Q. A streamlined body falls through air from a height $h$ on the surface of a liquid. If $d$ and $D(D > d)$ represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is
A) $\sqrt{\frac{2h}{g}}$
B) $\sqrt{\frac{2h}{g}.\frac{D}{d}}$
C) $\sqrt{\frac{2h}{g}.\frac{d}{D}}$
D) $\sqrt{\frac{2h}{g}}\ \left( \frac{d}{D - d} \right)$
Explanation
(d) Upthrust – weight of body = apparent weight
$VDg - Vdg = Vda$
Where $a =$ retardation of body $\therefore a = \left( \frac{D - d}{d} \right)g$
The velocity gained after fall from $h$ height in air, $v = \sqrt{2gh}$
Hence, time to come in rest
$t = \frac{v}{a} = \frac{\sqrt{2gh} \times d}{(D - d)g} = \sqrt{\frac{2h}{g}} \times \frac{d}{(D - d)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.