Question

A ball of mass m and density $\rho$ is immersed in a liquid of density $3\rho$ at a depth h and released. To what height will the ball jump up above the surface of liquid? (neglect the resistance of water and air).

• A. h
• B. $\frac{h}{2}$
• C. 2h
• D. 3h

Answer 1

The correct option is C 2h

Volume of ball $V=\frac{m}{\rho }$
Acceleration of ball inside the liquid
$a=\frac{F_{net}}{m}=\frac{\text{upthrust - weight}}{m}$
or $a=\frac{\left(\frac{m}{\rho }\right)\left(3\rho \right)\left(g\right)-mg}{m}=2g$ (upwards)
$\therefore$ Velocity of ball while reaching at surface
$v=\sqrt{2ah}=\sqrt{4gh}$
$\therefore$ The ball will jump to a height
$H=\frac{v^2}{2g}=\frac{4gh}{2g}=2h$