A water drop of radius 1 $\mu$m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 $\times$ 10$-$5 Nsm$-$2 and its density is negligible as compared to that of water 106 gm$-$3. Terminal velocity of the water drop is :
(Take acceleration due to gravity = 10 ms$-$2)
Solution
<p>$6\pi \eta rv = mg$</p>
<p>$6\pi \eta rv = {4 \over 3}\pi {r^3}\rho g$</p>
<p>or $$v = {2 \over 9}{{\rho {r^2}g} \over \eta } = {2 \over 9} \times {{{{10}^3} \times {{({{10}^{ - 6}})}^2} \times 10} \over {1.8 \times {{10}^{ - 5}}}}$$</p>
<p>$= 123.4 \times {10^{ - 6}}$ m/s</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
This question is part of PrepWiser's free JEE Main question bank. 183 more solved questions on Properties of Solids and Liquids are available — start with the harder ones if your accuracy is >70%.