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(C) The initial surface area of the drop is $A_i = 4 \pi r^2$. The initial surface energy is $U_i = S 	imes 4 \pi r^2$.Since the volume remains const

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$4 \pi r^2 S$

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(C) The initial surface area of the drop is $A_i = 4 \pi r^2$. The initial surface energy is $U_i = S 	imes 4 \pi r^2$.Since the volume remains constant, the volume of the large drop equals the sum of the volumes of the $8$ small droplets: $\frac{4}{3} \pi r^3 = 8 	imes \frac{4}{3} \pi (r')^3$.This simplifies to $r^3 = 8(r')^3$, so $r = 2r'$, which means $r' = r/2$.The final surface area of $8$ droplets is $A_f = 8 	imes 4 \pi (r')^2 = 8 	imes 4 \pi (r/2)^2 = 8 	imes 4 \pi (r^2/4) = 8 \pi r^2$.The final surface energy is $U_f = S 	imes 8 \pi r^2$.The work done is equal to the change in surface energy: $W = \Delta U = U_f - U_i = S(8 \pi r^2 - 4 \pi r^2) = 4 \pi r^2 S$.

$A$ spherical drop of radius $r$ is divided into $8$ equal droplets. If the surface tension is $S$, then the work done in the process will be

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