Question
Asked Dec 23, 2019
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The viscous force on an oil drop is measured to be equal to 3.0 x 10-13 N when the drop is falling through air with a speed of 4.5 x 10-4 m/s. If the radius of the drop is 2.5 x 10-6 m, what is the viscosity of air?

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Expert Answer

Step 1

Given,

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-13 Viscous force,F = 3×10¬ N Air speed,v = 4.5×10¬ m/s Drop radius,r =2.5×10 m

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Step 2

Viscosity of air can be calculate using Stoke’s law which can be mathematically expressed as below,

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F-6πμην where, µ = Viscosity of air

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Step 3

By plugging in the g...

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F-6μν (3x101') = 6πμ (2.5 x10") (4.5x10") 3x10-13 μ . 2.12×108 μ= 1.41 x105 N.sim?

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