Question

A wooden ball of density $D$ is immersed in water of density $d$ to a depth $h$ below the surface of water and then released. Upto what height will the ball jump out of water( measured from surface of water) ?

• A. $\frac{d}{D}h$
• B. $(\frac{d}{D}-1)h$
• C. $h$
• D. $\text{Zero}$

Answer 1

The correct option is B $(\frac{d}{D}-1)h$

Let the volume of the ball be $V$.
Force on the ball due to upthrust $=Vdg$
Net upward force $=Vdg-VDg$
$\therefore$ upward acceleration is given by
$VDa=Vdg-VDg$
$\therefore a=\left(\frac{d-D}{D}\right)g$
Velocity on reaching the surface, $v=\sqrt{2ah}$
When ball leaves the surface, let maximum height reached be $H.$
Then,
$v=\sqrt{2gH}$
$H=\frac{ah}{g}=\left(\frac{d-D}{D}\right)h=(\frac{d}{D}-1)h$
Hence, the correct answer is option (b).