An oil drop falls through air and achieves a steady terminal velocity of 1.0 mm s ⁻¹ . The coefficient of viscosity of air is 2.0 10⁻⁵ N s m ⁻² and the density of the oil is 1000 kg m ⁻³ . Assuming the effect of buoyant force is negligible, the radius of the oil drop is (Take g = 10 m s ⁻² )

Step-by-step solution

Terminal velocity is given by v = 2 r^2 g 9 (neglecting the density of air). Rearranging for the radius r gives r^2 = 9 v 2 g . Substituting the given values: v = 1.0 mm s ⁻¹ = 10⁻³ m s ⁻¹ = 2.0 10⁻⁵ N s m ⁻² = 1000 kg m ⁻³ g = 10 m s ⁻² r^2 = 9 2.0 10⁻⁵ 10⁻³ 2 1000 10 r^2 = 18 10⁻⁸ 2 10^4 = 9 10⁻¹² m ^2 r = 3 10⁻⁶ m = 3 m Answer: 3 m