Two rain drops of same radius ' r ' falling with same terminal velocity ' V ' merge and form bigger drop of radius ' R '. The terminal velocity of big drop is

Step-by-step solution

The terminal velocity of a spherical drop of radius r falling through a viscous medium is given by Stokes' law: V = 2 9 r^2 ( - ) g From this relation, the terminal velocity is directly proportional to the square of the radius of the drop: V r^2 For the smaller drop of radius r , the terminal velocity is V . For the bigger drop of radius R , let the terminal velocity be V' . Taking the ratio of their terminal velocities: V' V = R^2 r^2 V' = V R^2 r^2 Answer: VR^2 r^2