Question

A mercury drop of radius R is sprayed into n droplets of equal size. Calculate the energy expanded if surface tension of mercury is T.

Answer 1

If drop of radius R is sprayed into n droplets of equal radius r, then as a drop has only one surface, the initial surface area will be $4\pi R^2$ while final area is $n\left(4\pi r^2\right)$. So the increase in area .
$\Delta S=n\left(4\pi r^2\right)-4\pi R^2$
So energy expended in the process
$W=T\Delta S=4\pi T[n{r}^2-R^2]$____(i)
Now since the total volume of n droplets is same as that of initial drop, i.e.
$\frac{4}{3}\pi r^3=n\left[\right(4/3\left)\pi r^3\right]$
Or $r=\frac{R}{n^{1/3}}$----(ii)
Putting the value of r from Eq (ii) in Eq (i)
$W=4\pi R^{2}T\left[\right(n{)}^{1/3}-1]$