Q.A steel ball with radius 0.3 mm is falling with velocity of 2 m/s at a time t, through a tube filled with glycerin, having coefficient of viscosity 0.833 Ns/m². Determine viscous force acting on the steel ball at that time.

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 …