Explanation (Step-by-Step)

Initial energy stored in capacitor:

\[ U_0 = \frac{1}{2}CV_0^2 \]

In LC oscillation, total energy remains constant.

Energy in inductor at time t:

\[ U_L = U_0 \sin^2(\omega t) \]

Where angular frequency:

\[ \omega = \frac{1}{\sqrt{LC}} \]

Given 25% energy transferred to inductor:

\[ U_L = \frac{U_0}{4} \]

\[ U_0 \sin^2(\omega t) = \frac{U_0}{4} \]

Cancel \(U_0\):

\[ \sin^2(\omega t) = \frac{1}{4} \]

\[ \sin(\omega t) = \frac{1}{2} \]

Smallest positive solution:

\[ \omega t = \frac{\pi}{6} \]

Substitute \(\omega\):

\[ \frac{t}{\sqrt{LC}} = \frac{\pi}{6} \]

\[ t = \frac{\pi\sqrt{LC}}{6} \]

Final Answer matches Option D

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