Question

in millikan's Oil Drop experiment what is the terminal velocity of a drop of radius is 2 into 10 to the power minus 5 metre and density is 1.2 into 10 power 3 kg per metre cube take the viscosity of air at temperature of the experiment to be 1.8 into 10 power minus 5 Newton second per metre square how much is the viscous force on the drop at that speeds neglect points of the Drop due to air​

Answer 1

The viscous force on the drop is F = 3.9 × 10^-10 N

Explanation:

Given data:

Viscous force = 3.9 × 10^–10 N

Terminal speed = 5.8 cm/s  = 5.8 x 10^-2 m/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

Solution:

The formula of 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

Thus the viscous force on the drop is F = 3.9 × 10^-10 N

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