Figure 1

Consider filling a glass partially with water, covering it with a solid non-porous plate and then turning it upside down. The plate is not hinged to the glass. There is always some air trapped in the glass. Assume h << H.

  1. Is $P_1 = P_{atm}$? If yes, why does the water not fall down?

  2. What happens when $h \approx H$?

  • $\begingroup$ Is the class partially filled with water, a rigid cover added and then inverted? $\endgroup$
    – Eric S
    Commented Feb 27, 2022 at 18:13
  • $\begingroup$ @Eric yes, that was my initial thought. I guess if water is not able to pass at all, that adds hinge force on the cover. So, seepage of water should be permissible. $\endgroup$ Commented Feb 27, 2022 at 20:04
  • $\begingroup$ Please edit the question so it includes your focus on water seeping out of the cardboard to glass interface. $\endgroup$
    – J. Ari
    Commented Feb 27, 2022 at 20:06
  • $\begingroup$ @J.Ari I have edited the question. I think it now clarifies my question better. $\endgroup$ Commented Feb 27, 2022 at 20:16

2 Answers 2


First we need to assume that there is not a perfect seal at the bottom of the container so that fluid can seep in and out.

The pressure in the h section will drop until the difference between it and the atmospheric pressure is enough to support the column of water below it.

  1. $P_1 = P_{atm}$?

No. Consider what happens as you stretch H. At about 10 m or so (depending on salt / fresh water) $P_1 = 0$ and the h is a vacuum unable to lift any more than 10 m. This is the principle of operation of the mercury barometer with the advantage of a much more convenient tube length due to the density of mercury (13.6 times that of water, if my school memory is correct).

  1. What happens when $h \approx H$?

The same principle applies until $P_1$ reaches zero.


if we assume the glass is cylinder, the amount of h that can hold any water would be roughly 1/1000 0f the H.

Because if we let the water freely run out of the bottom of the glass the volume of the entrapped air will increase 1000times and its presser will be 1/1000atm. or equal to 1 cm heigh water at the bottom of the glass.

Any h larger than that would not hold any water.

  • 1
    $\begingroup$ I read the question and assumed it was a different scenario where a partially filled glass is covered with a rigid cover and then inverted. $\endgroup$
    – Eric S
    Commented Feb 27, 2022 at 18:15
  • $\begingroup$ water is incomprehensible. so if it can't be drained out nothing is going to happen. $\endgroup$
    – kamran
    Commented Feb 27, 2022 at 21:03
  • 1
    $\begingroup$ Where is the '1000' coming from? Why do we need to assume the glass is a cylinder? $\endgroup$
    – Transistor
    Commented Feb 27, 2022 at 23:34
  • $\begingroup$ @transistor, cylinder can substitute height for volume. 1000 is a test variable. i will edit my answer later and elaborate. $\endgroup$
    – kamran
    Commented Feb 28, 2022 at 2:08
  • $\begingroup$ Since there's a maximum limit on the height of the water column I don't think your "ratio" there is correct. $\endgroup$ Commented Feb 28, 2022 at 15:50

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