Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = $1.8\times {10}^{-4}$ poise, g = $9.9\times 10m{s}^{-2}$ and density of water = $1000kg{m}^{-3}$

r = 0.02 cm = $2\times {10}^{-4}m$

$\eta =1.8\times {10}^{-4}poise$

= $1.8\times {10}^{-5}$decapoise

g= 9.9 $m/s^2$

$\rho =1000kg/m^3$

To find terminal velocity of rain drops, the force acting on the drops are,

(i) The weight $\left(\frac{4}{3}\right)\pi r^3\rho g$ downward.

(ii) Force of buoyancy $\left(\frac{4}{3}\right)\pi r^3\rho g$ upward.

(iii) Force of viscosity $6\pi \eta rv$ upward.

Because , $\sigma$ of air is very small, the force of buoyancy may be neglected.

Thus, $6\pi \eta rv=\left(\frac{4}{3}\right)\pi r^3\rho g$

or v =$2r^2\frac{\pi g}{9\eta }$

= $2\times (0.02\times {10}^{-2}{)}^2\times 1000\times \frac{\left(9.9\right)}{9}\times (1.8\times {10}^{-5})$

= 5m /sec.