A small spherical ball of radius 0.1 mm and density $$10^4$$ 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 ball does not change and it continue to fall with same constant velocity inside water, then the value of $$h$$ will be ______ m. (Given $$g = 10$$ m s$$^{-2}$$, viscosity of water $$= 1.0 \times 10^{-5}$$ N-s m$$^{-2}$$).

A small spherical ball falls freely through height h, then enters water and continues at the same constant velocity (terminal velocity). We need to find h. The radius of the ball is $$r = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}$$, its density is $$\rho = 10^4 \text{ kg/m}^3$$, the density of water is $$\sigma = 10^3 \text{ kg/m}^3$$, the viscosity of water is $$\eta = 1.0 \times 10^{-5} \text{ N·s/m}^2$$, and the acceleration due to gravity is $$g = 10 \text{ m/s}^2$$.

At terminal velocity, the net downward force (weight minus buoyancy) equals the viscous drag force: $$\frac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi \eta r v_t$$. Solving for $$v_t$$ gives $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$.

Substituting the values into this expression yields $$v_t = \frac{2 \times (10^{-4})^2 \times (10^4 - 10^3) \times 10}{9 \times 1.0 \times 10^{-5}}$$. Calculating step by step: $$r^2 = (10^{-4})^2 = 10^{-8} \text{ m}^2$$, $$\rho - \sigma = 10^4 - 10^3 = 9000 \text{ kg/m}^3$$, hence the numerator is $$\text{Numerator} = 2 \times 10^{-8} \times 9000 \times 10 = 2 \times 10^{-8} \times 9 \times 10^4 = 18 \times 10^{-4}$$ and the denominator is $$\text{Denominator} = 9 \times 10^{-5}$$, giving $$v_t = \frac{18 \times 10^{-4}}{9 \times 10^{-5}} = \frac{18}{9} \times \frac{10^{-4}}{10^{-5}} = 2 \times 10 = 20 \text{ m/s}$$.

As the ball falls freely from rest through height h, using $$v^2 = u^2 + 2gh$$ with $$u = 0$$, we have $$v_t^2 = 2gh$$ and therefore $$h = \frac{v_t^2}{2g} = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 \text{ m}$$. The answer is 20 m.