A water drop of radius l cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain surface energy upto first decimal place will be  $[Given\pi =3.14]$

Given, radius of water drop, $R=1cm$  Surface tension $T=75dyne/cm=0.07N/m$  Volume of  $1drop=Volumeof729$ droplets 

$\frac{4}{3}\pi R^3=729\times \frac{4}{3}\pi r^3\therefore R^3=729r^3$  Where r is the  radius of small droplets 

$\Rightarrow {\left({10}^{-2}\right)}^3=729r^3\Rightarrow {\left({10}^{-2}\right)}^3={\left(9\right)}^3{r}^3\Rightarrow r=\frac{1}{9}{\times }^{{10}^{-2}}m$  Work done or gain in surface energy is given by,  $W=T\Delta A$  Where $\Delta A$ is the change in surface area $\therefore W=T[n\times 4\pi r^2-4R^2]$

$=0.075[729\times {(\frac{1}{9}\times {10}^{-2})}^2-{\left({10}^{-2}\right)}^2]\times 4\pi$  $=0.075[9\times {10}^{-4}-{10}^{-4}]\times 4\pi =7.536\times {10}^{-4}J$