Sixty-four drops are joined together to form | vedclass

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

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$4C$

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(B) Let the radius of each small drop be $r$ and the radius of the bigger drop be $R$.Since the volume remains constant,the volume of the bigger drop is equal to the sum of the volumes of $64$ small drops:$\frac{4}{3} \pi R^3 = 64 	imes \frac{4}{3} \pi r^3$$R^3 = 64 r^3$$R = 4r$The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 r$.Therefore,the capacitance of the bigger drop $C'$ is:$C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (4r) = 4(4 \pi \epsilon_0 r) = 4C$.Thus,the capacitance of the bigger drop is $4C$.

Sixty-four drops are joined together to form a bigger drop. If each small drop has a capacitance $C$,a potential $V$,and a charge $q$,then the capacitance of the bigger drop will be

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