If ‘n’ identical drops each charged to a potential V coalesce to a single bigger drop, the potential of the bigger drop is
Answer & explanation
Correct answer: option 2
$\text{ If charge on each drop is q then total charge on bigger drop is } nq.$
$\text{Since Volume remain constant then } n\times \frac{4\pi}{3} r^3 = \frac{4\pi}{3} R^3 $
$ \Rightarrow R = n^{1/3} r$
$ \text{Potential of bigger drop is }V' = \frac{k.nq}{R} = n^{2/3} \frac{kq}{r} = n^{2/3}V$