1. Given data

• Radius of raindrop: $R = 0.2\text{ mm} = 2 \times 10^{-4}\text{ m}$
• Density of water: $\rho_w = 1000\,\text{kg m}^{-3}$
• Density of air: $\rho_a = 1.2\,\text{kg m}^{-3}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$
• Coefficient of viscosity of air: $\eta = 1.8 \times 10^{-5}\,\text{N s m}^{-2}$
• Height of cloud: $h = 2000\text{ m}$

The question asks for the terminal speed of the raindrop.

2. Use Stokes' law for terminal velocity

For a small spherical drop, neglecting buoyancy as instructed, terminal velocity is given by

$v_t = \frac{2 R^2 \rho_w g}{9\eta}$

(If buoyancy were included, we would use $\rho_w - \rho_a$, but the problem says to neglect buoyancy.)

3. Substitute the values

First compute $R^2$:

$R^2 = (2 \times 10^{-4})^2 = 4 \times 10^{-8}$

Now,

$v_t = \frac{2 \times (4 \times 10^{-8}) \times 1000 \times 10}{9 \times 1.8 \times 10^{-5}}$

Simplify numerator:

$2 \times 4 \times 10^{-8} \times 1000 \times 10 = 8 \times 10^{-8} \times 10^4 = 8 \times 10^{-4}$

Simplify denominator:

$9 \times 1.8 \times 10^{-5} = 16.2 \times 10^{-5} = 1.62 \times 10^{-4}$

So,

$v_t = \frac{8 \times 10^{-4}}{1.62 \times 10^{-4}} = \frac{8}{1.62} \approx 4.94\,\text{m s}^{-1}$

4. Check whether terminal speed is attainable within 2000 m

Since the question gives the cloud height, we should verify whether the drop can attain terminal speed before reaching the ground. For such a small drop, terminal speed is reached very quickly, so the given height is more than sufficient.

Thus the terminal speed is indeed

$v_t \approx 4.94\,\text{m s}^{-1}$

5. Option matching

• A: $250.6\,\text{m s}^{-1}$ ❌
• B: $43.56\,\text{m s}^{-1}$ ❌
• C: $4.94\,\text{m s}^{-1}$ ✅
• D: $14.4\,\text{m s}^{-1}$ ❌

Therefore, the correct option is:

$\boxed{\text{C}}$