A stream-line body with relative density d, falls into air from a height h, on the surface of a liquid of relative density $d_{2}$ where d₂>d₁. The time of immersion of the body into liquid will be

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$t = \sqrt{(\frac{2 h_{1}}{g})} \times \frac{d_{1}}{d_{2} - d_{1}}$

$t = \sqrt{(\frac{2 h_{1}}{g})}$

$t = \sqrt{(\frac{2 h_{1}}{g})} \times \frac{d_{1}}{d_{2}}$

$t = \sqrt{(\frac{2 h_{1}}{g})} \times \frac{d_{2}}{d_{1}}$

answer is A.

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Detailed Solution

Velocity of the body when it strikes the liquid surface
$V = \sqrt{2 {gh}_{1}} upward thrust = {Vd}_{2} g - {Vd}_{1} g = (Retardation produced by upthrust) a = \frac{F}{m} = \frac{{Vd}_{2} g - {Vd}_{1} g}{{Vd}_{1}} = (\frac{d_{2} - d_{1}}{d_{1}}) g t = \frac{v}{a} = \sqrt{2 {gh}_{1}} \times (\frac{d_{1}}{d_{2} - d_{1}}) \frac{1}{g} = \sqrt{(\frac{2 h_{1}}{g})} (\frac{d_{1}}{d_{2} - d_{1}})$