Q 12-02-137JEE MainJEE Main 2021 (25 Feb, Shift 1)Easy
512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is $V$ in Volt. Find $V$.
Numerical value type. Enter your answer.
Answer: 128
Volume is conserved: $R^3 = 512r^3 \Rightarrow R = 8r$. Charge $Q = 512q$.
$$V = \frac{kQ}{R} = \frac{512}{8}\cdot\frac{kq}{r} = 64\times2 = 128\ \text{V}$$
(In general $V = n^{2/3}v$.)
Solution by Sreeraj P, M.Sc Physics