- Home
- Standard 11
- Physics
A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is $r$ and angular velocity of rotation is $\omega $ , then the difference in the heights of the liquid at the centre of the vessel and the edge is
$\frac{{{r^2}{\omega ^2}}}{{2g}}$
$\frac{{{\omega ^2}}}{{2g{r^2}}}$
$\sqrt {2gr\omega } $
none of these
Solution
Using Bemaulli's theorem
$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}+\rho \mathrm{gh}_{2}$
Here, ${{\rm{h}}_1} \simeq {{\rm{h}}_2}$
$\therefore {{\rm{P}}_1} + \frac{1}{2}\rho {\rm{v}}_1^2 = {{\rm{P}}_2} + \frac{1}{2}\rho {\rm{v}}_2^2$
or ${\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2} \rho\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right)}$
Here. $v_{1}=0 . \mathrm{v}_{2}=\mathrm{r} \omega \quad$ and
$\mathrm{P}_{1}-\mathrm{P}_{2}=\mathrm{h} \rho \mathrm{g}$
$\therefore $ ${\operatorname{hpg}=\frac{1}{2} \rho(10)^{2}} $
or ${h=\frac{r^{2} \omega^{2}}{2 g}}$
