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 s 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ____________s.

A capacitor $$C$$ charged to voltage $$V_0$$ is disconnected from the battery and connected across an inductor $$L$$. We need to find the time $$t$$ at which 25% of the initial energy in the capacitor is transferred to the inductor.

We recall the LC oscillation equations. In an LC circuit, the charge on the capacitor oscillates as:

$$ q(t) = Q_0\cos(\omega t) $$

where $$Q_0 = CV_0$$ is the initial charge, and the angular frequency of oscillation is:

$$ \omega = \frac{1}{\sqrt{LC}} $$

Next, we write the expressions for the energies. The initial energy stored in the capacitor is:

$$ U_0 = \frac{1}{2}CV_0^2 = \frac{Q_0^2}{2C} $$

At time $$t$$, the energy in the capacitor is:

$$ U_C = \frac{q^2}{2C} = \frac{Q_0^2\cos^2(\omega t)}{2C} = U_0\cos^2(\omega t) $$

By conservation of energy (no resistance, so no energy loss), the energy in the inductor is:

$$ U_L = U_0 - U_C = U_0 - U_0\cos^2(\omega t) = U_0\sin^2(\omega t) $$

Applying the condition that 25% of the initial energy is in the inductor gives:

$$ U_L = 0.25 \times U_0 $$

$$ U_0\sin^2(\omega t) = 0.25 \times U_0 $$

$$ \sin^2(\omega t) = 0.25 $$

$$ \sin(\omega t) = 0.5 $$

(Taking the first positive solution)

Finally, solving for $$t$$ yields:

$$ \omega t = \frac{\pi}{6} $$

since $$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.5$$.

$$ t = \frac{\pi}{6\omega} = \frac{\pi}{6 \times \frac{1}{\sqrt{LC}}} = \frac{\pi\sqrt{LC}}{6} $$

Therefore, the correct answer is Option (3): $$\frac{\pi\sqrt{LC}}{6}$$.