A streamlined body falls through air from a height $h$ on the surface of a liquid. If $d$ and $D (D > d)$ represent the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is:

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$\sqrt{\frac{2 h}{g}} \cdot \frac{d}{D}$

$\sqrt{\frac{2 h}{g}} \cdot \frac{D}{d}$

$\sqrt{\frac{2 h}{g}} \cdot (\frac{d}{D - d})$

$\sqrt{\frac{2 h}{g}}$

answer is D.

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Detailed Solution

Up thrust – Weight of body = apparent weight

$V D g - V d g = V d a$

Where $a =$ retardation of body $= (\frac{D - d}{d}) g$

The velocity gained after fall from height $h$ in air, $v = \sqrt{2 g h}$

Hence, time taken to come to rest

$t = \frac{v}{a} = \frac{\sqrt{2 g h} d}{(D - d) g} = \sqrt{\frac{2 h}{g}} \times \frac{d}{(D - d)}$

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