c
(c) Initially potential difference across both the capacitor is same hence energy of the system is
\({U_1} = \frac{1}{2}C{V^2} + \frac{1}{2}C{V^2} = C{V^2}\)\(……(i)\)
In the second case when key \(K\) is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes \(3C\), while potential difference across \(A\) is \(V\) and potential difference across \(B\) is \(\frac{V}{3}\) hence energy of the system now is
\({U_2} = \frac{1}{2}\,(3C){V^2} + \frac{1}{2}\,(3C)\,{\left( {\frac{V}{3}} \right)^2}\)\( = \frac{{10}}{6}\,C{V^2}\)\(……(ii)\)
So, \(\frac{{{U_1}}}{{{U_2}}} = \frac{3}{5}\)