Eight drops of mercury of equal radii and pos | vedclass

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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant during the combination of drops:V

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$2$

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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant during the combination of drops:Volume of big drop $= 8 	imes$ Volume of small drop$\frac{4}{3} \pi R^3 = 8 	imes \frac{4}{3} \pi r^3$$R^3 = 8 r^3 \Rightarrow R = 2r$The capacity of a spherical conductor is given by $C = 4 \pi \varepsilon_0 r$.For the big drop,the capacity $C'$ is:$C' = 4 \pi \varepsilon_0 R$Substituting $R = 2r$:$C' = 4 \pi \varepsilon_0 (2r) = 2(4 \pi \varepsilon_0 r) = 2C$.

Eight drops of mercury of equal radii and possessing equal charge combine to form a big drop. If the capacity of each drop is $C$,then the capacity of the big drop is: (in $C$)

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