Question

A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5cm$ and the angular speed of rotation is $\omega rad{s}^{-1}$. The difference in the height, $h$ (in $cm$) of liquid at the center of the vessel and at the side will be:

• A. $\frac{5{\omega }^2}{2g}$
• B. $\frac{2{\omega }^2}{25g}$
• C. $\frac{25{\omega }^2}{2g}$
• D. $\frac{2{\omega }^2}{5g}$

Answer 1

The correct option is C

$\frac{25{\omega }^2}{2g}$

Step 1. Given data:

Radius of vessel is $5cm$

Angular speed of rotation is $\omega rad{s}^{-1}$

Step 2. Solving for height $h$

Considers a mass $m$ of water, at a distance $x$ from its axis of rotation .The centripetal force will act on liquid $f_x=m\cdot x\cdot {\omega }^2$ radially outwards and its weight force $f_y=mg$,

The net work done along$x$ axis due to centripetal force is,

$$\begin{gathered} ={\int }_0^{R}m.x.{\omega }^{2}dx \\ ={\left[\frac{m{x}^2{\omega }^2}{2}\right]}_0^R \\ =\frac{m{R}^2{\omega }^2}{2} \end{gathered}$$

The net work done along$y$ axis is against the potential energy $mgh$

Where$h$ is the height of liquid rising due to rotation and$g$ is acceleration due to gravity

Equating the two equations, we get

$\frac{m{R}^2{\omega }^2}{2}=mgh$

$\Rightarrow$ $\begin{array}{rcl}h& =& \frac{R^2{\omega }^2}{2g}\\ & =& \frac{5^2{\omega }^2}{2g}\end{array}$

$\Rightarrow$ $h=\frac{25{\omega }^2}{2g}$

Hence, option C is correct