A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength $\frac{\mathrm{81}\mathrm{\pi }}{7}\times {\mathrm{10}}^5 \mathrm{V} {\mathrm{m}}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2\times {\mathrm{10}}^{-3 } \mathrm{m} {\mathrm{s}}^{-1}$. Given  $g=9.8 \mathrm{m} {\mathrm{s}}^{-2}$, viscosity of the air $1.8\times {\mathrm{10}}^{-5}{ \mathrm{N} \mathrm{s} \mathrm{m}}^{-2}$ and the density of oil  $=900 \mathrm{kg} {\mathrm{m}}^{-3}$, the magnitude of $q$ is

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 $\left.=1.8\times {10}^{-5}\mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}\right]$