Answer
5.66 V
Work Step by Step
Energy of empty capacitor= Energy of capacitor filled with the dielectric
$\implies \frac{1}{2}C_{0}(V_{0})^{2}=\frac{1}{2}CV^{2}$
But $C=\kappa C_{0}$ which gives
$\frac{1}{2}C_{0}(V_{0})^{2}=\frac{1}{2}\times\kappa\times C_{0}\times V^{2}$
Dividing both sides by $\frac{C_{0}}{2}$, we get
$V_{0}^{2}=\kappa\times V^{2}$
$\implies V=\sqrt {\frac{V_{0}^{2}}{\kappa}}=\frac{V_{0}}{\sqrt {\kappa}}=\frac{12.0\,V}{\sqrt {4.50}}=5.66\,V$
