Explain Stokes Law.

A small spherical body of radius $r$ and density $\rho$ moves with the terminal velocity $v$ in a fluid of coefficient of viscosity $\eta$ and density $\sigma .$ What will be the net force on the body?

A raindrop with radius $\mathrm{R}=0.2 \mathrm{mm}$ falls from a cloud at a height $\mathrm{h}=2000 \mathrm{m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water $f_{\mathrm{w}}=1000 \mathrm{kg}{ \mathrm{m}}^{-3}$ and Density of air $f_{\mathrm{a}}=1.2 \mathrm{kg}{ \mathrm{m}}^{-3}, \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2$ Coefficient of viscosity of air $\left.=1.8\times {10}^{-5}\mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}\right]$

A water drop of radius $1\mu \mathrm{m}$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is $1.8\times {10}^{-5} \mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}$ and its density is negligible as compared to that of water ${10}^6 \mathrm{g} {\mathrm{m}}^{-3}$. Terminal velocity of the water drop is

(Take acceleration due to gravity $=10{ \mathrm{m} \mathrm{s}}^{-2}$)