W = q(VB - VA)
=1 10 E.d-6× ×× ×× ×× ×
=1 10 50 4-6× × ×× × ×× × ×× × ×
= 200 10 = 2 10 = 0.2 10 J-6 -4 -3× × ×× × ×× × ×× × ×
Q. No. 77 The potential in an electric field has the form V = a(x2 + y
2 + z
2). The modulus of the
electric field at a point (x, y, z) is
Option 1 (((( ))))2a x + y + z3/2
2 2 2
Option 2 2a x + y + z
2 2 2
Option 3 a x + y + z
2 2 2
Option 4 2a
x + y + z2 2 2
Correct Answer 2
Explanation (((( ))))V = a x + y + z2 2 2
dV v vE = - i + j+ k = - 2ax i +2ay j+2azk
dx y z
→ ∧ ∧ ∧ ∧ ∧ ∧→ ∧ ∧ ∧ ∧ ∧ ∧→ ∧ ∧ ∧ ∧ ∧ ∧→ ∧ ∧ ∧ ∧ ∧ ∧ ∂ ∂∂ ∂∂ ∂∂ ∂ ∂ ∂∂ ∂∂ ∂∂ ∂
E =2a x + y +z→→→→
2 2 2
Q. No. 78 A conducting sphere of radius R is charged to a potential of V volt. Then the electric
field at a distance r(>R) from the centre of the sphere would be
Option 1 RV
r2
Option 2 V
r
Option 3 rV
R2
Option 4 R V
r
2
3
Correct Answer 1
Explanation Conducting sphere
If r > R
1 Q QV = . =k.
4 R Rπ∈π∈π∈π∈0
V.RR =
k
QE = k.
r2
VR VRE = k. E =
kr r⇒⇒⇒⇒
2 2
Q. No. 79 How should three charges q, 2q and 8q be arranged on a 9cm long line such that the
potential energy of the system is minimum?
Option 1 q at a distance of 3cm from 2q
Option 2 q at a distance of 5cm from 2q
Option 3 2q at a distance of 7cm from q
Option 4 2q at a distance of 9cm from q
Correct Answer 1
Explanation
2q 8q 16qU = k + +
x 9 - x q
2 2 2
T
If U minimum
dU= 0
dx
2q d 1 4 8k. + + = 0
dx x q- x 9
2
(((( )))) (((( ))))d d
x + 4. q- x + 0 = 0dx dx
-1 -1
(((( ))))1 4
- + = 0q- xx
2
1 2= x = 3cm
x q- x⇒⇒⇒⇒
q at distance 3cm form 2q
Q. No. 80 If 3 charges are placed at the vertices of equilateral triangle of charge ‘q’ each. What is
the net potential energy, if the side of equilateral is cm∆∆∆∆ ℓ ℓ ℓ ℓ
Option 1 1 q
4πεπεπεπε
2
0 ℓℓℓℓ
Option 2 1 2q
4πεπεπεπε
2
0 ℓℓℓℓ
Option 3 1 3q
4πεπεπεπε
2
0 ℓℓℓℓ
Option 4 1 4q
4πεπεπεπε
2
0 ℓℓℓℓ
Correct Answer 3
Explanation
q q qU= k. +k. +k.
2 2 2
ℓ ℓ ℓℓ ℓ ℓℓ ℓ ℓℓ ℓ ℓ
3qU= k.
2
ℓℓℓℓ
1 3qU= .
4π∈π∈π∈π∈
2
0 ℓℓℓℓ
Q. No. 81 If identical charges (-q) are placed at each corner of a cube of side b, then electric
potential energy due to charge (+q) which is placed at centre of the cube will be
Option 1 8 2 q
4 bπεπεπεπε
2
0
Option 2 -8 2 q
bπεπεπεπε
2
0
Option 3 -4 2 q
bπεπεπεπε
2
0
Option 4 -4 q
3 bπεπεπεπε
2
0
Correct Answer 4
Explanation There are 8 vertices having charge (-q) at each corner.
1 -q. + qU= 8 .
4 3b
2
××××π∈π∈π∈π∈0
-4qU=
3 bπ∈π∈π∈π∈
2
0
Q. No. 82 Three charges Q, +q and +q are placed at the vertices of an equilateral triangle of side l
as shown in the figure. If the net electric energy of the system is zero, then Q is equal
to
Option 1 q
-2
Option 2 -q
Option 3 +q
Option 4 Zero
Correct Answer 1
Explanation
2k.Qq k.qU= +
2
ℓ ℓℓ ℓℓ ℓℓ ℓ
U = 0
2kQq k.q= -
2
ℓ ℓℓ ℓℓ ℓℓ ℓ
qQ = -
2
Q. No. 83 A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of V
volt, its energy will be
Option 1 qV
Option 2 mqV
Option 3 qV
m
Option 4 q
mV
Correct Answer 1
Explanation Energy of particle = eV
Q. No. 84 Two equal charges q are placed at a distance of ‘2a’ and a third charge -2q is placed at
the midpoint. The potential energy of the system is
Option 1 q
8 aπεπεπεπε
2
0
Option 2 6q
8 aπεπεπεπε
2
0
Option 3 7q-
8 aπεπεπεπε
2
0
Option 4 9q
8 aπεπεπεπε
2
0
Correct Answer 3
Explanation
-k.2q k.2q k.qU= - +
a a 2a
2 2 2
-k.4q k.qU= +
a 2a
2 2
k.q 1 -7qU= -4 + =
a 2 8 a
π∈π∈π∈π∈
2 2
0
Q. No. 85 A proton and an -particle are situated at r distance apart. At very large distance apart∝∝∝∝
when released, the kinetic energy of proton will be
Option 1 2ke
r
2
Option 2 8 ke
5 r
2
Option 3 ke
r
2
Option 4 8ke
r
2
Correct Answer 2
Explanation Initially and proton are at r distance apart.∝∝∝∝
2eU =k. ,K.E = 0
r
2
i i
Uf = 0 (particles are for away)
Acc., to conservation of energy
K.Ef = Vi
2keK.E +K.E =
r∝∝∝∝
2
P
1 2keK.E + 4 m v =
2 rαααα××××
22
P p
(((( ))))V1 1 2ke
m v + 4m =2 2 16 r
××××
2 2p2
p p p
Acc., to conservation of Momentum
m v = m vα αα αα αα αp p
m v = 4m vα αα αα αα αp p
vv =
4αααα
p
1 1 2ke 1 8 ke 8 kem v 1+ = m v = K.E =
2 4 r 2 5 r 5 2
⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒
2 2 22 2
p p p p p
Q. No. 86 Two positively charged particles X and Y are initially far away from each other and at
rest. X begins to move towards Y with some initial velocity. The total momentum and
energy of the system are p and E.
Option 1 If Y is fixed, both p and E are conserved.
Option 2 If Y is fixed, E is conserved, but not p.
Option 3 If both are free to move, p is conserved but not E.
Option 4 If both are free, E is conserved, but not p.
Option 5
Option 6
Correct Answer 2
Explanation Conceptual
As Y is fixed and is moving, so p cannot be conserved. E is conserved.××××
Q. No. 87 Two identical particles of same mass each m are having same magnitude of charge Q.
One particle is initially at rest on a frictionless horizontal plane and the other particle is
projected directly towards the first particle from a very large distance with a velocity v.
The distance of closest approach of the particle will be
Option 1 1 4Q
4 mvπεπεπεπε
2
20
Option 2 1 2Q
4 mvπεπεπεπε
2
20
Option 3 1 Q
4 m vπεπεπεπε
2
2 20
Option 4 1 4Q
4 m vπεπεπεπε
2
2 20
Correct Answer 1
Explanation Acc. To conservation of Momentum
mv + 0 =2mv′′′′
vv =
2′′′′
Acc., to conservation of energy
1 mv k.Q2 =
2 4 r××××
2 2
4kQr =
mv
2
2
Q. No. 88 A bullet of mass 2g is having a charge of 2 C. Through what potential difference must it µµµµ
be accelerated, starting from rest, to acquire a speed of 10m/s ?
Option 1 50 kV
Option 2 5 V
Option 3 50 V
Option 4 5 kV
Option 5
Option 6
Correct Answer 1
Explanation F = qE
ma = qE
qE Va = E =
m
ℓℓℓℓ
qVa =
mℓℓℓℓ
v -u qV=
2 m
2 2
ℓ ℓℓ ℓℓ ℓℓ ℓ
mv -muV =
2q
2 2
2 10 100 1V = = 10
22 2 10
× ×× ×× ×× ×××××
× ×× ×× ×× ×
-35
-6
V = 50KV
Q. No. 89 Water is an excellent solvent because its molecules are
Option 1 Neutral
Option 2 Highly polar
Option 3 Non-polar
Option 4 Anodes
Correct Answer 2
Explanation Due to electronegativity difference, therefore is not dipole moment on water and can
attract other charged particles.
Q. No. 90 Electric potential at equatorial point of a small dipole with dipole moment p(At, r
distance from the dipole) is
Option 1 Zero
Option 2 p
4 rπεπεπεπε 20
Option 3 p
4 rπεπεπεπε 30
Option 4 2p
4 rπεπεπεπε 30
Correct Answer 1
Explanation
AP =PB = a + x2 2
(((( ))))-k.q
V Pot. at point P due to A =a + x
12 2
(((( ))))k.q
V Pot. at point P due to charge at B =a + x
22 2
Vp = V1 + V2 = 0
Q. No. 91 An electric dipole is fixed at the origin of coordinates. Its moment is directed in the
positive x-direction. A positive charge is moved from the point (r, 0) to the point (-r, 0)
by external agent. In this process, the work done by the agent is
Option 1 Positive and inversely proportional to r
Option 2 Positive and inversely proportional to r2.
Option 3 Negative and inversely proportional to r
Option 4 Negative and inversely proportional to r2.
Correct Answer 4
Explanation
2kP
E =r
3
F = QE
2kPQF =
r3
dW = F .dr→ →→ →→ →→ →
2K P QdW = dr
r
→→→→→→→→
3
4kPQ 4kPQ 4kPQW = + = - -
r r r
-r
2 2 2r
(((( ))))8kPQ
W = + -ver
2
1W
r∝∝∝∝
2
Q. No. 92 The distance between H and Cl ions in HCl molecule is 1.28 A. What will be the −−−−
o+
potential due to this dipole at a distance of 12 A on the axis of dipole. o
Option 1 0.13 V
Option 2 1.3 V
Option 3 13 V
Option 4 130 V
Correct Answer 1
Explanation Potential due to dipole at axis of dipole
kP 9 10 1.6 10 1.28 10V = =
r 12 12 10
× × × × ×× × × × ×× × × × ×× × × × ×
× ×× ×× ×× ×
9 -19 -10
2 -20
= 0.13 V
Q. No. 93 The potential at a point due to an electric dipole will be maximum and minimum when
the angles between the axis of the dipole and the Electric field vector are respectively
Option 1 900 and 180
0
Option 2 00 and 90
0
Option 3 900 and 0
0
Option 4 1800 and 0
0
Correct Answer 4
Explanation U= -PE cos θθθθ
= 0 cos0 =1θθθθ ⇒⇒⇒⇒0 0
U = -PE (minimum)
=180 cos180 = -1θθθθ ⇒⇒⇒⇒0 0
U = PE (maximum)
Q. No. 94 An electric dipole is placed at origin and is directed along the x-axis. At a point P far
away from the dipole the electric field is parallel to y -axis. OP makes an angle withθθθθ x axis, then
Option 1 tan = 3θθθθ
Option 2 tan = 2θθθθ
Option 3 1tan =
2θθθθ
Option 4 = 45θθθθ 0
Correct Answer 2
Explanation
2kPcos
E =r
θθθθx 3
kPsinE =
r
θθθθx 3
E cos = E sinθ θθ θθ θθ θx y
2kPcos kPsin=
r r
θ θθ θθ θθ θ2 2
3 3
tan = 2θθθθ2
tan = 2θθθθ
Q. No. 95 Two short dipoles each of dipole moment p are placed at origin. The dipole moment of
one dipole is along x axis, while that of other is along y axis. The electric field at point
(a, 0) is given by
Option 1 2p
4 aπεπεπεπε0
Option 2 p
4 aπεπεπεπε 30
Option 3 5 p
4 aπεπεπεπε 30
Option 4 zero
Correct Answer 3
Explanation
2kP kP
E = ,E =a a
x y3 3
E = E +E2x y
E = E +E2x y
kPE = 5
a3
Q. No. 96 A dipole is kept in front of a conducting sphere containing a total a charge Q. If the
dipole is released from rest it reaches the sphere in time t1. If the same experiment is
repeated with an insulating sphere with same charge distribution Q uniformly over its
surface the dipole reaches in time t2. Then
Option 1 t2 > t2
Option 2 t1 < t2
Option 3 t1 = t2
Option 4 No definite relation exists
Correct Answer 2
Explanation In case of dipole and charged sphere, attractive force increases, but in case of dipole
and insulated sphere there will be no distribution of charges. Hence t1 < t2.
Q. No. 98 A positive charge is fixed at the origin of coordinates. An electric dipole. Which is free
to move and rotate, is placed on the positive x-axis. Its moment is directed away from
the origin. The dipole will :
Option 1 Move towards the origin
Option 2 Move away from the origin
Option 3 Rotate by
2
ππππ
Option 4 Rotate by ππππ
Correct Answer 1
Explanation
-q charge is near the +Q as compare to +q, so attractive force will be greater than
repulsive force. Therefore, dipole moves towards the origin.
Q. No. 99 The electric dipole is situated in an electric field as shown in adjacent figure. The dipole
and the electric field are both in the plane of the paper. The dipole is rotated about an
axis perpendicular to the plane of the paper about its axis at a point A in anti-clockwise
direction. If the angle of rotation is measured with respect to the direction of the
electric field, then the torque for different values of the angle of rotation will beθθθθ
represented in fig. given below by the (clockwise torque + ve)
Q. No. 97 There exists a non-uniform electric field along x-axis as shown in figure. The field
increases at a uniform rate along positive x-axis. A dipole is placed inside the field as
shown. For the dipole which one of the following statement is correct
Option 1 Dipole moves along positive x-axis and undergoes a clockwise rotation
Option 2 Dipole moves along negative x-axis after undergoing a clockwise rotation
Option 3 Dipole moves along positive x-axis after under going an anticlockwise rotation
Option 4 Dipole moves along negative x-axis and undergoes an anticlockwise rotation
Correct Answer 3
Explanation Dipole rotate due to torque is anticlockwise along -ve axis, then comes in equilibrium
and moves along the +ve x-axis.
Option 1 Curve (1)
Option 2 Curve (2)
Option 3 Curve (3)
Option 4 Curve (4)
Correct Answer 2
Explanation
(((( ))))= qE sin θθθθττττ
Torque
Torque is function of θθθθ
Q. No. 100 When an electric dipole p is placed in a uniform electric field E then at what angle
→ →→ →→ →→ →
between P and E the value of toruqe will be maximum at.→ →→ →→ →→ →
Option 1 900
Option 2 00
Option 3 1800
Option 4 450
Correct Answer 1
Explanation =PE sin τ θτ θτ θτ θ
= 90θθθθ 0
(((( ))))= PE maxττττ
Q. No. 101 An electric dipole is placed in an electric field generated by a point charge
Option 1 The net electric force on the dipole must be zero
Option 2 The net electric force on the dipole may be zero
Option 3 The torque on the dipole due to the field must be zero
Option 4 The torque on the dipole due to the field may be zero
Correct Answer 4
Explanation Electric dipole is placed in non-uniform electric field. Therefore, net electric force will
exerted on the dipole but torque may or may not be zero depending upon orientation
of dipole.
Q. No. 102 An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is
placed in a uniform electric field E. If its dipole moment is along the direction of the
field, the force on it and its potential energy are respectively
Option 1 q. E and p . E
→ → →→ → →→ → →→ → →
Option 2 Zero and minimum
Option 3 q. E and maximum
→→→→
Option 4 2q. E and minimum
→→→→
Correct Answer 2
Explanation Dipole is placed in uniform electric field, so net force will be zero as dipole is directed
along the field [Fnet = 0]
U= -pE cos θθθθ
= 0θθθθ 0
U = -PE (minimum)
Q. No. 103 An electric dipole is situated in an electric field of uniform intensity E whose dipole
moment is p and moment of inertia is I. If the dipole is displaced slightly from the
equilibrium position, then the angular frequency of its oscillations
Option 1 pE
I
1
2
Option 2 pE
I
3
2
Option 3 I
pE
1
2
Option 4 p
IE
1
2
Correct Answer 1
Explanation
= pE sin τ θτ θτ θτ θ
(((( ))))sin angle is very smallθ ≈ θθ ≈ θθ ≈ θθ ≈ θ
= pE ...(i)τ θτ θτ θτ θ
= k τ θτ θτ θτ θ
K = PE
Restoring factor=
Inertial factorωωωω
k PE= =
m Iωωωω
Q. No. 104 The electric flux from a cube of edge ℓℓℓℓ is φφφφ . What will be value if edge of cube is
made 2ℓℓℓℓ and charge enclosed havled.
Option 1
2
φφφφ
Option 2 2φφφφ
Option 3 4φφφφ
Option 4 φφφφ
Correct Answer 1
Explanation Cube of edge = φφφφℓℓℓℓ
Q=⇒⇒⇒⇒ φφφφ
∈∈∈∈0
QIn q =
2enclosed
q Q= = =
2 2
φφφφ′ ′′ ′′ ′′ ′φφφφ ⇒⇒⇒⇒ φφφφ
∈ ∈∈ ∈∈ ∈∈ ∈enclosed
0 0
Q. No. 105 A uniformly charged and infinitely long line having a lines charge density λλλλ is placed at
a normal distance y from point O. Consider an imaginary sphere of radius R with O as
centre and R > y. Electric flux through the surface of the sphere is
Option 1 Zero
Option 2 2 Rλλλλ
εεεε0
Option 3 2 R - yλλλλ
εεεε
2 2
0
Option 4 2 R + yλλλλ
εεεε
2 2
0
Correct Answer 3
Explanation For R > y
Length inside the sphere
= 2 R - y2 2
Charge inside the sphere = 2 R - yλλλλ 2 2
2 R - yChange enclose= =
λλλλφφφφ
∈ ∈∈ ∈∈ ∈∈ ∈
2 2
0 0
Q. No. 106 A long string with a charge of λλλλ per unit length passes through an imaginary cube of
edge a. The maximum flux of the electric field through the cube will be
Option 1 aλλλλ
∈∈∈∈0
Option 2 2 aλλλλ
∈∈∈∈0
Option 3 6 aλλλλ
∈∈∈∈
2
0
Option 4 3 aλλλλ
∈∈∈∈0
Correct Answer 4
Explanation Flux will be maximum when charge enclosed will be maximum. i.e., string is along
diagonal of cube.
Q = 3 aλ ×λ ×λ ×λ ×enclosed
3 a=
λλλλφφφφ
∈∈∈∈Max
0
Q. No. 107 A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is
constructed with its centre at the periphery of the ring. The flux of electric field
through the sphere is
Option 1 Q
3∈∈∈∈0
Option 2 2Q
3∈∈∈∈0
Option 3 Q
2∈∈∈∈0
Option 4 3Q
4 ∈∈∈∈0
Correct Answer 1
Explanation
Q 2
Charge enclosed = r2 r 3
ππππ××××
ππππ
Q=
3
Q =
3φφφφ
∈∈∈∈0
Q. No. 108 A conducting spherical shell of radius R carries a charge Q. A point charge +Q is placed
at the centre. The electric field E varies with distance r (from centre of the shell) as
Option 1
Option 2
Option 3
Option 4
Correct Answer 1
Explanation
Electric field inside the sphere is due to +q charge maximum at centre and varies
inverse proportional with radius. Electric field outside the sphere is zero.
Q. No. 109 Which of the following graphs shows the variation of electric field E due to a hollow
spherical conductor of radius R as a function of distance from the centre of the sphere
Option 1
Option 2
Option 3
Option 4
Correct Answer 1
Explanation For hollow spherical conductor
E = 0 (Inside the sphere)
QE = k. (at the surface of sphere)
R2
QE = k. (outside the sphere)
R2
Q. No. 110 The surface density on the copper sphere is .σσσσ The electric field strength on the
surface of the sphere is
Option 1 σσσσ
Option 2
2
σσσσ
Option 3
2
σσσσ
∈∈∈∈0
Option 4 σσσσ
∈∈∈∈0
Correct Answer 4
Explanation 1 QE (at surface of sphere) = .
4 Rπ∈π∈π∈π∈ 20
Q= Q = 4 R
4 Rσσσσ ⇒⇒⇒⇒ π σπ σπ σπ σ
ππππ
2
2
E =σσσσ
∈∈∈∈0
Q. No. 111 Two conducting spheres of radii r1 and r2 are charged to the same surface charge
density. The ratio of electric fields near their surface is
Option 1 r
r
2122
Option 2 r
r
2221
Option 3 r
r
1
2
Option 4 1 : 1
Correct Answer 4
Explanation =σ σσ σσ σσ σ1 2
Q Q=
R R
1 22 21 1
Qas E = k.
R2
E 1=
E 1
1
2
Q. No. 112 A hollow metallic sphere of radius 10cm is given a charge of 3.2 10 C.×××× -9 The electric
intensity at a point 4cm from the centre is
Option 1 9 10 NC×××× -9 -1
Option 2 288 NC-1
Option 3 2.88 NC-1
Option 4 zero
Correct Answer 4
Explanation as r < R
Inside the hollow sphere, electric field is zero
Q. No. 113 The electric field due to a uniformly volume charged sphere of radius R as a function of
the distance from its centre is represented graphically by
Option 1
Option 2
Option 3
Option 4
Correct Answer 2
Explanation Outside the solid sphere
Q 1E = k. E
r r⇒⇒⇒⇒ ∝∝∝∝
2 2
At the surface
QE = k. (maximam)
R2
Inside Q
E = k. (E r)R
∝∝∝∝3
Q. No. 114 A solid sphere of radius R has a uniform distribution of electric charge in its volume. At
a distance x from its centre for x < R, the electric field is directly proportional to
Option 1 1
x2
Option 2 1
x
Option 3 x
Option 4 x2
Correct Answer 3
Explanation
Inside the solid sphere
QE = k. .x
R2
E x∝∝∝∝
Q 4Q = x
4 3R
3
′′′′ × π× π× π× π
ππππ
3
3
QQ = .x
R′′′′ 3
3
Q QE = k. = k. x
x R
′′′′2 3
Q. No. 115 An insulated sphere of radius R has charge density ρρρρ . The electric field at a distance r
from the centre of the sphere (r < R)
Option 1 r
3
ρρρρ
∈∈∈∈0
Option 2 R
3
ρρρρ
∈∈∈∈0
Option 3 rρρρρ
∈∈∈∈0
Option 4 Rρρρρ
∈∈∈∈0
Correct Answer 1
Explanation r < R
1 QE = . .x
4 Rπ∈π∈π∈π∈ 30
Multiply and divide by 3 in denominator
Q.xE =
43 R
3× π ∈× π ∈× π ∈× π ∈3
0
Q=
4R
3
ρρρρ
ππππ 3
rE =
3
ρρρρ
∈∈∈∈0
Q. No. 116 Which of the following is discontinuous across a charged conducting surface ?
Option 1 Electric potential
Option 2 Electric intensity
Option 3 both electric potential and intensity
Option 4 none of these
Correct Answer 2
Explanation Electric Intensity is discontinuous across a charged conducting surface.
Q. No. 117 Consider two points 1 and 2 in a region outside a charged sphere. Two points are not
very far away from the sphere. If E and V represent the electric field vector and the
electric potential, which of the following is not possible
Option 1 E = E ,V = V→ →→ →→ →→ →
1 2 1 2
Option 2 E E ,V V→ →→ →→ →→ →
≠ ≠≠ ≠≠ ≠≠ ≠1 2 1 2
Option 3 E E ,V = V→ →→ →→ →→ →
≠≠≠≠1 2 1 2
Option 4 E = E ,V V→ →→ →→ →→ →
≠≠≠≠1 2 1 2
Correct Answer 4
Explanation Outside the charged sphere
E = E ,V V→ →→ →→ →→ →
≠≠≠≠1 1 2
Q. No. 118 A system consists of uniformly charged sphere of radius R and a surrounding medium
filled by a charge with the volume density = , where is a positive constant andr
ααααρ αρ αρ αρ α
r is the distance from the centre of the sphere. The charge of the sphere for which
electric field intensity E outside the sphere is independent of r is
Option 1
2
αααα
∈∈∈∈0
Option 2 2
α∈α∈α∈α∈0
Option 3 2 Rπαπαπαπα 2
Option 4 Rαααα 2
Correct Answer 3
Explanation Volume charge density of medium =
r
ααααρρρρ
Let a small ring element in medium having charge ‘dq’
dq = 4 r drρ πρ πρ πρ π 2
dq = 4 r dr
2
αααα× π× π× π× π 2
dq= 4 rdrπαπαπαπα
q = 4 rdrπαπαπαπα ∫∫∫∫r
2
R
4q = r -R
2
παπαπαπα
2 2
1 Q 1 q
E = . + .4 4r rπ∈ π∈π∈ π∈π∈ π∈π∈ π∈
p 2 20 0
1 Q 1 1E = . + 2 r -R
4 4r r × × πα× × πα× × πα× × πα π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2 2p 2 2
0 0
1 Q 2 RE = . + 1-
4 4r r
παπαπαπα
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2
p 2 20 0
As Ep is independent of r
1 Q 2 R. - . = 0
4 4r r
παπαπαπα
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2
2 20 0
1 Q 2 R. = .
4 4r r
παπαπαπα
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2
2 20 0
Q = 2 Rπ ∝π ∝π ∝π ∝ 2
Q. No. 119 Two infinitely long parallel wires having linear charge densities and respectivelyλ λλ λλ λλ λ1 2
are placed at a distance of R metres. The force per unit length on either wire will be
1k =
4
πεπεπεπε 0
Option 1 2k
R
λ λλ λλ λλ λ1 22
Option 2 2k
R
λ λλ λλ λλ λ1 2
Option 3 k
R
λ λλ λλ λλ λ1 22
Option 4 k
R
λ λλ λλ λλ λ1 2
Correct Answer 2
Explanation
Electric field at pt. A on X2Y2 due to X1Y1
2E =
4 R
λλλλ
π∈π∈π∈π∈1
0
F = qE
q = 1λ ×λ ×λ ×λ ×2
Force per unit length
2F =
4 R
λ λλ λλ λλ λ
π∈π∈π∈π∈1 2
0
Q. No. 120 An electron moves round a circular path of radius 0.1m about an infinite linear charge
of density +1 C /m. The speed of the electron will be µµµµ
Option 1 5.6 10 m/ s×××× 3
Option 2 2.8 10 m/ s×××× 5
Option 3 5.6 10 m/ s×××× 7
Option 4 2.8 10 m/ s×××× 7
Correct Answer 3
Explanation
Electron moves round a circular path, force due to infinite line on e
– provide it
necessary centripetal force.
mv= qE
r
2
mv 1= q .
r 2 r
λλλλ××××
π∈π∈π∈π∈
2
0
2q 9 10 2 1.6 10 1 10v = =
4 m 9.1 10
λ × × × × × ×λ × × × × × ×λ × × × × × ×λ × × × × × ×
π∈π∈π∈π∈ ××××
9 -19 -62
-310
v = 32 10××××2 14
v = 4 2 10 = 5.6 10 m/ s× ×× ×× ×× ×7 7
Q. No. 121 A long thin rod lies along the x-axis with one end at the origin. It has a uniform charge
density C /m. Assuming it to infinite in length, the electric field point x = -a onλλλλ
the x-axis will
Option 1
a
λλλλ
πεπεπεπε0
Option 2
2 a
λλλλ
πεπεπεπε0
Option 3
4 a
λλλλ
πεπεπεπε0
Option 4 2
a
λλλλ
πεπεπεπε0
Correct Answer 3
Explanation
For infinite length
Electric field at a point along
=4 r
λλλλ
πεπεπεπε0
r = a
E =4 a
λλλλ
πεπεπεπε0
Q. No. 122 Two infinite plane parallel sheets separated by a distance d have equal and opposite
uniform charge densities . Electric field at a point between the sheets isσσσσ
Option 1 Zero
Option 2 σσσσ
εεεε0
Option 3
2
σσσσ
εεεε0
Option 4 2σσσσ
εεεε0
Correct Answer 2
Explanation
Electric field at point between the sheets
= EI - EII
E =2
σσσσ
∈∈∈∈I
0
-E =
2
σσσσ
∈∈∈∈II
0
= + =2 2
σ σ σσ σ σσ σ σσ σ σ
∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈0 0 0
Q. No. 123
An infinite plane with uniformly distributed positive charge has surface charge density
. A small metallic sphere S of mass m and charge + Q is attached to a thread andσσσσ
tied to a point P on the sheet AB. The angle which PS makes with the plane AB is given
by
Option 1 Q
tanmg
σσσσ ∈∈∈∈
-1
0
Option 2 Qcot
2 mg
σσσσ
∈∈∈∈
-1
0
Option 3 Qtan
2 mg
σσσσ
∈∈∈∈
-1
0
Option 4 Qcot
mg
σσσσ ∈∈∈∈
-1
0
Correct Answer 3
Explanation
At equilibrium of Pendulum
T sin =Feθθθθ
T cos =mgθθθθ
Fe QEtan = =
mg mgθθθθ
tan =2 mg
θσθσθσθσθθθθ
∈∈∈∈0
= tan2 mg
θσθσθσθσθθθθ
∈∈∈∈
-1
0
Q. No. 124 The electric potential due to an infinite sheet of positive charge density at a pointσσσσ
located at a perpendicular distance Z from the sheet is (Assume V0 to be the potential
at the surface of sheet)
Option 1 V0
Option 2 ZV -
σσσσ
εεεε0
0
Option 3 ZV +
2
σσσσ
εεεε0
0
Option 4 ZV -
2
σσσσ
εεεε0
0
Correct Answer 4
Explanation
-dV
E =dr
dV = -Edr
-dV = .dr
2
σσσσ
∈∈∈∈0
(dr = Z)
- ZdV =
2
σσσσ
∈∈∈∈0
Potential at point P ‘V’ = V0 + dV
Z= V -
2
σσσσ
∈∈∈∈0
0
Q. No. 125 A large metal surface has uniform charge density . An electron of mass m andσσσσ
charge e leaves the surface at A with speed u and returns to point B. Disregard gravity.
The maximum value of AB is
Option 1 u m
e
εεεε
σσσσ
20
Option 2 u e
m
εεεε
σσσσ
20
Option 3 u e
mε σε σε σε σ
2
0
Option 4 u e
m
σσσσ
εεεε
2
0
Correct Answer 1
Explanation
F = qE
eF =
2
σσσσ
∈∈∈∈0
eW.D = AB
2
σσσσ××××
∈∈∈∈ 0
1 emu = AB
2 2
σσσσ××××
∈∈∈∈
2
0
muAB =
e
∈∈∈∈
σσσσ
20
Q. No. 126 If two conducting spheres are separately charged and then brought in contact
Option 1 The total energy of the two spheres is conserved.
Option 2 The total charge on the two spheres in conserved.
Option 3 Both the total energy and the total charge are conserved.
Option 4 The final potential is always the mean of the original potential of the two spheres.
Correct Answer 2
Explanation Change neither be created nor destroyed.
Q. No. 127
A point charge +q is placed at the centre of a conducting spherical shell of inner radius
a and outer radius b. What charge will appear on the outer surface of the shell
Option 1 Zero
Option 2 q
Option 3 -q
Option 4 2q
Correct Answer 2
Explanation
Net electric field has to be zero.
qE .A =→ →→ →→ →→ →
φφφφ∈∈∈∈
enclosed
0
Q= 0
∈∈∈∈enclosed
0
Qenclosed = 0
q+Q = 0 Q = -q⇒⇒⇒⇒
As charge -q is thieve on inner surface of shell so charge +q will appear on the outer
urface
Q. No. 128 A point charge Q is placed inside a conducting spherical shell of inner radius 3R and
outer radius 5R at a distance R from the centre of the shell. The electric potential at
the centre of the shell will be
Option 1 1 Q.
4 Rπ∈π∈π∈π∈0
Option 2 1 5Q.
4 6Rπ∈π∈π∈π∈0
Option 3 1 13Q.
4 15Rπ∈π∈π∈π∈0
Option 4 1 7Q.
4 9Rπ∈π∈π∈π∈0
Correct Answer 3
Explanation
Q Q Q
V = k. -k. +R 3R 5R
0
13QV = k
15R
0
Q. No. 129 A point charge q is placed at a distance r from the centre of an uncharged conducting
sphere of radius R(< r). The potential at any point on the sphere is
Option 1 Zero
Option 2 1 q
4 rπ∈π∈π∈π∈0
Option 3 1 qR
4 rπ∈π∈π∈π∈ 20
Option 4 1 qR
4 Rπ∈π∈π∈π∈
2
0
Correct Answer 2
Explanation
Since in sphere shell, potential at centre and surface is always same.
Potential due to q at 'O'∴∴∴∴
q 1 q= k. = .
r 4 rπ∈π∈π∈π∈0
Q. No. 130 Figure shows two concentric, conducting shells of radii r and 2r. The outer shell is given
a charge Q. The amount of charge that will appear on inner surface of outer cell if
inner cell is grounded
Option 1 Q
-2
Option 2 Q+
2
Option 3 -2Q
Option 4 +2Q
Correct Answer 2
Explanation
k.Q k.qV = + = 0
2r rs
-Qq =
2
Since charge given to outer shell is Q charge on inner surface of outershell will be
Q+ so as not electric field will be zero.
2
Q. No. 131 In a region with a uniform electric field, the number of lines of force per unit area is E.
If a spherical metallic conductor is placed in the area, the field inside the conductor
will be
Option 1 Zero
Option 2 E
Option 3 more than E
Option 4 less than E
Correct Answer 1
Explanation Since no. of lines of force passing per unit area is equal to no. of lines of force leaving
through the spherical metallic conductor, so net electric field inside will be zero
Q. No. 132 When two uncharged metal balls of radius 0.09 mm each collide, one electron is
transferred between them. Then potential difference between them would be
Option 1 16 Vµµµµ
Option 2 16pV
Option 3 32 Vµµµµ
Option 4 32pV
Correct Answer 3
Explanation k(e) k(e) 2keV = V - V = - =
r r r∆∆∆∆ 1 2
2 9 10 1.6 10V = = 32 V
9 10
× × × ×× × × ×× × × ×× × × ×∆ µ∆ µ∆ µ∆ µ
××××
9 -19
-5
Q. No. 133 A hollow sphere of radius 2R is charged to V volt and another small sphere of radius R
Vis charged to volt. Then the smaller sphere is placed inside the bigger sphere
2
without changing the net charge on each sphere. The potential difference between the
two spheres would be
Option 1 3V
2
Option 2 V
4
Option 3 V
2
Option 4 V
Correct Answer 2
Explanation
V Q
= k.2 R
2
QV = k
2R
1
VRQ =
2K2
2RVQ =
K1
K(Q +Q )
V =2R
1 21
4RV VRK +
k.5VR 52k 2k= = = V
2R 2k.2R 4
kQ kQV = +
2R R
1 22
2RV VRk k
k 2k= +
2R 2R
V 3= V + = V
2 2
3V 5V - V = - V
2 42 1
VV - V =
42 1
Q. No. 134 If two conducting spheres are separately charged and then brought in contact
Option 1 The total energy of the two spheres is conserved.
Option 2 The total charge on the two spheres is conserved.
Option 3 Both the total energy and the total charge are conserved.
Option 4 The final potential is always the mean of the original potential of the two spheres.
Correct Answer 2
Explanation Charge neither be created nor be destroyed.
Q. No. 136 In a parallel-plate capacitor, the region between the plates is filled by a dielectric slab.
The capacitor is charged from a cell and then disconnected from it. The slab is now
taken out.
Option 1 The potential difference across the capacitor is reduced
Option 2 The potential difference across the capacitor is increased
Option 3 The energy stored in the capacitor is reduced
Option 4 No work is done by an external agent in taking the slab out
Correct Answer 3
Explanation When capacitor is disconnected and slow is now taken out,
V = constant
but Q decreases
1 U = QV
2∴∴∴∴
U will also decreases
Q. No. 137 A parallel plate capacitor is connected to a battery. The plates are pulled apart with a
uniform speed. If x is the separation between the plates, the time rate of change of
electrostatic energy of capacitor is proportional to
Option 1 x-2
Option 2 x
Option 3 x-1
Option 4 x2
Correct Answer 1
Explanation 1U= CV
2
2
V = constant
1 AU = V
2 x
∈∈∈∈ 20
(((( ))))dU 1 d= AV x
dx 2 dx∈∈∈∈ 2 -1
0
(((( ))))dU 1 1
= AV -1dx 2 x
∈∈∈∈ 20 2
dU 1 1= - AV
dx 2 x∈ ×∈ ×∈ ×∈ ×2
0 2
dU 1 dU OR x
dx 2 dx∝ ∝∝ ∝∝ ∝∝ ∝ -2
Q. No. 138 Charge Q on a capacitor varies with voltage V as shown in the figure, where Q is taken
along the X-axis and V along the Y-axis. The area of triangle OAB represents
Option 1 Capacitance
Option 2 Capacitive reactance
Option 3 Magnetic field between the plates
Option 4 Energy stored in the capacitor
Correct Answer 4
Explanation
Area of Triangle OAB
= Area under V-Q graph
1= OB AB
2× ×× ×× ×× ×
1= QV
2××××
= U = Energy stored in Capacitor
Q. No. 139 The capacity of a parallel plate air capacitor is 10 F and it is given a charge 40 C.µ µµ µµ µµ µ
The electrical energy stored in the capacitor in ergs is
Option 1 80 10×××× 6
Option 2 800
Option 3 8000
Option 4 20000
Correct Answer 2
Explanation C =10 F, Q = 40 Cµ µµ µµ µµ µ
1 Q 1 1600 10U = = = 8 10 J = 800 ergs
2 C 2 10 10
××××× ×× ×× ×× ×
××××
2 -12-5
-6
Q. No. 140 Two condensers of capacity 0.3 F and 0.6 F respectively are connected in series.µ µµ µµ µµ µ
The combination is connected across a potential of 6 volts. The ratio of energies stored
by the condensers will be
Option 1 1
2
Option 2 2
Option 3 1
4
Option 4 4
Correct Answer 2
Explanation
0.3 0.6
C = = 0.2 F0.9
××××µµµµeff
Change supplied by Battery QB = Ceff. V
Q = 0.2 10 6 = 1.2 C× × µ× × µ× × µ× × µ-6B
1 Q 1 1.2 1.2 10Energy stored in C 'U ' = =
2 C 2 0.3 10
× ×× ×× ×× ×××××
××××
2 -12
1 1 -61
U = 2.4 10 J×××× -61
1 Q 1 1.2 1.2 10Energy stored in C 'U ' = =
2 C 2 0.6 10
× ×× ×× ×× ×××××
××××
2 -12
2 2 -62
U =1.2 Jµµµµ2
U 2=
U 1
1
2
Q. No. 141 In a parallel plate capacitor, the distance between the plates is d and potential
difference across plates is V. Energy stored per unit volume between the plates of
capacitor is :
Option 1 Q
2V
2
2
Option 2 1 V
2 dεεεε
2
0 2
Option 3 1 V
2 dεεεε
2
20
Option 4 1 V
2 dεεεε
2
0 2
Correct Answer 2
Explanation
1
U = E2
∈∈∈∈ 20
VE =
d
1 VU =
2 d∈∈∈∈
2
0 2
Q. No. 142 In a parallel plate capacitor of plate area A, plate separation d and charge q, the force
of attraction between the plates is F
Option 1 F A∝∝∝∝ 2
Option 2 F q∝∝∝∝ 2
Option 3 F d∝∝∝∝
Option 4 1F
d∝∝∝∝
Correct Answer 2
Explanation E =
σσσσ
∈∈∈∈0
QE =
A∈∈∈∈0
QF = QE = F Q
A⇒⇒⇒⇒ ∝∝∝∝
∈∈∈∈
22
0
Q. No. 143 A spherical condenser has inner and outer spheres of radii a and b respectively. The
space between the two is filled will air. The difference between the capacities of two
condenser, formed when outer sphere is earthed and when inner sphere is earthed
will be
Option 1 Zero
Option 2 4 aπεπεπεπε0
Option 3 4 bπεπεπεπε0
Option 4 b4 a
b - a
πεπεπεπε
0
Correct Answer 3
Explanation When outer sphere is earthed
Q Q
V = -4 a 4 bπ ∈ π ∈π ∈ π ∈π ∈ π ∈π ∈ π ∈0 0
abC = 4 .
b - aπ∈π∈π∈π∈0
When Inner sphere is earthed
Indented charge on inner sphere
-a bQ = Q, C = 4
b b- a′ ′′ ′′ ′′ ′ π∈π∈π∈π∈
2
0
Q b- aV =
4 ab
′′′′ π∈π∈π∈π∈ 0
b abC - C = 4 - = 4 b
b - a b - a
′′′′ π∈ π∈π∈ π∈π∈ π∈π∈ π∈
2
0 0
Q. No. 144 Eight drops of mercury of equal radii possessing equal charges combine to form a big
drop. Then the capacitance of bigger drop compared to each individual small drop is
Option 1 8 times
Option 2 4 times
Option 3 2 times
Option 4 32 times
Correct Answer 3
Explanation Q = 8Q′′′′
(((( ))))V = 8 V′′′′2/3
V = 4V′′′′
Q 8QC = = = 2C
V 4V
′′′′′′′′
′′′′
Q. No. 145 Two conducting spheres of radii 5 cm and 10 cm are given a charge of 15 C each.µµµµ
After the two spheres are joined by a conducting wire, the charge on the smaller
sphere is
Option 1 5 Cµµµµ
Option 2 10 Cµµµµ
Option 3 15 Cµµµµ
Option 4 20 Cµµµµ
Correct Answer 2
Explanation
R = 5 10×××× -21 R = 10 10 m×××× -2
2
Q = 15 10 C×××× -6
when two spheres are joined by a conducting wire, charge will flow from high
potential to low potential.
Q R - Q RCharge flow x =
R +R
1 2 2 1
1 2
15 10 10 10 -15 10 5 10x =
15 10
× × × × × ×× × × × × ×× × × × × ×× × × × × ×
××××
-6 -2 -6 -2
-2
(((( ))))150 - 75 10x = = 5 C
15
××××µµµµ
-2
New charge on smaller sphere = Q -x = 15 - 5=10 Cµµµµ
Q. No. 146 Two capacitors of capacitances 3 F and 6 F are charged to a potential of 12V each.µ µµ µµ µµ µ
They are now connected each other, with the positive plate of each joined to the
negative plate of the other. The potential difference across each will be
Option 1 6 volt
Option 2 4 volt
Option 3 3 volt
Option 4 Zero
Correct Answer 2
Explanation Charge on C (3 F) is Q = 3 12 = 36 Cµ µ × µµ µ × µµ µ × µµ µ × µ1 1
Charge on C (6 F) is Q = 6 12 = 72 Cµ µ × µµ µ × µµ µ × µµ µ × µ2 2
C V - C VCommon Potential V =
C + C
1 1 2 2f
1 2
72- 36 36
= = = 4V9 9
Q. No. 147 A parallel plate capacitor carries a charge q. The distance between the plates is
doubled by application of a force. The work done by the force is
Option 1 Zero
Option 2 q
C
2
Option 3 q
2C
2
Option 4 q
4C
2
Correct Answer 3
Explanation Q QW = F.d = d =
2 A 2C××××
∈∈∈∈
2 2
0
A= C
d
∈∈∈∈0
1 QW.D =
2 C
2
Q. No. 148 A parallel plate capacitor of capacity C0 is charged to a potential V0
i) The energy stored in the capacitor when the battery is disconnected and the
separation is doubled E1
ii) The energy stored in the capacitor when the charging battery is kept connected and
E
the separation between the capacitor plates doubled is E . Then value is E
12
2
Option 1 4
Option 2 3
2
Option 3 2
Option 4 1
2
Correct Answer 1
Explanation AOriginal Capacitance 'C ' =
d
∈∈∈∈00
Case I : When separation between plates is doubled
AC =
2d
∈∈∈∈′′′′ 0
C =2
∈∈∈∈′′′′ 0
1 Q QE = . = 2
2 C 2C
′′′′
2 2
1
E1 = 2E
Case II : V0 = Constant
CC =
2′′′′′′′′ 0
1 1 1 EE = C V = C V =
2 2 2 2
′′′′′′′′
2 22 0 0 0
E 2E= = 4
EE
2
1
2
Q. No. 149 A parallel plate capacitor has an electric field of 105 V/m between the plates. If the
charge on the capacitor plate is 1 C, the force on each capacitor plate isµµµµ
Option 1 0.5 N
Option 2 0.05 N
Option 3 0.005 N
Option 4 None of these
Option 5
Option 6
Correct Answer 2
Explanation E = 10 V / m, Q = 1 Cµµµµ5
F = QE = 1 10 10 = 10 N× ×× ×× ×× ×-6 5 -1
F 1Force on each capacitor plate = = = 0.05 N
2 20
Q. No. 150 An air capacitor of capacity C = 10 F is connected to a constant voltage battery ofµµµµ
12 V. Now the space between the plates is filled with a liquid of dielectric constant 5.
The charge that flows now from battery to the capacitor is
Option 1 120 Cµµµµ
Option 2 699 Cµµµµ
Option 3 480 Cµµµµ
Option 4 24 Cµµµµ
Correct Answer 3
Explanation C =10 F, V = 12V, K = 5µµµµ
(((( ))))Q = CV = 120 C when no dielectric inserted µµµµB When dielectric slab inserted between plates
C =kC = 50 F′′′′ µµµµ
Q = C V = 50 12 = 600 C′ ′′ ′′ ′′ ′ × µ× µ× µ× µB
Extra charge that flow from battery after inserting dielectric slab
= Q -Q = 600 C-120 C = 480 C′′′′ µ µ µµ µ µµ µ µµ µ µB B
Q. No. 151 A capacitor of capacity C is connected with a battery of potential V in parallel. The
distance between its plates is reduced to half at once, assuming that the charge
remains the same. Then to charge to capacitance upto the potential V again, the
energy given by the battery will be
Option 1 CV
4
2
Option 2 CV
2
2
Option 3 3CV
4
2
Option 4 CV2
Correct Answer 4
Explanation
d
d =2
′′′′
AC = 2 C = 2C
d
∈∈∈∈′ ′′ ′′ ′′ ′⇒⇒⇒⇒0
Q = constant
1U = C V
2′′′′ 2
1U = 2CV = CV
2×××× 2 2
Q. No. 152 In a parallel-plate capacitor, the region between the plates is filled by a dielectric slab.
The capacitor is charged from a cell and then disconnected from it. The slab is now
taken out.
Option 1 The potential differdence across the capacitor is reduced
Option 2 The potential difference across the capacitor is increased
Option 3 The energy stored in te capacitor is reduced
Option 4 No work is done by an external agent in taking the slab out
Correct Answer 2
Explanation
When battery is disconnected
C = kc′′′′
Q =Q′′′′
VV =
k′′′′
UU =
k′′′′
When battery is disconnected and slab is taken out
C increases
Charge (Q) same
Potential increases
Energy stored increases
Q. No. 153 The capacity of a condenser is 4 10 farad and its potential is 100 volt. The×××× -6
energy released on discharging it fully will be
Option 1 0.02 joule
Option 2 0.04 joule
Option 3 0.025 joule
Option 4 0.05 joule
Correct Answer 1
Explanation 1 1U = CV = 4 10 10 = 2 10 J
2 2× × × ×× × × ×× × × ×× × × ×2 -6 4 -2
U = 0.02 J
Q. No. 154 Three capacitors are connected to d.c. source of 100 volts as shown in the adjoining
figure. If the charge accumulated on plates of C1, C2 and C3 are qa, qb, qc, qd, qe and qr
respectively, then
Option 1 100
q +q +q = coulombs9
b d r
Option 2 qb + qd + qr = 0
Option 3 qa + qc + qe = 50 coulbombs
Option 4 qb = qd = qf
Correct Answer 4
Explanation 12C F,V = 100 V
13= µµµµeff
1200Q C .V = 10 V
13= ×××× -6
eff
Since all three capacitors are connected in series charge stroed in each will be same.
Q. No. 155 An infinite number of identical capacitors each of capacitance 1 F are connected asµµµµ
in adjoining figure. Then the equivalent capacitance between A and B is
Option 1 1 Fµµµµ
Option 2 2 Fµµµµ
Option 3 1 F
2µµµµ
Option 4 ∞∞∞∞
Correct Answer 2
Explanation
1 1 1
C =1+ + + + ......2 4 8
eq
= 1 + 0.5 + 0.25 + 0.125
= 2 Fµµµµ
Q. No. 156 Two identical capacitors are joined in parallel, charged to a pot. V, separated and then
connected in series, i.e., the positive plate of one is connected to the negative of the
other :
Option 1 The charge on the plates connected together are destroyed
Option 2 The charge on free plates are enhanced
Option 3 The energy stored in the system increases
Option 4 The P.D. between the free plates is 2V
Correct Answer 2
Explanation
Potential difference between the free
Plates is 2V.
Q. No. 157 A finite ladder is constructed by connecting serveral sections of 2 F, 4 F capacitorµ µµ µµ µµ µ
combinations as shown in the fiture. It is terminated by a capacitor of capacitance C.
What value should be chosen for C such that the equivalent capacitance of the ladder
between the points A and B becomes independent of the number of sections in
between
Option 1 4 Fµµµµ
Option 2 2 Fµµµµ
Option 3 18 Fµµµµ
Option 4 6 Fµµµµ
Correct Answer 1
Explanation
4C
C = +2C + 4
6C + 8C =
C + 4
C2 + 4C = 6C + 8
C2 - 2C - 8 = 0
C2 - 4C + 2C - 8 = 0
C(C - 4) + 2(C - 4) = 0
(C + 2)(C - 4) = 0
C = 4 Fµµµµ
Q. No. 158 A capacitor capacitance C =1 F can with stand maximum voltage V = 6kVµµµµ1 1
(((( ))))kilo volt and another capacitor of capacitance C = 3 F can withstand maximum− µ− µ− µ− µ2
voltage V2 = 4 kV. When the two capacitors are connected in series, the combined
system can withstand a maximum voltage of
Option 1 4 kV
Option 2 6 kV
Option 3 8 kV
Option 4 10 kV
Correct Answer 3
Explanation =1 F, max. voltage can given to C (V 6 VC ) = kµµµµ 1 11
= 3 F, max. voltage can given to C (V 4 VC )= kµµµµ 2 22
Q1 = C1V1 = 6mc
Q2 = C2N2 = 12mc
Both capacitors will store 6mc charge.
3C = F
4µµµµeff
Q = Ceff. V
Q 6 10V = = = 8kV
3C10
4
××××
××××
-3
-6eff
Q. No. 159 Four identical square plates of side a are arranged as shown. The equivalent capacity
between points A and B is
Option 1 3 a
2d
εεεε 20
Option 2 3 a
4d
εεεε 20
Option 3 3 a
5d
εεεε 20
Option 4 a
2d
εεεε 20
Correct Answer 3
Explanation
A
C = C =d
∈∈∈∈01 2
AC =
2d
∈∈∈∈03
A A 3 AC = + =
d 2d 2 d
∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈0 0 023
A 3 A 3 A
3 Ad 2 d 2 dC = = =5A 3 5 d
1 +2d 2
∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈×××× ∈∈∈∈
∈∈∈∈
0 0 0
0eff
0
Q. No. 160 Three identical capacitors are combined differently. For the same voltage to each
combination, the one that stores the greatest energy is
Option 1 Two in parallel and the third in series with it
Option 2 Three in series
Option 3 Three in parallel
Option 4 Two in series and third in parallel with it
Correct Answer 3
Explanation In series
Total voltage = V
CTotal Capacitance =
3
1 1 CU = CV = V
2 2 3××××2 2
S
1U = CV
6
2S
In Parallel
Total voltage = V
Total Capacitor = 3C
1U = 3CV
2×××× 2
p
3U = CV
2
2p
We can provide same potential to all three capacitors in case of pure parallel
combination.
Q. No. 161 Five identical plates are connected across a battery as follows. If the charge on plate 1
be + q, then the charges on the plates 2, 3, 4 and 5 are
Option 1 -q, +q, -q, + q
Option 2 -2q, +2q, -2q, + q
Option 3 -q, +2q, -2q, + q
Option 4 None of these
Correct Answer 2
Explanation
Charge on plato 1 = +q
Hence change on plates 2, 3, 4 and 5 are
-2q, +2q, -2q, + q
Q. No. 162 Four plates, each of area A and each side are placed parallel to each other at a
distance d. A battery is connected between the combinations 1 and 3 and 2 and 4. The
modulus of charge on plate 2 is
Option 1 2 A
Ed
εεεε0
Option 2 3 AE
d
εεεε0
Option 3 2 AE
3d
εεεε0
Option 4 AE
d
εεεε0
Correct Answer 1
Explanation
Ceff = 3C
QB = 3 CE
Charge on plate 2 2
= Q = 3CE3 3
2 × ×× ×× ×× ×B
= 2CE
2 AE=
d
∈∈∈∈0
Q. No. 163 Two capacitors of capacitance 2 F and 3 F are joined in series. Outer plate of firstµ µµ µµ µµ µ
capacitor is at 1000 volt and outer plate of second capacitor is earthed (grounded).
Now the potential on inner plate of each capacitor will be
Option 1 700 Volt
Option 2 200 Volt
Option 3 600 Volt
Option 4 400 Volt
Correct Answer 4
Explanation
Acc. of Kirchhoff’s II
nd Law
q q1000 - - - 0 = 0
2 3
5q1000 =
6
6 1000q = = 1200 C
5
××××µµµµ
Pot. Differnence across 2 Fµµµµ
1200= = 600 V
2V1
VA - VB = V1
1000 - VB = 600
VB = 400V
Q. No. 164 The capacity of the capacitors are shown in the adjoining fig. The equivalent
capacitance between the points A and B and the charge on the 6 F will beµµµµ
Option 1 27 F; 540 Cµ µµ µµ µµ µ
Option 2 15 F; 270 Cµ µµ µµ µµ µ
Option 3 6 F; 180 Cµ µµ µµ µµ µ
Option 4 15 F; 90 Cµ µµ µµ µµ µ
Correct Answer 3
Explanation
9 18
C = = 6 C9+18
××××µµµµeff
QB = Ceff.V
= 6 C 90 = 540 Cµ × µµ × µµ × µµ × µ
Charge on C2
CQ =
C + C
22
1 2
6= 540 = 180 C
18× µ× µ× µ× µ
Q =180 Cµµµµ2
Q. No. 165 Five capacitors are conneted as shown in the digram. If the p.d. between A and B is
22 V, the emf of the cell is
Option 1 26 V
Option 2 42 V
Option 3 38 V
Option 4 46 V
Correct Answer 4
Explanation
C1 and C2 are in series
12 10 60C = = F
12+10 11
××××µµµµ12
C4 and C5 are in series
3 7 21C = = = 2.1 F
3+ 7 10
××××µµµµ45
C45 and C3 are in parallel
C = 2.9+2.1 = 5 Fµµµµ345
New circuit will be
Since both capacitors are in series. So, charge will be same on both.
Q = C1V1 = C2V2
6022 = 5 V
11× ×× ×× ×× × 2
60 2V = = 24 V
5
××××2
[VB - 0 = 24V]
VB = 24V
VA - VB = 22V
VA - 24 = 22V
VA = 46 V
Q. No. 166 Find the equivalent capacitance between X and Y
Option 1 3 Fµµµµ
Option 2 4 Fµµµµ
Option 3 5 Fµµµµ
Option 4 6 Fµµµµ
Correct Answer 3
Explanation
3 3
C = + = 3 F2 2
µµµµ2
C = 6 Fµµµµ3
C = 4 Fµµµµ4
New circuit is
1 1 1 1= + +
C 2 3 6123
Ceq = C123 + C4
3+ 2 +1
=6
= 1 F + 4 Fµ µµ µµ µµ µ
C =1 Fµµµµ123 C = 5 Fµµµµeq
Q. No. 167 Find the equivalent capacitance between X and Y
Option 1 C
Option 2 2C
Option 3 3C
Option 4 4C
Correct Answer 2
Explanation
C C C C
C = + + + = 2C2 2 2 2
eq
Q. No. 168 Find the equivalent capacitance between X and Y
Option 1 C
10
Option 2 10C
3
Option 3 3C
10
Option 4 9C
Correct Answer 3
Explanation
1 1 1 1 1 1
= + + + +C C 2C 3C 2C Ceq
1 1 1 1= 1 + + + +1
C 2 3 2
1 1 10=
C C 3
eq
3CC =
10eq
Q. No. 169 Find the equivalent capacitance between X and Y
Option 1 C
Option 2 5C
Option 3 2C
Option 4 3C
Correct Answer 4
Explanation
Ceq = C + C + C
Ceq = 3C
Q. No. 170 Find the equivalent capacitance between X and Y
Option 1 6C
Option 2 5C
Option 3 3C
Option 4 2C
Correct Answer 4
Explanation
Ceq (across X and Y) = 2C
Q. No. 171 Find the equivalent capacitance between X and Y
Option 1 C
Option 2 4C
Option 3 6C
Option 4 0
Correct Answer 1
Explanation
In outer circuit, ratio of capacitauces is both arms are same
i.e., VB = VC i.e., circuit is balanced
Now
Ceq = C
Q. No. 172 Find the equivalent capacitance between X and Y
Option 1 C
Option 2 2C
Option 3 3C
Option 4 2C
3
Correct Answer 4
Explanation
2C C
C =3C
××××eq
2CC =
3eq
Q. No. 173 In the given figure switch is closed. Find the change in energy stored in the capacitors.
Option 1 CV
2
20
Option 2 CV
4
20
Option 3 CV
8
20
Option 4 CV
12
20
Correct Answer 2
Explanation
When S Open→→→→
CC =
2eq
QB = Ceq.V0
CVQ =
2
0B
1U = Q V
2i B 0
1U = CV
4
2i 0
When S is closed
Ceff = C
QB = CV0
1U = CV
2
21 0
Change in energy U = U -U∆∆∆∆ f i
1= CV
4
20
Q. No. 174 A parallel plate capacitor is filled by a dielectric whose permittivity varies with the
applied voltage according to the law = V where = 1V . The capacitor withoutε α αε α αε α αε α α -1r
dielectric is charged to voltage V0 = 156 volt is connected in parallel with first nolinar
uncharged capacitor. What is final voltage across the capacitors.
Option 1 6 volt
Option 2 30 volt
Option 3 12 volt
Option 4 4 volt
Correct Answer 3
Explanation = Vε αε αε αε αr
= 1Vαααα -1
V0 = 156V
It C is capacitance without dielectric
Q1 = CV
Capacitance with dielectric = CVαααα 2
Initial charge Q0 = CV0
Acc. to conservation of energy
Q0 = Q1 + Q2
CV = CV + CVαααα 20
V + V - V = 0αααα 20
-1 1 - 4 VV =
2
± α± α± α± α
αααα
0
-1 1+ 4 1 156 -1 625V = =
2 2
± × × ±± × × ±± × × ±± × × ±
-1 25V =
2
±±±±
V = 12V
Q. No. 175 Condenser A has a capacity of 15 F when it is filled with a mdium of dielectricµµµµ
constant 15. Another condenser B has a capacity of 1 F with air between the plates.µµµµ
Both are charged separately by a battery of 100V. After charging, both are connected
in parallel without the battery and the dielectric medium being removed. The common
potential now is
Option 1 400 V
Option 2 800 V
Option 3 1200 V
Option 4 1600 V
Correct Answer 2
Explanation Charge on Capacitor A
Q = 15 10 C×××× -41
Charge on Capacitor B
Q = 1 10 100 = 10 C× ×× ×× ×× ×-6 -42
15 10Capacity of capacitor A after removing dielectric = = 1 F
15
××××µµµµ
-6
Now, the capacitors are connected in parallel
C = 2 Fµµµµeq
15 10 +1 10Common Potential =
2 10
× ×× ×× ×× ×
××××
-4 -4
-6
= 800 V
Q. No. 176 An uncharged capacitor with a solid dielectric is connected to a similar air capacitor
charged to a potential of V0. If the common potential after sharing of charges becomes
V, then the dielectric constant of the dielectric must be
Option 1 V
V
0
Option 2 V
V0
Option 3 (((( ))))V - V
V
0
Option 4 (((( ))))V - V
V
0
0
Correct Answer 3
Explanation Common potential = V
CV +kC 0V =
C +kC
××××0
(((( ))))CV V
V = k +1 =k +1 C V
⇒⇒⇒⇒0 0
Vk = -1
V
0
V - Vk =
V
0
Q. No. 177 The capacitance of a parallel plate condenser is C1 (fig. a). A dielectric of dielectric
constant K is inserted as shown in figure (b) and (c). If C2 and C3 are the capacitances in
figure (b) and (c), then
Option 1 Both C2 and C3 > C1
Option 2 C3 > C1 but C2 < C1
Option 3 Both C2 and C3 < C1
Option 4 C1 = C2 = C3
Correct Answer 1
Explanation
AC =
d
∈∈∈∈01
(((( ))))
A A4k .
d dC =A
2 k +1d
∈ ∈∈ ∈∈ ∈∈ ∈
∈∈∈∈
0 0
20
k +1 A
C =2 d
∈∈∈∈ ⇒⇒⇒⇒
0
3
Since k 1≥≥≥≥
Both C2 and C3 > C1
Q. No. 178 The expression for the capacity of the capacitor formed by compound dielectric placed
between the plates of a parallel plate capacitor as shown in figure, will be (area of
plate = A)
Option 1 A
d d d+ +
K K K
εεεε
0
1 2 3
1 2 3
Option 2 A
d +d +d
K +K +K
εεεε
0
1 2 3
1 2 3
Option 3 (((( ))))A K K K
d d d
εεεε0 1 2 3
1 2 3
Option 4 AK AK AK+ +
d d d
εεεε
1 2 30
1 2 3
Correct Answer 1
Explanation
1 1 1 1
= + +A A AC
k k kd d d
∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈∈ ∈ ∈0 0 0eq1 2 3
1 2 3
AC =
d d d+ +
k k k
∈∈∈∈
0eq
1 2 3
1 2 3
Q. No. 179 The space between the plates of a parallel plate capacitor is filled completely with a
dielectric substance having dielectric constant 4 and thickness 3mm. The distance
between the plates in now increased by inserting a second sheet of thickness 5mm
and dielectric constant K. If the capacitance of the capacitor so formed is one-half of
the original capacitance, the value of K is
Option 1 10
3
Option 2 20
3
Option 3 5
3
Option 4 15
3
Correct Answer 2
Explanation Initially
4. A
C =3 10
∈∈∈∈
××××
01 -3
Finally
4. A k A
3 10 5 10C =4 k
A3 10 5 10
∈ ∈∈ ∈∈ ∈∈ ∈××××
× ×× ×× ×× ×
∈ +∈ +∈ +∈ + × ×× ×× ×× ×
0 0-3 -3
2
0 -3 -3
(((( ))))4k. A
C =20 10 3 10 k
∈∈∈∈
× + × ×× + × ×× + × ×× + × ×
02 -3 -3
CC =
2
12
(((( ))))2k 1
=3 1020 10 +3 10 k ××××× × ×× × ×× × ×× × ×
-3-3 -3
6 10 k = 20 10 +3 10 k× × ×× × ×× × ×× × ×-3 -3 -3
203k = 20 k =
3⇒⇒⇒⇒
Q. No. 180 In a parallel-plate capacitor, the plates are kept vertical. The upper half of the sapace
between the plates is filled with a dielectric with dielectric constant K and the lower
half with a dielectric with dielectric constant 2K. The ratio of the charge density on the
upper half of the plates to the charge density on the lower half of the plates will be
equal to
Option 1 1
Option 2 2
Option 3 1
2
Option 4 3
2
Correct Answer 3
Explanation
k A
C =d
∈∈∈∈01
2k AC =
d
∈∈∈∈02
Pot. Difference across both capacitors are same.
(E1d = E2dV1) = V2
d = dk 2k
σ σσ σσ σσ σ× ×× ×× ×× ×
∈ ∈∈ ∈∈ ∈∈ ∈1 1
0 0
1=
2
σσσσ
σσσσ1
1
Q. No. 181 In a parallel-plate capacitor, the region between the plates is filled by a dielectric also.
The capacitor is connected to a cell and the slab is taken out.
Option 1 Some charge is drawn from the cell
Option 2 Some charge is returned to the cell
Option 3 The potential difference across the capacitor is reduced
Option 4 No work is done by an external agent in taking the slab out
Correct Answer 2
Explanation Battery still connected So, V = const.
Capacitance decreases as dielectric slab is taken out hence charge shotred decreases
i.e, some charge is returned to the cell.
Q. No. 182 If we increase ‘d’ of a parallel plate condenser to ‘2d’ and fill wax to the whole empty
space between its two plate, then capacitance increase from 1PF to 2PF. What is the
dielectric constant of wax.
Option 1 2
Option 2 4
Option 3 6
Option 4 8
Correct Answer 2
Explanation AInitial Capacitance 'C ' = = 1PF
d
∈∈∈∈01
ANew Capacitance 'C ' = k. = 2PF
2d
∈∈∈∈02
C k = = 2 k = 4
C 2⇒⇒⇒⇒2
1
Q. No. 183 A capacitor is connected across another charged capacitor. The energy in the two
capacitors will :
Option 1 Be equal to energy in the initial capacitor
Option 2 Be less than that in the initial capacitor
Option 3 Be more than that in the initial capacitor
Option 4 Be more or less depending on the relative capacities of the two capacitors
Correct Answer 2
Explanation During charging the uncharged capacitor some amount of energy stored in first
capacitor dissipiated in the form of Heat.
Q. No. 184 A small sphere of mass m and having charge q is suspended by a light thread
Option 1 Tension in thread must reduce if another charged sphere is placed vertically
below it
☐
Option 2 Tension in thread is greater that weight mg if another charged sphere is
held in same horizontal line in which first sphere stays in equilibrium
☒
Option 3 Tension in thread is always equal to weight mg ☐
Option 4 Tension may increase to double its original value if another charge is below ☒
it
Explanation
Net force acting on sphere
= mg + Fe
T > mg
T = 2mg if Fe = mg
Q. No. 186 A deuteron and an alpha particle are placed in uniform electric field. The forces acting
on them are F1 and F2 and their acceleration are a1 and a2 respectively.
Option 1 F1 = F2 ☐
Option 2 a1 = a2 ☒
Option 3 F F≠≠≠≠1 2 ☒
Option 4 a a≠≠≠≠1 2 ☐
Explanation qdenteron = +e , md = 2 a.m.u
q = +2eαααα m = 4 a.m.uαααα
eE 1= =
F F
F=
2eF E 2αααα
1 d
2
F F≠≠≠≠1 2
q E eEa = a = =
m 2amu
d1 d
d
q E 2eEa = a = =
m 4a.m.u
αααααααα
αααα2
a1 = a2
Q. No. 187 The electric field in a region is directed outward and is proportional to the distance r
from the origin. Taking electric potential to be zero at origion
Option 1 It is uniform in the region ☐
Option 2 It is propotional to r ☐
Option 3 It is proportional to r2
☒
Option 4 It decreases as one goes away from origin ☒
Explanation E r∝∝∝∝
E = kr
dv = - E dr→ →→ →→ →→ →
dv = -kr.dr
v = -k r dr∫∫∫∫
-krv =
2
2
v r∝∝∝∝ 2
Potential decrease as one goes away from origin
Q. No. 188 Ring with uniform charge Q and radius R is placed in y-z plane with its centre at origin.
Then
Option 1 The electric field at origin is maximum ☐
Option 2 kQThe potential at origin is
R
☒
Option 3 kQThe field at point (x, 0, 0) is
R + X2 2
☐
Option 4 qMaximum value of electric field will be
6 3 Rπ∈π∈π∈π∈ 20
☒
Explanation Electric field due to ring with uniform charge ‘Q’ at a point along its axis.
(((( ))))1 Qx
E = .4 R + xπ∈π∈π∈π∈
x 3/22 20
at origin (x = 0)
E = 0
RElectric field is maximum at
R±±±±
QE =
6 3 Rπ∈π∈π∈π∈ 20
Q. No. 189 Two metallic spheres have same radii. One of them is solid and other is hollow. They
are charged to same potential. The charge on former is q1 and one later is q2. We have
Option 1 q1 = q2 ☒
Option 2 Electric field inside both of them is zero. ☒
Option 3 Electrostatic potential in both the spheres at an inside point is same as on
surface
☒
Option 4 Charge in both is effectively concentrated at the centre for field strength at
an external point.
☒
Explanation
v1 = v2
Q Qk. = k. Q = Q
R R⇒⇒⇒⇒1 2
1 2
Electric field inside zero
Potential constant from centre to surface.
Q. No. 190 A spherical charged conductor has surface charge density. The electric field on its
surface is E and electric potential of the conductor is V. Now the radius of the sphere is
halved keeping the charge to be constant. The new value of electric field and potential
would be
Option 1 4E ☒
Option 2 2V ☒
Option 3 2E ☐
Option 4 4V ☐
Explanation For spherical conductor
1 1E and V
rr
∝ ∝∝ ∝∝ ∝∝ ∝
2
rIf r =
2′′′′
E = 4E and V = 2V′ ′′ ′′ ′′ ′
Q. No. 191 With regards to Gauss’s law and electric flux which of the following statements are
correct.
Option 1 Electric flux through closed surface is equal to total flux due to all charges
enclosed within that surface.
☒
Option 2 Gauss’s law is applicable only when there is symmetrical distribution of
charges
☐
Option 3 Electric field appearing in the Gauss’s law is resultant electric field due to all
the charges present inside as well outside the given closed surface
☒
Option 4 Electric field calculated by Gauss’s law is the field due to only those charges
which are enclosed inside the Gaussian surface.
☐
Explanation Q=φφφφ
∈∈∈∈enclosed
0
Electric field resultant of electric field due to all charges present inside as well outside
the closed surface.
For eg :
= +E E E + E→ → → →→ → → →→ → → →→ → → →
p 1 2 3
Q. No. 192 In uniform electric field equipotential surfaces must
Option 1 Be plane surfaces ☒
Option 2 Be normal to the direction of the field ☒
Option 3 Be spaced that surfaces having equal difference in potential are equally
spaced
☒
Option 4 Have decreasing potential along field ☒
Explanation Uniform electric field
Er
=dV
-d
→→→→
Equipotential surface.
Surfaces I, II and III are equispaces
V1 = V2 = V3
Q. No. 193 When the separation between two charges in increased
Option 1 Electric potential energy increases ☐
Option 2 Electric potential energy decreases ☐
Option 3 Force between them decreases ☒
Option 4 Electric potential energy may increase or decreases ☒
Explanation
d increases interaction of charges of two plates decrease, so, force decrease.
Electric potential energy may increases or decreases depending upon which factor is
constant Q or V
If Q = constant
U =1
Q2C
2
U decreases
If U = constant
U =1
CV2C
2
U increases
Q. No. 194 Two point charges 2q and 8q are placed at distance d apart. A third charge -q is placed
at d
distance from 2q online joining the charges 2q and 8q. Then3
Option 1 Electric potential energy of the system is maximum ☒
Option 2 Electric potential energy of system is minimum ☐
Option 3 Charge -q is in unstable equilibrium ☒
Option 4 Charge -q is in stable equilibrium ☐
Explanation
2q 8q 16qU = -k. -k. +k.
x d- x d
2 2 2
(((( ))))dU d -1 4 8
= k.2q - +dx dx x d- x d
2
(((( ))))
dU 1 4= k.2q -
dx x d- x
2
2 2
For maximum
dU= 0
dx
(((( ))))
1 4 d- = 0 x =
3x d- x⇒⇒⇒⇒
2 2
To check the equilibrium status
(((( ))))
d U d dU d 1 4= = -
dx dx dxdx x d- x
2
2 2 2
(((( ))))
-2 8= +
x d- x3 3
dat x =
3
d U -2 8 -27= + = = -ve
dx d ddd-
27 3
2
2 3 3 3
i.e., charge -q is in unstable equilibrium
Q. No. 195 A particle A of mass m and charge q moves directly towards a fixed particle B which
has charge q. The speed of A is v when it is far away from B. The minimum separation
between the particles is proportional to
Option 1 q2
☒
Option 2 1
v
☐
Option 3 1
m
☒
Option 4 1
v2
☒
Explanation
For minimum separation
1 qmv = k.
2 r
22
k.q 2kqr = =
1 mvmv2
2 2
22
1 1r q , h , r
m v∝ ∝ ∝∝ ∝ ∝∝ ∝ ∝∝ ∝ ∝2
2
Q. No. 196 In uniformly charge dielectric sphere a very thin tunnel has been made along the
diameter shown in figure. A particle with charge -q having mass m is released from rest
at one end of the tunnel for the situation described, mark out the correct statements
Option 1 Charge particle will perform SHM about the centre of sphere as mean
position
☒
Option 2 2 mR
The time period of the particle is 2Qq
πεπεπεπεππππ
30
☒
Option 3 QqSpeed of the particle crossing the mean position is
4 Rmπεπεπεπε0
☒
Option 4 Particle will perform oscillations but not SHM ☐
Explanation F = qE
1 QF = q. . .x
4 Rπ∈π∈π∈π∈ 30
F = kx
Qk =
4 Rπ∈π∈π∈π∈
2
30
m 4 mRT = 2 = 2
k Qq
π∈π∈π∈π∈π ππ ππ ππ π
30
2=
T
ππππωωωω
V 2=
R T
ππππ
2 R QqV =
2 4 mR
ππππ
ππππ π∈π∈π∈π∈ 30
QqV =
4 mRπ∈π∈π∈π∈0
Q. No. 197 (((( )))) (((( ))))An electric charge q = 20 10 C is placed at a point 1,2,4 . At the point (3, 2 ,1)×××× -9
the electric
Option 1 Field will increase by a factor K if the space between the points is filled with
a dielectric of dielectric constant K
☐
Option 2 Field will be along y axis ☐
Option 3 Potential will be 49.9 volt ☒
Option 4 Field will have no y component. ☒
Explanation q = 20 10 C×××× -9
r = r - r→ →→ →→ →→ →
2 1
(((( )))) (((( ))))= 3-1 +0 + 1 - 42 2
r = 4 +9
r = 13 m
QV = k.
r
9 10 20 10 180V = =
3.613
× × ×× × ×× × ×× × ×9 9
V = 50V
dVE = -
dr
→→→→
E = 50 2 i - 3k→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
E = 100 i - 150k N/ C→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧
No. of component of field
Q. No. 198 In the circuit shown in steady state
Option 1 Charge across 4 F capacitor is 20 Cµ µµ µµ µµ µ
☒
Option 2 Charge across 4 F capacitor is 10 Cµ µµ µµ µµ µ ☐
Option 3 Potential difference across 4 F capacitor is 5 voltµµµµ ☒
Option 4 Potential difference across 4 F capacitor is 10 voltµµµµ ☐
Explanation
Q Q
20 - -10 - = 04 4
Q10 - = 0 Q = 20 C
2⇒⇒⇒⇒ µµµµ
Pot. Diff across 4 Fµµµµ
20 x20 - = = 15V
4 x
VA - VB = 20 -15
= 5V
Q. No. 199 The separation between the plates of parallel plate capacitor is made double while it
remains connected to cell
Option 1 the cell absorbs some energy ☒
Option 2 the electric field in the region between the plates becomes half ☒
Option 3 the charge on the capacitor become half ☒
Option 4 some work has to be done by an external agency on the plates ☒
Explanation
d d′′′′ >>>>
C C′′′′ <<<<
Q Q′′′′ <<<<
(((( ))))Q Q goes back to cell′′′′< →< →< →< →
QE =
A ∈∈∈∈0
E < E′′′′∴∴∴∴
1U = CV
2
2
U C∝∝∝∝
i.e. U < U′′′′
Q. No. 200 A dielectric slab is inserted between the plates of an isolated charged capacitor. Which
of the following quantities will change.
Option 1 The electric field in the capacitor ☒
Option 2 The charge on the capacitor ☐
Option 3 The potential difference between the plates ☒
Option 4 The stored energy in the capacitor ☒
Explanation
V =E d0 0
= E0 - Ep
V = E0(d - t) + E(t)
U1 Q
=2 C
2
Q = Const
C increases
U decreases
Q. No. 201 Two identical parallel plate capacitors are connected in one case in parallel and in the
other in series. In each case the plates of one capacitor are brought closer by a
distance a and the plates of the other are moved apart by distance a. Then
Option 1 Total capacitance first system increase ☒
Option 2 Total capacitance of first system decreases ☐
Option 3 Total capacitance of second system remains constant ☒
Option 4 Total capacitance of second system decreases ☐
Explanation (I)
A A
= , C =d- a d+ a
C∈ ∈∈ ∈∈ ∈∈ ∈0 0
21
1 1= A +
d- a d+ aC
∈∈∈∈
0p
2C
d= A
d - a∈ ×∈ ×∈ ×∈ ×p 0 2 2
Capacitance of system increases
(II)
A A
=d- a d+
Ca
∈ ∈∈ ∈∈ ∈∈ ∈××××0 0
s
CA
=2d
∈∈∈∈s
0
No change in capacitance of system
Q. No. 202 Two conducting spheres of unequal radii are given charges such that they have the
same charge density. If they are brought in contact
Option 1 Some heat will be produced ☒
Option 2 Charge will flow from larger to smaller sphere ☒
Option 3 Charge will flow from smaller to larger sphere ☐
Option 4 No charge will be exchanged between the spheres ☐
Explanation
Q
=4 Rππππ
σσσσ11
21
Q
=4 Rππππ
σσσσ22
22
Q R =
Q=
R
σ σσ σσ σσ σ ⇒⇒⇒⇒
2
1 11 2
2 2
Large sphere has more charger, so when they brought in contact. Charge will flow
from larger to smaller sphere.
Larger sphere has more capacitance so, charge will flow until it reaches common
potential.
(((( ))))1 C C
Loss in energy = V - V2 C +C
21 21 2
1 2
= +ve
Q. No. 203 In a parallel plate capacitor, the region between the plates is filled by a dielectric slab.
The capacitor is connected to a cell and the slab is taken out then which of the
following are wrong statements
Option 1 Some charge is returned to cell ☐
Option 2 Some charge is drawn from the cell ☒
Option 3 The potential difference across the capacitor is reduced ☒
Option 4 No work is done by external agent in taking out the slab ☒
Explanation
A
C =d
∈∈∈∈0
Capacitance will decrease so charge return back to cell.
Voltage/P.D remains constant and battery is connected.
There is attractive forces acting on slab, so there must be work done by external agent
in taking out the slab.
Passage Text Three charges are placed as shown. The magnitude of q1 is 2 micro coulomb, but value
of q2 and its sign are not known. Charge q2 is 4 micro coulomb and the net force on q3
is entirely in the negative x-direction.
Q. No. 204 As per the condition given in the problem the sign of q1 and q2 will be
Option 1 +, +
Option 2 +, -
Option 3 -, +
Option 4 -, -
Correct Answer 3
Explanation
q = 2 C, q = ?, q = 4 Cµ µµ µµ µµ µ1 2 3
Net force on q3 is in negative x-direction
q1 = -ve
q3 = +ve
q2 = +ve
Q. No. 205 The magnitude of q2 is
Option 1 27C
64µµµµ
Option 2 27C
32µµµµ
Option 3 13C
32µµµµ
Option 4 7C
64µµµµ
Correct Answer 2
Explanation Passsage - I
Fy = 0
cos F = F sin × θ× θ× θ× θBC AB
F =F tan × θ× θ× θ× θBC AB
(((( )))) (((( ))))1 q q 1 q q 4
. = .4 4 34 10 3 10
××××π∈ π∈π∈ π∈π∈ π∈π∈ π∈×××× ××××
1 3 2 32-2 -20 0
9 10 3 27q = = C
3216 10 4
× ×× ×× ×× ×µµµµ
× ×× ×× ×× ×
-4
2 -4
Q. No. 206 The magnitude of the force acting on q3 is
Option 1 25.2 N
Option 2 32.2 N
Option 3 56.2 N
Option 4 18.2 N
Correct Answer 3
Explanation F3 = q3 Ex
1 q 4 q 3
= 44 5 516 10 9 10
× × ×× × ×× × ×× × ×
π∈π∈π∈π∈ × ×× ×× ×× ×
1 2-4 -4
0
2 10 4 27 10 3
= 4 10 9 10 +5 516 10 32 9 10
× ×× ×× ×× ×× × × × ×× × × × ×× × × × ×× × × × ×
× × ×× × ×× × ×× × ×
-6 -6-6 9
-4 -9
= 56.2 N
Passage Text Four charges +q, +q, -q and -q are placed respectively at the corners A, B, C and D of a
square of side L arranged in the given order. E and F are midpoints of sides BC and CD
respectively. O is the centre of the square.
Q. No. 207 The electric field at point O is
Option 1 q
2 Lπεπεπεπε 20
Option 2 q
3 Lπεπεπεπε 20
Option 3 3 q
Lπεπεπεπε 20
Option 4 2 q
Lπεπεπεπε 20
Correct Answer 4
Explanation
E = E + E→ → →→ → →→ → →→ → →
AC A C
q q= 2k. + 2k.
L L2 2
1E = 4k.
L
→→→→
AC 2
LOA = OB = OC = OD =
2
E = E + E→ → →→ → →→ → →→ → →
BD B D
4kq=
L2
4 2kqE = E +E =
L
→→→→2 2AC BD 2
4 2kq
=4 Lπ∈π∈π∈π∈ 2
0
2qE =
L
→→→→
π∈π∈π∈π∈ 20
Q. No. 208 The electric potential at the point O is
Option 1 2 q
Lπεπεπεπε0
Option 2 3 q
Lπεπεπεπε0
Option 3 q
Lπεπεπεπε0
Option 4 Zero
Correct Answer 4
Explanation q q q qV = k. +k. -k. -k.
OA OB OC ODcentre
Vcentre = 0
Q. No. 209 Work done in carrying a charge q0 from O to F is
Option 1 2 qq
Lπεπεπεπε0
0
Option 2 qq 1-1
L 5
πεπεπεπε
0
0
Option 3 qq 1+1
L 5
πεπεπεπε
0
0
Option 4 zero
Correct Answer 2
Explanation
1 q 1 qV = 2 . - .
L4 4L 2L +4
π∈ π∈π∈ π∈π∈ π∈π∈ π∈
F 10 02 2
2
q 1
= -1L 5
π∈π∈π∈π∈ 0
WOF = q0(VF - V0)
qq 1W = -1
L 5
π∈π∈π∈π∈
0OF
0
Passage Text A particle can oscillate harmonically in electric field if the restoring torque is
proportional to its angular displacement from equilibrium position. Take A point
particle of mass m which is attached to one end of a massless rigid non-conducting rod
of length L. Another point particle of same mass is attached to other end of the rod.
Two particles are given charges q and -q. This arrangement is held in a region of
uniform electric field E such that rod makes small angle with the field direction asθθθθ
shown in figure. Then answer the following question.
Q. No. 210 The magnitude of the torque acting on the rod is
Option 1 qE sinl θθθθ
Option 2 qE cosl θθθθ
Option 3 (((( ))))qE 1 - sinl θθθθ
Option 4 (((( ))))qE 1 - cosl θθθθ
Correct Answer 1
Explanation
(((( ))))= q AN Eττττ
= qE sinτ θτ θτ θτ θℓ ℓ ℓ ℓ
(((( ))))= q E sinτ θτ θτ θτ θℓ ℓ ℓ ℓ
= q E sinτ θτ θτ θτ θ ℓ ℓ ℓ ℓ
Q. No. 211 What will be time taken by the rod to align itself parallel to electric field from extreme
position once it is free to rotate.
Option 1 mt =
qEππππ
ℓℓℓℓ
Option 2 2m
2 qE
ππππ ℓℓℓℓ
Option 3 m
2 2qE
ππππ ℓℓℓℓ
Option 4 qE
2 m
ππππ2
ℓℓℓℓ
Correct Answer 3
Explanation MLMoment of Inertia of dipole =
2
2
PE =
Iωωωω
q E =
m
2
ωωωω2
ℓℓℓℓ
ℓℓℓℓ
2qE =
mωωωω
ℓℓℓℓ
mT = 2
2qEππππ
ℓℓℓℓ
Time period of oscillation of dipole from parallel to extreme position
T mT = =
4 2 2qE
ππππ′′′′
ℓℓℓℓ
Q. No. 212 The dipole with dipole moment p is placed in a uniform electric field E such that
→ →→ →→ →→ →
p is perpendicular to E . The work done to turn the dipole through an angle of 180→ →→ →→ →→ →
0
is work done to turn the dipole through an angle of 1800 is
Option 1 Zero
Option 2 qE
Option 3 qE
2
Option 4 2 qE
Correct Answer 1
Explanation W = PE(cos 400 - cos 270
0)
W = 0
Q. No. 213 Figure shows a charge arrangement known as quadrupole. P is the point on the axis at
a distance r >>> a. Which of the following statement is wrong.
Option 1 The dipole moment of quadrupole is 2qa
Option 2 The electric potential at P varies as r-3
Option 3 For a single dipole the electric potential at a point on its axis at distance r from centre
varies as r-2
.
Option 4 For a single charge the electric potential at a distance r from it varies as r-1
.
Correct Answer 1
Explanation Net dipole moment of quadrupole is zero.
Passage Text Read the following passage and answer the following questions at the end.
Some cells walls in human body have a layer of negative charge on the inside surface
and a layer of positive charge of equal magnitude on the outside surface. Suppose that
the surface charge densities are 5 10 Cm . The cell wall is 5 10 m thick and its± × ×± × ×± × ×± × ×-4 -2 -9
material has dielectric constant of k = 5.4. A typical cell in human body is spherical and
has the volume 10-16
m-3
. Then
Q. No. 214 The electric field between the inside and outside layers is
Option 1 106 Vm
-1
Option 2 10-6
Vm-1
Option 3 107 Vm
-1
Option 4 103 Vm
-1
Correct Answer 3
Explanation 5 10E = =
8.85 10
σ ×σ ×σ ×σ ×
∈∈∈∈ ××××
-4
-120
10 Vm≈≈≈≈ 7 -1
Q. No. 215 The potential difference between inside and outside walls is
Option 1 0.5 volt
Option 2 0.05 volt
Option 3 0.025 volt
Option 4 5 mV
Correct Answer 2
Explanation V = E.d
=10 5 10× ×× ×× ×× ×7 -9
= 0.05 V
Q. No. 216 Which wall is at higher potential
Option 1 Inner
Option 2 Outer
Option 3 Both at same potential
Option 4 None of these
Correct Answer 2
Explanation Outer wall is at higher potential as inner wall is -v charged and outer wall is +v
charged.
Q. No. 217 What is the potential energy store in a cell
Option 1 1.4 10 J×××× -9
Option 2 1.4 10 J×××× -12
Option 3 1.4 10 J×××× -13
Option 4 3.2 10 J×××× -9
Correct Answer 3
Explanation 1U = k. E volume
2∈ ×∈ ×∈ ×∈ ×2
0
1U = 5.4 8.854 10 10 10
2× × × × ×× × × × ×× × × × ×× × × × ×-12 14 -16
= 1.4 10 J×××× -13
Passage Text We have an isolated conducting spherical shell of radius 10cm. Some positive charge is
given to it so that resulting electric field has maximum intensity of 1.8 10 N / C. The×××× 6
same amount of negative charge is given to another isolated conducting spherical shell
of radius 20cm. Now first shell is placed inside second so that both are concentric.
Answer the following questions.
Q. No. 218 What is electric potential at any point inside the first shell.
Option 1 18 10 V×××× 4
Option 2 9 10 V×××× 4
Option 3 4.5 10 V×××× 4
Option 4 1.8 10 V×××× 4
Correct Answer 2
Explanation R1 = 10cm, R2 = 20cm
QE =k.
R1 2
1
Q1.8 10 = 9 10
100 10× × ×× × ×× × ×× × ×
××××
6 9
-4
Q = 2 10 C×××× -6
QV = k.
R2
9 10 2 10=
20 10
× × ×× × ×× × ×× × ×
××××
9 -6
-2
= 9 10 V×××× 9
Q. No. 219 What is electric field intensity just inside the outer shell.
Option 1 4.5 10 N/ C×××× 5
Option 2 9 10 N / C×××× 5
Option 3 4.5 10 N/ C×××× 4
Option 4 6 10 N/ C×××× 4
Correct Answer 1
Explanation QE =k.
R2 2
2
9 10 2 10 18
= = 104400 10
× × ×× × ×× × ×× × ×××××
××××
9 -65
-4
= 4.5 10 N/ C×××× 5
Q. No. 220 What is electrostatic energy stored in the system
Option 1 1.0 J
Option 2 1.8 J
Option 3 0.09 J
Option 4 0.045 J
Correct Answer 3
Explanation 1U = E
2∈∈∈∈ 2
0
(((( ))))1U = 8.854 10 4.5 10
2× × × ×× × × ×× × × ×× × × ×
2-12 5
= 0.09 J
Q. No. 221 What will happen if both the spheres are connected by a conducting wire.
Option 1 Charge on both spheres will be positive
Option 2 Nothing will happen
Option 3 Some part of energy stored in system will convert into heat
Option 4 Entire amount of energy will convert into heat
Correct Answer 4
Explanation Both spheres have charge of equal magnitude and opposite polarity.
When both spheres connected using conducting wire, charge on both spheres vanish
i.e., electrostatic potential energy disappear and convert into heat.
Passage Text Three large plates A, B and C are placed parallel to each other and charges are given as
shown below
Then answer the following questions
Q. No. 222 The charge that appears on the left surface of plate B is
Option 1 -3C
Option 2 3C
Option 3 6C
Option 4 5C
Correct Answer 3
Explanation Net charge -3C + 4C + 5C = 6C
3C, 3C distributed to extreme faces on plates
Q. No. 223 The charge on the inner surface of plate C if the plate B is earthed
Option 1 -3C
Option 2 3C
Option 3 6C
Option 4 5C
Correct Answer 4
Explanation
Q. No. 224 The charge on left surface of B if B and C are both earthed
Option 1 -3C
Option 2 3C
Option 3 6C
Option 4 5C
Correct Answer 2
Explanation
Q. No. 225 Statement-1 : Charge is quantized because only integral number of electrons can be
transferred.
Statement-2 : There is no possibility of transfer of some fraction of electron.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 226 Statement-1 : If the Gaussian surface does not encloses any charge, then electric
intensity at any point on Gaussian surface must be zero.
Statement-2 : No net charge is enclosed by Gaussian surface, so net flux passing
through the surface is zero.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 227 Statement-1 : Electric field on the surface of a conductor is more at the sharp corners.
Statement-2 : Surface charge density on conductor’s surface is inversely proportional
to radius of curvature.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 228 Statement-1 : Positive charge always moves from a higher potential point to lower
potential point if left free in electric field.
Statement-2 : Electric potential is vector quantity.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 3
Explanation Statement-1 is True; Statement-2 is False.
Q. No. 229 Statement-1 : Conductors having equal positive charge and volume must have same
potential.
Statement-2 : Potential depends on the charge, volume and shape of conductor.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 230 Statement-1 : A capacitor can be given only a limited quantity of charge.
Statement-2 : Charge stored by a capacitor depends on shape size of the plates of the
capacitor and the surrounding medium.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 231 Statement-1 : A dielectric is inserted between the plates of an isolated fully charged
capacitor. The dielectric completely fills the space between the plates. The magnitude
of the electrostatic force on either metal plate decreases as it were before the
insertion of the dielectric.
Statement-2 : Due to insertion of the dialectic in an isolated parallel plate capacitor
electrostatic potential energy of the capacitor decreases.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 232 Statement-1 : If electric potential is constant in a certain region of space, the electric
field in that region must be zero.
Statement-2 : Electric field intensity is equal to negative gradient of potential.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 233 Statement-1 : A small test charge is initially at rest at a point in an electrostatic field of
an electric dipole. When released it will move along the line of force passing through
that point.
Statement-2 : Tangent at a point on a line of force gives the direction of the electric
field at that point.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 234 Statement-1 : The work done by the electric field of a nucleus in moving an electron
around it in a complete orbit is greater if the orbit is elliptical than if it were circular.
Statement-2 : Electric field is conservative.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 235 Statement-1 : Two equipotential surfaces can not cut each other.
Statement-2 : Two equipotential surfaces are parallel to each other.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 3
Explanation Statement-1 is True; Statement-2 is False.
Q. No. 236 Statement-1 : Electric field intensity at a point in space can be determined if electric
potential at the point is known.
Statement-2 : If electric potential at a distance r from a point charge is V, magnitude of
Velectric field intensity at the points is
r
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 237 Statement-1 : The electric field in a region around the charge is uniform.
Statement-2 : The equi-potential surface of the electric field of a point charge is a
sphere with charge at its centre.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 4
Explanation Statement-1 is False, Statement-2 is True.
Q. No. 238 Statement-1 : A charge stored by the plates of capacitor, electrostatics potential
energy of the system is more if a metallic ball is introduced in between plates of the
capacitor.
Statement-2 : Potential at any point on equatorial line of dipole is zero.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 239 Statement-1 : Capacitance of a capacitor having liquid dielectric may decrease if
temperature of the dielectric is increased.
Statement-2 : For liquid dielectrics having polar molecules dielectric constant
decreases with rise in temperature.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 240 Statement-1 : Electrostatics energy for a system of charge does not obey the principle
of superposition.
Statement-2 : Electrostatic potential energy density is proportional to square of
magnitude of electric field intensity.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 241 Statement-1 : Electric field is always normal to the surface of a conductor.
Statement-2 : The potential at every point on the surface of the conductor is same.
Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Option 2 Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
Option 3 Statement-1 is True; Statement-2 is False.
Option 4 Statement-1 is False, Statement-2 is True.
Correct Answer 1
Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
Q. No. 242 Consider the situation shown below. The switch S is open for a long time and then
closed. Then match the columns
No. Column A Column B Column C Id of Additional
Answer
1 Charge flown
through battery
when switch is
closed
1CE
2
2
2 Change in energy
stored in capacitor
1CE
4
2
3 Heat developed in
system
1CE
8
2
4 Work done by
battery
CE
2
Explanation (((( ))))A S;B Q;C Q;D P→ → → →→ → → →→ → → →→ → → →
(a) When switch is open
C
C =2
eq
CE
Q =2
When switch is closed
Ceq = C
Q = CE
CE
Extra charge flow when switch is closed is 2
(b) When switch open
1 Q 1 C E 2
U = =2 C 2 4 C
× ×× ×× ×× ×2 2 2
i
When switch is closed
1
U = CE2
21
1
Change in energy stored = U -U = CE4
2f i
(c) Heat developed in system = Change in energy stored
1
= CE4
2
(d) 1
Work done by battery = CE2
2
Q. No. 243 In case of an isolated parallel plate capacitor there is effect on its capacity when a
dielectric is introduced or plate separation is changed. Match Column-1 with column-2
for the statements in Column-1.
No. Column A Column B Column C Id of Additional
Answer
1 When the plates of
parallel plate
capacitor are pulled
apart keeping
charge constant
Work done by
external agent is
negative
2 When the plates of
parallel plate
capacitor are pulled
apart keeping it
potential constant
Work done by
battery is positive
3 When a dielectric
slab is gradually
inserted between
the plates of
parallel plate
capacitor and its
potential is kept
constant
Electric potential
energy of the
system decreases
4 When a dielectric
slab is gradually
inserted between
the plates of an
Work done by
external agent is
positive
isolated parallel
plate capacitor
Explanation (((( ))))A - S; B - S, R; C -P,Q;D -R,P
(a) When separation between plates increases capacitance decreases
as Q = const.
1 Q
U =2 C
2
U increases
i.e., W.D by external agent is positive.
(b) at = V const.
W.D by external agent is positive
(c) When dielectric slab is gradually inserted at V = const.
W.D is negative
As Electric field inside plates pull the slab inward
(d) W.D is negative as field pull the slab inside and net electric field decreases inside
the plates. So, Potential energy of system decreases.
Q. No. 244 Different shaped charged bodies and their corresponding electric fields are mentioned
Match the Column 1 with Column 2.
No. Column A Column B Column C Id of Additional
Answer
1 Spherical charged
conductor
At centre electric
field is zero
2 Infinite plane sheet
of charge
Electric field is
uniform
3 Uniformly charged
ring
Electric field is
discontinuous at
the surface
4 Sphere with
uniform volume
distribution of
charge
At the surface
electric field is
continuous and
maximum
Explanation (A-R, P; B-Q; C-P; D-P, S)
(a) E. field inside the spherical conductor is field and discontinuous at the surface.
(b)
Electric field is uniform
(c) In uniformly charged ring (at any point along axis)
(((( ))))Qx
E = k.
x +R
x 32 2
at x = 0 E = 0⇒⇒⇒⇒
(d) at centre r = 0 E = 0⇒⇒⇒⇒
at surface (((( ))))Q
r = R E = k. maximumR
⇒⇒⇒⇒2
Q. No. 245 A solid sphere of radius R has a charge Q distributed in its volume with charge density
= kra, where k and a are constant and r is the distance from its centre. If the electric
R 1field are r = is times that at r =R. What is the value of a.
2 8
Correct Answer 0002
Is Integer Type ☒
Explanation = krρρρρ a
EE =
8
Rr
Rr =
2
1 q 1 Q 1. = .
4 4 8r R××××
π∈ π∈π∈ π∈π∈ π∈π∈ π∈enclosed
2 20 0
1 4 q 1 Q 1. = .
4 4 8R R
××××××××
π∈ π∈π∈ π∈π∈ π∈π∈ π∈enclosed
2 20 0
Q = 32 qenclosed
q = k a 4 r dr = 4 k r drπ ππ ππ ππ π∫ ∫∫ ∫∫ ∫∫ ∫R/2 R/2
r 2 2+aenclosed
0 0
4 k R=
a+3 2
ππππ
a +3
Q= 2 2 = 32
q⇒⇒⇒⇒a +3 a +3
enclosed
a+ 3 = 5 a = 2⇒⇒⇒⇒
Q. No. 246 Two point charges 4 C and -10 C are placed 10cm apart in air. An electric slab ofµ µµ µµ µµ µ
large length and breadth but of thickness 5cm is placed between them. Calculate the
force (in newton) of attraction between the charges if the relative permittivity of the
dielectric is 9.
Correct Answer 0009
Is Integer Type ☒
Explanation
(((( ))))1 q q
F = .4 d- t + t kπ∈π∈π∈π∈
1 22
0
(((( ))))9 10 4 10 10
=
10 10 - 5 10 + 5 10 9
× × × ×× × × ×× × × ×× × × ×
× × ×× × ×× × ×× × ×
9 -12
2-2 -2 -2
(((( ))))
36 10 36 10= = = 9N
400 105 10 +15 10
× ×× ×× ×× ×
××××× ×× ×× ×× ×
-2 -2
-4-2 -2
Q. No. 247 The linear charge density on a dielectric ring of radius R is varying with angle
as = cos where is constant. What is potential at the centre O of the2
θθθθ θ λ λ λθ λ λ λθ λ λ λθ λ λ λ
0 0
ring.
Correct Answer 0000
Is Integer Type ☒
Explanation
d = Rdθ λ θθ λ θθ λ θθ λ θ
= cos2
θθθθλ λλ λλ λλ λ0
1 ddV = .
4 R
θθθθ
π∈π∈π∈π∈0
cos .Rd2dV =
4 R
θθθθλ θλ θλ θλ θ
π∈π∈π∈π∈
0
0
V = dV∫∫∫∫ = = cos d4 2
ππππ
λ θλ θλ θλ θθθθθ
π∈π∈π∈π∈
2
0
00
∫
sin2V =
14
2
ππππθθθθ
λλλλ
π∈π∈π∈π∈
2
0
0
0
V = 0
Q. No. 248 A potential difference is applied to the plates of a capacitor filled with an insulator
with stored energy as U. The capacitor is disconnected and the insulator is pulled out
now. The work done in pulling out the insulator against the electric field is 4U. What is
dielectric constant of the insulator.
Correct Answer 0005
Is Integer Type ☒
Explanation Initially
Capacitor with Insulator
1 QU =
2 kC
2
0
When Insulator pulled out
1 Q5U = U =
2 C′′′′
2
0
1 Q 1 Q5 =
2 kC 2 C× ×× ×× ×× ×
2 2
0 0
k = 5
Q. No. 249 Two concentric spherical conducting shells of radius R and 2R are carrying q and 2q
respectively. Both are now connected by a conducting wire. Find the change in electric
potential on the outer shell.
Correct Answer 0000
Is Integer Type ☒
Explanation
Electric potential at spherical shell A
QV = k.
RA
Electric Potential at spherical shell B
2qV = k.
2RB
qV = k.
RB
VA - VB = 0
Q. No. 250 The kinetic energy of a charged particle decreases by 10 joule as it moves from a point
at potential 100 volt to a joint where potential is 200 volt.. What is charge on particle
(((( ))))in 10 Cx ×××× -1
Correct Answer 0001
Is Integer Type ☒
Explanation k.E = q V∆ ∆∆ ∆∆ ∆∆ ∆
(((( ))))= q 10 Cθ ×θ ×θ ×θ × -1
(((( )))) (((( ))))10J = 9 10 200 -10× ×× ×× ×× ×-1
10 = 9 10 100× ×× ×× ×× ×-1
q = 1
q = 1 10 C×××× -1
Q. No. 251 Two identical particles each having charge of 2 10 C and mass of 10gm are kept×××× -4
at a separation of 10cm and then released. What should be speed of the particle in
(((( )))) x 10 m/ s when separation becomes large.×××× 2
Correct Answer 006
Is Integer Type ☐
Explanation
1 Q
2 mv = k.2 d
2
Qmv = k.
d
2
9 10 4 10 1v =
10 10 10
× × ×× × ×× × ×× × ×××××
××××
9 -82
-1 -3
v = 36 10××××2 4
v = 600 m/s
v = 6 10 m/ s×××× 2
Q. No. 252 A capacitor with stored energy 4 joule is connected with an identical capacitor with no
electric field in between. What is energy stored in two capacitors in joule.
Correct Answer 0002
Is Integer Type ☐
Explanation Case I
Case II
C1 = C2 = C
U = 4J
U1 + U2 = U
When uncharged capacitor connected with charged capacitor
‘U1 = U2’ as both
Capacitors are identical
U = 4J = 2U2
U2 = 2J
Energy stored in two capacitors is 2J
Q. No. 253 An electric dipole consists of charges 2.0 10 C separated by a distnace of ± ×± ×± ×± × -7
2.0 10 m. It is placed near a long line charge of linear charge density×××× -3
4.0 10 C / m as shown in figure, such that the negative charge is at a distance×××× -4
of 2.0 cm from the line charge. Find the force acting on the dipole in newton.
Correct Answer 0006
Is Integer Type ☒
Explanation
F = q(E1 - E2)
2k 2kF = q -
r r
λ λλ λλ λλ λ 1 2
1 1F = 2kq -
r r
∝∝∝∝ 1 2
F = 2 9 10 2 10 4 10× × × × × ×× × × × × ×× × × × × ×× × × × × ×9 -7 -4
F = 6N
Q. No. 254 A charge of 1 C is given to one plate of a parallel-plate capacitor of capacitanceµµµµ
0.1 C and a charge of 2 C is given to the other plate. Find the potential difference,µ µµ µµ µµ µ
in volts developed between the plates.
Correct Answer 0005
Is Integer Type ☒
Explanation
1+2
Q = =1.5 C2
′′′′ µµµµ
Q = 0.5 Cµµµµ
C = 0.1 Cµµµµ
Q 0.5 CV = =
C 0.1 F
µµµµ
µµµµ
V = 5V
Q. No. 255 A capacitor having a capacitance of 100 F is charged to potential difference of 50V. µµµµ
The charging battery is disconnected and a dielectric slab of dielectric constant 2.5 is
inserted. What charge in millicoloumb would have produced this potential difference
in absence of the dielectric slab.
Correct Answer 0002
Is Integer Type ☒