There are 27 drops of a conducting fluid. Each has a radius r and they are charged to a potential V₀. These are combined to form a bigger drop. Its potential will be:

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V₀

3V₀

9 V₀

29V₀

answer is C.

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Detailed Solution

$\begin{array}{l} V_{o} = \frac{1}{4 {πε}_{0}} [\frac{q}{r}]; Q = 27 q; R^{3} = 27 r^{3} \Rightarrow R = 3 r \\ V^{'} = \frac{1}{4 {πε}_{0}} [\frac{Q}{R}] = \frac{1}{4 {πε}_{0}} \frac{27 q}{3 r} = 9 V_{0} \end{array}$

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