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A ball of radius $r$ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\eta$, the value of $h$ is given by
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The correct answer is:
$\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$
Velocity of ball when it strikes the water surface $v=\sqrt{2 g h}$...(i)
Terminal velocity of ball inside the water $v=\frac{2}{9} r^2 g \frac{(\rho-1)}{\eta}$ ....(ii)
Equating (i) and (ii) we get $\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(\rho-1)$
$\Rightarrow \quad h=\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$
Terminal velocity of ball inside the water $v=\frac{2}{9} r^2 g \frac{(\rho-1)}{\eta}$ ....(ii)
Equating (i) and (ii) we get $\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(\rho-1)$
$\Rightarrow \quad h=\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$
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