Explanation

(d) Resistance of the bulb
$R = \frac{V^{2}}{P} = \frac{(100)^{2}}{50} = 200\ \Omega$
Current through bulb$(I) = \frac{V}{R}$
$= \frac{100}{200} = 0.5\ A$
In a circuit containing inductive reactance $(X_{L})$ and resistance (R), impedance (Z) of the circuit is
$Z = \sqrt{R^{2} + \omega^{2}L^{2}}$ …(i)
Here, $Z = \frac{200}{0.5} = 400\ \Omega$
Now, $X_{L}^{2} = Z^{2} - R^{2}$
$= (400)^{2} - (200)^{2}$
$(2\pi fL)^{2} = 12 \times 10^{4}$
$L = \frac{2\sqrt{3} \times 100}{2\pi \times 50}$
$= \frac{2\sqrt{3}}{\pi} = 1.1\ H$