A spherical ball of radius $1\times10^{-4}\ \text{m}$ and density $10^5\ \text{kg/m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change, then the value of $h$ is approximately: (The coefficient of viscosity of water is $9.8\times10^{-6}\ \text{N s/m}^2$)
Answer: (B) $2518\ \text{m}$
The ball must enter the water at its terminal velocity. The ball is 100 times denser than water, so the buoyancy is neglected here (as the options intend):
$$v_t = \frac{2r^2\rho g}{9\eta} = \frac{2\times10^{-8}\times10^5\times9.8}{9\times9.8\times10^{-6}} = \frac{2000}{9}\approx 222\ \text{m/s}$$
Free fall: $v_t^2 = 2gh$
$$h = \frac{(2000/9)^2}{2\times9.8} \approx 2519\ \text{m} \approx 2518\ \text{m}$$
(Including buoyancy would reduce $v_t$ by 1% and give about $2470\ \text{m}$.)
Solution by Sreeraj P, M.Sc Physics