Q. In identical mercury droplets charged to the same potential $V$ coalesce to form a single bigger drop. The potential of new drop will be
A) $\frac{V}{n}$
B) $nV$
C) $nV^{2}$
D) $n^{2/3}V$
Explanation
(d) $V_{bigdrop} = n^{2/3}\ V_{small}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.