A 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, radius of rain drop \(=\mathrm{r}\) Since, rain drop starts falling from rest, hence its initial speed, \(u=\mathbf{0}\) Final velocity of rain drop is equal to the terminal velocity \(v\), which is given by \(v=\frac{2 g r^2(\rho-\sigma)}{9 \eta}\) where, \(\rho \rightarrow\) density of the rain drop \(\sigma \rightarrow\) density of air \(\eta \rightarrow\) coefficient of viscosity \(\therefore\) According to the work-energy theorem, work done by all the forces on the drop = change in its kinetic energy \(\begin{array}{rlr} W & =\frac{1}{2} m v^2-\frac{1}{2} m u^2=\frac{1}{2} m v^2 & {[\because u=0]} \\ & =\frac{1}{2} \times \frac{4}{3} \pi r^3 \cdot \rho\left(\frac{2 g r^2(\rho-\sigma)}{9 \eta}\right)^2 & {\left[\because \rho=\frac{m}{V}\right]} \\ k & =\frac{8 \pi \rho g^2(\rho-\sigma)^2 r^7}{243 \eta^2} \Rightarrow W=k r^7 & \text { or } W \propto r^7 \end{array}\)