If ' n ' identical water drops assumed spherical each charged to a potential energy U coalesce to form a single drop, the potential energy of the single drop is(Assume that drops are uniformly charged):

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$n^{5 / 3} U$

$n^{4 / 3} U$

$n^{1 / 3} U$

$n^{2 / 3} U$

answer is D.

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Detailed Solution

Let q is charge and a is radius of single drop. U = $\frac{3 k q^{2}}{5 a}$
charge on big drop = nq.
Let Radius of big drop is R.

$\frac{4}{3} π R^{3} = n . \frac{4}{3} π a^{3}$

$R = a n^{1 / 3}$
P.E. of big drop = $\frac{3}{5} \frac{k {(q n)}^{2}}{R} = \frac{3}{5} \frac{k . q^{2} n^{2}}{a n^{1 / 3}} = U n^{\frac{5}{3}}$

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