A water drop of radius 1 μ m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1 . 8 × 10 - 5 N s m - 2 and its density is negligible as compared to that of water 10 6 g m - 3 . Terminal velocity of the water drop is (Take acceleration due to gravity = 10 m s - 2 )

Step-by-step solution

Terminal velocity is given by   v T = 2 r 2 ρ - σ g 9 η , where, r is radius of falling body, ρ is density of falling body and σ is density of fluid. ⇒ v T = 2 × 10 - 12 × 10 6 × 10 - 3 × 10 9 × 1 . 8 × 10 - 5 ⇒ v T = 123 . 4 × 10 - 6   m   s - 1