Answer 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 × 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 × (ρ – ρ0) g / 9η = 2 × (2 × 10-5)2 (1.2 × 103 – 0 ) × 9.8 / (9 × 1.8 × 10-5) = 5.8 × 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 × 3.14 × 1.8 × 10-5 × 2 × 10-5 × 5.8 × 10-2 = 3.9 × 10-10 N Hence, the viscous force on the drop is 3.9 × 10–10 N. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-10 Mechanical Properties of Fluids



Q.29:- 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.

 

 

Answer:-



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 × 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 × (ρ – ρ0) g / 9η
= 2 × (2 × 10-5)2 (1.2 × 103 – 0 ) × 9.8 / (9 × 1.8 × 10-5)
= 5.8 × 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 × 3.14 × 1.8 × 10-5 × 2 × 10-5 × 5.8 × 10-2
= 3.9 × 10-10 N
Hence, the viscous force on the drop is 3.9 × 10–10 N.