If $n$ drops of a liquid, each with surface energy E, joined to form a single drop, then

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Some energy will be released in the process

Some energy will be absorbed in the process

The energy released or absorbed will be $E (n - n^{2 / 3})$

The energy released or absorbed will be $n E (2^{2 / 3} - 1)$

answer is A, C.

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Detailed Solution

Let S=surface tension=surface energy per unit area
r=radius of each small drop
R=radius of a single drop
$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} o r R = r n^{1 / 3}$
Initial surface energy, $E_{i} = n \times 4 π r^{2} \times S = n E$
Final surface energy, $E_{f} = 4 π R^{2} S = 4 π r^{2} n^{2 / 3} S = n^{2 / 3} E$
Therefore, energy released $= E_{i} - E_{f} = E (n - n^{2 / 3})$