A streamlined body falls through air from a h | vedclass

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(D) Let $V$ be the volume of the body. The velocity of the body just before entering the liquid is $v = \sqrt{2gh}$.When the body enters the liquid,th

Vedclass · Accepted Answer

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

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(D) Let $V$ be the volume of the body. The velocity of the body just before entering the liquid is $v = \sqrt{2gh}$.When the body enters the liquid,the forces acting on it are its weight $W = Vdg$ (downwards) and the buoyant force $F_B = VDg$ (upwards).The net force $F_{net}$ acting on the body inside the liquid is $F_B - W = VDg - Vdg = V(D - d)g$.The retardation $a$ of the body is given by $a = \frac{F_{net}}{m} = \frac{V(D - d)g}{Vd} = \left( \frac{D - d}{d} \right)g$.The body comes to rest when its velocity becomes zero. Using $v = at$,we get $t = \frac{v}{a}$.Substituting the values,$t = \frac{\sqrt{2gh}}{\left( \frac{D - d}{d} \right)g} = \frac{\sqrt{2gh}}{g} \cdot \frac{d}{D - d} = \sqrt{\frac{2h}{g}} \cdot \frac{d}{D - d}$.

$A$ 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 the liquid respectively,then the time after which the body will be instantaneously at rest is:

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