Fluid Mechanics

JEE Main level, single correct, medium

185 of 192

Question 185 of 192

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0×10−5m2.0\times {10}^{-5}m and density 1.2×103kgm−31.2\times {10}^{3}kg{m}^{-3}? Take viscosity of liquid =1.8×10−5Nsm−2.=1.8\times {10}^{-5}Ns{m}^{-2}. (Neglect buoyancy due to air).

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Answer: (B) 3.9×10−10N3.9\times {10}^{-10}N

Use the concept the viscous force is the same as the weight of the body,
⇒F=mg=ρ×(43πr3)×g=3.9×10−10N\Rightarrow F=mg=\rho \times \left(\dfrac{4}{3}\pi {r}^{3}\right)\times g=3.9\times {10}^{-10}N.

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