We know that when a vacuumised tube is inverted on a fluid then the fluid rises up to fill the tube so that Pascal's law is follower. An example would be that of the mercury barometer in which mercury rises to fill the test tube up-to ($\approx$)$76$ cm. Clearly the height $h_m$ that a fluid attains is given by:

$$h \approx \frac {P_{\text{atm}}}{\rho g}$$

Now the question is

  • what might happen if the height $h_{\text{tube}}$ of the tube is less than $h_m$ (i.e., $h_{\text{tube}} \lt h_m$)?

Clearly when the pressure difference arises then a force would act and I think that there might be some movement of the tube and I think it might topple so an additional requirement that I am putting is that the tube is surrounded by a pipe (a kind of supporting structure) of about the same radius (actually a bit larger) so that it doesn't topple.

So what does happen?

Diagram related to the Question

Diagram of the structure

What I think might happen (though I don't have much to argue for it)

I think that since there is a lower pressure on the bottom of the tube as compared to surrounding therefore the fluid would try to flow in and in due process make the tube fly up. But looking at it from different perspective it does not seem to be possible as then it would act as a perpetual motion machine, which isn't a thing of reality. So a proper question is

  • Why this isn't the case?

  • What happens in reality then?

  • 1
    $\begingroup$ What do you think happens and why? You should let us know your attempt and reasoning because "what happens?" isn't really a conceptual question. $\endgroup$
    – Bill N
    Commented Jan 16, 2020 at 16:30
  • $\begingroup$ @BillN I have modified the question now. $\endgroup$
    – user249968
    Commented Jan 16, 2020 at 17:21

3 Answers 3


You can actually do an experiment with water to see what happens. (I just did it.) Get a large bowl of water and a cylindrical container like a flower vase or a graduated cylinder. Fill the cylinder with water and cover the opening with a small card of some type. Invert it into the bowl of water. You'll probably get a small air bubble but that's not a significant problem. See what happens. In my result, the glass flower vase did not shoot up.

Now, the height of a 1 atm water column would be $13.6 \times 76$ cm $= 10.3$ meters. My vase was about 20 cm tall. Why didn't it shoot up? The weight of the container was much larger than the fluid-pressure-generated force pushing upward on the inverted bottom of the vase.

If I used a tube/container that is less dense than water, I might have gotten a different result. That would all depend on the thickness of the tube wall and the thickness of the closed end of the tube.

If oscillations began they would definitely not be perpetual due to surface tension and friction effects.


If the surface of the fluid reaches the top of the tube, then the top of the tube applies a pressure on the fluid "surface". In other words, if the tube is fixed the pressure in the fluid at the top surface is $P_{atm} - \rho g h_{tube}$ .

If the tube can move vertically it will move up, until equilibrium is reached when this pressure in the fluid at the top of the tube exactly offsets the weight of the tube itself.


I discussed this question with John Rennie sir over here and he explained it quite wonderfully. This is a summary of that discussion.

Even when we place the inverted tube filled with mercury whose height is less than $76 cm$ then also the pressure at the bottom of the tube is equal to $P_{\text {atm}}$. Why? Because the glass in the tube acts as a spring and applies pressure on the mercury inside this pressure gets transmitted to the bottom of the tube which is equal to $P_{\text {atm}}$. The pressure applied by the tube is equal to $P_{\text {atm}} - \rho g h$.

One interesting thing to note is that if the tube isn't strong enough then the stress generated on it would cause the tube to break or crack!


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