A raindrop with radius $R = 0.2$ mm falls from a cloud at a height $h = 2000$ m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyancy may be neglected, then the terminal speed attained by the raindrop is: [Density of water $f_w = 1000\ \text{kg m}^{-3}$ and density of air $f_a = 1.2\ \text{kg m}^{-3}$, $g = 10\ \text{m/s}^2$, coefficient of viscosity of air $= 1.8\times10^{-5}\ \text{N s m}^{-2}$]
Answer: (C) $4.94\ \text{m s}^{-1}$
$$v_t = \frac{2R^2(\rho - \sigma)g}{9\eta} = \frac{2\times(0.2\times10^{-3})^2\times(1000 - 1.2)\times10}{9\times1.8\times10^{-5}}$$
$$v_t = \frac{7.99\times10^{-4}}{1.62\times10^{-4}} \approx 4.94\ \text{m s}^{-1}$$
(Ignoring the air density entirely gives $4.94$ m/s as well; the height of the cloud is not needed.)
Solution by Sreeraj P, M.Sc Physics