Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 ~cm / s coalesce to form a bigger drop. The terminal velocity of bigger drop is _ _ _ _ cm / s .

Step-by-step solution

At terminal velocity, gravitational force equals Stokes drag: mg = 6 rv For a sphere with mass m = 4 3 r^3 , this gives v = 2r^2 g 9 Therefore, v r^2 When 64 drops coalesce: 64 4 3 r₁^3 = 4 3 r₂^3 , so r₂ = 4r₁ New terminal velocity: v₂ v₁ = ( r₂ r₁ )^2 = 16 v₂ = 16 10 = 160 cm/s