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(C) The volume of the liquid remains constant during the splitting process.When a large drop of radius $R$ splits into $n$ small drops of radius $r$,

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energy is absorbed

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(C) The volume of the liquid remains constant during the splitting process.When a large drop of radius $R$ splits into $n$ small drops of radius $r$, the total volume is conserved: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$, which implies $R^3 = nr^3$ or $r = R / n^{1/3}$.The total surface area of the large drop is $A_1 = 4 \pi R^2$.The total surface area of $n$ small drops is $A_2 = n 	imes 4 \pi r^2 = n 	imes 4 \pi (R / n^{1/3})^2 = n^{1/3} 	imes 4 \pi R^2$.Since $n > 1$, $n^{1/3} > 1$, therefore $A_2 > A_1$. The surface area increases.Surface energy is given by $U = T 	imes A$, where $T$ is surface tension.Since the surface area increases, the surface energy of the system increases.To increase the surface energy, work must be done on the system, meaning energy is absorbed from the surroundings.

$A$ big liquid drop splits into $n$ similar small drops under isothermal conditions. Then, in this process:

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