A small spherical drop fall from rest in viscous liquid. Due to...

Viscous force acting on spherical drop $F_v=6 \pi \eta r v \therefore$ Terminal velocity, $v=\frac{2}{9} \frac{r^2(\sigma-\rho) g}{\eta}$ where, $\eta=$ coefficient of viscosity of liquid $\sigma=$ density of material of spherical drop $\rho=$ density of liquid Power imparted by viscous force = Rate of production of heat $$\begin{aligned} P & =\frac{d H}{d t}=F_v \cdot v=6 \pi \eta r v^2 \\ & =6 \pi \eta r \cdot\left(\frac{2}{9} \frac{r^2(\sigma-\rho) g}{\eta}\right)^2 \\ \Rightarrow \quad \frac{d H}{d t} & \propto r^5\end{aligned}$$