JEE Main 27-Aug-2021 Shift 1 Solved Paper

Given, radius of oil drop, r=2.0×10−5m
Density of oil drop, ρ=1.2×103kgm−3
Viscosity of liquid, η=1.8×10−5Nsm−2
To neglect buoyancy, consider the density of air, ρa = 0
Viscous force acting on drop can be given as
F=6πη‌rV ...(i)
Here, v is terminal velocity.
The terminal velocity of drop can be given as
v=

2r2(ρ−ρa)g

9η

⇒v=

2×(2.0×10−5)2(1.2×103−0)×9.8

9×1.8×10−5

=5.807×10−2ms−1
Substituting all values in Eq. (i), we get =3.94×10−10N
≃3.9×10−10N