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

Answer 1

Answer: (B)

From Bernoulli’s theorem,
$P_A+\frac{1}{2}d{v}_A^2+dg{h}_A$
$=P_{{}_B}+\frac{1}{2}d{v}_B^2+dg{h}_B$
Here, $h_A=h_B$
$\therefore P_A+\frac{1}{2}d{v}_A^2=P_B+\frac{1}{2}d{v}_B^2$
$P_A-P_B=\frac{1}{2}d\left[v_B^2-v_A^2\right]$
Now, $v_A=0,v_B=r\omega$ and $P_A-P_B=hdg$
$\therefore hdg=\frac{1}{2}d{r}^2{\omega }^2$ or $h=\frac{r^2{\omega }^2}{2g}$