A small mercury drop is charged such that its surface charge density is 2 $\mathrm{\mu C}/{\mathrm{m}}^2$. Exactly 125 such drops are combined to form a big drop. What is the surface charge density of the big drop?

For one drop, if the radius is r and the charge is q, then
$\mathrm{\sigma }=\frac{\mathrm{q}}{4{\mathrm{\pi r}}^2}=2\mathrm{\mu C}/{\mathrm{cm}}^2$
When 125 drops are combined, the charge on it becomes 125 q and if its radius is R, then
$125\times \frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}{\mathrm{\pi R}}^3\Rightarrow \mathrm{R}=5\mathrm{r}$
Charge density of the big drop is
${\mathrm{\sigma }}_1=\frac{125\mathrm{q}}{4{\mathrm{\pi R}}^2}=\frac{125\mathrm{q}}{4\mathrm{\pi }\times 25{\mathrm{r}}^2}=5\left(\frac{\mathrm{q}}{4{\mathrm{\pi r}}^2}\right)=(5\times 2)\mathrm{\mu C}/{\mathrm{cm}}^2$