Question

In Millikan's oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10⁻⁵ m and density 1.2 x 10 ³kg m⁻³. Take the viscosity of air at the temperature of the experiment to be 1.8 x 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 1

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Answer:
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Viscous force = 3.9 × 10^–10 N

Terminal speed = 5.8 cm/s

Density of the uncharged drop, 

ρ = 1.2 × 10^3 kg m^–3

Radius of the given uncharged drop, 

r = 2.0 × 10^–5 m

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

Terminal velocity (v) is given by the relation:

v = 2r^2 × (ρ – ρ0) g / 9η
= 2 × (2 × 10^-5)^2 (1.2 × 10^3 – 0 ) × 9.8 /
(9 × 1.8 × 10^-5)
= 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.

#Be Brainly❤️

Answer 2

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Viscous force = 3.9 × 10^–10 N

Terminal speed = 5.8 cm/s

Density of the uncharged drop, 

ρ = 1.2 × 10^3 kg m^–3

Radius of the given uncharged drop, 

r = 2.0 × 10^–5 m

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

Terminal velocity (v) is given by the relation:

v = 2r^2 × (ρ – ρ0) g / 9η
= 2 × (2 × 10^-5)^2 (1.2 × 10^3 – 0 ) × 9.8 /
(9 × 1.8 × 10^-5)
= 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.✔✔✔


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