A metallic sphere of radius $1.0 \times {10}^{-3} \mathrm{m}$ and density $1.0 \times {10}^4 \mathrm{kg}/{\mathrm{m}}^3$ enters 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 = $1.0 \times {10}^{-3} \mathrm{N} - \mathrm{s}/{\mathrm{m}}^2, \mathrm{g} = 10 \mathrm{m}/{\mathrm{s}}^2 \mathrm{and} \mathrm{density} \mathrm{of} \mathrm{water} = 1.0 \times {10}^3 \mathrm{kg}/{\mathrm{m}}^3.$

detailed solution

Correct option is B

The velocity attained by the sphere in falling freely from a height h isv = 2gh----------(i)This is the terminal velocity of the sphere in water. Hence by Stokes's law, we havev = 29r2(ρ-σ)gηwhere r is the radius of the sphere, ρ is the density of the material of the sphereσ( = 1.0×103 kg/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