Pure R, L and C Circuits question from MHT CET 2026 (17 Apr, Shift 2)

Answer: (a)

1. $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$, so $\frac{X_C}{X_L} = \frac{1}{\omega^2 LC} \propto \frac{1}{\omega^2}$.
2. Initially $\frac{X_C}{X_L} = 1$. Doubling $\omega$ gives $\frac{X_C'}{X_L'} = \frac{1}{2^2} = \frac{1}{4}$.
3. So the answer is (a) $\frac{1}{4}$.