A stream-lined body falls through air from a height h on surface of liquid. Let d and D d

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9-1.Fluid Mechanics

hard

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

$\sqrt{\frac{2h}{g}}$

$\sqrt{\frac{2h}{g} \frac{D}{d}}$

$\sqrt{\frac{2h}{g} \frac{d}{D}}$

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

Solution

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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