Physics - Fluid Question with Solution | TestHub

Let $n$ be the number of small drops of radius $r$.

Since the total volume remains constant during coalescence, the volume of the large drop is equal to the sum of the volumes of the small drops.

$V=n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

The initial surface area of the $n$ drops is:

$A_i=n\times 4\pi r^2$

Substituting $n=\frac{3V}{4\pi r^3}$:

$A_i=\left(\frac{3V}{4\pi r^3}\right)\times 4\pi r^2=\frac{3V}{r}$

The final surface area of the single large drop is:

$A_f=4\pi R^2=\frac{3V}{R}$

The change in surface area is:

$\Delta A=A_i-A_f=3V\left[\frac{1}{r}-\frac{1}{R}\right]$

Since $r<R$, the surface area decreases ($\Delta A>0$), which means surface energy is released.

The energy released is given by:

$E=T\Delta A=3VT\left[\frac{1}{r}-\frac{1}{R}\right]$

Answer: $E=3VT\left[\frac{1}{r}-\frac{1}{R}\right]$ is released