Question

As shown in the figure, a liquid is at the same levels in two arms of a U-tube of uniform cross-section when at rest. If the U-tube moves with an acceleration 'f' towards the right,the difference in liquid heights between the two arms of the U-tube will be,(acceleration due to gravity=g)

• A. $\frac{f}{g}a$
• B. $\frac{g}{f}a$
• C. a
• D. 0

Answer 1

Answer: (A)

$\frac{f}{g}a$

Answer 2

Given pressures P 1​ and P 2​ at the bottom of the left and right ends of the tube respectively, the force F can be expressed as the difference between the pressures multiplied by the cross-sectional area A , which is F =(P 1​− P 2​)A =ρghA.

The mass of the liquid in the horizontal portion is represented as m =ρLA. Substituting this into F =ma gives us ρghA =ρLAa.

Hence, the answer is f /g =a.

The correct answer is option (A): $\frac{f}{g}a$