Sixty-four conducting drops,each of radius 0. | vedclass

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(B) Let $r$ be the radius of each smaller drop and $R$ be the radius of the bigger drop.Since the volume remains constant,the volume of the bigger dro

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$4: 1$

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(B) Let $r$ be the radius of each smaller drop and $R$ be the radius of the bigger drop.Since the volume remains constant,the volume of the bigger drop is equal to the sum of the volumes of $64$ smaller drops:$\frac{4}{3} \pi R^3 = 64 	imes \frac{4}{3} \pi r^3$$R^3 = 64 r^3 \implies R = 4r$.Let $q$ be the charge on each smaller drop and $Q$ be the charge on the bigger drop.Since charge is conserved,$Q = 64q$.The surface charge density $\sigma$ is defined as $\sigma = \frac{	ext{Charge}}{	ext{Area}} = \frac{q}{4\pi r^2}$.For the smaller drop,$\sigma_s = \frac{q}{4\pi r^2}$.For the bigger drop,$\sigma_B = \frac{Q}{4\pi R^2} = \frac{64q}{4\pi (4r)^2} = \frac{64q}{4\pi (16r^2)} = 4 	imes \frac{q}{4\pi r^2}$.Therefore,the ratio $\frac{\sigma_B}{\sigma_s} = \frac{4 	imes \sigma_s}{\sigma_s} = 4:1$.

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 the surface charge density of the bigger drop to that of the smaller drop will be ............

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