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 $v^{'}$ is the velocity of ball on reaching the surface after being released at depth $h$ is
$v=\sqrt{2 a s}=\left[2\left(\frac{p_{0}-\rho}{\rho}\right) g h\right]^{1 / 2}$
If $h$ ' is the vertical distance reached by ball above the surface of water, then
$ h^{'} =\frac{v^{2}}{2 g}=\frac{2\left(\rho_{0}-\rho\right)}{\rho} g h \times \frac{1}{2 g} $
$ =\left(\frac{\rho_{0}-\rho}{\rho}\right) h=\left(\frac{\rho_{0}}{\rho}-1\right) h$