Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

If each drop has a charge ‘q’ and radius ‘r’. Then from conservation of charge, charge on the big drop is nq = 27 q (n = 27)
from conservation of volume $\frac{4}{3}{\mathrm{\pi r}}^3\mathrm{n}=\frac{4}{3}{\mathrm{\pi R}}^3$
R = n^1/3 r
Now potential of the small drop $\mathrm{V}=\frac{\mathrm{q}}{4{\mathrm{\pi ϵ}}_0\mathrm{r}}=220\mathrm{V}$
Potential of the big drop,
$\mathrm{V}=\frac{\mathrm{nq}}{4\mathrm{\pi }{\in }_0\mathrm{R}}=\frac{\mathrm{nq}}{4\mathrm{\pi }{\in }_0{\mathrm{n}}^{1/3}\mathrm{r}}={\mathrm{n}}^{2/3}\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0\mathrm{r}}$
V = (27)^2/3 x 220V
= 9 × 220 = 1980 V