A capacitor of unknown capacitance is connected with a battery of V volts. The charge stored in it is 200 μC. When the potential of the capacitor is reduced by 100 volts, the charge stored in it becomes 150 μC. The unknown capacitance is

0.5 μF

The correct answer is Option (1) → 0.5 μF

Charge stored on capacitor: $Q = CV$

Case 1: $Q_1 = 200 \,\mu C$, potential $= V$

$C = \frac{Q_1}{V} = \frac{200 \times 10^{-6}}{V}$

Case 2: $Q_2 = 150 \,\mu C$, potential $= V - 100$

$C = \frac{Q_2}{V - 100} = \frac{150 \times 10^{-6}}{V - 100}$

Equating both expressions for $C$:

$\frac{200 \times 10^{-6}}{V} = \frac{150 \times 10^{-6}}{V - 100}$

$\frac{200}{V} = \frac{150}{V - 100}$

$200(V - 100) = 150V$

$200V - 20000 = 150V$

$50V = 20000$

$V = 400$

Now, $C = \frac{200 \times 10^{-6}}{400}$

$C = 0.5 \times 10^{-6}$ F

$C = 0.5 \,\mu F$

Answer: The unknown capacitance is $0.5 \,\mu F$.