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Home » 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.
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.
CBSE | Class 11 | Mechanical Properties of Fluids | NCERT | Physics
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

According to the question, the acceleration due to gravity is

g = 9.8 m/s2
The radius of the uncharged drop is

r = 2.0 × 10-5 m
The density of the uncharged drop is

ρ = 1.2 × 103 kg m-3
The viscosity of air is

η =  1.8 × 10-5 Pa s
To avoid taking into account air buoyancy, we set the density of air to zero.Therefore terminal velocity (v) is :

v=\frac{2{{r}^{2}}g\rho }{9\eta }

v=\frac{2{{(2\times {{10}^{-5}})}^{2}}\times 9.8\times 1.2\times {{10}^{3}}}{9\times 1.8\times {{10}^{-5}}}

v=5.8c{{m}^{-1}}

And the expression of the viscous force on the drop is as follows :
F = 6πηrv
F = 6 x 3.14 x 1.8 × 10-5x2 x 10-5 x  5.8 10-2
f = 3.91 x 10-10 N

i

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