Question

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. $\sqrt{\frac{2 h}{g}}$
• B. $\sqrt{\frac{2 h}{g} \cdot \frac{D}{d}}$
• C. $\sqrt{\frac{2 h}{g} \cdot \frac{d}{D}}$
• D. $\sqrt{\frac{2 h}{g}}\left(\frac{d}{D-d}\right)$

Answer 1

Answer: (D)

Upthrust - weight of body = apparent weight
$V D g-V d g=V d a$
where $a=$ retardation of body
$\therefore a=\left(\frac{D-d}{d}\right) g$
The velocity gained after fall from $h$ height in air, $v=\sqrt{2 g h}$ Hence, time to come in rest,
$t=\frac{v}{a}=\frac{\sqrt{2 g h} \times d}{(D-d) g}=\sqrt{\frac{2 h}{g}} \times \frac{d}{(D-d)}$