Question

A small wooden ball of density $\rho$ is immersed in water of density $\sigma$ to depth $h$ and then released. The height $H$ above the surface of water up to which the ball will jump out of water is

• A. $\frac{\sigma h}{\rho }$
• B. $(\frac{\sigma }{\rho }-1)h$
• C. $h$
• D. $zero$

Answer 1

Answer: (B)

$(\frac{\sigma }{\rho }-1)h$

Net upward force acting on the body $F=\sigma Vg-\rho Vg$
$\Rightarrow ma=\sigma Vg-\rho Vg$
$\Rightarrow \left(\rho V\right)a=Vg(\sigma -\rho )$
$\Rightarrow a=g(\frac{\sigma }{\rho }-1)$
Velocity $v$ after height $h=\sqrt{2g(\frac{\sigma }{\rho }-1)h}$
Conserving mechanical energy at max height h' above surface,
$mg{h}^{\prime }=\frac{1}{2}m\left(2g(\frac{\sigma }{\rho }-1)h\right)$
$\Rightarrow h^{\prime }=(\frac{\sigma }{\rho }-1)h$