A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8×10-5 kg/(m-s), what will be the terminal velocity of the drop? (density ofwater = 1.0×103 kg/m2 and g = 9.8 m/s2) Density of air can be neglected.

By Stokes'law , the terminal velocity of a water drop of radius r is given byv = 29r2(ρ-σ)gηwhere ρ is the density of water, σ is the density of air and η the coefficient of viscosity of air. Here σ is negligible and r = 0.0015 mm = 1.5 x 10-3 mm: 1.5 x 10-6 m. Substituting the values:v = 29×(1.5×10-6)2×(1.0×103)×9.81.8×10-5   = 2.72 ×10-4 m/s