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### JEE Main 2021 (Online) 1st September Evening Shift

A capacitor is connected to a 20 V battery through a resistance of 10$$\Omega$$. It is found that the potential difference across the capacitor rises to 2 V in 1 $$\mu$$s. The capacitance of the capacitor is __________ $$\mu$$F. Given : $$\ln \left( {{{10} \over 9}} \right) = 0.105$$
A
9.52
B
0.95
C
0.105
D
1.85

## Explanation

Given, the peak voltage of the battery, V0 = 20V

The voltage of the battery, V = 2V

Time, t = 1 $$\mu$$s = 1 $$\times$$ 10$$-$$6s

Resistance of the capacitor, R = 10 $$\Omega$$

As we know that,

V = V0(1 $$-$$ e$$-$$t/RC)

Substituting the values in the above equation, we get

2 = 20(1 $$-$$ e$$-$$t/RC)

$$\Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)$$

$$\Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95$$ $$\mu$$F

$$\therefore$$ The capacitance of the capacitor is 0.95 $$\mu$$F.
2

### JEE Main 2021 (Online) 1st September Evening Shift

Two resistors R1 = (4 $$\pm$$ 0.8) $$\Omega$$ and R2 = (4 $$\pm$$ 0.4) $$\Omega$$ are connected in parallel. The equivalent resistance of their parallel combination will be :
A
(4 $$\pm$$ 0.4) $$\Omega$$
B
(2 $$\pm$$ 0.4) $$\Omega$$
C
(2 $$\pm$$ 0.3) $$\Omega$$
D
(4 $$\pm$$ 0.3) $$\Omega$$

## Explanation

Given,

R1 = (4 $$\pm$$ 0.8) $$\Omega$$

R2 = (4 $$\pm$$ 0.4) $$\Omega$$

Equivalent resistance when the resistors are connected in parallel is given by

$${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} \Rightarrow {1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 4}$$

$${R_{eq}} = 2\,\Omega$$

Now, $${{\Delta {R_{eq}}} \over {R_{eq}^2}} = {{\Delta {R_1}} \over {R_1^2}} + {{\Delta {R_2}} \over {R_2^2}}$$

Substituting the values in the above equation, we get

$${{\Delta {R_{eq}}} \over 4} = {{0.8} \over {16}} + {{0.4} \over {16}} \Rightarrow \Delta {R_{eq}} = 0.3\,\Omega$$

$$\therefore$$ The equivalent resistance in parallel combination is $${R_{eq}} = (2 \pm 0.3)\Omega$$.
3

### JEE Main 2021 (Online) 1st September Evening Shift

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at $$-$$10$$^\circ$$C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 k$$\Omega$$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 $$\times$$ 105 J kg$$-$$1, specific heat of ice = 2 $$\times$$ 103 J kg$$-$$1 and density of ice = 103 kg/m3
A
0.353 s
B
35.3 s
C
3.53 s
D
70.6 s

## Explanation

Given, the length of the water pipe, L = 1 m

The cross-sectional area of the water pipe, A = 1 cm2 = 10$$-$$4 m2

The temperature of the ice = $$-$$ 10$$^\circ$$C

Current passing in the conductor, I = 0.5 A

Resistance of the conductor, R = 4 k$$\Omega$$

The latent heat of fusion for ice, Lf = 3.33 $$\times$$ 105 J/kg

The density of the ice, d = 1000 kg/m3

The specific heat of the ice, cp, ice = 2 $$\times$$ 103 J/kg

Heat required to melt the ice at 10$$^\circ$$C to 0$$^\circ$$C

Q = mcp$$\Delta$$T + mLf $$\Rightarrow$$ Q = dVcp$$\Delta$$T + dVLf

= 1000 $$\times$$ 10$$-$$4 $$\times$$ 2 $$\times$$ 103 $$\times$$ (10) + 1000 $$\times$$ 10$$-$$4 $$\times$$ 3.33 $$\times$$ 105 ($$\because$$ V = A $$\times$$ L)

= 35300 J

According to the Joule's law of heating,

H = I2Rt

$$\Rightarrow$$ 35300 = (0.5)2(4000) (t)

$$\Rightarrow$$ t = 35.3 s

Thus, the minimum time required to melt the ice is 35.3 s.
4

### JEE Main 2021 (Online) 1st September Evening Shift

A cube is placed inside an electric field, $$\overrightarrow E = 150{y^2}\widehat j$$. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :

A
3.8 $$\times$$ 10$$-$$11 C
B
8.3 $$\times$$ 10$$-$$11 C
C
3.8 $$\times$$ 10$$-$$12 C
D
8.3 $$\times$$ 10$$-$$12 C

## Explanation

Given, the side of the cube, s = 0.5 m

Electric field, E = 150 y2$$\widehat j$$

The direction of electric field is as shown in the below figure,

At bottom surface, y = 0

As we know that, the expression of electric flux,

$$\phi$$ = E . A cos$$\theta$$

Here, E is the electric field passing through the cube and A is the surface area of the cube.

Substituting the values in the above equations, we get

$$\phi$$ = 150y2 . (0.5 $$\times$$ 0.5) $$\times$$ cos180$$^\circ$$

= 150(0)2 . (0.25) $$\times$$ ($$-$$1) = 0

Hence, the electric flux is zero at the bottom surface.

At the top surface, y = 0.5 m

Electric field, E = 150 y2$$\widehat j$$ = 150(0.5)2 = 37.5 N/C

Electric flux at the top surface,

$$\phi$$ = E . A cos$$\theta$$

= (37.5) . (0.5 $$\times$$ 0.5) cos0$$^\circ$$

= 9.375 N / C - m2

By using the Gauss's law, $$\phi = {{{Q_{in}}} \over {{\varepsilon _0}}}$$

Here, Qin = net charge enclosed in the cube and $${{\varepsilon _0}}$$ = permittivity of the free space.

Substituting the values in the above equation, we get

$$9.375 = {{{Q_{in}}} \over {8.85 \times {{10}^{ - 12}}}}$$

Qin = 8.3 $$\times$$ 10-11 C

The charge inside the cube is 8.3 $$\times$$ 10-11 C.

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