Question

Calculate the amount of energy needed to break a drop of water $2$mm in diameter into ${10}^9$ droplets of equal size, taking the surface tension of water $7.3\times {10}^{-2}N{m}^{-1}$

Answer 1

Given, $Diameter=2mm,droplets={10}^9,Tension=0.073N/m$

Since the liquid drop breaks into ${10}^9$ droplets of equal mass so the volume of the big drop is equal to the volume of each droplet multiplied by ${10}^9$

$R^3=\frac{4}{3}\pi R^3={10}^9\times \frac{4}{3}\pi r^3\Rightarrow R^3=({10}^{3}r{)}^3={10}^r$

The initial energy of the liquid drop $E_1=TA$ where T is the tension and A is the area

$E_1=T\times 4\pi R^2$

Final energy of the ${10}^9$ droplets $E_2={10}^{9}Ta$

Where T is the tension and a is the area.

$E_2={10}^9\times T\times 4\pi r^2$

NOw, Change in energy $E_2-E_1={10}^{9}T\times 4\pi r^2-T4\pi R^2=T\times 4\pi [{10}^9{r}^2-R^2]$

$\Rightarrow T\times 4\pi \left[{10}^9\right(\frac{R}{{10}^3}{)}^2-R^2]=T\times 4\pi [{10}^3{R}^2-R^2]=3996\pi T{R}^2$

Now put the value of T and R

$3996\times 3.14\times (1\times {10}^{-3}{)}^2\times 0.073=915\times {10}^{-6}J/m^2$