JEE Advanced 2021 · Paper 2, Physics question 11

The lamp. It takes 500 W at 100 V, so $I=5$ A and its resistance is $R=20\ \Omega$.

Capacitor voltage. The resistor and capacitor voltages are $90^\circ$ apart:
$$V_C=\sqrt{200^2-100^2}=100\sqrt3\text{ V},\qquad X_C=\frac{100\sqrt3}{5}=20\sqrt3\ \Omega.$$

Capacitance. With $\omega=100\pi$ and $\pi\sqrt3\approx5$,
$$C=\frac1{\omega X_C}=\frac1{100\pi\times20\sqrt3}=\frac1{2000\pi\sqrt3}\approx\frac1{10^4}\text{ F}=100\ \mu\text{F}.$$