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There are 27 drops of a conducting fluid. Each drop has radius $r$ , and each of them is charged to the same potential $V_{1}$ . They are then combined to form a bigger drop. The potential of the bigger drop is $V_{2}$ . The ratio $V_{1} / V_{2}$ , is
(Ignore the change in density of the fluid on combining the drops.)

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Detailed Solution

Let $q$ be the charge on each drop, then $V_{1} = k q / r$ . Let $R$ be radius of the bigger drop. Then

$\begin{aligned} \frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3} or R = 3 r \\ V_{2} = \frac{k (27 q)}{R} = \frac{27 k q}{3 r} = 9 V_{1} or \frac{V_{2}}{V_{1}} = 9 \end{aligned}$

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