Fluid Mechanics

JEE Main level, single correct, medium

181 of 192

Question 181 of 192

A water drop of radius 1μm1\mu m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×10−5Nsm−21.8\times {10}^{-5}Ns{m}^{-2} and its density is negligible as compared to that of water 106gm−3{10}^{6}g{m}^{-3}. Terminal velocity of the water drop is
(Take acceleration due to gravity =10ms−2=10{ms}^{-2})

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Answer: (B) 123.4×10−6ms−1123.4\times {10}^{-6}{ms}^{-1}

Terminal velocity is given by vT=2r2(ρ−σ)g9η{v}_{T}=\dfrac{2{r}^{2}\left(\rho -\sigma \right)g}{9\eta }, where, rr is radius of falling body, ρ\rho is density of falling body and σ\sigma is density of fluid.
⇒vT=2×10−12×106×10−3×109×1.8×10−5\Rightarrow {v}_{T}=\dfrac{2\times {10}^{-12}\times {10}^{6}\times {10}^{-3}\times 10}{9\times 1.8\times {10}^{-5}}
⇒vT=123.4×10−6ms−1\Rightarrow {v}_{T}=123.4\times {10}^{-6}m{s}^{-1}

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