Eight drops of mercury of equal radii combine | vedclass

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(A) Let $R$ and $r$ be the radii of the bigger and each smaller drop, respectively.Volume of the bigger drop $= 8 	imes$ Volume of a smaller drop.$\f

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$2$

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(A) Let $R$ and $r$ be the radii of the bigger and each smaller drop, respectively.Volume of the bigger drop $= 8 	imes$ Volume of a smaller drop.$\frac{4}{3} \pi R^{3} = 8 	imes \frac{4}{3} \pi r^{3}$$R^{3} = 8r^{3} \Rightarrow R = 2r$ ... $(i)$As the capacitance of a spherical conductor is given by $C = 4 \pi \varepsilon_{0} r$, we have $C \propto r$.Therefore, the ratio of the capacitance of the bigger drop to the smaller drop is:$\frac{C_{	ext{bigger}}}{C_{	ext{smaller}}} = \frac{R}{r} = \frac{2r}{r} = 2$.Thus, the capacitance of the bigger drop is $2$ times that of each smaller drop.

Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is (in $\times$)

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