Eight charged water droplets, each with a radius of 1 mm and a charge of $10^{–10}\; C$, coalesce to form a single drop. Calculate the potential of the bigger drop.

Let R and r be the radii of the bigger drop and one droplet respectively.
Volume of 8 droplets = Volume of bigger drop
$8 \times \large\frac{4}{3} $$\pi r^3=\large\frac{4}{3} \pi$$R^3$
=>$R= (8)^{1/3}r=2r$
or $R= 2 \times 1 \times 10^{-3} m= 2 \times 10^{-3} m$
Charge on bigger drop, $Q = 8 × 10^{–10} C$
Potential of the bigger drop $=\large\frac{1}{4 \pi \in_0} \frac{Q}{R}$
=>$3.6 \times 10^{4} volt$
Hence A is the correct answer.