Question

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

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

Answer 1

Answer: (B)

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

Answer 2

From Bernoulli?s theorem, $P_{A}+\frac{1}{2}dv^{2}_{A}+dgh_{A}$ $=P_{_B}+\frac{1}{2}dv^{2}_{B}+dgh_{B}$ Here, $h_{A}=h_{B}$ $\therefore P_{A}+\frac{1}{2}dv^{2}_{A}=P_{B}+\frac{1}{2}dv^{2}_{B}$ $P_{A}-P_{B}=\frac{1}{2}d\left[v_{B}^{2}-v^{2}_{A}\right]$ Now, $v_{A}=0, v_{B}=r\omega$ and $P_{A}-P_{B}=hdg$ $\therefore hdg=\frac{1}{2}dr^{2} \omega^{2}$ or $h=\frac{r^{2}\omega^{2}}{2g}$