Question

Calculate the work done in breaking a mercury drop of radius $1mm$ into $1000$ droplets of the same size. Surface tension of mercury is $525\times {10}^{-3}N/m$

Answer 1

Here,

$T=525\times {10}^{-3}N{m}^{-1}$

$R=1\times {10}^{-3}$

Now, the surface area of the bigger drop.

$=4\pi R^2$

$=4\times 3.14\times (1\times {10}^{-3}{)}^2$

then,

$S_1=0.1256\times {10}^{-4}{m}^2$

Now, the total initial volume $=\frac{4}{3}\pi R^3$

and the volume smaller drops $=\frac{\frac{4}{3}\pi R^3}{1000}$

consider $r$ be the radius of smaller drops,

Hence, volume of droplets $=\frac{4}{3}\pi r^3$

and $\frac{4}{3}\pi r^3=\frac{\frac{4}{3}\pi R^3}{1000}$

therefore, $r=\frac{R}{10}$

Now, the surface area of small drop $=4\pi r^2=4\pi (R/10{)}^2$

$=4\times 3.14\times (1\times {10}^{-4}{)}^2$

$=12.56\times {10}^{-8}{m}^2$

Total surface area vof $1000$ drops $S_2=1000\times 12.56\times {10}^{-8}$

$=1.256\times {10}^{-4}{m}^2$

So, change of surface area $=S_2-S_1$

$\Delta S=1.256\times {10}^{-4}-0.1256\times {10}^{-4}=1.13\times {10}^{-4}$

and, we know that

$workdone=T\Delta S$

$=525\times {10}^{-3}\times 1.13\times {10}^{-4}$

$=593.25\times {10}^{-7}=5.9\times {10}^{-5}$

Hence, the work done in breaking a mercury drop of radius $1mm$ into $1000$ droplets of same size will be $5.9\times {10}^{-5}J$