Question

A wooden ball of density $\rho$ is immersed in water of density ${\rho }_0$ to depth $h$ and then released. The height $H$ above the surface of water upto which the ball jump out of water is

• A. zero
• B. h
• C. $\frac{{\rho }_{0}h}{\rho }$
• D. $\left(\frac{{\rho }_0}{\rho }-1\right)h$

Answer 1

Answer: (D)

Let $V$ be the volume of wooden ball. The mass of ball is $m=V\rho$
Upward acceleration,
$a=\frac{\text{ upward thrust }-\text{ weight of ball }}{\text{ mass of ball }}$
$=\frac{V{\rho }_{0}g-V\rho g}{V\rho }=\frac{\left({\rho }_0-\rho \right)g}{\rho }$
If $v^{{}^{\prime }}$ is the velocity of ball on reaching the surface after being released at depth $h$ is
$v=\sqrt{2as}={\left[2\left(\frac{p_0-\rho }{\rho }\right)gh\right]}^{1/2}$
If $h$ ' is the vertical distance reached by ball above the surface of water, then
$h^{{}^{\prime }}=\frac{v^2}{2g}=\frac{2\left({\rho }_0-\rho \right)}{\rho }gh\times \frac{1}{2g}$
$=\left(\frac{{\rho }_0-\rho }{\rho }\right)h=\left(\frac{{\rho }_0}{\rho }-1\right)h$