Rain drop of radius ‘r’ is falling through air starting from rest. The work done by all the forces on the drop, when it attains terminal
velocity is proportional to 

Given, raindrop radius equals r
Since a raindrop begins to fall at rest, its initial speed is u=0.
The terminal velocity v, which is determined by, is equal to the final velocity of a raindrop.

$\mathrm{v}=\frac{2{\mathrm{gr}}^2\left(\mathrm{\rho }-\mathrm{\sigma }\right)}{9\mathrm{\eta }}$

Thus by the work energy theorem,

$\mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2$

as u =0,

$\mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{W}=\frac{1}{2}\times \frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\rho }{\left(\frac{2{\mathrm{gr}}^2(\mathrm{\rho }-\mathrm{\sigma })}{9\mathrm{\eta }}\right)}^2\left(\because \mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{v}}\right)$

$\mathrm{k}=\frac{8{\mathrm{\pi \rho g}}^2{(\mathrm{\rho }-\mathrm{\sigma })}^2{\mathrm{r}}^7}{243{\mathrm{\eta }}^2}$

$\mathrm{W}={\mathrm{kr}}^7$

$\mathrm{W}\propto {\mathrm{r}}^7$

Hence the correct answer is ${\mathrm{r}}^7.$