Q 12-01-097JEE MainJEE Main 2022 (26 Jun, Shift 2)Easy
Sixty four conducting drops each of radius $0.02\ \text{m}$ and each carrying a charge of $5\ \mu\text{C}$ are combined to form a bigger drop. The ratio of surface density of the bigger drop to the smaller drop will be
Answer: (B) $4 : 1$
Volume conservation: $R = 64^{1/3}r = 4r$. Charge $Q = 64q$.
$$\frac{\sigma_{\text{big}}}{\sigma} = \frac{64q/(4\pi\cdot16r^2)}{q/(4\pi r^2)} = 4$$
Solution by Sreeraj P, M.Sc Physics