A capacitor has a reactance of 100 Ω at 50 Hz. Its reactance at 125 Hz would be

40 Ω

The correct answer is Option (3) → 40 Ω

Given:

• Initial reactance, $X_{C1} = 100 \ \Omega$ at $f_1 = 50 \ \text{Hz}$
• New frequency, $f_2 = 125 \ \text{Hz}$

Formula for capacitive reactance:

$X_C = \frac{1}{2 \pi f C}$

So, reactance is inversely proportional to frequency:

$\frac{X_{C2}}{X_{C1}} = \frac{f_1}{f_2}$

$X_{C2} = X_{C1} \cdot \frac{f_1}{f_2} = 100 \cdot \frac{50}{125} = 100 \cdot \frac{2}{5} = 40 \ \Omega$