Sixty four conducting drops each of radius 0. | vedclass

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Question

Let $R=$ radius of combined drop

$r=$ radius of smaller drop

Volume will remain same

$\frac{4}{3} \pi R ^{3}=64 	imes \frac{4}{3} \pi r ^{3}

Vedclass · Accepted Answer

$4: 1$

Vedclass · Answer

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

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 ............

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