Q. 285.0( 7 Votes )

# In Millikan’s oil

Answer :

Radius of the drop = 2 × 10-5 m

Density of the drop = 1.2 × 103 kg/m3

Viscosity of the air, μ = 1.8 × 10-5 Pa.s

*Consider density of air to be zero in order to neglect buoyancy force create by air on drop.

Terminal velocity is given by the relation, Where,

r = radius

ρ = highest density = density of drop

ρ0 = Lowest density = density of air

g = acceleration due to gravity = 9.81 m/s

μ = viscosity  = 5.8 × 10-2 m/s

= 5.8 cm/s

Hence, the terminal speed of the drop = 5.8 cm/s

The viscous force on the drop is given by, Where,

μ = viscosity

r = radius

v = velocity  = 3.9 × 10-10 N

Hence, the viscous force on drop = 3.9 × 10-10 N

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