   Chapter 9, Problem 63P

Chapter
Section
Textbook Problem

The viscous force on an oil drop is measured to be equal to 3.0 × 10−13 N when the drop is falling through air with a speed of 4.5 × 10−4 m/s. If the radius of the drop is 2.5 × 10−5 m, what is the viscosity of air?

To determine

The viscosity of air.

Explanation

Stokes’ law gives the resistive force on a very small spherical object falling slowly through a fluid of viscosity η with radius r is Fr=6πrvη. Now this expression is rearranged for the viscosity of air , if all other values are known.

Given info: The viscos force on an oil drop is 3.0×1013N, the speed of oil drop is 4.5×104m/s, and radius of the oil drop is 2.5×106m.

The formula for the viscosity of air is,

η=Fr6πrv

• Fr is viscos force.
• r is radius of the oil drop.
• v is speed of oil drop.

Substitute 3.0×1013N for Fr, 4.5×104m/s for v, and 2

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