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 ............
JEE MAIN 2022, Medium
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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
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{\frac{ Q }{4 \pi R ^{2}}}{\frac{ q }{4 \pi r ^{2}}}=\frac{ Q }{ q } \cdot \frac{ r ^{2}}{ R ^{2}}$
$=64 \frac{ r ^{2}}{16 r ^{2}}=4$
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{4}{1}$
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