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= ⇒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
F=6π×1.8×10−5×2.0×10−5×5.807×10−2
=3.94×10−10N ≃3.9×10−10N