A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

A bulb and capacitor are in series across an AC supply. When a dielectric is placed between the capacitor plates:

The capacitance increases ($$C' = \kappa C$$ where $$\kappa > 1$$).

The capacitive reactance decreases: $$X_C = \frac{1}{\omega C}$$, so larger C means smaller $$X_C$$.

The total impedance of the series circuit decreases: $$Z = \sqrt{R^2 + X_C^2}$$ (R is the resistance of the bulb).

Since $$Z$$ decreases, the current increases: $$I = \frac{V}{Z}$$.

More current through the bulb means the glow increases.

The correct answer is Option 1: increases.