A streamline body with relative density $\rho_{1}$ falls into air from a height $h_{1}$ on the surface of a liquid of relative density$\rho_{2}$, where$p_{2} p_{1}$. The time of immersion of the body into the liquid will be [image:10_image31.emf]

Question

A streamline body with relative density $\rho_{1}$ falls into air from a height $h_{1}$ on the surface of a liquid of relative density$\rho_{2}$, where$p_{2} > p_{1}$. The time of immersion of the body into the liquid will be [image:10_image31.emf]

Dr. Abubacker · Accepted Answer

(d) If $V$ is the volume of the body, its weight$= V\ ho_{1}g$. Velocity gained by body when it falls from a height $h_{1} = \sqrt{2gh_{1}}$ The weight of liquid displaced by the body as body starts immersing into the liquid $= V\ ho_{2}g$. The net retarding force on the body when it starts going in the liquid $F = V(ho_{2} - ho_{1})$g
$	herefore$ Retarding, $a = \frac{F}{Vho_{1}} = \left\lbrack \frac{V\left( ho_{2} - ho_{1} ight)g}{V\ ho_{1}} ightbrack$
The time of immersion of the body is that time in which the velocity of the body becomes zero. Using the relation $v = u + at$, we have $v = 0,\ u = \sqrt{2gh_{1}}$,
$a = - \frac{V\left( ho_{2} - ho_{1} ight)g}{V\ ho_{1}} = - \left( \frac{ho_{2} - ho_{1}}{ho_{1}} ight)g;$
We have $= 0 = \sqrt{2gh_{1}} - \left( \frac{ho_{2} - ho_{1}}{ho_{1}} ight)gt$
Or $t = \sqrt{\frac{2h_{1}}{g}} 	imes \left( \frac{ho_{1}}{ho_{2} - ho_{1}} ight)$

Q. A streamline body with relative density $\rho_{1}$ falls into air from a height $h_{1}$ on the surface of a liquid of relative density$\rho_{2}$, where$p_{2} > p_{1}$. The time of immersion of the body into the liquid will be

Explanation

Reviewed by Dr. Abubacker

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