'n' identical small drops are charged to V volt each. They coalesce to form a bigger drop. The potential of the bigger drop will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( n^{\frac{2}{3}}\,V \)
When the drops coalesce, the total volume and the total charge are conserved. Let each small drop have radius \( r \) and charge \( q \), so \( V = kq/r \).
\( \displaystyle n \times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3} \;\Rightarrow\; R = n^{1/3}\,r \)
\( Q = nq \)
\( \displaystyle V' = \frac{kQ}{R} = \frac{k\,nq}{n^{1/3}\,r} = n^{2/3}\,\frac{kq}{r} = n^{2/3}\,V \)