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(C) The surface area of the initial liquid drop is $A_1 = 4 \pi R^2$. Its surface energy is $E = T \cdot A_1$,where $T$ is the surface tension. When t

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(C) The surface area of the initial liquid drop is $A_1 = 4 \pi R^2$. Its surface energy is $E = T \cdot A_1$,where $T$ is the surface tension. When the drop splits into $n = 512$ droplets of radius $r$,the volume remains conserved: $\frac{4}{3} \pi R^3 = 512 	imes \frac{4}{3} \pi r^3$. This implies $R^3 = 512 r^3$,so $R = 8r$ or $r = \frac{R}{8}$. The total surface area of the $512$ droplets is $A_2 = 512 	imes 4 \pi r^2$. Substituting $r = \frac{R}{8}$,we get $A_2 = 512 	imes 4 \pi \left(\frac{R}{8}ight)^2 = 512 	imes 4 \pi 	imes \frac{R^2}{64} = 8 	imes 4 \pi R^2 = 8 A_1$. The final surface energy $E_f = T \cdot A_2 = T \cdot (8 A_1) = 8 (T \cdot A_1) = 8E$.

$A$ 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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