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Let's say you have a pot with some water in it (as shown). Then you take an empty glass and put it upside down (its mouth on the flat pot bottom) in the pot. What can happen with this experiment?

Four possibilities are shown below:

A Level of water in the pot and the glass are same

B No water inside the glass. The level of water inside the pot rises

C Some water inside the glass. The level of water inside the pot rises but below the one in B

D Level of water inside the Glass is above the level of water in the pot.

My smart Physicist friend tells me that ALL FOUR situations can happen under different set of conditions.

Is he right? Can you explain? State your assumptions

No partial answers please.

BTW I did the experiment but saw only one result. Hmmm

enter image description here

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  • $\begingroup$ Oh, I thought it always results in rot13(o), now this is interesting to know that all 4 are possible. $\endgroup$ – athin Jul 31 at 14:10
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I'm not a physicist so this may be completely wrong...

A: vacuum

B: high air pressure; effect of gravity negligible in comparison

C: low air pressure; weight of excess water outside glass compresses air inside glass

D: pot of cold water on a hot day; water cools down air inside glass reducing its pressure

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  • $\begingroup$ This looks pretty good to me! $\endgroup$ – Don Thousand Jul 31 at 14:24
  • $\begingroup$ Good answers. I'm stuck on B, since this lets you get an arbitrarily small amount of water inside the glass, but the gas inside the glass would have to be ideally incompressible for there to truly be no water inside. This is the closest I can think of, though. $\endgroup$ – Nuclear Hoagie Jul 31 at 14:28
  • $\begingroup$ If you immersed the glass quickly enough in case b (maybe around mach 1 (dammit, I'm a nerd, not a doctor)), you could make the air incompressible inside the glass, then create a tight enough seal against the bottom of the pot that no water could enter. You'd probably end up with some of the water outside the pot though. $\endgroup$ – Charles Bamford Jul 31 at 22:35
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If you realise this experiment at home you get result:

C

To respond to this question, you need some notions of physics and more specifically about pressure. The ambient air exercise pressure on every surface , it's the atmospheric pressure. Likewise water exercise pressure on any oject in it.

When you put your "empty" glass in your pot, your glass isn't fully empty: there is some air in it. As such,when outside water, the pressure inside and outside your glass is the same. When you put your glass in your pot, the pressure inside your glass and at the entry of it is different: there is some water pressure added. Which will make some water go into it. This will push the air in the glass into a smaller space and augment it's pressure. The water will stop moving in when the pressure at the bottom of the glass outside and inside of it is the same . enter image description here enter image description here

Bonus question:

To obtain the other results you need some specific environements.

For Case A: you could do the experiment in an environnement without any air: the only pressure left would be the pressure from water. As such the level of water inside and outside the glass would be the same.

For Case D: As for case A you do your experiment in the void. But after that you move your pot in an environement with some air. There is still no air in the glass, but there is an additionnal air pressure on the outside which push more water in.

For Case B: It is the opposite of case D; do the experiment in a normal environnement then move it to the void. Note that you need to have sufficiently few water in the pot or you will get answer C anyway.

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  • $\begingroup$ Thanks, but I am expecting more explanation on "Specific Environments " to get other answers $\endgroup$ – DrD Jul 31 at 13:25
  • $\begingroup$ BTW @Pierre Schneegans, when I did my experiment I got another answer -- other than yours $\endgroup$ – DrD Jul 31 at 13:47
  • $\begingroup$ @DEEM answer edited. What answer do you get when you do the experiment? $\endgroup$ – Pierre Schneegans Jul 31 at 14:33
  • $\begingroup$ rot 13 N vs lbh qbag tb rknpgyl crecnaqvphyne lbh yrg gur nve sebz gur tynff rfpncr, juvpu unccrarq gb zr $\endgroup$ – DrD Jul 31 at 14:35
  • $\begingroup$ Hello @PierreSchneegans and welcome to PuzzlingSE! That a good and thorough answer, looking forward to reading more of your answers. I'd invite you to take the Tour and check out the Help Center or Chat if you got any questions. $\endgroup$ – Christoph Jul 31 at 14:46
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A: water level equal in both the glass and the pot

the result is is between B and D so a combined technique could be used. or you could start at D and leak air into the class until you reach this result.

B: water level lower in the glass than in the pot

air is slightly compressible, inserting the glass straight downwards will give this result.

C: glass empty.

chill the glass so that it is full of cold air, this will expand pushing the water out bottom water as the glass warms up.

D: water level higher in the glass than in the pot.

warm the glass and fill it with steam before putting it in the pot. as it cools the steam will condensce sucking the water level up,

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  • $\begingroup$ Very smart thinking @Jasen +1 $\endgroup$ – DrD Aug 1 at 11:27
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For readability's sake, I will present these out of order.

B:

The glass must be lowered into the pot perfectly straight, so that the air cannot escape through the sides. This is not easy to do by hand, but is completely plausible.

C:

This is the one you likely have done. It happens because the glass is lowered down at an angle, so that some of the air can escape, but not enough to have the water level be above that of the pot.

D:

Lower the glass into the pot at a flat angle. For the best results, lower it sideways, then flip the glass so that it is vertical against the bottom of the pot. This is because it's letting enough air escape that the water level can't drop without forming a vacuum.

A:

There are two methods to do this. First, you can lower the glass at a precise angle such that enough air is let out that the water level matches that of the pot. The easier method, however, is to drill a hole in the bottom of the glass. This will create a way for the air to escape, and the water level will rise to match that of the pot as the air flows out naturally.

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  • 2
    $\begingroup$ $B$ is not correct. Even if no air escapes, gases are compressible. $C$ is also wrong, for similar reasons. $A,D$ is not allowed by the problem conditions. $\endgroup$ – Don Thousand Jul 31 at 14:24
  • $\begingroup$ The problem simply states that the glass must be lowered into the pot, not that it must be lower straight down. $\endgroup$ – Bewilderer Jul 31 at 14:27
  • $\begingroup$ Even if that is the case, $B,C$ are still wrong. $\endgroup$ – Don Thousand Jul 31 at 16:35
  • $\begingroup$ it says in the question that the glass must be lowered straight down, there is not much room in the pot to tilt the glass anyway. $\endgroup$ – Jasen Aug 1 at 9:01
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I would say option C) is the answer. It occurs in the most common situation, to me the other options occur only in different pressure environments/depends on atmospheric pressure as well as the temperature.

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