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Suppose that the are of the plates is 6.35 m sq and height is negligible. Both the plates could be made to slide. In the initial case the whole apparatus was empty. Door 2 was in closed position and door 1 was open such that when the water from the top of the tunnel was poured no water could leak from the door 2 side. After filling the apparatus the top was sealed and the door 1 was closed. After that door 2 was opened.

Pressure at the top surface of the door 1 is indicated as green and that to the bottom as red.

So the pressure green will be :

Pressure= Density of water X Acceleration due to gravity X height

    = 1000 X 10 X 15
    = 150000 N/m sq

But what height should be considered while finding the pressure red?

  • $\begingroup$ Are you assuming the doors have negligible thickness? If so, remember that the layer of water that is in contact with atmosphere is at 1atm of pressure and that the other parts of the same body of water that are at the same height will be at equal pressure. $\endgroup$ Jul 21, 2020 at 12:46
  • $\begingroup$ The answer depends on the details of the experiment. If door 1 is tightly sealed and no liquid leaks through it, the pressure on the "red" side of door 1 will be independent of the pressure on the "green" side of door 1. $\endgroup$ Jul 21, 2020 at 17:24

1 Answer 1


Pressure red will be one atmosphere plus whatever extra is caused by the difference in height between the bottom of door 1 and the water level at door 2. Pressure green is perhaps more subtle. After door 1 is closed but before the top was sealed, it was σgh + 1 atm. I can't envision any mechanism that would change that when the top is sealed. Your sketch shows a little air trapped in the top of the cylinder. This would tend to maintain the air pressure at that point.


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