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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 dr

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$12$

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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 $64$ small droplets: $\frac{4}{3} \pi R^3 = 64 	imes \frac{4}{3} \pi r^3$. Solving for $r$, we get $R^3 = 64r^3$, which implies $r = \frac{R}{4}$. The work done $W$ is equal to the change in surface energy: $W = T 	imes \Delta A$, where $\Delta A$ is the increase in surface area. Initial surface area $A_i = 4 \pi R^2$. Final surface area $A_f = 64 	imes (4 \pi r^2) = 64 	imes 4 \pi (\frac{R}{4})^2 = 64 	imes 4 \pi 	imes \frac{R^2}{16} = 16 \pi R^2$. Change in area $\Delta A = A_f - A_i = 16 \pi R^2 - 4 \pi R^2 = 12 \pi R^2$. Therefore, the work done $W = T 	imes 12 \pi R^2 = 12 \pi TR^2$.

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ surface tension of water). (in $\pi TR^2$)

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