Question 169 of 280

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, $h\left(incm\right)$ of liquid at the Centre of vessel and at the sides of the vessel will be :

A$\dfrac{2{\omega }^{2}}{25g}$B$\dfrac{5{\omega }^{2}}{2g}$C$\dfrac{25{\omega }^{2}}{2g}$D$\dfrac{2{\omega }^{2}}{5g}$

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Answer: (C) $\dfrac{25{\omega }^{2}}{2g}$

$\begin{matrix}\rho \left(dr\right){\omega }^{2}r=\rho g\left(dh\right)\end{matrix}$
$\begin{matrix}{\omega }^{2}{\int }_{0}^{R}rdr=g{\int }_{0}^{h}dh\end{matrix}$
$\dfrac{{\omega }^{2}{R}^{2}}{2}=gh$
$h=\dfrac{{\omega }^{2}{R}^{2}}{2g}=\dfrac{25{\omega }^{2}}{2g}cm$.