Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μ C are combined to form a bigger drop. The ratio of surface charge density of bigger drop to the smaller drop will be: (a) 1: 4 (b) 4: 1 (c) 1: 8 (d) 8: 1

Answer: (b). Volume conserved: 4/3π R³ = 64·4/3π r³ ⇒ R = 4r; charge Q = 64q. σbigσsmall = (Q/R²)/(q/r²) = 64/16 = 4 So the answer is (b) 4: 1.

Sixty four conducting drops each of radius 0.02 m and each carrying a… | JEE Main 2022 (26 Jun, Shift 2)

Q74JEE Main 2022 · 26 JUN · SHIFT 2Medium

Electric Potential & Equipotentials

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $\mu$ C are combined to form a bigger drop. The ratio of surface charge density of bigger drop to the smaller drop will be:

Show answer and solution

Answer: (b)

Volume conserved: $\frac{4}{3}\pi R^3 = 64\cdot\frac{4}{3}\pi r^3 \Rightarrow R = 4r$ ; charge $Q = 64q$ .
$\frac{\sigma_{big}}{\sigma_{small}} = \frac{Q/R^2}{q/r^2} = \frac{64}{16} = 4$
So the answer is (b) $4 : 1$ .

Report a mistake

Electric Potential & Equipotentials question from JEE Main 2022 (26 Jun, Shift 2)

More questions on Electric Potential & Equipotentials