Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is........$times$
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(c) Volume of $8$ small drops $=$ Volume of big drop
$8 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}$ $==>$ $ R = 2r$
As capacity is $r$, hence capacity becomes $2$ times.
$8 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}$ $==>$ $ R = 2r$
As capacity is $r$, hence capacity becomes $2$ times.
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