Question

A small wooden ball of density $\rho$ is immersed in water of density $\sigma$ to a 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

The correct option is B $(\frac{\sigma }{\rho }-1)h$

If v is the volume of the ball,

ma = FB = mg
$\rho va=\sigma vg-\rho vg$
$a=(\frac{\sigma }{\rho }-1)g$
if v is the velocity at the surface,
$v^2=u^2+2as$
$v^2=2(\frac{\sigma }{\rho }-1)hg$
maximum height reached is
$H=\frac{v^2}{2g}$
$H=(\frac{\sigma }{\rho }-1)h$