wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The terminal velocity of a raindrop of radius 0.3 mm falling through the air of viscosity 1.8×105 Ns/m2 is equal to (Consider density of raindrop as 103kg/m3,acceleration due to gravity g=9.8m/s2and Neglect density of air)

A
10.9 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
8.3 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9.2 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7.6 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 10.9 m/s

Given
r=0.3 mm=0.3×103m,ρ=103 kg/m3,g=9.8 ms2 η=1.8×105Ns/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)2×103×9.89×1.8×105=10.9 m/s


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Viscosity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon