The terminal velocity of a raindrop of radius 0.3 mm falling through the air of viscosity 1.8×10−5 Ns/m2 is equal to (Consider density of raindrop as 103kg/m3,acceleration due to gravity g=9.8m/s2and Neglect density of air)
Given
r=0.3 mm=0.3×10−3m,ρ=103 kg/m3,g=9.8 ms−2 η=1.8×10−5Ns/m2
Weight of raindrop =mg=ρVg=43πr3ρg
From stokes law
Drag force on the spherical raindrop=6πηrvt
Here vt is terminal velocity of raindrop
For equilbrium
Weight of raindrop= Drag force on the spherical raindrop
43πr3ρg=6πηrvt
vt=2r2ρg9η
Where ρ is the density of water and η the viscosity of air.
Substituting
We get vt=2×(0.3)2×10−3×9.89×1.8×10−5=10.9 m/s