Q. No. 1 One metallic sphere A is given positive charge where ...

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××××


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

×××× ××××


-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


∧ ∧∧ ∧∧ ∧∧ ∧

-1

Option 3 2.7 i - 3.6 j kNC

∧ ∧∧ ∧∧ ∧∧ ∧

-1


∧ ∧∧ ∧∧ ∧∧ ∧

-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


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


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


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


Option 3 Q


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


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

Qq

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

× × × ×× × × ×× × × ×× × × ×


-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


1 2E = .

4 r0

λλλλ


V - V = -E dr

b

b a r

a

∫∫∫∫

1 2= . dr

4 r∫b

0a

λλλλ


bV - V = ln

2 ab a

0

λλλλ


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


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


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


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

∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +


343

∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧ + ++ ++ ++ +


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


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


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


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


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


Option 3 p


Option 4 2p


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


∈∈∈∈

2

0


∈∈∈∈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


2

p 2 20 0

As Ep is independent of r

1 Q 2 R. - . = 0

4 4r r

παπαπαπα


2

2 20 0

1 Q 2 R. = .

4 4r r

παπαπαπα


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


Option 3 k

R


Option 4 k

R


Correct Answer 2

Explanation

Electric field at pt. A on X2Y2 due to X1Y1

2E =

4 R

λλλλ


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××××


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




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 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





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


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 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





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


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




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





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


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


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


Option 1 C

Option 2 5C

Option 3 2C

Option 4 3C

Correct Answer 4

Explanation

Ceq = C + C + C

Ceq = 3C


Option 1 6C

Option 2 5C

Option 3 3C

Option 4 2C

Correct Answer 4

Explanation

Ceq (across X and Y) = 2C


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


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


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


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


F 10 02 2

2

q 1

= -1L 5


WOF = q0(VF - V0)

qq 1W = -1

L 5


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


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.


Statement-1.


Statement-1.



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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



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.


Statement-1.


Statement-1.



Correct Answer 4


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



Correct Answer 4


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



Correct Answer 4


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.


Statement-1.


Statement-1.



Correct Answer 4


Q. No. 235 Statement-1 : Two equipotential surfaces can not cut each other.

Statement-2 : Two equipotential surfaces are parallel to each other.


Statement-1.


Statement-1.



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


Statement-1.


Statement-1.



Correct Answer 4


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.


Statement-1.


Statement-1.



Correct Answer 4


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


Statement-1.


Statement-1.



Correct Answer 1


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.


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.


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


cos .Rd2dV =

4 R

θθθθλ θλ θλ θλ θ


0

0

V = dV∫∫∫∫ = = cos d4 2

ππππ

λ θλ θλ θλ θθθθθ


2

0

00

∫

sin2V =

14

2

ππππθθθθ

λλλλ


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 ☒


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 ☒

Explanation

Initially

C = 100 Fµµµµ0

V0 = 50V

Q = C0V0

= 5 10 C× µ× µ× µ× µ3

When dielectric slab inserted (battery is disconnected)

New pot. diff

Q 5 10V = =

C 2.5 100 10

××××

× ×× ×× ×× ×

3

1 -6

V1 = 20V

Q1 = C1V1

=100 10 20× ×× ×× ×× ×-6

= 2000 10×××× -6

= 2 10 C×××× -3

Q1 = 2mC

Q. No. 1 One metallic sphere A is given positive charge where ...

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