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(C) Work done,$W = S 	imes \Delta A$,where $\Delta A$ is the change in surface area.Initial surface area of the big drop,$A_{	ext{initial}} = 4 \pi

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$4 \pi R^2 S$

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(C) Work done,$W = S 	imes \Delta A$,where $\Delta A$ is the change in surface area.Initial surface area of the big drop,$A_{	ext{initial}} = 4 \pi R^2$.Let the radius of each small drop be $r$. The total volume remains constant:$\frac{4}{3} \pi R^3 = 8 	imes \frac{4}{3} \pi r^3 \implies R^3 = 8r^3 \implies R = 2r$ or $r = R/2$.Final surface area of $8$ small drops,$A_{	ext{final}} = 8 	imes 4 \pi r^2 = 32 \pi (R/2)^2 = 32 \pi (R^2/4) = 8 \pi R^2$.Change in surface area,$\Delta A = A_{	ext{final}} - A_{	ext{initial}} = 8 \pi R^2 - 4 \pi R^2 = 4 \pi R^2$.Work done,$W = S 	imes \Delta A = S 	imes 4 \pi R^2 = 4 \pi R^2 S$.

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

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