Question

A small ball of density $\rho$ is immersed in a liquid of density $\sigma (>\rho )$ to a depth h and released. The height above the surface of water up to which the ball will jump is

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

Answer 1

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

$a=\left(\frac{\sigma -\rho }{\rho }\right)g$
$v=\sqrt{\frac{2(\sigma -\rho )gh}{\rho }}$
$\therefore H=\frac{v^2}{2g}=\left(\frac{\sigma -\rho }{\rho }\right)h$