Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.28

Question 10.28:
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.
Solution:
Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10–10 N
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, η = 1.8 X 10-5 Pa s
Density of air (ρ0) 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:
v = 2r2 X (ρ - ρ0) g / 9η
= 2 X (2 X 10-5)2 (1.2 X 103 - 0 ) X 9.8 / (9 X 1.8 X 10-5)
= 5.8 X 10-2 m/s
= 5.8 cm s-1
Hence, the terminal speed of the drop is 5.8 cm s–1.
The viscous force on the drop is given by:
F = 6πηrv
∴ F = 6 X 3.14 X 1.8 X 10-5 X 2 X 10-5 X 5.8 X 10-2
= 3.9 X 10-10 N
Hence, the viscous force on the drop is 3.9 × 10–10 N.

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