Correct Option (b) (\pi/4) \(\sqrt{(}\) LC)
Explanation:
As initially charge is maximum
\(q=q_{0} \cos \omega t\)
\(\Rightarrow \quad i=\frac{d q}{d t}=-\omega q_{0} \sin \omega t\)
Given, energy. stored in inductor \(\left(U=\frac{1}{2} L^{2}\right)\) is
equal to energy stored in capacitor \(\left(U=\frac{q^{2}}{2 C}\right)\).
\(\therefore \quad \frac{1}{2} L i^{2}=\frac{q^{2}}{2 C}\)
\(\Rightarrow \quad \frac{1}{2} L\left(\omega q_{0} \sin \omega t\right)^{2}=\frac{\left(q_{0} \cos \omega t\right)^{2}}{2 C}\)
But
\(\omega=\frac{1}{\sqrt{L C}}\)
\(\Rightarrow \quad \tan \omega t=1\)
\(\Rightarrow\)
\(\omega t=\frac{\pi}{4}\)
\(\Rightarrow\)
\(t=\frac{\pi}{4 \omega}=\frac{\pi}{4} \sqrt{L C}\)
