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Mechanical Properties of Fluids

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Question : 28 of 31
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In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×10−52.0 \times 10^{-5} m and density 1.2×103kg m−31.2 \times 10^{3} \text{kg m}^{-3}. Take the viscosity of air at the temperature of the experiment to be 1.8×10−5Pa1.8 \times 10^{-5} \text{Pa} 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,r = 2 \times 10^{-5} \text{m},
η=1.8×10−5Pa s,\eta = 1.8 \times 10^{-5} \text{Pa s},
ρ=1.2×103kg m−3\rho = 1.2 \times 10^{3} \text{kg m}^{-3}
When the buoyancy of the drop due to air is neglected, the terminal speed of the drop is given by
v=2r2ρg9ηv = \frac{2 r^{2} \rho g}{9 \eta}
=2×(2×10−5)2×1.2×103×9.89×1.8×10−5= \frac{2 \times (2 \times 10^{-5})^{2} \times 1.2 \times 10^{3} \times 9.8}{9 \times 1.8 \times 10^{-5}}
=5.81×10−2m s−1= 5.81 \times 10^{-2} \text{m s}^{-1}
=5.81cm s−1= 5.81 \text{cm s}^{-1}.
Viscous force on the drop,F=6πηrvF = 6 \pi \eta r v
=6×227×(1.8×10−5)×(2.0×10−5)×(5.8×10−2)= 6 \times \frac{22}{7} \times (1.8 \times 10^{-5}) \times (2.0 \times 10^{-5}) \times (5.8 \times 10^{-2})
=3.93×10−10N= 3.93 \times 10^{-10} \text{N}
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