Question

A liquid is kept in a cylindrical vessel, which is rotated along its axis. The liquid rises at the side. If the radius of the vessel is $5cm$ and the frequency of rotation is $4rev/s$, then the difference in the height of the liquid at the centre of the vessel and its sides is:
$A)8cm$
$B)\text{ 2}cm$
$C)\text{ 40}cm$
$D)\text{ 4}cm$

Answer 1

Hint: This problem can be solved by realizing the fact that the change in the potential energy of the water will be equal to the change in the rotational kinetic energy of the water. The water in the centre of the cylinder loses potential energy which is gained by the water in the sides to attain the rotational kinetic energy.

Formula used:
$KE={\displaystyle \frac{1}{2}}I{\omega }^2$
$PE=mgh$
$I=m{r}^2$

Complete step by step answer:
In this problem, we must realize that the water at the centre drops in height and relatively the water at the sides rise. Since, the water in the cylinder rotates it has some rotational kinetic energy. This rotational kinetic energy is attained due to the loss of the potential energy of the water due to the change in height.
The rotational kinetic energy $KE$ of a body of moment of inertia $I$ about the axis of rotation is given by
$KE={\displaystyle \frac{1}{2}}I{\omega }^2$ --(1)
where $\omega$ is the angular frequency of rotation.
The change in potential energy $PE$ of a body of mass $m$ due to a change $h$ in the height of its position is given by
$PE=mgh$ --(2)
Where $g=9.8m/s^2$ is the acceleration due to gravity.
The moment of inertia $I$ of a particle of mass $m$ about an axis is given by
$I=m{r}^2$ --(3)
Where $r$ is the perpendicular distance of the particle from the axis.
Now, let us analyze the question.
We will consider a particle of mass $m$ of the water at the side of the cylinder.
The axis in this case is the axis of the cylinder through its centre.
Therefore, the perpendicular distance of the particle from the axis, will be nothing but the radius $r=5cm=0.05m$ of the cylinder. $(\because 1cm=0.01m)$
The angular frequency of rotation is $\omega =4rev/s=4\times 2\pi rad/s=8\pi rad/s$ $(\because 1rev/s=2\pi rad/s)$
Let the change in the height of the position of the particle at the side be $h$.
Let the moment of inertia of this particle about the axis be $I$
Let the change in the rotational kinetic energy of this particle be $KE$ while the change in the potential energy be $PE$.
Using (3), we get
$I=m{r}^2$ --(4)
Using (1), we get that,
$KE={\displaystyle \frac{1}{2}}I{\omega }^2$ --(5)
Putting (4) in (5), we get,
$KE={\displaystyle \frac{1}{2}}\left(m{r}^2\right){\omega }^2$ --(6)
Using (2), we get,
$PE=mgh$ --(7)
As explained above, the magnitude of the change in the rotational kinetic energy will be equal to the magnitude of the change in the potential energy.
$\therefore KE=PE$ --(8)
Using (6) and (7) in (8), we get,
${\displaystyle \frac{1}{2}}\left(m{r}^2\right){\omega }^2=mgh$
$\therefore {\displaystyle \frac{1}{2}}{r}^2{\omega }^2=gh$
$\therefore h={\displaystyle \frac{{\omega }^2{r}^2}{2g}}$
Putting the values of the variables in the above equation, we get,
$h={\displaystyle \frac{{\left(8\pi \right)}^2{\left(0.05\right)}^2}{2\times 9.8}}={\displaystyle \frac{64{\pi }^2(25\times {10}^{-4})}{2\times 9.8}}=0.08m=8cm$ $(\because 1m=100cm)$
Hence, the required rise in height is $8cm$.

So, the correct answer is “Option A”.

Note:
In this problem, a very common mistake that could have been made by the students is not changing the units for the angular frequency. The angular frequency is given in revolutions per second, whereas its SI unit is radians per second. If the proper conversion is not done, then the angular frequency will not be the correct value that can be used in the calculation and the ultimate answer will be incorrect.