Question

A spherical of radius $3.0\times {10}^{-4}m$ and density ${10}^{4}Kg/m^3$ falls freely under the gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the balls does not change, Find $h$. Viscosity of water is $9.8\times {10}^{-6}N-s/m^2.[g=9.8m/s^2]$

Answer 1

Before entering the water the velocity of ball i s $\sqrt{2gh}$.

Now, if after entering the water this velocity does not change then, value should be equal to the terminal velocity.

$\begin{array}{c}\therefore \sqrt{2gh}=\frac{2}{9}\frac{r^2\left(p-\sigma \right)g}{n}\\ h=\frac{{\left[\frac{2}{9}\frac{r^2\left(p-\sigma \right)g}{n}\right]}^2}{2g}\\ =\frac{\frac{4}{81}\frac{r^4{\left(p-\sigma \right)}^2{g}^2}{n^2}}{2g}\\ =\frac{4}{81}\frac{r^2{\left(p-\sigma \right)}^2{g}^2}{n^{2}2g}\\ =\frac{2}{81}\frac{r^2{\left(p-\sigma \right)}^{2}g}{n^2}\\ =\frac{2}{81}\times \frac{{\left(3\times {10}^{-4}\right)}^4{\left({10}^4-{10}^3\right)}^2\times 9.8}{{\left(9.8\times {10}^{-6}\right)}^2}\\ =1.65\times {10}^{3}m\end{array}$