JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A water drop of radius \(1\,\mu m\) falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is \(1.8 \times 10^{-5}\,Nsm ^{-2}\) and its density is negligible as compared to that of water \(10^{6}\,gm ^{-3}\). Terminal velocity of the water drop is________ \(\times 10^{-6}\,ms ^{-1}\) (Take acceleration due to gravity \(=10\,ms ^{-2}\) )
- A \(145.4\)
- B \(118.0\)
- C \(132.6\)
- D \(123.4\)
Answer & Solution
Correct Answer
(D) \(123.4\)
Step-by-step Solution
Detailed explanation
\(6 \pi \eta r v _{ t }=\frac{4}{3} \pi r ^{3} \rho g\) \(v _{ t }=\frac{4}{3} \times \frac{\pi r^{3} \rho g }{6 \pi \eta r}\) \(v _{ t }=\frac{4}{3} \times \frac{\pi r ^{3} \rho g }{6 \pi \eta r}=\frac{2 \times 10^{-12} \times 10^{3} \times 10}{9 \times 1.8 \times 10^{-5}}\)…
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