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 $$f_w = 1000$$ kg m$$^{-3}$$ and Density of air $$f_a = 1.2$$ kg m$$^{-3}$$, g = 10 m/s$$^2$$, Coefficient of viscosity of air = $$1.8 \times 10^{-5}$$ N s m$$^{-2}$$]

For a very small spherical body falling slowly through a viscous fluid, Stokes established that the viscous drag acting opposite to the motion is

$$F_{\text{drag}} = 6\pi \,\eta\, r\, v$$

where $$\eta$$ is the coefficient of viscosity of the fluid, $$r$$ is the radius of the sphere and $$v$$ is its speed relative to the fluid.

At terminal speed the net force on the sphere becomes zero. The downward forces are its weight and (if considered) the upward buoyant force. Because the problem tells us to neglect buoyancy, the force balance at terminal speed $$v_t$$ is simply

$$\text{weight} = \text{viscous drag}$$

$$\rho_w \, V \, g = 6\pi \,\eta\, r\, v_t$$

where $$V$$ is the volume $$\left(\dfrac{4}{3}\pi r^3\right)$$ of the raindrop and $$\rho_w$$ is the density of water.

Substituting the volume of the sphere we have

$$\rho_w \left(\dfrac{4}{3}\pi r^3\right) g = 6\pi \eta r v_t$$

Dividing both sides by $$\pi r$$ and then simplifying, we get

$$\dfrac{4}{3}\,\rho_w \, r^2 \, g = 6 \eta v_t$$

Next, dividing both sides by $$6\eta$$ gives

$$v_t = \dfrac{4}{18}\,\dfrac{\rho_w\, r^2 g}{\eta}$$

Recognising that $$\dfrac{4}{18} = \dfrac{2}{9}$$, the standard Stokes-law expression for terminal speed when buoyancy is neglected becomes

$$v_t = \dfrac{2 r^{2} g \rho_w}{9 \eta}$$

More generally, if buoyancy is included it is $$v_t = \dfrac{2 r^{2} g (\rho_w - \rho_a)}{9 \eta}$$, and because $$\rho_a$$ is so much smaller than $$\rho_w$$ the same result is obtained numerically whether or not we subtract it. For completeness we shall keep the subtraction:

$$v_t = \dfrac{2 r^{2} g (\rho_w - \rho_a)}{9 \eta}$$

Now we substitute the numerical values. The radius is given as

$$r = R = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m} = 2.0 \times 10^{-4}\ \text{m}$$

First we calculate $$r^2$$ :

$$r^2 = \left(2.0 \times 10^{-4}\right)^2 = 4.0 \times 10^{-8}\ \text{m}^2$$

The difference in densities is

$$(\rho_w - \rho_a) = 1000\ \text{kg m}^{-3} - 1.2\ \text{kg m}^{-3} = 998.8\ \text{kg m}^{-3}$$

The factor $$2 r^{2} g (\rho_w - \rho_a)$$ therefore equals

$$2 \times (4.0 \times 10^{-8}) \times 10 \times 998.8$$

We multiply step by step:

$$2 \times 4.0 \times 10^{-8} = 8.0 \times 10^{-8}$$

Multiplying by $$10$$ gives

$$8.0 \times 10^{-7}$$

Finally, multiplying by $$998.8$$ gives

$$8.0 \times 998.8 \times 10^{-7} = 7990.4 \times 10^{-7} = 7.9904 \times 10^{-4}$$

Now we find $$9\eta$$ :

$$\eta = 1.8 \times 10^{-5}\ \text{N s m}^{-2}$$

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

Putting numerator and denominator together, the terminal speed is

$$v_t = \dfrac{7.9904 \times 10^{-4}}{1.62 \times 10^{-4}}$$

Dividing the powers of ten first we have $$10^{-4} / 10^{-4} = 1$$, so we simply divide the coefficients:

$$v_t = \dfrac{7.9904}{1.62}$$

Carrying out the division,

$$v_t \approx 4.93\ \text{m s}^{-1}$$

The tabulated options list 4.94 m s$$^{-1}$$, which is the same to two significant figures.

Hence, the correct answer is Option C.