Question

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10^-5 m and density 1.2 × 10³ kg m³? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10^-5 Pas. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.(Take g = 9.8 ms²)

Answer 1

Radius of the given unchanged drop, r = 2 × 10^-5 m

Density of the uncharged drop,

ρ = 1.2 × 10³ kg/m^-3

Viscosity of air, η = 1.8 × 10^-5 Pa S

The density of air (ρ_o) can be taken as zero in order to neglect the buoyancy of air.

Terminal velocity (v) is given by the relation:

V = 5.807 × 10^-2 m/s

V = 5.8 cm/s

The terminal speed of the drop is 5.8 cm/s

Viscous force, F = π 6 hrv

∴ F = 6 × 3.14 × 1.8 × 10^-5 × 2 × 10⁵ × 5.8 × 10²

⇒ F = 3.9 × 10^-10N

The viscous force on the drop is

= 3.9 × 10^-10 N.