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**Q.33. The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure [2019]What is the value of current at t = 4 s?(1) zero(2) 3 μA(3) 2 μA(4) 1.5 μAAns.** (1)

Slope of q-t graph gives current

I = dq/dt

At t = 4 s, slope is zero

Hence, current at 4 s will be zero.

Ans.

(1) 1.2 n C

(2) 0.3 n C

(3) 2.4 n C

(4) 0.9 n C

Ans.

Q = KCV

Q

⇒ Q

(1) 1/√2

(2) √3

(3) √2

(4) 1/√2

Ans.

m = I x πr

2m = I x π(r')

**(1) **

**(2) **

**(3) **

**(4) ****Ans.** (2)

The electric flux through the square surface is:

(1) Q/6∈

(3) Q/3∈

(4) Q/2∈

Ans.

Through whole cube = Q/∈o

Through one face = Q/6∈

(1) 5.4 µF

(2) 2.4 µF

(3) 3.6 µF

(4) 4.9 µF

Ans.

(1) The new equilibrium position is at a distance 2qE/K from x = 0

(2) The new equilibrium position is at a distance qE/2K from x = 0

(3) The total energy of the system is 1/2 mω

**(4) The total energy of the system is 1/2 mω ^{2}A^{2} + ** (4)

Ans.

Energy at the extreme

= 1/2 kA

After switching on electric field

New mean position ⟹ kx

x

So entreme position also shifts by qE/k

(1) 3F/4

(2) F/2

(3) 3F/8

(4) F

Ans.

when A and C are touched charge on both will be q/2

Then when B and C are touched

(2) 90°

(3) 30°

(4) 45°

Ans.

From (i) and (ii)

(1)

(2)

(3) CE

(4)

Ans.

In steady state, flow of current through capacitor will be zero.

The minimum number of capacitors required to achieve this is [2017]

(1) 24

(2) 32

(3) 2

(4) 16

Ans.

Following arrangement will do the needful:

8 capacitors of 1μF in parallel with four such branches in series.

(1) Maximum potential energy when the torque is maximum

(2) Zero potential energy when the torque is maximum

(3) Zero potential energy when the torque is minimum

(4) Minimum potential energy when the torque is maximum

PE = – PE cos θ

τ = PE sin θ

τ

PE = 0

(2) 16 mm

(3) 28 mm

(4) 20 mm

Energy of sphere = Q

= 16 × 10

(1) 589.4 V

(2) 589.5 V

(3) 589.2 V

(4) 589.6 V

Ans.

1 DV = E.d

0.8 = Ed (max)

DV = Edcosθ = 0.8 × cos 60

= 0.4

∴ V= 589.4V

Let the respective electric fluxes through the surfaces be Φ

(2)

(3)

(4)

Ans.

Eq = same in all Φ = Same

(1) Away from the wire

(2) Towards the wire

(3) Parallel to the wire along the current

(4) Parallel to the wire opposite to the current

**(1)****(2)****(3)****(4)****Ans.** (1)

Charge in the shaded region =

Total field at P =

For field to be independent of r : Q = 2πAa^{2}**Q.52. A magnetic dipole is acted upon by two magnetic fields which are inclined to each other a tan angle of 75°. One of the fields has magnitude of 15mT. The dipole attains stable equilibrium at an angle 30° with this field. The magnitude of the other field (in mT) is close to: [2016](1) 11(2) 1060(3) 36(4) 1Ans.** (1)

B

V(z) = 30 – 5z

V(z) = 35 – 10 |z| for |z| ≥ 1m

(1) ρ

(2) ρ

(3) ρ

(4) ρ

Ans.

after integral on RHS

We must obtain r

⇒ ρ ∝ 1/r

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