Question

A cylindrical vessel of radius $r$ containing a liquid is rotating about a vertical axis passing through the centre of circular base. If angular velocity of rotation of vessel is $\omega$, then the difference in the heights of liquid at the centre of vessel and the edge of vesse is:

• A. $\frac{r\omega }{2g}$
• B. $\sqrt{2gr\omega }$
• C. $\frac{r{\omega }^2}{2g}$
• D. $\frac{r^2{\omega }^2}{2g}$

Answer 1

The correct option is D $\frac{r^2{\omega }^2}{2g}$

$\begin{array}{c}{P}_A+\frac{1}{2}\rho {v_A}^2+\rho g{h}_A=P_B+\frac{1}{2}\rho {v_B}^2+\rho g{h}_B\\ here,h_A=h_B\\ \therefore P_A+\frac{1}{2}\rho {v_A}^2=P_B+\frac{1}{2}\rho {v_B}^2\\ P_A-P_B=\frac{1}{2}\rho \left[{v_B}^2-{v_A}^2\right]\\ Now,v_A=0,v_B=r\omega and{P}_A-P_B=h\rho g\\ h\rho g=\frac{1}{2}\rho r^2{\omega }^2\\ or,h=\frac{r^2{\omega }^2}{2g}\end{array}$

Hence,

option $\left(D\right)$ is correct answer.