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 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 $= 1.8 \times 10^{-5} \text{ Ns m}^{-2}$ ]

• A. $250.6 \text{ ms}^{-1}$
• B. $4.94 \text{ ms}^{-1}$
• C. $14.4 \text{ ms}^{-1}$
• D. $43.56 \text{ ms}^{-1}$

Hint:

The height from which the object falls is irrelevant for calculating the terminal speed itself, although it determines whether the object has enough time to reach it. For raindrops in air, the density of air is often negligible compared to the density of water.

Answer 1

The Correct Option is B

The terminal speed ($v_t$) of a spherical object falling through a viscous fluid is reached when the gravitational force is balanced by the viscous drag force.
The formula for terminal speed is given by Stokes' law:
$v_t = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$
Where:
$r$ = radius of the sphere
$\rho$ = density of the sphere (water)
$\sigma$ = density of the fluid (air)
$g$ = acceleration due to gravity
$\eta$ = coefficient of viscosity of the fluid (air)
Given values:
$R = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$.
$\rho = f_w = 1000 \text{ kg m}^{-3}$.
$\sigma = f_a = 1.2 \text{ kg m}^{-3}$.
$g = 10 \text{ m/s}^2$.
$\eta = 1.8 \times 10^{-5} \text{ Ns m}^{-2}$.
Since $\rho \gg \sigma$, we can approximate $(\rho - \sigma) \approx \rho = 1000 \text{ kg m}^{-3}$.
Now, substitute the values into the formula:
$v_t = \frac{2}{9} \frac{(0.2 \times 10^{-3})^2 (1000 - 1.2) \times 10}{1.8 \times 10^{-5}}$.
$v_t = \frac{2}{9} \frac{(4 \times 10^{-8}) (998.8) \times 10}{1.8 \times 10^{-5}}$.
$v_t = \frac{2 \times 4 \times 998.8 \times 10}{9 \times 1.8} \times \frac{10^{-8}}{10^{-5}}$.
$v_t = \frac{79904}{16.2} \times 10^{-3}$.
$v_t \approx 4932.3 \times 10^{-3} \text{ m/s}$.
$v_t \approx 4.93 \text{ m/s}$.
This value is very close to 4.94 ms$^{-1}$.