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.

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 $${R_{eq}} = (2 \pm 0.3)\Omega $$.

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, L_f = 3.33 $$\times$$ 10⁵ J/kg

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

The specific heat of the ice, c_{p, ice} = 2 $$\times$$ 10³ J/kg

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

Q = mc_p$$\Delta$$T + mL_f $$\Rightarrow$$ Q = dVc_p$$\Delta$$T + dVL_f

= 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.

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

Substituting the values in the above equation, we get

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

Q_in = 8.3 $$\times$$ 10^-11 C

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