Question

A liquid is kept in a cylindrical vessel, which is being rotated about the vertical axis through the centre of the circular base. The radius of the vessel is r and angular velocity of rotation is $\omega $, then, what is difference in the heights of liquid at the centre of vessel and the angle is-
(A) $\dfrac{{r\omega }}{{2g}}$
(B) $\dfrac{{{r^2}{\omega ^2}}}{{2g}}$
(C) $\sqrt {2gr\omega } $
(D) $\dfrac{{{w^2}}}{{2g{r^2}}}$

Answer 1

Apply the Bernoulli’s theorem which can be expressed as-
$p + \dfrac{1}{2}\rho {v^2} + \rho gh = c$ (here, $c$ is the constant)
Then, find out the difference between pressure at sides and pressure at centre.
We know that, velocity of liquid at the centre is zero.
Then, compare the equations.

Complete step by step answer:
Bernoulli’s Principle states that the summation of mechanical energy of moving fluid which comprises the gravitational potential energy of elevation, the energy which is associated with pressure of fluid and kinetic energy of fluid motion is constant.
$p + \dfrac{1}{2}\rho {v^2} + \rho gh = c$
where, $p$ is the pressure exerted by the fluid
$v$ is the velocity
$\rho $ is the density
$h$ is height of container
$c$ is the constant value
According to the question, when the cylindrical vessel is rotate at angular speed $\omega $ about its axis the, velocity of fluid at sides is ${v_s} = r\omega $
Apply the Bernoulli’s theorem at centre and sides of vessel we get
\[{P_s} + \dfrac{1}{2}\rho {v^2}_s + \rho g{h_s} = {P_c} + \dfrac{1}{2}\rho {v^2}_c + \rho g{h_c}\]
where, ${P_s} = $pressure at sides
${P_c} = $pressure at centre
${v_s}$= velocity at sides
${v_c} = $velocity at centre
${h_s}$= height at sides
${h_c}$= height at centre
${h_s} = $
$\therefore {P_c} - {P_s} = \dfrac{1}{2}\rho v_s^2 \Rightarrow \dfrac{1}{2}\rho {r^2}{\omega ^2} \cdots (1)$
Because, ${P_c} > {P_s}$, the liquid rises at sides of the vessel.
$\therefore {P_c} - {P_s} = \rho gh \cdots (2)$
From equation $(1)$ and $(2)$ we get
$\rho gh = \dfrac{1}{2}\rho {r^2}{\omega ^2}$
By transposition
$ \Rightarrow h = \dfrac{{{r^2}{\omega ^2}}}{{2g}}$

So, option (B) is the correct answer.

Note Bernoulli’s equation can be produced by conservation of energy when it is applied to fluid flow. The net work done is the result of change in fluid’s kinetic energy and gravitational potential energy. Bernoulli's equation can also be modified depending on the form of energy.