A metallic sphere of radius 1.0 × 10-3 m and density 1.0×104 kg/m3 enter a tank of water, after a free fall through a distance of h in the earth's gravitational field, if its velocity remains unchanged after entering water, determine the value of h. Given: coefficient of viscosity of water = 1.0 × 10-3 N-s/m2, g = 10 m/s2 and density of water = 1.0×103 kg/m3.

The velocity attained by the sphere in falling freely from a height h is                   v = 2gh-----(i)This is the terminal velocity of the sphere in water. Hence by Stoke's law, we have               v = 29r2(ρ-σ)gηwhere r is the radius of the sphere, ρ is the density of the material of the sphereσ(=1.0×103kg/m3) is the density of water and η is coefficient of viscosity of water.v = 2×(1.0×10-3)2(1.0×104-1.0×103)×109×1.0×10-3  = 20 m/sfrom equation (i), we haveh = v22g = 20×202×10 = 20 m