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:
$\text{A}\text{. }\dfrac{2g{{r}^{2}}}{{{\omega }^{2}}}$
$\text{B}\text{. }\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
$\text{C}\text{. }\dfrac{g{{r}^{2}}}{2{{\omega }^{2}}}$
$\text{D}\text{. }\sqrt{2gr\omega }$

Answer 1

Hint: Find the velocities of the liquid at the edge and at the centre of the vessel by using $v=\omega r'$. Then apply Bernoulli’s theorem at the edge and at the centre of the vessel. i.e. $P+\dfrac{1}{2}\rho {{v}^{2}}=\text{constant}$. Once you can see the pressure difference between the edge and the centre, use $\Delta P=\rho gh$ to find the difference in heights.
Formula used:
$v=\omega r'$
$P+\dfrac{1}{2}\rho {{v}^{2}}=\text{constant}$
$\Delta P=\rho gh$

Complete step by step solution:
It is given that a cylindrical vessel is filled with a liquid and the vessel is being rotated about a vertical axis passing through the centre of its circular base. The radius of the circular base is r and the angular velocity of the cylinder is $\omega $.
When the vessel rotates, due to friction between the edge (curved surface) of the cylindrical vessel and the liquid in contact with the edge, the liquid in contact also rotates along with the surface. And due to viscosity of the liquid, all the layers of the liquid also rotate. As a result, the whole liquid rotates with an angular velocity of $\omega $.
When the liquid is rotating, the velocity of the cylindrical layer of liquid with the same centre as the vessel is $v=\omega r'$, where r’ is the radius of this layer.
With this formula we know that the outermost layer i.e. the liquid at the edge will have the maximum velocity, which is equal to ${{v}_{e}}=\omega r$.
The liquid at the centre will have the minimum velocity, which is equal to ${{v}_{c}}=\omega (0)=0$.
According to Bernoulli’s theorem, $P+\dfrac{1}{2}\rho {{v}^{2}}=\text{constant}$.
Here, P is the pressure of a point, $\rho $ is the density of the liquid and v is the speed of the liquid at that point.
Let apply Bernoulli’s theorem at a point on the edge and at the centre.
Let the pressure and velocity at a point on the edge be ${{P}_{e}}$ and ${{v}_{e}}$.
Let the pressure and velocity at the centre be ${{P}_{c}}$ and ${{v}_{c}}$.
Hence, we get
${{P}_{e}}+\dfrac{1}{2}\rho v_{e}^{2}={{P}_{c}}+\dfrac{1}{2}\rho v_{c}^{2}$
But we know that ${{v}_{c}}=0$ and ${{v}_{e}}=\omega R$. Substitute these values in the above equation.
$\Rightarrow {{P}_{e}}+\dfrac{1}{2}\rho {{\omega }^{2}}{{r}^{2}}={{P}_{c}}+0$
$\Rightarrow {{P}_{c}}-{{P}_{e}}=\dfrac{1}{2}\rho {{\omega }^{2}}{{r}^{2}}$
This implies that the pressure at the centre is greater than the pressure at the edge, which makes the liquid rise at the edge.
Let the level of liquid at the edge be at a height h from the level of liquid at the centre.
In a liquid, the pressure difference for a height of h is given by $\Delta P=\rho gh$, where g is acceleration due to gravity.
And here, $\Delta P={{P}_{c}}-{{P}_{e}}$.
Hence, $\Rightarrow \Delta P=\dfrac{1}{2}\rho {{\omega }^{2}}{{r}^{2}}$
$\Rightarrow \rho gh=\dfrac{1}{2}\rho {{\omega }^{2}}{{r}^{2}}$
$\Rightarrow h=\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
Therefore, the difference in the heights of the liquid at the centre of the vessel and the edge is $h=\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$.
Hence, the correct option is B.

Note: If you observe the options carefully, then we do not even need the concept that we used. This is because only one option in the given four options has the correct dimension of height.
What we are finding in this question is difference in heights, which has the dimension of length. Then the expression that we get for h must also have the dimension of length. And only the expression in option B satisfies this condition.