Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.

From law of conservation of charge <br/><br/> $$ \begin{aligned} &q_{i}=\mathrm{q}_{\mathrm{f}} \Rightarrow 8 \times\left(4 \pi \mathrm{E}_{0} \mathrm{R}\right) \times 12=\left(4 \pi E R^{1}\right) \times \mathrm{V}_{f} \\\\ &\Rightarrow 96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \mathrm{R}^{1} \\\\ &\text { And, } 8 \times \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{1} 3 \\\\ &8=\left(\frac{R^{1}}{R}\right)^{3} \end{aligned} $$ <br/><br/> $$ \begin{aligned} & \mathrm{R}^{1}=2 \mathrm{R} \end{aligned} $$ <br/><br/> From (i) &amp; (ii), we get <br/><br/> So, $96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \times 2 \mathrm{R} \Rightarrow \mathrm{V}_{\mathrm{f}}=48$ Volt <br/><br/> $$ \begin{aligned} &V_{f}=\frac{1}{2} C_{f} V_{f}^{2}=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \mathrm{R}^{1}\right) \mathrm{V}_{f}^{2} \\\\ &=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \times 2 \mathrm{R}\right) \times 48^{2} \\\\ &=\left(\frac{1}{2} \times 4 \pi \varepsilon_{0} R \times 12^{2}\right) \times \frac{48^{2} \times 2}{12^{2}}=32 \,E \end{aligned} $$