A small spherical ball of radius r and density is dropped from a height h above the surface of a large tank of oil. The density of the oil is (where < ) and its viscosity is . It is observed that the velocity of the ball does not change upon entering the oil. Assuming the acceleration due to gravity is g and neglecting air resistance, the expression for the height h is

Step-by-step solution

The velocity of the ball just before entering the oil is obtained from the kinematics of free fall: v = 2gh Since the velocity of the ball does not change upon entering the oil, this entry velocity must be exactly equal to the terminal velocity of the ball in the oil. The terminal velocity v_T is given by Stokes' law: v_T = 2 9 r^2 ( - ) g Equating the two velocities: 2gh = 2 9 r^2 ( - ) g Squaring both sides: 2gh = 4 81 r^4 ( - )^2 g^2 ^2 Solving for h : h = 1 2g ( 4 81 r^4 ( - )^2 g^2 ^2 ) h = 2 81 r^4 ( - )^2 g