Q. 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 then ball jump out of water?
A) $\frac{d}{D}h$
B) $\left( \frac{d}{D} - 1 \right)h$
C) $h$
D) Zero
Explanation
(b) Let the volume of the ball be V . Force on the hall due to upthrust =Vdg
Net upward force = Vdg =VDg
If upward acceleration is a, then
$Vda = Vdg - VDg$
$\therefore$ $a = \left( \frac{d - D}{D} \right)g$
Velocity on reaching the surface, $v = \sqrt{2ah}$
Further $v = \sqrt{2ah}$
$\therefore$ $2ah = 2gH$
or $H = \frac{ah}{8} = \left( \frac{d - D}{D} \right)h = \left( \frac{d}{D} - 1 \right)h$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.