Question

$27$ small drops each having charge $q$ and radius $r$ coalesce to from big drop. How many times charge and capacitance will become?

• A. 44647
• B. 27, 3
• C. 27, 27
• D. 44623

Answer 1

Answer: (B)

27, 3

Answer 2

In coalescing into a single drop charge remains conserved. Also volume before and after coalescing remains same. Let $R$ and $r$ be the radii of bigger and each smaller drop respectively. In coalescing into a single drop, charge remains conserved. Hence, charge on bigger drop $=27 \times$ charge on smaller drop i.e., $q '=27 q$ Now, before and after coalescing, volume remains same. That is, $\frac{4}{3} \pi R^{3}=27 \times \frac{4}{3} \pi r^{3}$ $\therefore R=3 r$ Hence, capacitance of bigger drop $C'=4 \pi \varepsilon_{0} R=4 \pi \varepsilon_{0}(3 r)$ $=3\left(4 \pi \varepsilon_{0} r\right)$ $=3 C$