Question

A spherical ball of radius $1\times 10^{-4}m$ 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 the ball does not change, find $h$. The viscosity of water is $9.8\times 10^{-6}N-s/m^2$. (in $m$)

Answer 1

Answer: 20.40

Given: $r=1\times 10^{-4}m,g=9.8m/s^2$
$\rho =10^{4}kg/m^3,h=9.8\times 10^{-6}Ns/m^2$
$\sigma =1000kg/m^3$
Since ball is falling from height $h$ and when it enters water. Its velocity does not change, i.e. it has acquired terminal velocity.
$v_t=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$
$=\frac{2}{9}\times \frac{{\left(10^{-4}\right)}^2(10000-1000)9.8}{9.8\times 10^{-6}}$
$=\frac{2}{9}\times 10^{-2}\times 9000=20m/s$
and $v^2=2gh$
$h=\frac{v^2}{2g}=\frac{20\times 20}{2\times 9.8}$
$h=20.41m$