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27 drops of mercury of equal radius and possessing equal charge are allowed to combine to form a big drop. The ratio of potential of the big drop to that of a small drop is x. Find the value of x.

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answer is 9.

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Detailed Solution

$27 r^{3} = R^{3} \Rightarrow R = 3 r a n d Q = 27 q$

$\frac{V_{b}}{V_{s}} = \frac{Q}{R} \times \frac{r}{q} = \frac{27 q}{3 r} \times \frac{r}{q} = 9$

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