Question

The vertical limbs of U–tube are filled with a liqiuid of density ρ upto a height h on each side. The horizontal portion of the U–tube having length 2h contains an immisible liquid of density 2ρ . The U–tube is moved horizontally with an acceleration of g/2 parallel to the horizontal arm. In steady state the difference in the level of liquid in the vertical limbs is

(1) 2h/7

(2) 12h/5

(3) 4h/5

(4) 3h/5

Answer 1

(2) 12h/5

Fig. 29 (a) shows the given arrangement. Fig.(b) shows the liquid levels when U–tube has a horizontal acceleration a* = g/2. Let the liquid level fall by an amount x in limb M as shown in Fig(b).

M = Mass of liquid in horizontal part of U–tube

where a is area of cross–section of the limb of U–tube. When U–tube has a horizontal acceleratio a* = g/2. the liquid in horizontal part experiences a ficticous force F_fic directed towards limb L. Obvisouly

F_fic is equal to force F on bottom of limb L due to liquid pressure. Obviously

From Eqns. (1) and (2) we have

The difference in liquid level in the two limbs = 2x = 12h/5