Q.A spherical drop of oil falls at a constant speed of 4 cm/s in steady air. Calculate the radius of the drop. The density of the oil is 0.9 g/cm³, density of air is 0.0013 g/cm³ and the coefficient of viscosity of air is 1.8 × 10⁻⁴ poise, (g = 980 cm/s²)

Stokes' law gives the viscous drag force on a small sphere falling through a viscous fluid: $F_v = 6\pi\eta r v$, where r is the sphere's radius, v its speed, and η the fluid's coefficient of viscosity — the constant $6\pi$ being an empirically (and dimensionally) determined proportionality factor.

A sphere falling through a viscous medium experiences three forces: its weight $F_g$ (downward, constant), a buoyant upthrust $F_u$ (upward, constant), and the viscous drag $F_v$ (upward, growing as speed grows). Starting from rest, the sphere accelerates as long as $F_g$ exceeds $F_v + F_u$, but because $F_v$ keeps rising with speed, the net force shrinks and eventually vanishes once $F_g = F_v + F_u$ — from that point on the sphere falls at a constant speed, its terminal velocity. For a sphere of radius r and density ρ falling through a medium of density σ and viscosity η, this …