A large spherical drop of liquid of radius $R$ and volume $V$ is broken into $n$ smaller identical d

Question

A large spherical drop of liquid of radius $R$ and volume $V$ is broken into $n$ smaller identical drops, each of radius $r$. If $T$ is the surface tension of the liquid, the energy absorbed during this process is

Professor PC Thomas Classes · Answer

(a) : Total initial volume $V = \frac{4}{3} \pi R^3$. Total final volume is $n 	imes \frac{4}{3} \pi r^3 = V$. The energy absorbed is equal to the work done against surface tension to increase the surface area: $\Delta E = T 	imes \Delta A = T(n 	imes 4\pi r^2 - 4\pi R^2)$. We can rewrite $n 	imes 4\pi r^2$ as $(n 	imes \frac{4}{3}\pi r^3) 	imes \frac{3}{r} = V 	imes \frac{3}{r}$. Similarly, $4\pi R^2 = V 	imes \frac{3}{R}$. Therefore, $\Delta E = T \left( \frac{3V}{r} - \frac{3V}{R} ight) = 3VT \left( \frac{1}{r} - \frac{1}{R} ight)$.