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,

$P_{A} + \frac{1}{2}dv_{A}^{2} + dgh_{A} = P_{B} + \frac{1}{2}dv_{B}^{2} + dgh_{B}$

Here, $h_{A} = h_{B}$ $\therefore\mspace{6mu} P_{A} + \frac{1}{2}dv_{A}^{2} = P_{B} + \frac{1}{2}dv_{B}^{2}$

⇒ $P_{A} - P_{B} = \frac{1}{2}d\lbrack v_{B}^{2} - v_{A}^{2}\rbrack$

Now, $v_{A} = 0,\mspace{6mu} v_{B} = r\omega$ and $P_{A} - P_{B} = hdg$

$\therefore\mspace{6mu}\mspace{6mu} hdg = \frac{1}{2}dr^{2}\omega^{2}$ or $h = \frac{r^{2}\omega^{2}}{2g}$