Consider a U-tube containing some liquid. The two ends of the tube are connected to closed volumes of air with different initial pressures, such that there is some difference in the liquid levels inside the tube. Assume that the U-tube has uniform cross section, the liquid is incompressible, and there are no losses.

Now, if the U-tube is accelerated vertically, will there be any changes in the liquid level at one leg relative to that in the other, and why?

I initially thought they would change. When the tube is accelerated upwards, the liquid in the tube would tend to move from the leg with higher liquid level to the other leg. The increased pressure that drives it is given as $\rho d a$, where $\rho$ is the liquid density, $d$ is the initial difference in the liquid levels of the two legs, and $a$ is the acceleration of the U-tube. But then, I thought there should be an equal and opposing pressure increase from the leg with lower liquid level, which prevents the liquid from moving. But I am not sure.

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    $\begingroup$ What are your thoughts? $\endgroup$
    – jinawee
    Jan 15 '14 at 16:03
  • $\begingroup$ @jinawee, I have just added my thoughts in the above question. $\endgroup$
    – adipro
    Jan 20 '14 at 11:31
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    $\begingroup$ @adipro.you are right.as water moves from open end to close end there will be a pressure increase in the sealed side.but unlike liquids the pressure increase happens by volume reduction(compression) of air in the sealed side thereby changing liquid height in the open end. $\endgroup$ Jan 21 '14 at 4:03

The difference in water column changes depends on how fast you accelerate.the pressure in the sealed side is compensated by the difference in water column height on the open side of the tube.this pressure is the product of mass of liquid column and and acceleration due to gravity (9.8m/sec^2) at rest and area.when you accelerate the tube upwards the water column (differnce in height) will have more value for g depends on how fast you move.it reaches to an acceleration where the weight of liquid column overcome the pressure on the sealed side thereby changing liquid height. enter image description here

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    $\begingroup$ Your answer is wrong, heights will change no matter what is the acceleration of column at whatever rate, the equation is $P -P_a = h \rho g$ since column will be accelerated $g$ will be replaced by $a_{liquid}$ $\endgroup$ Jan 15 '14 at 18:10
  • $\begingroup$ But the other liquid surface on the tube will be equally submitted to the same acceleration force, right? $\endgroup$
    – cinico
    Jan 15 '14 at 18:19
  • $\begingroup$ Even at a constant velocity the height will change as the air pressure at the open end will change, the Pitot tube functions on this principle: en.wikipedia.org/wiki/Pitot_tube $\endgroup$
    – Fergus
    Jan 15 '14 at 18:53
  • $\begingroup$ @Fergus, what do you think would happen if the open end is connected to a vacuum? $\endgroup$
    – adipro
    Jan 20 '14 at 12:33
  • $\begingroup$ @adipro If the open end was connected to a vacuum there would be no liquid water in the tube ;) So let's assume something that remains liquid in a vacuum is used, I also agree with cinico, all the fluid is subjected to the same acceleration so I wouldn't expect any change in height on either side. (unless of course the fluid is compressible in which case each side would drop by the same distance) $\endgroup$
    – Fergus
    Jan 20 '14 at 23:26

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