64 small drops of mercury,each of radius r an | vedclass

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(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume is conserved,the volume of $64$ small drops equals

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$1:4$

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(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume is conserved,the volume of $64$ small drops equals the volume of the big drop: $64 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.This gives $R^3 = 64 r^3$,so $R = 4r$.The total charge of the big drop is $Q = 64q$.The surface charge density of a small drop is $\sigma_{small} = \frac{q}{4 \pi r^2}$.The surface charge density of the big drop is $\sigma_{Big} = \frac{Q}{4 \pi R^2} = \frac{64q}{4 \pi (4r)^2} = \frac{64q}{4 \pi (16r^2)} = \frac{4q}{4 \pi r^2}$.Therefore,the ratio is $\frac{\sigma_{small}}{\sigma_{Big}} = \frac{q / (4 \pi r^2)}{4q / (4 \pi r^2)} = \frac{1}{4}$.

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a big drop. The ratio of the surface charge density of each small drop to that of the big drop is:

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