A stream-lined body falls through air from a height $h$ on the surface of liquid. Let $d$ and $D$ denote the densities of the materials of the body and the liquid respectively. If $D > d$, then the time after which the body will be instantaneously at rest, is
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Velocity $u$ of the body when it enters the liquid is given by
$\mathrm{mgh}=\frac{1}{2} \mathrm{mu}^{2} \quad$ or $\quad \mathrm{u}=\sqrt{2 \mathrm{gh}}$
Let $\quad$ volume of the body $=V$
mass of the body $=\mathrm{Vd}$
weight of the body $=\mathrm{Vdg}$
Mass of liquid displaced $= VD$
Weight of liquid displaced $=\mathrm{VDg}$
Net upward force $=\mathrm{VDg}-\mathrm{Vdg}$
$=\mathrm{Vg}(\mathrm{D}-\mathrm{d})$
Retardation $=\frac{\text { net weight }}{\text { mass }}$
$=\frac{V(D-d) g}{V d}=\left(\frac{D-d}{d}\right) g$
Acceleration $a=-\left(\frac{D-d}{d}\right) g$
Final velocity, v in the liquid when the body is instantaneously at rest is zero. Let the time taken be $t.$
$\text { Now } \quad \mathrm{v}=\mathrm{u}+a \mathrm{t}$
$0=\sqrt{2 \mathrm{gh}}-\left(\frac{\mathrm{D}-\mathrm{d}}{\mathrm{d}}\right) \mathrm{gt},\left(\frac{\mathrm{D}-\mathrm{d}}{\mathrm{d}}\right) \mathrm{gt}=\sqrt{2 \mathrm{gh}}$
$\therefore \quad t=\left[\frac{d}{D-d}\right] \sqrt{\frac{2 h}{g}}$
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