Q65 / 75 · JEE Main 2026 (21 January Shift 2) · Physics
A capacitor $C$ is first charged fully with potential difference of $V_{0}$ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance $L$. In $t \mathrm{~s} \ 25 \%$ of the initial energy in the capacitor is transferred to the inductor. The value of $t$ is $\_\_\_\_$ s.
- A$\frac{\pi \sqrt{L C}}{3}$
- Correct answerB$\frac{\pi \sqrt{L C}}{6}$
- C$\frac{\pi \sqrt{L C}}{2}$
- D$\pi \sqrt{\frac{L C}{2}}$
Solution
In an LC circuit, energy oscillates sinusoidally.
Initial energy in capacitor is $E_0 = \frac{1}{2}CV_0^2$.
When 25% is transferred to the inductor, 75% remains in the capacitor.
Since energy in capacitor varies as $E_C(t) = E_0\cos^2(\omega t)$, we need $\cos^2(\omega t) = 0.75$.
This gives $\sin^2(\omega t) = 0.25$, so $\sin(\omega t) = 0.5$, yielding $\omega t = \frac{\pi}{6}$.
With $\omega = \frac{1}{\sqrt{LC}}$, we get $t = \frac{\pi\sqrt{LC}}{6}$ seconds.
Initial energy in capacitor is $E_0 = \frac{1}{2}CV_0^2$.
When 25% is transferred to the inductor, 75% remains in the capacitor.
Since energy in capacitor varies as $E_C(t) = E_0\cos^2(\omega t)$, we need $\cos^2(\omega t) = 0.75$.
This gives $\sin^2(\omega t) = 0.25$, so $\sin(\omega t) = 0.5$, yielding $\omega t = \frac{\pi}{6}$.
With $\omega = \frac{1}{\sqrt{LC}}$, we get $t = \frac{\pi\sqrt{LC}}{6}$ seconds.
Hence, the correct answer is option: 'B'.