A water drop of radius 1,mu m falls in a situ | vedclass

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(D) At terminal velocity,the viscous force is equal to the weight of the water drop (since buoyancy is neglected).$6 \pi \eta r v_t = \frac{4}{3} \pi

Vedclass · Accepted Answer

$123.4$

Vedclass · Answer

(D) At terminal velocity,the viscous force is equal to the weight of the water drop (since buoyancy is neglected).$6 \pi \eta r v_t = \frac{4}{3} \pi r^3 ho g$Rearranging for terminal velocity $v_t$:$v_t = \frac{2 r^2 ho g}{9 \eta}$Given values:$r = 1\,\mu m = 10^{-6}\,m$$\eta = 1.8 	imes 10^{-5}\,Nsm^{-2}$$ho = 10^3\,kgm^{-3}$$g = 10\,ms^{-2}$Substituting the values:$v_t = \frac{2 	imes (10^{-6})^2 	imes 10^3 	imes 10}{9 	imes 1.8 	imes 10^{-5}}$$v_t = \frac{2 	imes 10^{-12} 	imes 10^4}{16.2 	imes 10^{-5}}$$v_t = \frac{2 	imes 10^{-8}}{16.2 	imes 10^{-5}} = \frac{2}{16.2} 	imes 10^{-3} \approx 0.1234 	imes 10^{-3} = 123.4 	imes 10^{-6}\,ms^{-1}$

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