'n' identical small drops are charged to V volt each. They coalesce to form a bigger drop. The potential of the bigger drop will be

$n^{\frac{2}{3}} V$

The correct answer is Option (3) → $n^{\frac{2}{3}} V$

Let the radius of each small drop be $r$, charge $q$, and potential $V$. For a sphere,

$V = \frac{q}{4 \pi \epsilon_0 r} \Rightarrow q = 4 \pi \epsilon_0 r V$

Total charge after $n'$ drops coalesce:

$Q_\text{total} = n' q = n' (4 \pi \epsilon_0 r V) = 4 \pi \epsilon_0 n' r V$

Volume conservation: volume of big drop = sum of volumes of small drops:

$\frac{4}{3} \pi R^3 = n' \frac{4}{3} \pi r^3 \Rightarrow R^3 = n' r^3 \Rightarrow R = r n'^{1/3}$

Potential of big drop:

$V' = \frac{Q_\text{total}}{4 \pi \epsilon_0 R} = \frac{4 \pi \epsilon_0 n' r V}{4 \pi \epsilon_0 (r n'^{1/3})} = n'^{2/3} V$