A wooden ball of density $\sigma$ is released from the bottom of a tank which is filled with a liquid of density $\rho ;(\rho >\sigma )$ up to a height $h_1$. The ball rises in the liquid, emerges from its surface and attains a height $h_2$ in air. If viscous effects are neglected, the ratio $h_2/h_1$ is

Weight of the ball $W=mg=\sigma Vg$. Upthrust $U=\rho Vg$. Therefore, the net upward force acting on the ball is
$F=U-W=(\rho -\sigma )Vg$
Now, mass of the ball is $m=\sigma V$. Therefore, upward acceleration of the ball while it is rising in the liquid is
$a=\frac{F}{m}=\frac{(\rho -\sigma )Vg}{\sigma V}=\left(\frac{\rho -\sigma }{\sigma }\right)g$
Velocity of the ball on reaching the surface of water is
$v=\sqrt{2a{h}_1}$
This is the initial upward velocity of the ball in air. If it rises to a height $h_2$ in air, we have
$v=\sqrt{2g{h}_2}$
Equating (i) and (ii), we have

 $$\begin{gathered} a{h}_1=g{h}_2 \\ or \frac{h_2}{h_1}=\frac{a}{g}=\frac{\rho -\sigma }{\sigma }=\left(\frac{\rho }{\sigma }-1\right) \end{gathered}$$
Hence the correct choice is (2).