Answer :

Radius of the drop = 2 × 10-5 m


Density of the drop = 1.2 × 103 kg/m3


Viscosity of the air, μ = 1.8 × 10-5 Pa.s


*Consider density of air to be zero in order to neglect buoyancy force create by air on drop.


Terminal velocity is given by the relation,



Where,


r = radius


ρ = highest density = density of drop


ρ0 = Lowest density = density of air


g = acceleration due to gravity = 9.81 m/s


μ = viscosity




= 5.8 × 10-2 m/s


= 5.8 cm/s


Hence, the terminal speed of the drop = 5.8 cm/s


The viscous force on the drop is given by,



Where,


μ = viscosity


r = radius


v = velocity




= 3.9 × 10-10 N


Hence, the viscous force on drop = 3.9 × 10-10 N


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