Question

A wooden ball of density $\rho$ is immersed in a liquid of density $\sigma$ to a depth H and then released. The height h above the surface to which the ball rises will be:

• A. $h=H$
• B. $h=\frac{\sigma }{\rho }H$
• C. $h=\frac{\sigma -\rho }{\rho }H$
• D. $h=\frac{\sigma -\rho }{\sigma }H$

Answer 1

The correct option is C $h=\frac{\sigma -\rho }{\rho }H$

Net upward force acting on the ball $F=\sigma Vg-\rho Vg$
$\Rightarrow ma=\sigma Vg-\rho Vg$
$\Rightarrow \left(\rho V\right)a=Vg(\sigma -\rho )$
$\Rightarrow a=g(\frac{\sigma }{\rho }-1)$
Velocity $v$ as it reaches the water surface $=\sqrt{2aH}=\sqrt{2g(\frac{\sigma }{\rho }-1)H}$
Conserving mechanical energy at max height h and at surface,
$mgh=\frac{1}{2}m\left(2g\right(\frac{\sigma }{\rho }-1\left)H\right)$
$\Rightarrow h=\left(\frac{\sigma -\rho }{\rho }\right)H$