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  • A water drop of radius  falls in a situation where the effect of buoyant force is ne

A water drop of radius \mathrm{1 \mu \mathrm{m}} falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is \mathrm{1.8 \times 10^{-5} \mathrm{Nsm}^{-2}} and its density is negligible as compared to that of water \mathrm{10^{6} \mathrm{gm}^{-3}}. Terminal velocity of the water drop is :

\text { (Take acceleration due to gravity }=10 \mathrm{~ms}^{-2} \text { ) } 

Option: 1

\mathrm{145.4 \times 10^{-6} \mathrm{~ms}^{-1}}


Option: 2

\mathrm{118.0 \times 10^{-6} \mathrm{~ms}^{-1}}


Option: 3

\mathrm{132.6 \times 10^{-6} \mathrm{~ms}^{-1}}


Option: 4

\mathrm{123.4 \times 10^{-6} \mathrm{~ms}^{-1}}


Answers (1)

best_answer

\mathrm{\mathrm{V_{T}=\frac{2}{9} r^{2} \frac{(\sigma-\rho) g}{\eta }}}

            \mathrm{\rho=0}

\mathrm{V_{T}=\frac{2}{9}\times \frac{\left ( 1\times 10^{-6} \right )^{2}\times \left ( 10^{3} \right )\times 10}{1.8\times 10^{-5}}}

\mathrm{=123.4 \times 10^{-6} \mathrm{~m} / \mathrm{s}}

Hence option 4 is correct.

Posted by

Divya Prakash Singh

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