Question 180 of 192

A small spherical ball of radius $0.1mm$ 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 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 $=1.0\times {10}^{-5}N‐s{m}^{-2}$ ).

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Answer: 20

Using equation of motion, ${v}^{2}={u}^{2}+2as$
${v}^{2}=0+2\left(-g\right)\left(-h\right)$
$\Rightarrow {v}^{2}=2gh$
Initial velocity of ball is $v=\sqrt{2gh}=\sqrt{20h}$.
If after entering the water this velocity does not change, then this value should be equal to the terminal velocity.
Then, $\sqrt{2gh}=\dfrac{2}{9}\dfrac{{r}^{2}\left(\rho -\sigma \right)g}{\eta }$, here, $\rho$ is density of ball, $\sigma$ is density of water and $\eta$ is viscosity.
$\Rightarrow \sqrt{20h}=\dfrac{2}{9}\dfrac{{\left(0.1\times {10}^{-3}\right)}^{2}\left({10}^{4}-{10}^{3}\right)10}{1\times {10}^{-5}} \Rightarrow \sqrt{20h}=\dfrac{2}{9}\dfrac{{10}^{-8}\left(10-1\right){10}^{3}\times 10}{1\times {10}^{-5}}$
$\Rightarrow 20h=400$
$∴h=20m$