A capacitor of capacitance \(\frac{50}{\pi}\ \mu\mathrm{F}\) is connected to an ac source. Its capacitive reactance is 100 Ω. The frequency of the source is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 100 Hz
The capacitive reactance is \( X_C=1/\omega C \), with \( \omega=2\pi\nu \). So the frequency of the source is \( \nu=1/(2\pi X_CC) \).
Given:
\( X_C=100\ \Omega \)
\( C=\frac{50}{\pi}\times10^{-6}\ \mathrm{F} \)
\( \begin{aligned} \nu &= \frac{1}{2\pi X_CC} \\ &= \frac{1}{2\pi\times100\times\frac{50}{\pi}\times10^{-6}} \\ &= \frac{1}{10^{-2}} \\ &= 100\ \mathrm{Hz} \end{aligned} \)
NCERT (2026-27):
Example 7.4 – “Solution The capacitive reactance is”
\( \displaystyle X_C=\frac{1}{2\pi\nu C} \)\( \displaystyle =\frac{1}{2\pi(50\ \mathrm{Hz})(15.0\times10^{-6}\ \mathrm{F})}=212\ \Omega \)