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Terminal Velocity

In Millikan's...

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 m^–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10^–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

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Solution

Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10^–10N

Radius of the given uncharged drop, r = 2.0 × 10^–5 m

Density of the uncharged drop, ρ = 1.2 × 10³ kg m^–3

Viscosity of air,

Density of air can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g = 9.8 m/s²

Terminal velocity (v) is given by the relation:

Hence, the terminal speed of the drop is 5.8 cm s^–1.

The viscous force on the drop is given by:

Hence, the viscous force on the drop is 3.9 × 10^–10N.

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