Physics · Alternating Current

JEE Main 2026 — 21 January, Evening Shift — Question 28

A capacitor C is first charged fully with potential difference of $\mathrm{V}_{0}$ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L . In t s $25 \%$ of the initial energy in the capacitor is transferred to the inductor. The value of t is $\_\_\_\_$ s.

Option A:
$\frac{\pi \sqrt{L C}}{3}$
Option B:
$\frac{\pi \sqrt{\mathrm{LC}}}{6}$
Correct
Option C:
$\frac{\pi \sqrt{\text { LC }}}{2}$
Option D:
$\pi \sqrt{\frac{L C}{2}}$

Answer: B

Step-by-step solution

$\mathrm{U}_{\mathrm{c}_{\mathrm{f}}}=75 \% \mathrm{U}_{\mathrm{c}_{\mathrm{i}}}$ $\mathrm{Q}_{\mathrm{F}}^{2}=\frac{3}{4} \mathrm{Q}_{\mathrm{I}}^{2}$ $\mathrm{Q}_{\mathrm{i}} \cos \omega \mathrm{t}=\frac{\sqrt{3}}{2} \mathrm{Q}_{\mathrm{i}} \Rightarrow \mathrm{t}=\frac{\mathrm{T}}{12}$ $\mathrm{t}=\frac{\pi}{6} \sqrt{\mathrm{LC}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Alternating Current
Topic: L-C Circuits oscillations and Quality Factor, Resonance