An air bubble of radius ${10}^{-2} \mathrm{m}$ is rising up at a steady rate of $2\times {10}^{-3} {\mathrm{m}\mathrm{s}}^{-1}$ through a liquid of density $1.5\times {10}^3 \mathrm{k}\mathrm{g}\mathrm{ }{\mathrm{m}}^{-3}$ , the coefficient of viscosity neglecting the density of air, will be ($\mathrm{g}=10\mathrm{ }{\mathrm{m}\mathrm{s}}^{-2}$)

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