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(B) Let the radius of each small drop be $r$ and the charge be $q$. The potential of a small drop is $v = \frac{kq}{r}$.When $n = 8$ drops combine to

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(B) Let the radius of each small drop be $r$ and the charge be $q$. The potential of a small drop is $v = \frac{kq}{r}$.When $n = 8$ drops combine to form a big drop of radius $R$,the volume remains constant: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r = 8^{1/3} r = 2r$.The total charge on the big drop is $Q = nq = 8q$.The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left( \frac{kq}{r} ight) = 4v$.Therefore,the ratio of the potential of the big drop to the small drop is $\frac{V}{v} = 4:1$.

Eight small drops,each of radius $r$ and having same charge $q$,are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

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