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(C) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$.Since the volume remains constant,the volume of the large drop

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$8$

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(C) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of $512$ small droplets:$\frac{4}{3} \pi R^3 = 512 	imes \frac{4}{3} \pi r^3$$R^3 = 512 r^3$$R = 8r$ or $r = \frac{R}{8}$.The initial surface energy is $E = T 	imes A = T 	imes 4 \pi R^2$,where $T$ is the surface tension.The final surface energy $E'$ is the sum of the surface energies of $512$ droplets:$E' = 512 	imes (T 	imes 4 \pi r^2)$Substitute $r = \frac{R}{8}$ into the equation:$E' = 512 	imes T 	imes 4 \pi \left(\frac{R}{8}ight)^2$$E' = 512 	imes T 	imes 4 \pi 	imes \frac{R^2}{64}$$E' = 8 	imes (T 	imes 4 \pi R^2)$$E' = 8E$.

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

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