216 charged droplets, each of radius r and charge q, are combined to form a big drop. The ratio of potential of the big drop to the small droplet will be:

36

The correct answer is Option (3) → 36

When $n$ small droplets combine to form a big droplet, volume is conserved:

$n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \text{ implies } R = n^{1/3} r$

Total charge on big droplet: $Q = n q$

Potential of a sphere: $V = \frac{Q}{4 \pi \epsilon_0 R}$

Potential of small droplet: $V_\text{small} = \frac{q}{4 \pi \epsilon_0 r}$

Potential of big droplet: $V_\text{big} = \frac{n q}{4 \pi \epsilon_0 n^{1/3} r} = n^{2/3} \frac{q}{4 \pi \epsilon_0 r}$

Ratio:

$\frac{V_\text{big}}{V_\text{small}} = n^{2/3}$

Given $n = 216$, $n^{2/3} = (6^3)^{2/3} = 6^2 = 36$

Final Answer: $V_\text{big} : V_\text{small} = 36 : 1$