A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water fw = 1000 kg m$-$3 and Density of air fa = 1.2 kg m$-$3, g = 10 m/s2, Coefficient of viscosity of air = 1.8 $\times$ 10$-$5 Nsm$-$2]
Solution
At terminal speed<br><br>a = 0<br><br>F<sub>net</sub> = 0<br><br>mg = F<sub>v</sub> = 6$\pi$ $\eta$Rv<br><br>$v = {{mg} \over {6\pi \eta Rv}}$<br><br>$v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}$<br><br>$= {{2{\rho _w}{R^2}g} \over {9\eta }}$<br><br>$= {{400} \over {81}}$ m/s<br><br>= 4.94 m/s
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%.