A water drop of radius $$1 \mu$$m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is $$1.8 \times 10^{-5}$$ N s m$$^{-2}$$ and its density is negligible as compared to that of water $$10^6$$ g m$$^{-3}$$. Terminal velocity of the water drop is
(Take acceleration due to gravity $$= 10$$ m s$$^{-2}$$)

Given: radius $$r = 1 \, \mu$$m $$= 10^{-6}$$ m, coefficient of viscosity $$\eta = 1.8 \times 10^{-5}$$ N s m$$^{-2}$$, density of water $$\rho = 10^6$$ g m$$^{-3}$$ $$= 10^3$$ kg m$$^{-3}$$, $$g = 10$$ m s$$^{-2}$$. Buoyant force is negligible.

Write the formula for terminal velocity (neglecting buoyancy).

When buoyancy is negligible, at terminal velocity the drag force equals the weight:

$$6\pi\eta r v_t = \frac{4}{3}\pi r^3 \rho g$$

$$v_t = \frac{2r^2 \rho g}{9\eta}$$

Substitute the values.

$$v_t = \frac{2 \times (10^{-6})^2 \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}}$$

$$v_t = \frac{2 \times 10^{-12} \times 10^4}{16.2 \times 10^{-5}}$$

$$v_t = \frac{2 \times 10^{-8}}{16.2 \times 10^{-5}}$$

$$v_t = \frac{2}{16.2} \times 10^{-3} = 0.1234 \times 10^{-3}$$

$$v_t = 123.4 \times 10^{-6} \text{ m s}^{-1}$$

The correct answer is Option B.