A small cork ball of density $\sigma$ is immersed in water of density $\rho$ (> $\sigma$) to a depth $h$ and released. If $\rho =4\sigma$, the height up to which the ball will rise above the surface of water is

Let V be the volume of the ball. The upward force when the ball is completely immersed in water is 

$U=\rho Vg$

Downward force acting on the ball is its weight

$W=\sigma Vg$

$\therefore Net upward force is$

F = U$-$W = $\left(\rho -\sigma \right)$Vg 

Mass of the ball is $m=\sigma v$. Therefore, upward acceleration is 

$a=\frac{F}{m}=\frac{\left(\rho -\sigma \right)vg}{\sigma v}=\left(\frac{\rho -\sigma }{\sigma }\right)g \left(1\right)$

If u is the velocity of the ball when it emerges from water, then 

$$\begin{gathered} 0-u^2=-2ah \\ \Rightarrow u=\sqrt{2ah} \end{gathered}$$

If H is the maximum height attained,

$$\begin{gathered} 0-u^2=2\times \left(-g\right)H \\ \Rightarrow H=\frac{u^2}{2g}=\frac{2ah}{2g}=\frac{ah}{g} \end{gathered}$$

Using (1) in (2) we get

$H=\left(\frac{\rho -\sigma }{\sigma }\right)h$

Putting $\rho =4\sigma$, we get $H=3h$, which is choice (3)