Question

A closed tube is filled with liquid $\rho =800{\text{kg/m}}^3$. It is rotating about an axis shown in figure with an angular velocity $\omega =4\text{rad/sec}$ and length of the tube is $9\text{m}$. Find the pressure difference between the ends of the tube.

• A. $2\text{bar}$
• B. $1\text{bar}$
• C. $1.512\text{bar}$
• D. $1.728\text{bar}$

Answer 1

The correct option is D $1.728\text{bar}$

Given that,
Density of the liquid, $\rho =800{\text{kg/m}}^3$
Angular velocity of the tube, $\omega =4\text{rad/sec}$
Pressure difference between axis of rotation and a point at distance $x$ from the axis is given by
$\Delta P=\pm \frac{\rho {\omega }^2{x}^2}{2}$
[(+) ve shows that pressure increases in moving away from axis and (-) ve shows pressure decreases in moving towards the axis of rotation]
$\therefore P_A>P_B$ and $P_C>P_B$
$P_C-P_A=(P_C-P_B)-(P_A-P_B)$
$=\left[\frac{\rho {\omega }^2{\left(\frac{2L}{3}\right)}^2}{2}\right]-\left[\frac{\rho {\omega }^2{\left(\frac{L}{3}\right)}^2}{2}\right]$
$[\because AB=\frac{L}{3},BC=\frac{2L}{3}]$
$=\frac{\rho {\omega }^2}{2}[\frac{4L^2}{9}-\frac{L^2}{9}]$
$=\frac{\rho {\omega }^2}{2}\times \frac{3L^2}{9}=\frac{\rho {\omega }^2}{2}\times \frac{L^2}{3}=\frac{\rho {\omega }^2{L}^2}{6}$

$\Rightarrow P_C-P_A=\frac{800\times 4\times 4\times 9\times 9}{6}$
$=172,800\text{Pa}$
$\therefore P_C-P_A=1.728\text{bar}$