Question

A small spherical ball of radius $0.1\, mm$ 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 $g =10 ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5}$ $N - sm ^{-2}$ ).

Answer 1

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