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 $5 \,cm$ and the angular speed of rotation is $\omega\, rad \,s^{-1}$. The difference in the height, $h ($ in $cm$ ) of liquid at the centre of vessel and at the side will be:

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

Answer 1

Answer: (A)

Applying pressure equation from A to B
$P_{0}+\rho \cdot \frac{R \omega^{2}}{2} \cdot R-\rho g h=P_{0}$
$\frac{\rho R^{2} \omega^{2}}{2}=\rho g h$
$h =\frac{ R ^{2} \omega^{2}}{2 g }=(5)^{2} \frac{\omega^{2}}{2 g }=\frac{25}{2} \frac{\omega^{2}}{ g }$