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

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Electrostatics

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 density of bigger drop to the smaller drop will be :

Answer: (B)

When 64 small drops combine, volume is conserved:

\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}

, giving

R = 4 r

. The total charge is

Q = 64 q

. Surface charge density is

σ = \frac{charge}{area}

. Thus,

\frac{σ _{bigger}}{σ _{smaller}} = \frac{Q / ( 4 π R ^{2} )}{q / ( 4 π r ^{2} )} = \frac{Q}{q} \cdot \frac{r ^{2}}{R ^{2}} = 64 \cdot \frac{r ^{2}}{( 4 r ) ^{2}} = 64 \cdot \frac{1}{16} = 4

. So the ratio is

4 : 1

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option