Question

Difference in height of the liquid columns in the limbs of a rotating $U-$ tube as shown in the figure is:
(Neglect the width of each limb and take $g=10{\text{m/s}}^2$)

• A. $1\text{m}$
• B. $0.8\text{m}$
• C. $0.5\text{m}$
• D. $0.6\text{m}$

Answer 1

The correct option is B $0.8\text{m}$

Let the difference in heights between the liquid columns in the limbs be $x^{\prime }$ and atmospheric pressure be $p_0$.


For a rotating fluid, the pressure difference can be given as
$\Delta P=\frac{1}{2}\rho {\omega }^2{x}^2...\left(1\right)$
where $x$ is the distance between two points in a liquid, measured $\perp$ from the axis of rotation.
The pressure will increase in a direction away from the axis of rotation.
$\Rightarrow p_B>p_A$
For pressure at $A$,
$p_A=p_0+\rho g{h}_2...\left(2\right)$ (due to variation in depth)
For pressure at $B$:
$p_B=p_A+\frac{1}{2}\rho {\omega }^2{L}^2...\left(3\right)$
[$\because$ we can take $AB=x\approx L=2\text{m}$]

Also, $p_B=p_0+\rho g{h}_1...\left(4\right)$ (due to variation in depth)

From Eq.$\left(2\right),\left(3\right)\&\left(4\right)$,
$p_0+\rho g{h}_1=p_0+\rho g{h}_2+\frac{1}{2}\rho {\omega }^2{L}^2$
$\Rightarrow h_1-h_2=\frac{1}{2}\frac{{\omega }^2{L}^2}{g}$
Then, difference in heights of the liquid in the limbs
$\Rightarrow x^{\prime }=h_1-h_2=\frac{1}{2}\times \frac{2^2\times 2^2}{10}$
$\therefore x^{\prime }=0.8\text{m}$