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Name: Chris H.
Status: student
Age: 17
Location: N/A
Country: N/A
Date: 12/10/2002

I have a bottle with two holes (one in the top one in the bottom), I am performing an experiment showing how when the hole in the top is covered (air hole) the water will not flow out the small hole in the bottom. As there becomes less and less water in the bottle the time delay between covering the hole at the top of the bottle and water coming out the bottom increases - why?

Dear Chris,

The water column over the hole in the bottom of the bottle is held up by the air pressure on the bottom of the column being greater than the air pressure on top. The water over the parts of the bottom of the bottle which are solid glass is held up by the glass pushing up on the water. The force per unit area must be the same as the air pressure on the water above the hole when the water is not flowing out.

If the hole on the top is open, the air pressure is (almost) the same on the top and bottom of the water, so the water will fall with the acceleration of gravity, ignoring viscosity, surface tension, and friction between the water and the sides of the bottom hole.

However, when the top hole is closed, the air above the water must expand to fill the increasing volume above the water. This, of course, reduces the pressure of the air (pV = constant at constant temperature). Notice that a 10% increase in the volume V produces a 10% decrease in the pressure p. (Plug in some numbers if this is not obvious to you)

When the jar is almost filled with water, a 10% change in the volume of the water occurs very quickly since the volume is small. Conversely, when the jar is almost empty of water, losing the same amount of water produces a much smaller percentage change in the volume (though the same magnitude of volume change). Therefore it takes a longer time after the top hole is covered for the water to stop flowing when the jar is almost full of water then when there is less water in the jar.

Best, Dick Plano

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