In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $$2.0 \times 10^{-5}$$ m and density $$1.2 \times 10^3$$ kg m$$^{-3}$$? Take viscosity of liquid = $$1.8 \times 10^{-5}$$ N s m$$^{-2}$$. (Neglect buoyancy due to air).

In Millikan’s oil-drop experiment each drop very quickly reaches its terminal (steady) velocity. At this stage the forces on the uncharged drop balance exactly, so the downward gravitational force is matched by the upward viscous (Stokes) force. Neglecting the tiny buoyant force of the air, we may therefore write

$$\text{Viscous force } F_v \;=\; \text{Weight of the drop } mg.$$

We first determine the mass $$m$$ of the spherical oil drop. The formula for the volume of a sphere is

$$V \;=\;\frac{4}{3}\,\pi r^{3},$$

where $$r$$ is the radius. Substituting $$r = 2.0 \times 10^{-5}\,\text{m}$$, we get

$$V \;=\;\frac{4}{3}\,\pi\,(2.0 \times 10^{-5})^{3}.$$ Now, $$\left(2.0 \times 10^{-5}\right)^{3} = 2^{3} \times 10^{-15} = 8 \times 10^{-15}.$$ So

$$V \;=\;\frac{4}{3}\,\pi \,\left(8 \times 10^{-15}\right) = \frac{32}{3}\,\pi \times 10^{-15}$$ $$\qquad = 10.666\ldots\,\pi \times 10^{-15}$$ $$\qquad \approx 33.51 \times 10^{-15}\;\text{m}^{3}$$ $$\qquad = 3.351 \times 10^{-14}\;\text{m}^{3}.$$

The mass is obtained from $$m = \rho V,$$ where the density of oil is $$\rho = 1.2 \times 10^{3}\,\text{kg m}^{-3}.$$ Hence

$$m \;=\; \left(1.2 \times 10^{3}\right)\left(3.351 \times 10^{-14}\right) = 4.021 \times 10^{-11}\;\text{kg}.$$

The weight of the drop is $$mg,$$ with $$g = 9.8\;\text{m s}^{-2}:$$

$$mg \;=\; (4.021 \times 10^{-11})(9.8) = 3.94 \times 10^{-10}\;\text{N}.$$

Because the drop is falling at terminal velocity, this weight is opposed by an equal viscous drag, so the magnitude of the viscous force is

$$F_v \;=\; 3.9 \times 10^{-10}\;\text{N}.$$

Hence, the correct answer is Option B.