Question

A ball of mass $m$ and density $p$ is immersed in a liquid of density $3p$ at a depth $h$ and is 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. $3h$
• D. $2h$

Answer 1

The correct option is D $2h$

$\begin{array}{c}\Rightarrow \left(3p\times v\times g\right)h-Mgh=\frac{1}{2}m{v}^2\\ \Rightarrow 3mgh-mgh=\frac{1}{2}m{v}^2\\ \Rightarrow 4mgh=v^2.............\left(i\right)\\ \therefore h^{\prime }=\frac{v^2}{2g}=\frac{4gh}{2g}=2h\end{array}$

$\therefore$ Option $D$ is correct.