Question

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 density of bigger drop to the smaller drop will be :

• A. $1: 4$
• B. $4: 1$
• C. $1: 8$
• D. $8: 1$

Answer 1

Answer: (B)

Let $R =$ radius of combined drop $r =$ radius of smaller drop
Volume will remain same
$\frac{4}{3} \pi R ^{3}=64 \times \frac{4}{3} \pi r ^{3}$
$R =4 r$
$Q =64 q$;
$q : $ charge of smaller drop
$Q$ : Charge of combined drop
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{\frac{ Q }{4 \pi R ^{2}}}{\frac{ q }{4 \pi r ^{2}}}=\frac{ Q }{ q } \cdot \frac{ r ^{2}}{ R ^{2}} $
$=64 \frac{ r ^{2}}{16 r ^{2}}=4$
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{4}{1}$