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

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

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

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

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

answer is D.

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Detailed Solution

Upthrust – weight of body = apparent weight

$VDg - Vdg = Vda,$

Where a = retardation of body

$∴ a = (\frac{D - d}{d}) g$
The velocity gained after fall from h height in air, $v = \sqrt{2 gh}$
Hence, time to come in rest, $t = \frac{v}{a} = \frac{\sqrt{2 gh} \times d}{(D - d) g} = \sqrt{\frac{2 h}{g}} \times \frac{d}{(D - d)}$

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