A liquid is kept in a cylindrical vessel whic | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Using Bemaulli's theorem

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2

Vedclass · Accepted Answer

$\frac{{{r^2}{\omega ^2}}}{{2g}}$

Vedclass · Answer

Using Bemaulli's theorem

$\mathrm{P}_{1}+\frac{1}{2} ho \mathrm{v}_{1}^{2}+ho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} ho \mathrm{v}_{2}^{2}+ho \mathrm{gh}_{2}$

Here,  ${{m{h}}_1} \simeq {{m{h}}_2}$

$	herefore {{m{P}}_1} + \frac{1}{2}ho {m{v}}_1^2 = {{m{P}}_2} + \frac{1}{2}ho {m{v}}_2^2$

 or    ${\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2} ho\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}ight)}$

Here. $v_{1}=0 . \mathrm{v}_{2}=\mathrm{r} \omega \quad$ and

$\mathrm{P}_{1}-\mathrm{P}_{2}=\mathrm{h} ho \mathrm{g}$

$	herefore $ ${\operatorname{hpg}=\frac{1}{2} ho(10)^{2}} $

or ${h=\frac{r^{2} \omega^{2}}{2 g}}$

Similar Questions

A .......... $m/s$ speed the velocity head of water is equal to pressure head of $40\, cm$ of $Hg$ .

An $L-$ shaped glass tube is just immersed in flowing water towards tube as shown. If speed of water current is $V$, then the height $h$ upto which water rises will be

A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D (D >> d)$ as shown in figure. If it is burning at the rate of $2\, cm/hour$ then the top of the candle will

Water falls from a tap, down the streamline

The height of water in a tank is $H$. The range of the liquid emerging out from a hole in the wall of the tank at a depth $\frac {3H}{4}$ form the upper surface of water, will be