1. Circuit elements and impedance

   The circuit is a series combination of:

   • resistance $R$ (the bulb),
   • capacitance $C$.

   For a series $RC$ circuit, the capacitive reactance is $X_C = \frac{1}{\omega C}.$

   Hence the total impedance is $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}.$

2. Effect of increasing $\omega$

   As angular frequency $\omega$ increases, $X_C = \frac{1}{\omega C}$ decreases.

   Therefore, $Z = \sqrt{R^2 + X_C^2}$ also decreases.

3. Current in the circuit

   Since the source amplitude $V_0$ is fixed, the current amplitude is $I_0 = \frac{V_0}{Z}.$

   Because $Z$ decreases when $\omega$ increases, $I_0$ increases.

4. Brightness of the bulb

   The bulb is purely resistive, so its average power is $P = I_{\text{rms}}^2 R.$

   Since current increases, the power dissipated in the bulb increases. Therefore the bulb glows brighter.

5. Checking options

   • A: the bulb glows dimmer. ❌ False
   • B: the bulb glows brighter. ✅ True
   • C: total impedance of the circuit is unchanged. ❌ False
   • D: total impedance of the circuit increases. ❌ False

Therefore, the correct option is B.