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A small body with relative density $d_{1}$ falls in air from a height 'h' on to the surface of liquid of relative density $d_{2}$ where $d_{2} > d_{1}$ . The time elapsed after entering the liquid to the instant when it comes to instantaneous rest inside liquid:

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2 h g d 2 d 1

2 h g d 1 ∣ d 2

2 h g d 1 d 2 − d 1

2 h g d 2 − d 1 d 1

answer is B.

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Detailed Solution

The speed with which the body enters water is $u = 2 g h$

Retardation is given by ${vd}_{1} = {αvd}_{2} g - {vd}_{1} g$

$α = - (\frac{d_{2} - d_{1}}{d_{1}}) g$

Using the equation of motion

$0 = \sqrt{2 gh} - (\frac{d_{2} - d_{1}}{d_{1}}) gt$
$t = 2 h g d 1 d 2 − d 1$

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