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

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