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:

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The free surface tilts at angle \(\tan\phi=a_0/g\); multiply by the limb separation \(l\).
Updated On: Jul 2, 2026
  • \(\dfrac{a_0 l}{g}\)
  • \(\dfrac{l}{g}\)
  • \(\dfrac{gl}{a_0}\)
  • \(l\)
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The Correct Option is A

Solution and Explanation

Step 1: In the frame of the accelerating tube, a pseudo-force acts on the liquid opposite to the acceleration. The liquid surface tilts so that it is perpendicular to the net effective gravity.

Step 2: The effective gravity has a vertical part \(g\) (downward) and a horizontal part \(a_0\) (backward). The free surface makes an angle \(\phi\) with the horizontal where \[\tan\phi=\frac{a_0}{g}.\] Step 3: The two vertical limbs are separated by horizontal distance \(l\). The height difference between the liquid levels in the two limbs is \[\Delta h=l\tan\phi=l\cdot\frac{a_0}{g}.\] Step 4: Therefore \[\Delta h=\frac{a_0 l}{g}.\] \[\boxed{\Delta h=\dfrac{a_0 l}{g}}\]
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