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(C) Let the radius of the large drop be $R$ and its capacitance be $C = 4\pi\epsilon_0 R = 1\,\mu F$.Let the radius of each small drop be $r$. Since t

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$0.5$

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(C) Let the radius of the large drop be $R$ and its capacitance be $C = 4\pi\epsilon_0 R = 1\,\mu F$.Let the radius of each small drop be $r$. Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $8$ small drops:$\frac{4}{3}\pi R^3 = 8 	imes \frac{4}{3}\pi r^3$$R^3 = 8r^3 \implies R = 2r$ or $r = R/2$.The capacitance of each small drop is $c = 4\pi\epsilon_0 r$.Substituting $r = R/2$,we get $c = 4\pi\epsilon_0 (R/2) = C/2$.Given $C = 1\,\mu F$,the capacitance of each small drop is $c = 1/2 = 0.5\,\mu F$.

$A$ spherical drop of capacitance $1\,\mu F$ is broken into eight drops of equal radius. Then,the capacitance of each small drop is ......$\mu F$.

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