A big liquid drop splits into n similar small | vedclass

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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 $r$,t

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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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