Question

The terminal velocity of a raindrop of radius $0.3mm$ falling through the air of viscosity $1.8\times {10}^{-5}Ns/m^2$ is equal to $($Consider density of raindrop as ${10}^{3}kg/m^3,$acceleration due to gravity $g=9.8m/s^2$and Neglect density of air$)$

• A. $10.9m/s$
• B. $8.3m/s$
• C. $9.2m/s$
• D. $7.6m/s$

Answer 1

The correct option is A $10.9m/s$

Given
$r=0.3mm=0.3\times {10}^{-3}m,\rho ={10}^{3}kg/m^3,g=9.8m{s}^{-2}$ $\eta =1.8\times {10}^{-5}Ns/m^2$


Weight of raindrop $=mg=\rho Vg=\frac{4}{3}\pi r^3\rho g$
From stokes law
Drag force on the spherical raindrop$=6\pi \eta r{v}_t$
Here $v_t$ is terminal velocity of raindrop

For equilbrium
Weight of raindrop= Drag force on the spherical raindrop

$\frac{4}{3}\pi r^3\rho g=6\pi \eta r{v}_t$
$v_t=\frac{2r^2\rho g}{9\eta }$

Where $\rho$ is the density of water and $\eta$ the viscosity of air.

Substituting

We get $v_t=\frac{2\times (0.3{)}^2\times {10}^{-3}\times 9.8}{9\times 1.8\times {10}^{-5}}=10.9m/s$