ElectrostaticsGauss'sLawandCapacitors-Exercise-3

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Q. No. 1 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 Mass of A and mass of B still remain equal Mass of A increases Mass of B decreases Mass of B increases 4 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 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

A body can be negatively charged by Giving excess of electrons to it Removing some electrons from it Giving some protons to it. Removing same neutrons from it 1 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 : 1 2

Option 1 Option 2 Option 3

3 1

Option 4

3

Correct Answer Explanation

4







F is resultant of F AC and F BC → F2

2 2 = FAC +FBC +2 FACFBC cos θ



F = F2 +F2 +2 F2 cos 1200



F =F

→  F AC  → F2

 →  =  F BC  

 →   =  F AB    

2 2 = FCB +FAB +2 FABFCB cos 600

= 3F → F1

=



F2

Q. No. 4 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

1 3

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 d x= 2 d x= 2 d x= 2 2 d x= 2 3 3

EP =

F=

1 qx . 4 π∈0 a2 + x 2

(

1 Qqx . 4 π∈0 a2 + x2

(

)

)

3/2

F = maximum, when x= a= x=

Q. No. 5 Option 1 Option 2

3/2

dF =0 dx

a 2 d 2 d 2 2

Point charges +4q, -q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a respectively, then Only -q is in stable equilibrium None of the charges are in equilibrium

Option 3 Option 4 Correct Answer Explanation

All the charges are in unstable equilibrium All the charges are in stable equilibrium 3

Net force on –q FAO = FOB 1 4q2 1 4q2 . 2 = . 4 π∈0 a 4 π∈0 a2 →



F OA = F OB -q is in unstable Equilibrium Force on charges at O and B are also same ∴ System is in equilibrium (unstable)

Q. No. 6 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 24 cm from +9c 12 cm from +9e 24 cm from +e 12 cm from +e 2

Charge q will be in equilibrium If force due to +qe and +e will same qe.q q.e k. 2 = k. x ( 16 - x )2 2

2

 3  1    =   x   16 - x  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 Option 3 Option 4 Correct Answer Explanation

2:1 1:3 3:1 3

Net force on A FA = k.

2q2

4q2

= k.

3q2

4a2 a2 FA = TAB ( Tensionin string AB ) Net force on C FC = k.

a2

+ k.

8q2 a2

+ k.

TBC = FC = k.

4q2 4a2

9q2 a2

TAB 1 = TBC 3

Q. No. 8

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Two particles, each of mass 10 g and having charge of 1 µC are in equilibrium on a Horizontal table at a distance of 50cm. The minimum coefficient of friction between the particles and the table is 0.18 0.54 0.36 0.72 3 As particles are in equilibrium, therefore electrostatic force balance frictional force. µmg = Fe µmg = k. µ=

q2

d2 9 × 109 × 1 × 10-6 × 1 × 10-6

25 × 10-2 × 10 × 10-3 × 10 µ = 0.36

Q. No. 9 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

A thin insulator rod is placed between two unlike point charges q1 and -q2. How will the forces acting on the charges change. Total force acting on q1 will increase Total force acting on charge q1 will decrease Total force acting on q1 will remain unaffected None of these 2 F F′ = , when insulator rod placed b / w two unlike point charges, total force acting on K either charges will decrease.

Q. No. 10

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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? 1 2 2 1 3

2 4 Initially

1 k 2π m When charge provided to balls ρ0 =

kℓ =

1 q2 1 q2 . ⇒ k = . 4 π∈0 ( 2l)2 4 π ∈0 4l3

FR = k1 ∆x k ( ℓ + ∆x ) k ( ℓ + ∆x ) -

1 q2 . = k1 ∆ x 4 π∈0 ( ℓ + ∆x )2 k.4ℓ 3

( 2ℓ + ∆x )2

= k1 ∆x

-2  4 ℓ 3  ∆x   k  ℓ + ∆x - 2  1+   = k 1 ∆x x   4ℓ     ∆x   k  ℓ + ∆x - ℓ  1-   = k1 ∆x ℓ    K1= 2k

k ρ = 1 = 2 ρ0 k

Q. No. 11

Option 1 Option 2

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 2 4

Option 3 Option 4 Correct Answer Explanation

2 1 2

F=

qq k . 1 22 4 π∈0 ( 20 )

q1q2 k F . = 2 4 π ∈0  d - t + t k  2   q1q2 1 . 4 π ∈0  d - t + t k  2 F′   = q1q2 1 F . 4 π∈0 ( 20 )2

F′ =

2

20   1   =  20 - 8 + 8 k  2 20 1 = 12 + 8 k 2 12 + 8 k = 20 2 8 k = 20 2 -12 20 2 -12 k= 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 ℓ. A total charge Q is distributed on the block such that maximum elongation of the spring at equilibrium is equal to x. Value of Q is

Option 1

2ℓ 4πε0k ( ℓ + x )

Option 2

2 ( ℓ + x ) 4πε0k ( ℓ + x )

Option 3

2 ( ℓ + x ) 4πε0k ( ℓ )

Option 4

(ℓ + x)

Correct Answer Explanation

2

4πε0k ( x )

For maximum elongation Charge distributed to both balls θ q1 = q2 = 2 Restoring force = Electrostatic force

kx =

1 θ2 1 . × 4 π∈0 4 ( ℓ + x )2 2

θ = 16π∈0 kx ( ℓ + x )

θ = 2 ( ℓ + x ) 4 π∈0 kx Q. No. 13

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 10cm 8cm 5cm 16cm 2

Equivalent ⇒ To equivalent dielectric separation Fair = Fdielectric qq qq k. 12 2 = k. 1 22 r k.x 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 × 109 N-m2 / c2   4 πε0 

Option 1

12 × 10 4

Option 2

24 × 10 4

Option 3

36 × 10 4

Option 4

48 × 10 4 3

Correct Answer Explanation

  1  5 × 10-9 5 × 10-9 5 × 10-9  E= + - - - - - - - 4 π∈0  1 × 10-2 2 2 × 10-2 2 4 × 10-2 2    1 5 × 10-9  1 1 1   E= × 1+ 4 π∈0 10-4  ( 2 )2 ( 4 )2 ( 8 )2 

(

) (

) (

)

   1  1 1 1 1 -  E = 195 × 10 4   1 + + + +   ( 2 )2 ( 8 )2 ( 32 )2    ( 4 )2 ( 16 )2    

    1 45 × 104 E = 45 × 104   1 4 1 -   16 

  1 1 1 +  + + 2 2  ( 4 ) ( 16 ) 

E = 48 × 104 -12 × 104 = 36 × 10 4 N / C

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 Explanation

4

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

 volt  Electric field intensity at a point at a distance 60cm from charge is 2   then  metre  charge will be

Option 1

8 × 10 -11 C

Option 2

8 × 1011 C

Option 3

4 × 1011 C

Option 4

4 × 10 -11 C 1

Correct Answer Explanation

E = 2 Vm-1 , d = 60 cm = 60 × 10-2 m q E = k. 2 d E.d2 2 × 3600 × 10-4 q= = = 800 × 10-13 = 8 × 10-11 C k 9 × 109

Q. No. 17

Electric field strengths due to a point charge of 5µC at a distance of 80cm from the charge is

Option 1

8 × 10 4 N / C

Option 2

7 × 10 4 N / C

Option 3

5 × 10 4 N / C

Option 4

4 × 10 4 N / C 2 q E = k. 2 d 9 × 109 × 5 × 10-6 9 × 5 × 103 = = 2 6400 × 10-4 80 × 10-ℓℓ

Correct Answer Explanation

(

)

= 0.007 × 107 = 7 × 104 N / C

Q. No. 18

A positive point charge 50 µC is located in the plane xy at a point with radius vector →













r 0 = 2 i + 3 j . Evaluate electric field vector E at a point whose radius vector is r = 8 i - 5 j where r0 and r are expressed in meters. Option 1

∧  ∧ 1.8 i 2.6 j  kNC-1   

Option 2

∧  ∧ 1.4 i -2.6 j  kNC-1   

Option 3

∧  ∧ 2.7 i 3.6 j  kNC-1   

Option 4

∧  ∧ 2.8 i -1.5 j  kNC-1    3

Correct Answer Explanation



E=

1 . 4 π ∈0

q → → 3 r-r0

→ →   r- r0  

 ∧ ∧  ∧ ∧ ∧ ∧ r - r 0 =8 i-5 j -2 i+3 j  =6 i-8 j    

→ →

→ → r-r0 →

E=

= 62 + 82 = 10m

9 × 109 × 50 × 106  ∧ ∧  ×6 i-8 j    ( 10 )3

∧  ∧ E =  2.7 i - 3.6 j  kNC-1  



Q. No. 19 Option 1

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 q

Option 2

b2 q

Option 3

2b2 32q

Option 4 Correct Answer Explanation

b2 Zero 4 In a regular body having equal charges at all vertices net electric field at centre is Zero.

Q. No. 20

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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. n q 4 πε 0 r2 n q 4πε πε 0 ℓ 2 1 q 4 πε 0 r2 1 q 4πε πε 0 ℓ 2 3 1 q E= . 2 4 π ∈0 r

Q. No. 21

Two very long line charges of uniform charge density λ and - λ are placed along same 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

λ 2πε 0 a

Option 2

λ 4 πε 0 a

Option 3

λ 3πε 0 a

Option 4

λ 8πε 0 a

Correct Answer Explanation

1

→ E1



λ ; Net - electric field at 0 4 π ∈0 a λ λ = 2× = 4 π ∈0 a 2π ∈0 a

= E2 =

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 3a x = only 2 2

Option 4 Correct Answer Explanation

Let at point P near change Q at a distance x net electric field is zero. Q 2Q k. 2 = k. x ( a + x )2 2

2 1  2     =   x   a + x  a+ x = 2 x

a=x

(

x=

a 2 -1

x=

(

2 -1

)

)

2 +1 a

Position of pt. P from origin. = a-

(

)

2 +1 a

=- 2a Q. No. 23

In the figure distance of the point from A, where the electric field is zero is

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

20cm 10cm 33cm None of these 3

Let at point P from A, net Electric field is zero k.

10 × 10-6 x2

= k.

20 × 10-6

( 80 - x )2 2

2 1  2     =  x   80 - x  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 Option 2 Option 3 Option 4 Correct Answer Explanation

Between -q and +2q charges Towards +2q on the line drawn Away from the line towards 2q Away for the line towards -q 4 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 × 10-7 m lies in an electric field of 6.28 × 105 V / m. The surrounding medium is air whose coefficient of viscosity is 1.6 × 10-5 N- s / m2 . If the particle moves with a horizontal uniform velocity of 0.02 m/s, the number of electrons

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

on it is 10 20 30 40 3 Particle moves with uniform velocity in an electric field when force due to electric field balance with drag force due to air. Fe = 6πηrν qE = 6 πη rν

ne = 6πηrν

( q = ne )

6πηrν e 6 × 3.14 × 1.6 × 10-5 × 5 × 107 × 0.02 n= 1.6 × 10-19 n = 30 n=

Q. No. 26 Option 1 Option 2

The displacement of a charge Q in the electric field E = e1i+ e2 j+ e3k is r = ai+bj. The work done is Q ( ae1 + be2 ) Q

( ae1 )2 + ( be2 )2

Option 3

Q ( e1 + e2 ) a2 +b2

Option 4

Q e12 + e22 ( a+b )

Correct Answer Explanation

1 →







E = e1 i + e2 j+ e3 k →





d = a i +b j V E= d → →

V = E . d = e1a + e2b W = Q.V = Q ( ae1 + be2 )

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 × 104 N / C. Then the magnitude of the charge is Option 1

5 × 10 -4 C

Option 2

5 × 10 -10 C

Option 3

-18 × 10-6 C

Option 4

-5 × 10-9 C 2 Charge particle is stationary if mg + qE = 0 qE = -mg

Correct Answer Explanation

q=

-mg −0.003 × 10-3 × 9.8 = E 6 × 10 4

q = -4.9 × 10-10 C Magnitude of q = 4.9 × 10-10 C in 5 × 10-10 C Q. No. 28 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 29 Option 1 Option 2 Option 3

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 East South West North 4 Cathode rays deflect Towards North

An electron and a proton are kept in a uniform electric field. The ratio of their acceleration will be Unity Zero mp

me Option 4

me mp

Correct Answer Explanation

3

qE m Both have same charge But mass of proton is greater than electron mp > me 1 a∝ m ae mp = ap me a=

Q. No. 30 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

A drop of 10-6 kg water carries 10-6C charge. What electric field should be applied to balance its weight (assume g = 10 m/s2) 10V m upward

10V m downward 0.1V m downward 0.1V m upward 1 qE = mg

E=

Q. No. 31

mg 10-6 × 10 = = 10 V / m ( upward) q 10-6

The acceleration of an electron in an electric field of magnitude 50V/CM, if e/m value

of the electron is 1.76 × 1011 C / kg,is Option 1

8.8 × 1014 m / sec2

Option 2

6.2 × 1013 m / sec2

Option 3

5.4 × 102 m / sec2 Zero 1 50 E = 50 V / cm = -2 = 50 × 102 V / m 10 e = 1.76 × 1011 C / kg m eE = ma eE a = = 1.76 × 1011 × 5 × 103 m

Option 4 Correct Answer Explanation

= 88 × 1013 = 8.8 × 1014 m / s2

Q. No. 32 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

An electron enters an electric field with its velocity in the direction of the electric lines of force. Then The path of the electron will be a circle The path of the electron will be a parabola The velocity of the electron will decrease The velocity of the electron will increase 3 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 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 34 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 qEy2 qE2y qEy q2Ey 3 u=0 qE a= m 2qE V2 -u2 = 2as ⇒ V2 = y m 1 K.E = mv 2 = qEy ⇒ K.E = qEy 2 If an electron has an initial velocity in a direction different from that of an uniform electric field, the path of the electron is A straight line A circle An ellipse A parabola 4

x x ν= ⇒ t= t ν eE a= m 1 2 S = at 2 1 eE x2 y= . 2 m ν2 y ∝ x2 Path of electron will be Parabola. Q. No. 35

A pendulum bob of mass 80mg and carrying a charger of 2 × 10-8 C is at rest in a horizontal uniform electric field of 20,000 V/m. The tension in the thread of the pendulum is

Option 1

2.2 × 10-4 N

Option 2

4.4 × 10-4 N

Option 3

8.8 × 10-4 N

Option 4

17.6 × 10-4 N 3

Correct Answer

Explanation

m = 80 × 10-6 kg g = 9.8 m/s2 E = 20,000 V/m T = Resultant of mg and qE when pendulum bob is at rest T=

( mg )2 + ( qE )2

T=

( 80 × 10

-6

× 10

2

-4 2

) + ( 4 × 10 )

= 64 × 10-8 +16 × 10-8

= 80 × 10-14 = 4 5 × 10-4 = 8.8 × 10 -14 N

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 -6 s

Option 2

3 × 10 -7 s

Option 3

3 × 10 -8 s

Option 4

3 × 10 -9 s 4 E = 105 N/C

Correct Answer Explanation

S = 8cm = 8 × 10-2 m u=0 eE a= m 1 S =ut + at2 2 1 eE 2 S= t 2m

1 1.6 × 10-19 × 105 2 8 × 10-2 = × t 2 9.1 × 10-31 t2 =

8 × 9.1 × 10-33

8 × 10-15 t2 = 9 × 10-18 t = 3 × 10-9 sec

Q. No. 37

A particle of specific charge (q/m) enters into uniform electric field E along the centre qE line, with velocity v = 2q . After how much time it will collide with one of the plates md (figure)

Option 1 Option 2

Not possible d 2V

Option 3

md qE

Option 4

2md qE

Correct Answer Explanation

3 v = 2q

qE md

Sx = ℓ ,S y =

d 2

qE m uy = 0 a=

Vy2 - Vy2 = 2aSy Vy = 2aSy Vy = 2 ×

qE d × m 2

qE d m Vy = uy + at Vy = at Vy =

qE t= = m qE a m md t= qE Vy

Q. No. 38

An electron moving with the speed 5 × 106 m / s is shooted parallel to the electric field of intensity 1 × 103 N / C. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest

(

for an instant mass of e = 9 × 10-31 kg. charge = 1.6 × 10-19 C Option 1 Option 2 Option 3

7m 0.7mm 7cm

)

Option 4 Correct Answer Explanation

0.7cm 3 u = 5 × 106 m / s E = 1 × 103 N / C V=0 eE a= m V2- u2 = -2as u2 = 2as S=

u2 u2m 25 × 1012 × 9.1 × 10-31 = = 2a 2eE 2 × 1.6 × 10-19 × 1 × 103

S = 71.09 × 10-3 = 7.109 × 10 -2 S = 7cm

Q. No. 39

Option 1

Option 2

Option 3

Option 4 Correct Answer Explanation

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 T time period changes to T. The ratio is equal to T0

1

qE  2   g+ m     g    3

 2  g     g + qE  m  1

 2  g     g + qE  m  None of these 3

qE m g′ = g + a

a=

qE m L T0 = 2π g

g′ = g +

T = 2π T = 2π

T = T0

ℓ g′ ℓ  qE  g+  m 

g qE g+ m 1

 2 T  g  =  T0  qE  g+ m 

Q. No. 40

Option 1

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 Q 12 3πε 0 a2

Option 2

Q 16 3πε 0 a2

Option 3

Q 8 2πε 0 a2

Option 4 Correct Answer Explanation

Q 8 2ε 0 a2 1

θ dθ = dx 6 dER = dEcos θ dθ OP = × = 2 AP 4 π∈0 ( AP )

yθ dθ

(

6 × 4 π∈0 a y

2

3 2 2 +x

)

E = ∫ dEcos θ a/2

yθ = 6 × 4 π ∈0 a

∫ (y + x ) dx

2

2 3/2

-a/ 2

x = y tan θ,dx = y sec2 θ dθ =

2

=

y sec2 θ dθ

∫ ( y + x ) ∫ y sec θ dx

1 y

E=

E=

2

2 3/2

3

2

1

∫ cos θ dθ = y2 sin θ θ

6 × 4 π∈0 y a2 + 4y 2 θ

6 × 4 π ∈0 E=

=

3 2 a 2

θ

12π 3 ∈0 a2

Q. No. 41

In the figure shown, if the linear charge density is λ , then the net electric field at O will be

Option 1 Option 2

Zero kλ R 2kλ R 2kλ R 1

Option 3 Option 4 Correct Answer Explanation

Segment 1



Ex = →

Ey = → E1

λ  ∧ - i  4 π∈0 R   λ  ∧ - j 4 π∈0 R   →



= Ex+ E y =

λ  ∧ ∧ - i- j  4 π∈0 R  

... (i)

Segment 2

λ ∧ ∧  j- i  4 π∈0 R   Segment 3 → E2

=



E3 = →

...(ii)

λ ∧ i 2π∈0 R







E = E1 +E2 + E3

=

Q. No. 42

Option 1 Option 2 Option 3 Option 4 Correct Answer

λ  ∧ ∧ λ ∧ ∧ λ ∧ i =0 - i-y+  j- i  + 4 π∈0 R   4π∈0 R   2π∈0 R

10µC charge is uniformly distributed over a thin ring radius 1m. A particle (mass = 0.9 gram, charg e -1 µC) is placed on the axis of ring. It is displaced towards centre of ring, then time period of oscillations of particle 0.6sec 0.2sec 0.3sec 0.4sec 1

Explanation

Q = 10 × 10-6 C

q= 1 × 10-6 C m = 0.9 × 10-3 kg R = 1m T = 2π

T = 2π

T=

Q. No. 43 Option 1

4 π ∈0 mR3 Qq 0.9 × 10-3 × ( 1)

9 × 109 × 10 × 10-6 × 10-6

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 Qq Qq 4 π ∈0 r 2

Option 3

Qq 2

8π ∈0 r2 Option 4

Qq 2

Correct Answer Explanation

= 2π 10-2

2π = 0.6 sec 10

8 π ∈0 r2 Option 2

3

4 π ∈0 r2 3

dθ k.dθ.q = 2 2 r k.θqdl T dθ = 2πr3 dl k.θqdl T = r 2πr3 kθ q T= 2 πr 2

2T sin

T=

Q. No. 44

Option 1 Option 2

θq 2

8 π ∈0 r 2

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 Zero

qQ 4 πε 0

2 -1 2R

Option 3

2qQ 2 πε 0 R

Option 4

qQ 4 πε 0 1

Correct Answer Explanation

2 +1 2R

Potential at point A k.2q k.2q VA = + R R 2 k.q VA = 2+ 2 R Potential at point B k.q VB = 2+ 2 R A change θ moves from B to A V -V WBA = B A = 0 [ as VA = VB ] θ

(

)

(

)

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 Option 2 Option 3

E and 0 0 and V E 2 and

V 2

Option 4 Correct Answer Explanation

E V and 2 2

1

VD = VA + VB + VC k.Q k.2Q k.Q = + a a a VD = 0 →



E A = E C =E EAC = E2 + E2 + 2E2 ×

-1 2

EAC = E EB = 2E Net Electric field at D = EB - EAC = E

Q. No. 46

Two charges of 4µC each are placed at the corners A and B of an equilateral triangle of  1 N-m2  side length 0.2 m in air. The electric potential at C is  = 9 × 109   4 πε C2  0 

Option 1

9 × 10 4 V

Option 2

18 × 10 4 V

Option 3

36 × 10 4 V

Option 4

72 × 10 4 V 3

Correct Answer Explanation

VC = VCA + VCB = k.

4 × 10-6

+k.

4 × 10-6

2 × 10-1 2 × 10-1 = 2 × 9 × 109 × 2 × 10-5 = 36 × 10 4 V

Q. No. 47

Electric charges of + 10µC + 5µC,-3µC and + 8µC are placed at the corners of a square of side 2m. The potential at the centre of the square is

Option 1 Option 2

1.8V

Option 3

1.8 × 105 V

Option 4

1.8 × 10 4 V 3

Correct Answer Explanation

1.8 × 106 V

OA = OB = OC = OD = 1m Vcentre = VOA + VOB + VOC + VOD Vcentre =

1  10 × 10-6 5 × 10-6 3 × 10-6 8 × 10-6  + +   4 π ∈0  1 1 1 1 

= 9 × 109 × 10-6 [15 + 5] = 180 × 103 = 1.8 × 105 V

Q. No. 48

10 × 10-9 are placed at each of the four corners of a square of side of 8cm. 3 The potential at the intersection of the diagonals is

Option 1

150 2 volt

Option 2

1500 2 volt

Option 3

900 2 volt 900 volt

Option 4 Option 5 Option 6 Correct Answer Explanation

Equal charg es +

2

10 × 10-9 C 3 8 d = OA = OB = OC = OD = × 10-2 m 2 1 q VCatte = 4 × . 4 π∈0 d q = qA = qB = qC = qD =

4 × 9 × 109 × =

10 × 10-9 3

8 × 10-2 2

= 1.5 2 × 103 = 1500 2 volt

Q. No. 49

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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. rn r(n - 1) ( n -1) r r ( n -1 ) n 2 q E = k. 2 r q k. 2 E r = q V k. ( n-1 ) r q V = k. ( n-1 ) r E 1 = V ( n -1 ) r

V = ( n-1) r E Q. No. 50

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Two electric charges 12µC and - 6µC are placed 20 cm apart in air. There will be a point 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 - 6µC charge is 0.10 m 0.15m 0.20m 0.25m 3

Potential due to change at A + Potential due to change at B = 0 k.12 × 10 -6 k.6 × 10 -6 =0 20 + x x 20 + x = 2x ⇒ x = 20cm x = 0.2m

Q. No. 51 Option 1 Option 2

Two unlike charges of magnitude q are separated by a distance 2d. The potential at a point midway between them is Zero 1 4πε πε0

Option 3

1 q . 4πε0 d

Option 4

1 2q . 4 πε 0 d2

Correct Answer Explanation

1

Potential at point P = Potential at Point P due to charge at A + Potential at point P due to charge at B q q V = -k. + k. = 0 a a

Q. No. 52 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Three charges 2q, -q, -q are located at the vertices of an equilateral triangle. At the centre of the triangle The field is zero but potential is non-zero The field is non-zero but potential zero Both field and potential are zero Both field and potential are non-zero. 2

Vcentre = VOA + VOB + VOC

2q q q -k. -k. = 0 a a a Vcentre = 0 q EB = EC = k. 2 a = k.

EBC = EB2 + E2C + 2EBEC cos1200 = k. E A = k.

2q a2

E centre = E A + EBC

Ecentre = k.

3q a2

( +ve )

q a2

Q. No. 53

Two points are at distances a and b(a > b) from a long string of charge per unit length λ. The potential difference between the points is proportional to

Option 1

b a

Option 2

b2

Option 3 Option 4 Correct Answer Explanation

a2 b a b In a 4

1 dq 1 λ dx . = . 4 π∈0 r 4 π∈0 x 2 + y2 1 2λ E= . 4 π∈0 r dV =

b

Vb - Va = ∫ -Er dr a b

=

∫ 4π∈ . r dr 2λ

1

a

Vb - Va =

0

λ b ln   2π∈0  a 

b Vb - Va ∝ ln    a Q. No. 54

An arc of radius r carries charge. The linear density of charge is λ and the arc π subtends and angle at the centre. What is electric potential at the centre 3

Option 1

λ 4ε 0

Option 2

λ 8ε 0

Option 3

λ 12ε 0

Option 4

λ 16ε 0

Correct Answer Explanation

3

V = Kλ - α λ α= 3 π V = Kλ 3 1 λπ V= . 4 π∈0 3

V=

Q. No. 55 Option 1

λ 12 ∈0

A hemisphere of radius R is charged uniformly with surface density of charge σ. What will be the potential at centre ? σR 2∈0

Option 2

σ 4 ∈0

Option 3

σ 2∈0

Option 4

4σR 3∈0 1

Correct Answer Explanation

Potential at Centre Q V =k R

1 2πR2 σ . 4 π∈0 R σR V= 2 ∈0

V=

Q. No. 56

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

A solid sphere of radius R is charged uniformly through out the volume. At what 1 distance from its surface is the electric potential of the potential a the centre ? 4 8R 3 R 3 5R 3 2R 3 3

Vsurface =

1 Q . 4 π∈0 R

3 Vsurface 4 3 Q Vcentre = k. 2 R Vcentre =

Let Potential at point P =

1 × Vcentre 4

Q 1 3 kQ = × × R 4 2 R 1 3 8R = ⇒r= r 8R 3 Distance of point P from surface of sphere 5 r -R = R 3 k.

Q. No. 57 Option 1

In a hollow spherical shell potential (V) changes with respect to distance (r) from centre O, as shown in graph

Option 2

Option 3

Option 4

Correct Answer Explanation

2

In case of hollow sphere Vcentre = Vsurface and outside the surface Q V = k. r 1 i.e., V ∝ r

Q. No. 58 Option 1 Option 2 Option 3

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 Vn V n-1 1

Vn 3

Option 4

2 Vn 3

Correct Answer Explanation

4 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 3 4 πR = n × πr 3 3 3 1

R3 = nr 3 ⇒ R = n 3 . r

V′ = k.

Q nq = k. 1 R n3 . r

2

V ′ = n 3 . k.

q r

2

V′ = n 3 .V

Q. No. 59 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 60 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 61 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 62 Option 1 Option 2 Option 3 Option 4

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 straight line circle a parabola An ellipse 2 Conceptual Two copper spheres of same radii one hollow and other solid are charged to the same potential then both will hold same charge solid will hold more charge hollow will hold more charge hollow can not be charged 1 Q V = k. r If both spheres (hollow and solid) having same radii and potential. Charge on both will also same. 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 The charge on each sphere will be uniformly distributed over its surface. The potential on the part of the surface R which faces S is less than the potential on the other part. Each sphere has the same potential throughout its volume but this potential on one is same to that on the other sphere. The electrostatic force on the bigger sphere is same as that on the smaller sphere. 4 Both spheres are opposite charged so they exerts electrostatic force (equal) on each other. 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 : ∧

-8 i ∧

8i ∧

-16 i ∧

8 5i

Correct Answer Explanation

1 →





V = 4x 2 V, x = 1 i + 2 k -dV E= dx -d E= 4x 2 = -8x dx

( )





E ( 1,0,2 ) = -8 i N / C

Q. No. 63

Equipotential surfaces are shown in figure. Then the electric field strength will be

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

100 Vm-1 along X-axis 100 Vm-1 along Y-axis 200 Vm-1 at an angle 1200 with X-axis 50 Vm-1 at an angle 1200 with X-axis 3 d = 10 sin 300 = 5cm ∆V ( 20 -10 ) 1000 E= = = = 200V / m d 5 × 10-2 5 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 Zero Positive Negative Optimum 1 Equipotential surface, potential at every point is same. i.e., potential difference = 0 ( ∆V = 0 )

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

W = ( ∆V ) q

W=0

Q. No. 65 Option 1 Option 2 Option 3 Option 4 Correct Answer

At a certain distance from a point charge the electric field is 500 V/m and potential is 3000V. What is this distance 6m 12m 36m 144m 1

Explanation

E = 500 V/m V = 3000 V Q k. V 3000 = r =r ⇒ r = = 6m Q E 500 k. 2 r

Q. No. 66

Electric potential at any point is V = - 5x + 3y + 15 z, then the magnitude of the electric field is

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

3 2 4 2 5 2

7 4 E=-

dV  ∂ ∂ ∂ = -  ( -5x ) + ( 3y ) + dr ∂ x ∂ y ∂ z 

(

 15 z  

)

= 5- 3- 15 →

E =

Q. No. 67

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

( 5 )2 + ( -3 )2 + (

5

)

2

=7

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 Q 4 Q 2 Q 3Q 2 2 V 4  QE = mg ⇒ Q =  πR3 ρ  g d 3  3

  Q 1  R1  V2  R  R3 600 Q∝ ⇒   × = R  × V Q 2  R2  V1   2400 2 Q1 Q = 2 ⇒ Q2 = Q2 2 3

Q. No. 68

Option 1 Option 2 Option 3

100 V / m. x2 The Potential difference between the points x = 10cm and x = 20cm will be 15V 10V 5V The intensity of electric field in a region of space is represented by E =

Option 4 Correct Answer Explanation

1V 3 E=

100

V /m x2 dV = Edr 20

20

20

∫ dV = ∫ x

100

10

2

dx ⇒

10

∫x

100 2

dx

10 20

1 V = -100    x  10  1-2  V = -100   20  V = 5V Q. No. 69

Option 1 Option 2

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 4µC placed at A along Y - axis will be 0.12N

48 × 10-3 N

Option 3

24 × 10-4 N

Option 4

96 × 10-2 N 3

Correct Answer Explanation

VA = 56V, VB = 54.8V VA - VB = 50 - 54.8 = 1.2V ∆V 1.2 E= = d 0.2 × 10 - 2 E = 600 N/C F = qE

= 4 × 10-6 × 600 F = 24 × 10-4 N Q. No. 70 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 200V -200V 400V -400V 3 E = 100 Vm-1 t = 4m V = E.d V = 400V

Q. No. 71 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation







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 100V -100V 200V -200V 4 →





E = 50 i + 75 j N / C

 ∧ ∧ d =  - i - 2 j m  



→ →

V= E.d = -50 – 150 = -200V

Q. No. 72

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 : 0 4V 8V 10V 2

E= →

-dV dx ∧





E = 2 i + 2 j+ 2k →







dr = dx i + dy j+ dz k ( 1,1 ) → → ∆V = ∫ E .dr ( 0,0,0 ) ( 1,1,1 ) = - ∫ 2dx + 2dy + 2dz ( 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 Option 2 Option 3

2.5 volt/m -2.5 volt/m 2 volt / m 5 2 - volt / m 5 1

Option 4 Correct Answer Explanation

Eat R = 5m = =-

∆V ∆R

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 Option 2 Option 3 Option 4 Correct Answer Explanation

E1 > E2 > E3 > E4 > E5 E1 = E3 = E5 and E2 < E4 E2 = E4 = E5 and E1 < E3 E1 < E2 < E3 < E4 < E5 2 In region 1, 3, 5 → V = constant. Therefore E1 = E3 = E5 = 0 In region 2 → Slope is + ve dV E2 = dr In region 4 → Slope is - ve dV E4 = dr E4 > E2

Q. No. 75

Electric potential in an electric field is given as V =

K , (K being constant), if position r









vector r = 2 i + 3 j + 6 k , then electric field will be

Option 1

∧ K  ∧ ∧ 2 i + 3 j+ 6k    243

Option 2

∧ K  ∧ ∧ 2 i + 3 j+ 6k    343

Option 3

∧ K  ∧ ∧  3 i + 2 j+ 6k  243  

Option 4

K  ∧ ∧ ∧  6 i + 2 j + 3k  343  

Correct Answer Explanation

2 dV dr k V= r → k → E= 3 r

E=-



r

 ∧ ∧ ∧ . 2 i + 3 j+ 6 k  3    2 2+ 3 2+ 6 2   ( ) ( ) ( )  → k  ∧ ∧ ∧ E=  2 i + 3 j+ 6k  343   →

E=

k

Q. No. 76

An electric field of strength 50 Vm-1 exists along the negative direction of Y-axis. If 1 µC of positive charge is shifted from a point A(1m, -1m) to B(1m, 3m), the work done by agent is

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

0 -0.2 mJ +0.2 mJ +0.8 mJ 3 W = qV W = q(VB - VA)

= 1 × 10-6 × E.d = 1 × 10-6 × 50 × 4 = 200 × 10-6 = 2 × 10-4 = 0.2 × 10-3 J Q. No. 77

The potential in an electric field has the form V = a(x2 + y2 + z2). The modulus of the electric field at a point (x, y, z) is

Option 1

2a x 2 + y 2 + z2

Option 2

2a x 2 + y 2 + z2

Option 3

a x 2 + y 2 + z2

(

Option 4

)

3/2

2a 2

x + y 2 + z2 Correct Answer Explanation

2

(

V = a x2 + y2 + z2

)

∧ ∧ ∧  dV ∧ ∂v ∧ ∂v ∧   E =- i+ j+ k  = - 2ax i +2ay j+2azk   dx ∂y ∂z   





E = 2a x2 + y2 + z2  

Q. No. 78 Option 1 Option 2 Option 3

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 RV r2 V r rV R2

Option 4

R2 V r3

Correct Answer Explanation

1 Conducting sphere

If r > R 1 Q Q V= . = k. 4 π∈0 R R

V.R k Q E = k. 2 r VR VR E = k. 2 ⇒ E = 2 kr r R=

Q. No. 79 Option 1 Option 2 Option 3 Option 4 Correct Answer

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? q at a distance of 3cm from 2q q at a distance of 5cm from 2q 2q at a distance of 7cm from q 2q at a distance of 9cm from q 1

Explanation

 2q2 8q2 16q2  UT = k  + +  q   x 9 - x If U minimum dU =0 dx

2q2d  1 4 8  + + =0 dx  x q - x 9  d d ( x )-1 + 4. ( q- x )-1 + 0 = 0 dx dx 1 4 - 2+ =0 x ( q- x ) 1 2 = ⇒ x = 3cm x q- x q at distance 3cm form 2q k.

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 ℓ cm

Option 1

1 q2 4 πε 0 ℓ

Option 2

1 2q2 4 πε 0 ℓ

Option 3

1 3q2 4 πε 0 ℓ

Option 4

1 4q2 4 πε 0 ℓ

Correct Answer Explanation

3

q2 q2 q2 U= k. +k. +k. ℓ ℓ ℓ 2 3q U = k. ℓ 1 3q2 U= . 4 π∈0 ℓ

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 q2 4 πε 0b

Option 2

-8 2 q2 πε 0b

Option 3

-4 2 q2 πε0b

Option 4

-4 q2 3πε 0b

Correct Answer Explanation

4 There are 8 vertices having charge (-q) at each corner. 1 -q. + q U= 8 × . 4 π∈0 3 b 2

U=

-4q2 3 π∈0 b

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 -q +q Zero 1

Option 2 Option 3 Option 4 Correct Answer Explanation

-

2k.Qq k.q2 U= + ℓ ℓ U=0 2kQq k.q2 =ℓ ℓ q Q=2

Q. No. 83 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of V volt, its energy will be qV mqV q  V m

q mV 1 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

q2 8 πε 0 a

Option 2

6q2 8 πε 0 a

Option 3

7q2 8πε 0 a

Option 4

9q2 8 πε 0 a

Correct Answer Explanation

3

-k.2q2 k.2q2 k.q2 + a a 2a 2 2 -k.4q k.q U= + a 2a 2 k.q  1 -7q2 U= -4 + = a  2  8π∈0 a U=

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

2ke2 r

Option 2

8 ke2 5 r

Option 3

ke2 r

Option 4

8ke2 r 2

Correct Answer

Explanation

Initially ∝ and proton are at r distance apart.

2e2 ,K.Ei = 0 r Uf = 0 (particles are for away) Acc., to conservation of energy K.Ef = Vi Ui = k.

K.EP +K.E∝ =

2ke2 r

1 2ke2 K.EP + 4 × mp v 2α = 2 r Vp2 2ke2 1 1 mp vp2 + × 4mp = 2 2 16 r Acc., to conservation of Momentum mp vp = mα v α

(

)

mp vp = 4mα v α

vα =

vp 4

2 1 1 8 ke2 8 ke2  1  2ke mp vp2 1+  = ⇒ K.Ep = ⇒ mp vp2 = 2 r 2 5 r 5 2  4

Q. No. 86

Option 1 Option 2 Option 3 Option 4 Option 5 Option 6 Correct Answer Explanation

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. If Y is fixed, both p and E are conserved. If Y is fixed, E is conserved, but not p. If both are free to move, p is conserved but not E. If both are free, E is conserved, but not p.

2 Conceptual As Y is fixed and × is moving, so p cannot be conserved. E is conserved.

Q. No. 87

Two identical particles of same mass each m are having same magnitude of charge Q. One particle is initially at rest on a frictionless horizontal plane and the other particle is projected directly towards the first particle from a very large distance with a velocity v. The distance of closest approach of the particle will be

Option 1

1 4Q 2 4 πε 0 mv 2

Option 2

1 2Q 2 4 πε 0 mv 2

Option 3

1 Q2 4 πε 0 m2 v 2

Option 4

1 4Q 2 4 πε 0 m2 v 2

Correct Answer Explanation

1 Acc. To conservation of Momentum

mv + 0 =2mv′ v′ =

v 2

Acc., to conservation of energy

2× r=

Q. No. 88 Option 1 Option 2 Option 3 Option 4 Option 5 Option 6 Correct Answer Explanation

1 mv 2 k.Q 2 = 2 4 r 2 4kQ mv 2

A bullet of mass 2g is having a charge of 2µC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10m/s ? 50 kV 5V 50 V 5 kV

1 F = qE ma = qE qE  V  a = E =  m  ℓ qV a= mℓ 2 v -u2 qV = 2ℓ mℓ

V=

mv2 -mu2 2q

2 × 10 -3 × 100 1 = × 105 -6 2 2 × 2 × 10 V = 50KV V=

Q. No. 89 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Water is an excellent solvent because its molecules are Neutral Highly polar Non-polar Anodes 2 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 Option 2

Zero p 4 πε 0r 2

Option 3

p 4 πε 0r 3

Option 4

2p 4 πε 0r 3

Correct Answer Explanation

1

AP = PB = a2 + x2 V1 ( Pot. at point P due to A ) =

-k.q a2 + x2

V2 ( Pot. at point P due to charge at B ) =

k.q a2 + x2

Vp = V1 + V2 = 0

Q. No. 91

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 Positive and inversely proportional to r Positive and inversely proportional to r2. Negative and inversely proportional to r Negative and inversely proportional to r2. 4

E=

2kP

r3 F = QE 2kPQ F= 3 r → →

dW = F .dr →

dW =

2K P Q r

3



dr -r

4kPQ 4kPQ  4kPQ  W = + 2  = - 2 - 2 r r  r r

8kPQ

W=+ W∝

Q. No. 92

r2

( -ve )

1 r2 o

The distance between H+ and Cl− ions in HCl molecule is 1.28 A . What will be the o

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

potential due to this dipole at a distance of 12 A on the axis of dipole. 0.13 V 1.3 V 13 V 130 V 1 Potential due to dipole at axis of dipole kP 9 × 109 × 1.6 × 10-19 × 1.28 × 10-10 = r2 12 × 12 × 10 -20 = 0.13 V

V=

Q. No. 93 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 900 and 1800 00 and 900 900 and 00 1800 and 00 4 U= -PE cos θ

θ = 00 ⇒ cos 00 = 1 U = -PE (minimum) θ = 1800 ⇒ cos1800 = -1 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 θ with x axis, then

Option 1

tan θ = 3

Option 2 Option 3

tan θ = 2

Option 4 Correct Answer

tan θ = 0

θ = 45 2

1 2

Explanation

Ex = Ex =

2kPcos θ r3 kPsin θ

r3 Ex cosθ = Ey sin θ 2kPcos2 θ kPsin2 θ = r3 r3

tan2 θ = 2 tan θ = 2

Q. No. 95

Option 1 Option 2

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 2p 4 πε 0 a p 4 πε 0 a3

Option 3

5p 4 πε0a3

Option 4 Correct Answer Explanation

zero 3

Ex =

2kP 3

a

,E y =

kP a3

E = E2x +Ey E = E2x +Ey E= 5

Q. No. 96

kP a3

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

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

surface the dipole reaches in time t2. Then t2 > t2 t1 < t2 t1 = t2 No definite relation exists 2 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. 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 Option 2 Option 3 Option 4 Correct Answer Explanation

Dipole moves along positive x-axis and undergoes a clockwise rotation Dipole moves along negative x-axis after undergoing a clockwise rotation Dipole moves along positive x-axis after under going an anticlockwise rotation Dipole moves along negative x-axis and undergoes an anticlockwise rotation 3 Dipole rotate due to torque is anticlockwise along -ve axis, then comes in equilibrium and moves along the +ve x-axis.

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 : Move towards the origin Move away from the origin π Rotate by 2 Rotate by π 1

Option 1 Option 2 Option 3 Option 4 Correct Answer 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 θ will be represented in fig. given below by the (clockwise torque + ve)

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Curve (1) Curve (2) Curve (3) Curve (4) 2

τ

= qE sin θ

( Torque) Torque is function of θ

Q. No. 100

→ →

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation



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. 900 00 1800 450 1 τ = PE sin θ

θ = 900 τ = PE ( max ) Q. No. 101 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

An electric dipole is placed in an electric field generated by a point charge The net electric force on the dipole must be zero The net electric force on the dipole may be zero The torque on the dipole due to the field must be zero The torque on the dipole due to the field may be zero 4 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 Option 2 Option 3 Option 4 Correct Answer Explanation



→ →

q. E and p . E Zero and minimum →

q. E and maximum →

2q. E and minimum 2 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 θ

θ = 00 U = -PE (minimum) Q. No. 103

Option 1

Option 2

Option 3

Option 4

Correct Answer Explanation

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 1

 pE  2    I  3

 pE  2    I  1

 I 2    pE  1

 p 2    IE  1

τ = pE sin θ sin θ ≈ θ ( angle is very small) τ = pE θ ...(i) τ =k θ K = PE Restoring factor ω= Inertial factor

ω=

k PE = m I

Q. No. 104 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

The electric flux from a cube of edge ℓ is φ . What will be value if edge of cube is made 2ℓ and charge enclosed havled. φ 2 2φ φ 4φ φ

φ 1 Cube of edge ℓ = φ Q ⇒φ= ∈0 Q In qenclosed = 2 qenclosed Q φ φ′ = = ⇒ φ′ = ∈0 2 ∈0 2

Q. No. 105

A uniformly charged and infinitely long line having a lines charge density λ 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 Option 2

Zero 2λR ε0

Option 3

2λ R2 - y 2 ε0

Option 4

2λ R2 + y 2 ε0

Correct Answer Explanation

3 For R > y Length inside the sphere

= 2 R2 - y 2 Charge inside the sphere = 2λ R2 - y2

φ=

Q. No. 106 Option 1

Change enclose 2λ R2 - y 2 = ∈0 ∈0

A long string with a charge of λ per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be λa ∈0

Option 2 Option 3

2 λa ∈0 6 λa2 ∈0

Option 4 Correct Answer Explanation

3 λa ∈0 4 Flux will be maximum when charge enclosed will be maximum. i.e., string is along diagonal of cube. Q enclosed = λ × 3 a

φMax =

3λa ∈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 Explanation

1

Charge enclosed = =

Q 3

φ=

Q. No. 108

Q 2π × r 2πr 3

Q 3 ∈0

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 Explanation

1

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 Option 1

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 2

Option 3

Option 4

Correct Answer Explanation

1 For hollow spherical conductor E = 0 (Inside the sphere) Q E = k. 2 (at the surface of sphere) R Q E = k. 2 (outside the sphere) R

Q. No. 110

The surface density on the copper sphere is σ . The electric field strength on the surface of the sphere is σ σ 2 σ 2 ∈0

Option 1 Option 2 Option 3 Option 4

σ ∈0

Correct Answer Explanation

4 E (at surface of sphere) = σ=

Q ⇒ Q = 4 πR 2 σ 4 πR2

1 Q . 2 4π ∈0 R

E=

σ ∈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

r12 r22

Option 2

r22 r12

Option 3

r1 r2

Option 4 Correct Answer Explanation

1:1 4 σ1 = σ 2 Q1 Q 2 = R12 R12 as E = k.

Q R2

E1 1 = E2 1

Q. No. 112

A hollow metallic sphere of radius 10cm is given a charge of 3.2 × 10-9 C. The electric intensity at a point 4cm from the centre is

Option 1

9 × 10-9 NC -1 288 NC-1 2.88 NC-1 zero 4 as r < R Inside the hollow sphere, electric field is zero

Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 113 Option 1

Option 2

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 3

Option 4

Correct Answer Explanation

2 Outside the solid sphere Q 1 E = k. 2 ⇒ E ∝ 2 r r At the surface Q E = k. 2 (maximam) R Q Inside E = k. 3 (E ∝ r) R

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 1

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

x2 1 x x x2 3

Inside the solid sphere Q E = k. 2 .x R E∝x

Q 4 × πx 3 4 3 3 πR 3 Q Q ′ = 3 .x 3 R Q′ Q E = k. 2 = k. 3 x x R Q′ =

Q. No. 115 Option 1

An insulated sphere of radius R has charge density ρ . The electric field at a distance r from the centre of the sphere (r < R) ρr 3 ∈0

Option 2

ρR 3 ∈0

Option 3

ρr ∈0

Option 4

ρR ∈0

Correct Answer Explanation

1 r C1 C3 > C1 but C2 < C1 Both C2 and C3 < C1 C1 = C2 = C3 1

C1 =

∈0 A d

∈0 A ∈0 A . d d C2 = ∈0 A 2 ( k +1 ) d 4k

 k +1  ∈0 A ⇒ C3 =    2  d 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

ε0 A  d1 d2 d3   + +   K1 K 2 K 3 

Option 2

ε0 A  d1 + d2 + d3     K1 +K 2 +K 3 

Option 3

ε 0 A ( K1K 2K 3 ) d1d2d3

Option 4

 AK AK AK  ε0  1 + 2 + 3  d2 d3   d1 1

Correct Answer Explanation

1 1 1 1 = + + Ceq k ∈0 A k ∈0 A k ∈0 A 1 2 3 d1 d2 d3 ∈0 A C eq =  d1 d2 d3  k + k + k   1 2 3 Q. No. 179

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 10 3 20 3 5 3 15 3 2 Initially

C1 =

4. ∈0 A

3 × 10 -3 Finally

4. ∈0 A k ∈0 A × 3 × 10 -3 5 × 10 -3 C2 = k   4 ∈0 A  + -3 5 × 10 -3   3 × 10 4k. ∈0 A C2 = 20 × 10 -3 + 3 × 10 -3 × k

(

C2 =

)

C1 2

2k

( 20 × 10

-3

-3

+ 3 × 10 × k

)

=

1 3 × 10-3

6 × 10-3 k = 20 × 10 -3 + 3 × 10-3 k 20 3k = 20 ⇒ k = 3 Q. No. 180

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 1 2 1 2 3 2 3

k ∈0 A d 2k ∈0 A C2 = d Pot. Difference across both capacitors are same. C1 =

(E1d = E2dV1) = V2 σ1 σ ×d = 1 ×d k ∈0 2k ∈0 σ1 1 = σ1 2

Q. No. 181 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 182

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 183 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 184 Option 1 Option 2 Option 3 Option 4

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. Some charge is drawn from the cell Some charge is returned to the cell The potential difference across the capacitor is reduced No work is done by an external agent in taking the slab out 2 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. 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. 2 4 6 8 2 ∈ A Initial Capacitance 'C1 ' = 0 = 1PF d ∈0 A New Capacitance 'C2 ' = k. = 2PF 2d C2 k = = 2⇒k=4 C1 2 A capacitor is connected across another charged capacitor. The energy in the two capacitors will : Be equal to energy in the initial capacitor Be less than that in the initial capacitor Be more than that in the initial capacitor Be more or less depending on the relative capacities of the two capacitors 2 During charging the uncharged capacitor some amount of energy stored in first capacitor dissipiated in the form of Heat. A small sphere of mass m and having charge q is suspended by a light thread Tension in thread must reduce if another charged sphere is placed vertically below it 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 Tension in thread is always equal to weight mg 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 Option 1 Option 2 Option 3 Option 4 Explanation

Q. No. 187 Option 1 Option 2 Option 3 Option 4 Explanation

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. F1 = F2 a1 = a2 F1 ≠ F2

a1 ≠ a2 qdenteron = +e , md = 2 a.m.u qα = +2e mα = 4 a.m.u F1 Fd eE 1 = = = F2 Fα 2eE 2 F1 ≠ F2 qE eE a1 = ad = d = md 2amu q E 2eE a2 = aα = α = mα 4a.m.u a1 = a2

☐ ☒ ☒ ☐

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 It is uniform in the region It is propotional to r It is proportional to r2 It decreases as one goes away from origin E∝r E = kr

☐ ☐ ☒ ☒

→→

dv = - E dr dv = -kr.dr v = -k ∫ r dr -kr 2 2 v ∝ r2 Potential decrease as one goes away from origin v=

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 Option 2 Option 3 Option 4 Explanation

☐ ☒ ☐

The electric field at origin is maximum kQ The potential at origin is R kQ The field at point (x, 0, 0) is R 2 + X2

Maximum value of electric field will be

q 6 3π∈0 R2



Electric field due to ring with uniform charge ‘Q’ at a point along its axis. 1 Qx Ex = . 2 4π ∈0 R + x 2 3/2

(

)

at origin (x = 0) E=0 Electric field is maximum at ±

E=

Q. No. 189 Option 1 Option 2 Option 3 Option 4

R R

Q 6 3π∈0 R2

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 q1 = q2 Electric field inside both of them is zero. Electrostatic potential in both the spheres at an inside point is same as on surface Charge in both is effectively concentrated at the centre for field strength at an external point.

☒ ☒ ☒ ☒

Explanation

v1 = v2 Q Q k. 1 = k. 2 ⇒ Q1 = Q 2 R R Electric field inside zero Potential constant from centre to surface.

Q. No. 190

Option 1 Option 2 Option 3 Option 4 Explanation

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 4E 2V 2E 4V For spherical conductor

☒ ☒ ☐ ☐

1 1   E ∝ 2  and  V ∝  r r    r If r′ = 2 E′ = 4E and V ′ = 2V

Q. No. 191 Option 1 Option 2 Option 3 Option 4 Explanation

With regards to Gauss’s law and electric flux which of the following statements are correct. Electric flux through closed surface is equal to total flux due to all charges enclosed within that surface. Gauss’s law is applicable only when there is symmetrical distribution of charges 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 Electric field calculated by Gauss’s law is the field due to only those charges which are enclosed inside the Gaussian surface. Q φ = enclosed ∈0 Electric field resultant of electric field due to all charges present inside as well outside the closed surface. For eg :

☒ ☐ ☒ ☐

→ Ep

Q. No. 192 Option 1 Option 2 Option 3 Option 4 Explanation







= E 1+ E 2+ E 3

In uniform electric field equipotential surfaces must Be plane surfaces Be normal to the direction of the field Be spaced that surfaces having equal difference in potential are equally spaced Have decreasing potential along field Uniform electric field

☒ ☒ ☒ ☒



dV dr Equipotential surface. Surfaces I, II and III are equispaces V1 = V2 = V3 E =-

Q. No. 193 Option 1

When the separation between two charges in increased Electric potential energy increases



Option 2 Option 3 Option 4 Explanation

Electric potential energy decreases Force between them decreases Electric potential energy may increase or decreases

☐ ☒ ☒

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 1 U = Q2 2C U decreases If U = constant 1 U = CV2 2C U increases

Q. No. 194

Option 1 Option 2 Option 3 Option 4 Explanation

Two point charges 2q and 8q are placed at distance d apart. A third charge -q is placed d at distance from 2q online joining the charges 2q and 8q. Then 3 Electric potential energy of the system is maximum Electric potential energy of system is minimum Charge -q is in unstable equilibrium Charge -q is in stable equilibrium

☒ ☐ ☒ ☐

2q2 8q2 16q2 - k. + k. x d- x d dU d  -1 4 8 = k.2q2  +  dx dx  x ( d- x ) d  1 dU 4  = k.2q2  2  2 dx  x ( d- x )  For maximum dU =0 dx 1 4 d =0 ⇒ x= 2 2 3 x ( d- x ) U = -k.

To check the equilibrium status d2U d  dU  d  1 4  = =  2 2 dx  dx  dx 2 dx    x ( d- x )  -2 8 = 3+ x ( d- x ) 3

 d at  x =   3 2 d U -2 8 -27 = 3 + = 3 = -ve 2 3 dx d d  d d-   27  3  i.e., charge -q is in unstable equilibrium

Q. No. 195

Option 1 Option 2 Option 3 Option 4

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 q2 1 v 1 m 1

☒ ☐ ☒ ☒

v2 Explanation

For minimum separation 1 2 q2 mv = k. 2 r 2 k.q 2kq2 r= = 1 2 mv2 mv 2 1 1 r ∝ q2 , h ∝ , r ∝ 2 m v

Q. No. 196

Option 1

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 Charge particle will perform SHM about the centre of sphere as mean position

Option 2 The time period of the particle is 2π

Option 3 Option 4 Explanation

☒ ☒

2πε 0mR 3 Qq

Speed of the particle crossing the mean position is Particle will perform oscillations but not SHM F = qE 1 Q F = q. . 3 .x 4π ∈0 R

Qq 4πε 0Rm

☒ ☐

F = kx

Q2

k=

4π∈0 R3

T = 2π

m 4π ∈0 mR 3 = 2π k Qq

2π T V 2π = R T 2πR Qq V= 2π 4 π ∈0 mR3

ω=

V=

Q. No. 197

Option 1 Option 2 Option 3 Option 4 Explanation

Qq 4π∈0 mR

(

)

An electric charge q = 20 × 10-9 C is placed at a point ( 1,2,4 ) . At the point (3, 2 ,1) the electric Field will increase by a factor K if the space between the points is filled with a dielectric of dielectric constant K Field will be along y axis Potential will be 49.9 volt Field will have no y component. q = 20 × 10 -9 C

→ →

r = r2 - r1 =

( 3 -1 ) 2 +0 + (1 - 4 ) 2

r = 4 +9 r = 13 m Q V = k. r 9 × 109 × 20 × 109 180 V= = 3.6 13 V = 50V → dV E =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 4µF capacitor is 20 µC Charge across 4µF capacitor is 10 µC Potential difference across 4µF capacitor is 5 volt Potential difference across 4µF capacitor is 10 volt

Option 2 Option 3 Option 4

☒ ☐ ☒ ☐

Explanation

Q Q -10 - = 0 4 4 Q 10 - = 0 ⇒ Q = 20 µC 2 Pot.Diff across 4 µF 20 x 20 - = = 15V 4 x VA - VB = 20 -15 = 5V 20 -

Q. No. 199 Option 1 Option 2 Option 3 Option 4 Explanation

The separation between the plates of parallel plate capacitor is made double while it remains connected to cell the cell absorbs some energy the electric field in the region between the plates becomes half the charge on the capacitor become half some work has to be done by an external agency on the plates

☒ ☒ ☒ ☒

d′ > d C′ < C

Q′ < Q

( Q < Q ′ ) → goes back

to cell

Q A ∈0 ∴ E′ < E 1 U = CV2 2 U∝ C i.e. U′ < U E=

Q. No. 200 Option 1 Option 2 Option 3 Option 4 Explanation

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will change. The electric field in the capacitor The charge on the capacitor The potential difference between the plates The stored energy in the capacitor

☒ ☐ ☒ ☒

V0 = E0d = E0 - Ep V = E0(d - t) + E(t) 1 Q2 Q = Const U= 2 C C increases U decreases Q. No. 201

Option 1 Option 2 Option 3 Option 4 Explanation

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 Total capacitance first system increase Total capacitance of first system decreases Total capacitance of second system remains constant Total capacitance of second system decreases (I)

☒ ☐ ☒ ☐

∈0 A ∈ A , C2 = 0 d- a d+ a 1   1 Cp =∈0 A  +  d- a d + a 

C1 =

Cp =∈0 A ×

2d 2

d - a2 Capacitance of system increases (II)

∈0 A ∈0 A × d- a d+ a ∈ A Cs = 0 2d No change in capacitance of system Cs =

Q. No. 202 Option 1 Option 2 Option 3 Option 4 Explanation

Two conducting spheres of unequal radii are given charges such that they have the same charge density. If they are brought in contact Some heat will be produced Charge will flow from larger to smaller sphere Charge will flow from smaller to larger sphere No charge will be exchanged between the spheres

☒ ☒ ☐ ☐

σ1 =

Q1 4πR12

σ2 =

Q2 4πR22

2

Q 1  R1  = Q 2  R2  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  2 Loss in energy =  1 2  ( V1 - V2 ) 2  C1 + C 2 

σ1 = σ2 ⇒

= +ve

Q. No. 203

Option 1 Option 2 Option 3 Option 4 Explanation

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 Some charge is returned to cell Some charge is drawn from the cell The potential difference across the capacitor is reduced No work is done by external agent in taking out the slab

☐ ☒ ☒ ☒

C=

∈0 A d

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 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

As per the condition given in the problem the sign of q1 and q2 will be +, + +, -, + -, 3

q1 = 2µC, q2 = ?, q3 = 4µC Net force on q3 is in negative x-direction

q1 = -ve q3 = +ve q2 = +ve

Q. No. 205 Option 1 Option 2

The magnitude of q2 is 27 µC 64 27 µC 32

Option 3 Option 4 Correct Answer Explanation

13 µC 32 7 µC 64 2 Passsage - I

Fy = 0 cos × FBC = FAB sin θ FBC = FAB × tan θ 1 q1q3 1 q2q3 . = . -2 4π ∈0 4 × 10 4π ∈0 3 × 10-2

(

q2 =

Q. No. 206 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

)

(

)

2

×

4 3

9 × 10 -4 × 3 27 = µC 16 × 10 -4 × 4 32

The magnitude of the force acting on q3 is 25.2 N 32.2 N 56.2 N 18.2 N 3 F3 = q3 E x  1  q1 4 q2 3  = 4 × × ×   -4 -4 5 9 × 10 5   4π∈0  16 × 10  2 × 10 -6 4 27 × 10 -6 3 = 4 × 10 -6 × 9 × 109  + ×  × -4 -9 5 32 × 9 × 10 5  16 × 10 = 56.2 N

Passage Text

Q. No. 207 Option 1

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. The electric field at point O is q

2πε 0L2 Option 2

q 3πε 0L2

Option 3

3q

πε 0L2

Option 4 Correct Answer Explanation

2q

πε 0L2 4







EAC = EA + EC = 2k.

q 2

L

+ 2k.



EAC = 4k.

q L2

1 L2

OA = OB = OC = OD = →



L 2



EBD = EB + ED =

4kq L2



2 E = E2AC +EBD =

= →

E =

4 2kq L2

4 2kq 4π∈0 L2 2q

π∈0 L2

Q. No. 208 Option 1

The electric potential at the point O is

Option 2

3q πε0L

Option 3

q πε0L

Option 4 Correct Answer Explanation

Zero 4

2q πε0L

Vcentre = k.

q q q q +k. -k. -k. OA OB OC OD

Vcentre = 0

Q. No. 209

Work done in carrying a charge q0 from O to F is

Option 1

2 qq0 πε0L

Option 2

qq0  1  -1   πε 0L  5 

Option 3

qq0  1  +1   πε0L  5  zero 2

Option 4 Correct Answer Explanation

      1 q 1 q VF = 2  . . 1 4π ∈ L   4π ∈0 0 2 2  L  2 2 L +     4     q  1  = -1 π∈0 L  5  WOF = q0(VF - V0) qq0  1  WOF = -1 π∈0 L  5 

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 θ with the field direction as shown in figure. Then answer the following question.

Q. No. 210 Option 1

The magnitude of the torque acting on the rod is qEl sinθ

Option 2

qEl cos θ

Option 3

qEl (1 - sin θ )

Option 4

qEl ( 1 - cos θ )

Correct Answer

1

Explanation

τ = q ( AN) E τ = qEℓ sinθ τ = ( qℓ ) E sin θ τ = q Eℓ sinθ Q. No. 211 Option 1

What will be time taken by the rod to align itself parallel to electric field from extreme position once it is free to rotate.

mℓ qE

t=π

Option 2

π 2mℓ 2 qE

Option 3

π mℓ 2 2qE

Option 4

π qE 2 mℓ 2 3

Correct Answer Explanation

Moment of Inertia of dipole =

ω=

ω=

ML2 2

PE I

qℓ E

mℓ 2 2 2qE ω= mℓ

mℓ 2qE Time period of oscillation of dipole from parallel to extreme position T π mℓ T′ = = 4 2 2qE T = 2π

Q. No. 212

→ →

Option 1 Option 2



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 Zero qE

Option 3 Option 4 Correct Answer Explanation

qE 2 2 qE 1 W = PE(cos 400 - cos 2700) 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 Option 2 Option 3

The dipole moment of quadrupole is 2qa The electric potential at P varies as r-3 For a single dipole the electric potential at a point on its axis at distance r from centre varies as r-2. For a single charge the electric potential at a distance r from it varies as r-1. 1 Net dipole moment of quadrupole is zero.

Option 4 Correct Answer Explanation

Passage Text

Q. No. 214 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 -4 Cm-2 . The cell wall is 5 × 10 -9 m thick and its 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 The electric field between the inside and outside layers is 106 Vm-1 10-6 Vm-1 107 Vm-1 103 Vm-1 3

E=

σ 5 × 10-4 = ∈0 8.85 × 10-12

≈ 107 Vm-1 Q. No. 215 Option 1 Option 2 Option 3

The potential difference between inside and outside walls is 0.5 volt 0.05 volt 0.025 volt

Option 4 Correct Answer Explanation

5 mV 2 V = E.d = 107 × 5 × 10-9 = 0.05 V

Q. No. 216 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Which wall is at higher potential Inner Outer Both at same potential None of these 2 Outer wall is at higher potential as inner wall is -v charged and outer wall is +v charged.

Q. No. 217 Option 1

What is the potential energy store in a cell

Option 2

1.4 × 10 -12 J

Option 3

1.4 × 10 -13 J

Option 4

3.2 × 10-9 J 3 1 U = k. ∈0 E2 × volume 2 1 U = × 5.4 × 8.854 × 10-12 × 1014 × 10-16 2 = 1.4 × 10-13 J

Correct Answer Explanation

Passage Text

Q. No. 218 Option 1

1.4 × 10-9 J

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 × 106 N / C. The 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. What is electric potential at any point inside the first shell.

18 × 10 4 V

Option 2

9 × 10 4 V

Option 3

4.5 × 10 4 V

Option 4

1.8 × 104 V 2 R1 = 10cm, R2 = 20cm Q E1 = k. 2 R1

Correct Answer Explanation

1.8 × 10 6 = 9 × 109 ×

Q = 2 × 10-6 C Q V = k. R2

Q 100 × 10 -4

9 × 109 × 2 × 10 -6 20 × 10 -2 = 9 × 109 V =

Q. No. 219 Option 1

What is electric field intensity just inside the outer shell.

Option 2

9 × 105 N / C

Option 3

4.5 × 104 N / C

Option 4

6 × 104 N / C 1 Q E2 = k. 2 R2

Correct Answer Explanation

4.5 × 105 N / C

9 × 10 9 × 2 × 10 -6 18 = × 105 4 400 × 10 -4 5 = 4.5 × 10 N / C =

Q. No. 220 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

What is electrostatic energy stored in the system 1.0 J 1.8 J 0.09 J 0.045 J 3 1 U = ∈0 E2 2 2 1 U = × 8.854 × 10-12 × 4.5 × 105 2 = 0.09 J

(

)

Q. No. 221 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

What will happen if both the spheres are connected by a conducting wire. Charge on both spheres will be positive Nothing will happen Some part of energy stored in system will convert into heat Entire amount of energy will convert into heat 4 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

Q. No. 222 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Then answer the following questions The charge that appears on the left surface of plate B is -3C 3C 6C 5C 3 Net charge -3C + 4C + 5C = 6C 3C, 3C distributed to extreme faces on plates

Q. No. 223 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

The charge on the inner surface of plate C if the plate B is earthed -3C 3C 6C 5C 4

Q. No. 224 Option 1 Option 2 Option 3 Option 4 Correct Answer

The charge on left surface of B if B and C are both earthed -3C 3C 6C 5C 2

Explanation

Q. No. 225

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 226

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 227

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 228

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. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 : Positive charge always moves from a higher potential point to lower potential point if left free in electric field.

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 229

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 230

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 231

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Statement-2 : Electric potential is vector quantity. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 3 Statement-1 is True; Statement-2 is False. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True.

Q. No. 232

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 233

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 234

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 235 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. Statement-1 : Two equipotential surfaces can not cut each other. Statement-2 : Two equipotential surfaces are parallel to each other. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 3 Statement-1 is True; Statement-2 is False.

Q. No. 236

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 237

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 238

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 239

Option 1 Option 2 Option 3 Option 4

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 V electric field intensity at the points is r Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 4 Statement-1 is False, Statement-2 is True. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True.

Correct Answer Explanation

1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Q. No. 240

Statement-1 : Electrostatics energy for a system of charge does not obey the principle of superposition. Statement-2 : Electrostatic potential energy density is proportional to square of magnitude of electric field intensity. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

Q. No. 241 Option 1 Option 2 Option 3 Option 4 Correct Answer Explanation

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 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. Statement-1 is True; Statement-2 is False. Statement-1 is False, Statement-2 is True. 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Q. No. 242

Consider the situation shown below. The switch S is open for a long time and then closed. Then match the columns

No.

Column A

Column B

1

Charge flown through battery when switch is closed Change in energy stored in capacitor

1 2 CE 2

2 3 4 Explanation

1 2 CE 4 Heat developed in 1 2 CE system 8 Work done by CE battery 2 ( A → S;B → Q;C → Q;D → P )

Column C

Id of Additional Answer

(a) When switch is open C Ceq = 2 CE Q= 2 When switch is closed Ceq = C Q = CE

Extra charge flow when switch is closed is

CE 2

(b) When switch open 1 Q 2 1 C 2E2 2 Ui = = × × 2 C 2 4 C When switch is closed 1 U1 = CE2 2

1 Change in energy stored = Uf -Ui = CE2 4 (c) Heat developed in system = Change in energy stored 1 = CE2 4 1 (d) Work done by battery = CE2 2 Q. No. 243

No. 1

2

3

4

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. Column A Column B Column C Id of Additional Answer When the plates of Work done by parallel plate external agent is capacitor are pulled negative apart keeping charge constant When the plates of Work done by parallel plate battery is positive capacitor are pulled apart keeping it potential constant When a dielectric Electric potential slab is gradually energy of the inserted between system decreases the plates of parallel plate capacitor and its potential is kept constant When a dielectric Work done by slab is gradually external agent is inserted between positive the plates of an

Explanation

Q. No. 244 No. 1 2 3

4

Explanation

isolated parallel plate capacitor ( A - S; B - S, R; C -P,Q;D -R,P ) (a) When separation between plates increases capacitance decreases as Q = const. 1 Q2 U= 2 C 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. Different shaped charged bodies and their corresponding electric fields are mentioned Match the Column 1 with Column 2. Column A Column B Column C Id of Additional Answer Spherical charged At centre electric conductor field is zero Infinite plane sheet Electric field is of charge uniform Uniformly charged Electric field is ring discontinuous at the surface Sphere with At the surface uniform volume electric field is distribution of continuous and charge maximum (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 Ex = k. 3 x 2 +R2

(

)

at x = 0 ⇒ E = 0 (d) at centre r = 0 ⇒ E = 0

at surface r = R ⇒ E = k.

Q R2

( maximum)

Q. No. 245

Correct Answer Is Integer Type Explanation

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 1 field are r = is times that at r = R. What is the value of a. 2 8 0002



ρ = kr a E Er = R 8 R r= 2 1 qenclosed 1 Q 1 . = . 2× 2 4 π ∈0 4 π ∈ 8 r 0 R 1 4 × qenclosed 1 Q 1 . = . 2× 2 4π ∈0 4 π ∈ 8 R 0 R Q = 32 qenclosed R/2

qenclosed =



R/2

k ar 4πr 2dr = 4πk

0 a +3

=



r 2+adr

0

4πk  R  a+ 3  2  Q = 2a +3 ⇒ 2a +3 = 32

qenclosed a+3 = 5 ⇒ a = 2

Q. No. 246

Correct Answer Is Integer Type Explanation

Two point charges 4µC and -10µC are placed 10cm apart in air. An electric slab of 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. 0009



F=

=

1 q1q2 . 4π ∈0 d- t + t k

(

)

2

9 × 109 × 4 × 10 × 10-12

(10 × 10-2 - 5 × 10-2 + 5 × 10-2 9 )

2

=

Q. No. 247

Correct Answer Is Integer Type Explanation

36 × 10-2

( 5 ×10

-2

+15 × 10 -2

)

=

36 × 10-2 = 9N 400 × 10-4

The linear charge density on a dielectric ring of radius R is varying with angle θ θ as λ = λ 0 cos   where λ 0 is constant. What is potential at the centre O of the 2 ring. 0000



dθ = λ Rdθ θ λ = λ 0cos 2 1 dθ dV = . 4 π ∈0 R

θ λ 0 cos .Rdθ 2 dV = 4 π ∈0 R 2π

λ V = ∫ dV = 0 = 4π∈0

∫ cos 2dθ θ

0 2π

 θ λ 0  sin 2  V= 4 π ∈0  1   2 0 V=0

Q. No. 248

Correct Answer Is Integer Type Explanation

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. 0005



Initially

Capacitor with Insulator 1 Q2 U= 2 kC0

When Insulator pulled out 1 Q2 5U = U′ = 2 C0

1 Q2 1 Q2 5× × = 2 kC0 2 C0 k=5 Q. No. 249

Correct Answer Is Integer Type Explanation

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. 0000



Electric potential at spherical shell A Q VA = k. R Electric Potential at spherical shell B 2q VB = k. 2R q VB = k. R 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 x × 10-1 C Correct Answer Is Integer Type Explanation

)

0001



∆k.E = q∆V

( θ = q ×10 C ) 10J = ( 9 × 10 ) × ( 200 -10 ) -1

-1

10 = 9 × 10 -1 × 100 q=1 q = 1 × 10 -1 C

Q. No. 251

Two identical particles each having charge of 2 × 10 -4 C and mass of 10gm are kept at a separation of 10cm and then released. What should be speed of the particle in

( x ×10 Correct Answer Is Integer Type Explanation

2

)

m / s when separation becomes large.

006



Q 1  2  mv 2  = k. d 2  Q mv2 = k. d 9 × 10 9 × 4 × 10 -8 1 v2 = × -1 10 10 × 10 -3 2 4 v = 36 × 10 v = 600 m/s

v = 6 × 102 m / s Q. No. 252 Correct Answer Is Integer Type Explanation

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. 0002



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 -7 C separated by a distnace of 2.0 × 10 -3 m. It is placed near a long line charge of linear charge density 4.0 × 10 -4 C / m as shown in figure, such that the negative charge is at a distance of 2.0 cm from the line charge. Find the force acting on the dipole in newton.

Correct Answer Is Integer Type Explanation

0006



F = q(E1 - E2)  2kλ 2kλ  F = q r2   r1

1 1 F = 2kq ∝  -   r1 r2  F = 2 × 9 × 109 × 2 × 10-7 × 4 × 10-4 F = 6N Q. No. 254

Correct Answer Is Integer Type Explanation

A charge of 1 µC is given to one plate of a parallel-plate capacitor of capacitance 0.1 µC and a charge of 2µC is given to the other plate. Find the potential difference, in volts developed between the plates. 0005



1+2 = 1.5 µC 2 Q = 0.5 µC C = 0.1 µC Q 0.5 µC V= = C 0.1 µF V = 5V Q′ =

Q. No. 255

Correct Answer Is Integer Type

A capacitor having a capacitance of 100 µF is charged to potential difference of 50V. 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. 0002



Explanation

Initially C 0 = 100 µF V0 = 50V Q = C0V0 = 5 × 10 3 µC When dielectric slab inserted (battery is disconnected) New pot. diff Q 5 × 103 V1 = = C 2.5 × 100 × 10 -6 V1 = 20V Q1 = C1V1 = 100 × 10-6 × 20 = 2000 × 10-6 = 2 × 10-3 C Q1 = 2mC

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