A liquid drop having surface energy E is spre | vedclass

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(C) Let the radius of the original drop be $R$ and the radius of each small droplet be $r$.Surface energy $E = S 	imes A$,where $S$ is the surface te

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$8$

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(C) Let the radius of the original drop be $R$ and the radius of each small droplet be $r$.Surface energy $E = S 	imes A$,where $S$ is the surface tension and $A$ is the surface area.Initial surface area $A_1 = 4 \pi R^2$.Volume of the large drop = Volume of $512$ small droplets:$\frac{4}{3} \pi R^3 = 512 	imes \frac{4}{3} \pi r^3$$R^3 = 512 r^3 \implies R = 8r$.Final surface area $A_2 = 512 	imes (4 \pi r^2)$.Substituting $r = R/8$:$A_2 = 512 	imes 4 \pi \left(\frac{R}{8}ight)^2 = 512 	imes 4 \pi 	imes \frac{R^2}{64} = 8 	imes (4 \pi R^2) = 8 A_1$.Since surface energy is directly proportional to surface area $(E \propto A)$:$E_2 = 8 E_1 = 8 E$.

$A$ liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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