# A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m

Question:

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 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 $=1.8 \times 10^{-5} \mathrm{Nsm}^{-2}$ ]

1. $250.6 \mathrm{~ms}^{-1}$

2. $43.56 \mathrm{~ms}^{-1}$

3. $4.94 \mathrm{~ms}^{-1}$

4. $14.4 \mathrm{~ms}^{-1}$

Correct Option: , 3

Solution:

At terminal speed

$\mathrm{a}=0$

$\mathrm{F}_{\text {net }}=0$

$m g=F_{v}=6 \pi \eta R v$

$\mathrm{v}=\frac{\mathrm{mg}}{6 \pi \eta \mathrm{Rv}}$

$\mathrm{v}=\frac{\rho_{\mathrm{w}} \frac{4 \pi}{3} \mathrm{R}^{3} \mathrm{~g}}{6 \pi \eta \mathrm{R}}$

$=\frac{2 \rho_{w} R^{2} g}{9 \eta}$

$=\frac{400}{81} \mathrm{~m} / \mathrm{s}$

$=4.94 \mathrm{~m} / \mathrm{s}$