Question

A mercury drop of radius $1\text{cm}$ is sprayed into ${10}^6$ droplets of equal size. Calculate the energy expended ( in $\text{mJ}$ upto two decimals) if surface tension of mercury is $35\times {10}^{-3}\text{N/m}$.

Answer 1

When a drop of radius $R$ is sprayed into $n$ droplets of equal radii $r$, then the initial and final surface area are different
Change in area $\Delta A=n\left(4\pi r^2\right)-4\pi R^2$
So work required to increase surface energy in the process,
$W=T\Delta A=4\pi T(n{r}^2-R^2)\dots \dots \left(1\right)$
Now, the total mass of $n$ droplets is the same as that of initial drop, i.e.,
$\rho \frac{4}{3}\pi R^3=\rho n\left[\frac{4}{3}\pi r^3\right]$ or $r=\frac{R}{n^{\frac{1}{3}}}\dots \dots \left(2\right)$
So substituting the value of $r$ from Eqs. (2) in (1), we get
$W=4\pi R^{2}T\left[\right(n{)}^{\frac{1}{3}}-1]=4\times 3.14\times (1\times {10}^{-2}{)}^2\times 35\times {10}^{-3}[{10}^2-1]$
$=4.352\times {10}^{-3}\text{J}$