Mechanical Properties of Fluids

Question : 28

Total: 31

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×10–5 m and density 1.2×103kg m–3. Take the viscosity of air at the temperature of the experiment to be 1.8×10–5Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Solution:

Here, r=2×10−5m,
η=1.8×10−5 Pa s,
ρ=1.2×103 kg m−3
When the buoyancy of the drop due to air is neglected, the terminal speed of the drop is given by
v=

2r2ρg

9η

2×(2×10−5)2×1.2×103×9.8

9×1.8×10−5

=5.81×10−2 m s−1
=5.81 cm s−1.
Viscous force on the drop,F=6πηrv

=6×

×(1.8×10−5)×(2.0×10−5)×(5.8×10−2)

=3.93×10−10 N

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