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Step-by-Step Solution
Step 1: Understand the Concept of Terminal Velocity
When a raindrop falls through air, two main forces act on it:
The gravitational force $mg$ pulling it downward.
The viscous drag force exerted by air, which opposes the motion.
Eventually, these two forces balance each other, resulting in zero net acceleration. The constant speed at which this happens is called the terminal velocity.
Step 2: Express the Forces Mathematically
Let:
$m$ be the mass of the raindrop.
$g$ be the acceleration due to gravity.
$\eta$ be the coefficient of viscosity of air.
$R$ be the radius of the raindrop.
$v$ be the terminal velocity of the raindrop.
$\rho_w$ be the density of water (which makes up the raindrop).
The gravitational force on the raindrop is:
$$
F_g = mg
$$
Since the raindrop is assumed spherical, its mass $m$ can be written in terms of its volume and density:
$$
m = \rho_w \times \text{(volume)} = \rho_w \times \frac{4}{3}\pi R^3
$$
The viscous force (drag) acting on a sphere of radius $R$ moving at speed $v$ in a fluid with viscosity $\eta$ is given by Stokes’ Law:
$$
F_{\text{viscous}} = 6\pi \eta \, R \, v
$$
Step 3: Apply the Condition for Terminal Velocity
At terminal velocity, the net force is zero, so:
$$
F_g = F_{\text{viscous}}
$$
Substituting the expressions, we get:
$$
\rho_w \times \frac{4}{3}\pi R^3 \, g = 6\pi \eta \, R \, v
$$
Simplifying for $v$:
$$
v = \frac{\rho_w \cdot \frac{4}{3}\pi R^3 \, g}{6\pi \eta \, R}
$$
Cancel $\pi$ and rearrange terms:
$$
v = \frac{2\,\rho_w\,R^2\,g}{9\,\eta}
$$
Step 4: Substitute the Given Numerical Values
Given:
$\rho_w = 1000\ \text{kg\,m}^{-3}$
$R = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m}$
$g = 10\ \text{m\,s}^{-2}$
$\eta = 1.8 \times 10^{-5}\ \text{N\,s\,m}^{-2}$
Substitute into the formula:
$$
v = \frac{2 \times 1000 \times (0.2\times10^{-3})^2 \times 10 }{9 \times 1.8\times10^{-5}}
$$
Simplify the numerator and denominator carefully to get:
$$
v \approx 4.94\ \text{m\,s}^{-1}
$$
Step 5: Conclusion
Therefore, the terminal velocity of the raindrop is approximately
$$
4.94\ \text{m\,s}^{-1}.
$$
