Question

Difference in heights between the liquid columns in the vertical limbs of an accelerated $U$-tube as shown in the figure is:
[Take $g=10{\text{m/s}}^2$]

• A. $3\text{m}$
• B. $5\text{m}$
• C. $4\text{m}$
• D. $2\text{m}$

Answer 1

The correct option is D $2\text{m}$

Let the difference in heights of the liquid in the two limbs be $x$ and atmospheric pressure be $p_0$.


For pressure at $A$:
The difference in pressure in the liquid along horizontal direction is given as (due to horizontal acceleration)
$\Delta p=\rho a\Delta x...\left(i\right)$
Pressure will increase opposite to the direction of acceleration.
$\Rightarrow p_A>p_B$
From Eq.$\left(i\right)$,
$p_A-p_B=\rho aL$
$(\because \Delta x=L,p_B=p_0)$
$\Rightarrow p_A=p_0+\rho \times 10\times 2$
$\Rightarrow p_A=p_0+20\rho ...\left(ii\right)$
Also, due to vertical height, the pressure difference is
$\Delta p=\rho gh$
Pressure will increase in vertically downward direction
$\Rightarrow p_A=p_0+\rho gx$
[$\because h=x$, is the depth of $A$]
$\Rightarrow p_A=p_0+\rho \times 10\times x...\left(iii\right)$
From Eq.$\left(ii\right)$ and $\left(iii\right)$,
$p_0+20\rho =p_0+10\rho x$
$\Rightarrow 10\rho x=20\rho$
$\therefore x=2\text{m}$
The difference in the heights of the liquid in the two limbs is $2\text{m}$