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(C) When a large liquid drop of radius $R$ splits into $n$ smaller drops of radius $r$, the volume remains constant: $\frac{4}{3} \pi R^3 = n 	imes \

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energy is absorbed

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(C) When a large liquid drop of radius $R$ splits into $n$ smaller drops of radius $r$, the volume remains constant: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$, which implies $R = n^{1/3} r$.Since $n > 1$, the total surface area of the $n$ small drops $(A_{final} = n 	imes 4 \pi r^2)$ is greater than the surface area of the original large drop $(A_{initial} = 4 \pi R^2)$.Surface energy is directly proportional to the surface area $(U = T 	imes A)$.Since the total surface area increases, the system must absorb energy from the surroundings to perform the work required to increase the surface area.

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

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