Question

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 $\frac{h_2}{h_1}$ is

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

Answer 1

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

Weight of the ball,
$W=mg=\sigma Vg$,
where, $V$ is the volume of the ball.

Upthrust due to liquid acting on the ball,
$U=\rho Vg$,

Therefore, the net upward force acting on the ball is
$F=U-W=(\rho -\sigma )Vg$

Now, mass of the ball , $m=\sigma V$.

Therefore, upward acceleration of the ball while it is rising in the liquid is
$$a=\frac{F}{m}$$
$\Rightarrow a=\frac{(\rho -\sigma )Vg}{\sigma V}$

$\Rightarrow a=\frac{(\rho -\sigma )}{\sigma }g$

Velocity of the ball on reaching the surface of water is
$v=\sqrt{2a{h}_1}....\left(1\right)$

This is the initial upward velocity of the ball in air.

If it rises to a height $h_2$ in air, we have
$v=\sqrt{2g{h}_2}....\left(2\right)$

Equating equation $\left(1\right)$ and $\left(2\right)$, we have
$a{h}_1=g{h}_2$

$\Rightarrow \frac{h_2}{h_1}=\frac{a}{g}$

On subtituting the value of $a$,
$\Rightarrow \frac{h_2}{h_1}=\frac{\rho -\sigma }{\sigma }$

$\therefore \frac{h_2}{h_1}=(\frac{\rho }{\sigma }-1)$

Hence, option (b) is the correct answer