Question

A small spherical ball of radius $0.1mm$ and density $10^{4}kg{m}^{-3}$ falls freely under gravity through a 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 $g=10m{s}^{-2}$, viscosity of water $=1.0\times 10^{-5}$ $N-s{m}^{-2}$ ).

Answer 1

Answer: 20

Speed after falling through height $h$
Should be equal to terminal velocity
$\sqrt{2gh}=\frac{2}{9}\frac{r^2(d-\rho )g}{\eta }$
$\sqrt{2gh}=\frac{2}{9}\frac{10^{-8}(10000-1000)\times 10}{10^{-5}}$
$=\frac{2}{9}\times 10^{-8}\frac{9\times 10^4}{10^{-5}}=20$
$2\times 10\times h=400$
$h=20m$