MHTCET 11 Apr 2026 Shift 2 Paper

Concept:The brightness of the bulb depends on the current flowing through it, which is controlled by the capacitive reactance of the capacitor in the AC circuit.Explanation:For a capacitor connected to an AC source, the capacitive reactance is given by:$$X_C = \frac{1}{2\pi f C}$$where $f$ is the frequency of the AC source and $C$ is the capacitance.The bulb and capacitor are in series, so the same current flows through both.If $X_C$ increases, the impedance of the circuit increases, and the current decreases.A smaller current means the bulb glows with reduced brightness.When the capacitance $C$ is reduced, the denominator $2\pi fC$ becomes smaller, so $X_C$ increases.Hence, the current decreases and the brightness of the bulb is reduced.When the frequency $f$ is reduced, the product $2\pi fC$ also becomes smaller, so $X_C$ increases again.Thus, the current decreases and the brightness of the bulb is reduced in this case as well.Therefore, reducing capacitance and reducing frequency both cause the brightness to be reduced.Answer:Option A: is reduced, is reduced