A light bulb and an air core inductor are connected to an AC source.…

The correct answer is Option (4) → (B), (C) and (E) only

The bulb (resistance \( R \)) and the inductor are in series, so the current is

    \( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{\sqrt{R^{2} + X_L^{2}}}, \qquad X_L = 2\pi f L \)

The bulb glows brighter whenever the impedance falls, that is, when the inductive reactance is reduced.

(A) the frequency of an AC source increases — Incorrect. \( X_L = 2\pi f L \) rises, the current falls and the bulb dims.

(B) a copper rod is inserted inside the inductor — Correct. Copper is non-magnetic, and the eddy currents induced in it oppose the changing flux of the coil, so the effective inductance falls; \( X_L \) decreases and the bulb glows brighter.

(C) the frequency of an AC source decreases — Correct. \( X_L \) decreases, so the current rises.

(D) an iron rod is inserted inside the inductor — Incorrect. Iron is ferromagnetic with a large relative permeability, so \( L \) and \( X_L \) increase and the bulb dims.

(E) a capacitor of suitable capacitance is connected in series with an inductor such that \( X_L = X_C \) — Correct. This is resonance: the reactances cancel, \( Z = R \) is the least possible impedance and the current is the largest.

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.”