n small drops of same size are charged to V volt each. If they coalesce to form a single large drop, then its potential will be

Let R be the radius of big drop and r, radius of small drop.
$\therefore \frac{4}{3}{\mathrm{\pi R}}^3=\mathrm{n}\frac{4}{3}{\mathrm{\pi r}}^3\text{ or }\mathrm{R}=\mathrm{r}(\mathrm{n}{)}^{1/3}$
Total charge on big drop = n q
Potential of bigger drop
$\begin{array}{l}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{nq}}{\mathrm{r}(\mathrm{n}{)}^{1/3}}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{\mathrm{r}}\times {\mathrm{n}}^{2/3}={\mathrm{Vn}}^{2/3}\end{array}$