Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is

Let $R$ and $r$ be the radii of bigger and each smaller drop, respectively. Volume of bigger drop $=8 \times$ Volume of smaller drop $\Rightarrow \quad \frac{4}{3} \pi R^{3}=8 \times \frac{4}{3} \pi r^{3}$ or $R=2 r...(i)$ As, capacitance of a smaller spherical drop is $\begin{aligned} C &=4 \pi \varepsilon_{0} r \\ \Rightarrow \quad C & \propto r \end{aligned}$ So, $\quad \frac{C_{\text {bigger }}}{C_{\text {smaller }}}=\frac{r_{\text {bigger }}}{r_{\text {smaller }}}=\frac{R}{r}=\frac{2 r}{r}=2$[using Eq. (i)] So, capacitance of bigger drop is two times as compared to each smaller drop.