$A$ light bulb and an open coil inductor are connected to an $AC$ source through a key as shown in the figure. The switch is closed and after some time,an iron rod is inserted into the interior of the inductor. Does the glow of the light bulb $(a)$ increase,$(b)$ decrease,or $(c)$ remain unchanged? Give your answer with reasons.

(B) When the iron rod is inserted into the inductor,the magnetic permeability of the core increases,which in turn increases the self-inductance $(L)$ of the coil.
The inductive reactance of the coil is given by $X_L = \omega L$. As $L$ increases,the inductive reactance $X_L$ also increases.
The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb. As $X_L$ increases,the total impedance $Z$ of the circuit increases.
Since the source voltage $V$ is constant,the current in the circuit $I = V/Z$ decreases.
The glow of the light bulb depends on the power dissipated,$P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases,and therefore,the glow of the light bulb decreases.