Question

A U-tube containing liquid is accelerated horizontally with a constant acceleration \(a_0\). If the separation between the vertical limbs is \(l\), then the difference in the heights of the liquid in the two arms is:

• A. \(\dfrac{a_0 l}{g}\)
• B. \(\dfrac{l}{g}\)
• C. \(\dfrac{gl}{a_0}\)
• D. \(l\)

Hint:

The free surface tilts at angle \(\tan\phi=a_0/g\); multiply by the limb separation \(l\).

Answer 1

The Correct Option is A

Use pressure balance along a horizontal line inside the liquid.

Set up. Take the horizontal bottom tube joining the two limbs. Let the liquid density be $\rho$ and the height difference be $\Delta h$. Because the tube accelerates with $a_0$, the liquid needs a horizontal pressure gradient to supply the force.

Horizontal force balance. Consider a liquid column of length $l$ and area $A$ in the bottom tube. Net horizontal force is the pressure difference between the two ends times area: \[(P_1-P_2)A=(\rho A l)\,a_0.\] So \[P_1-P_2=\rho l a_0.\] Relate to heights. The pressure at the bottom of each limb equals $\rho g$ times the liquid height in that limb (open tops at atmospheric pressure). Thus \[P_1-P_2=\rho g\,\Delta h.\] Combine. \[\rho g\,\Delta h=\rho l a_0\;\Rightarrow\;\Delta h=\frac{a_0 l}{g}.\] \[\boxed{\Delta h=\dfrac{a_0 l}{g}}\]