School boy experiment with water

[TinselKoala]

So, from examining my diagram, we can see that the pressure in the system is set by the depth of the opening in the tube on the right -- that is, by the weight of the water column above the opening, or equivalently the water pressure at that depth. As soon as the pressure in the tube reaches that pressure, the water will be all displaced from the fixed end and the air will bubble out. Push the free end down faster and the air will bubble out faster, but not at any higher pressure. This is why it doesn't seem like your hand is pushing very much... it isn't.

But if you make your tank very deep and put the fixed end of the tube very deep in the tank, then you will feel how much pressure it takes to displace all that water.

[vineet_kiran]

@TinselKoala,

You have done an excellent analysis of the experiment with a very nice diagram.   But what I am thinking is totally different  which is about direction and area of application of forces.    I have  given my reply vide attachment showing one more experiment   in which you can totally absorb the buoyancy forces at  any depth.  I request you to kindly go through it and let me have your  comments.
with thanks and regards,
Vineet.K.

[TinselKoala]

Vineet, once again your assumptions and diagram do not correspond to reality. Now you are apparently failing to consider the stretch of the balloon. When you initially fill the balloon with water and seal it, you have stored energy in the stretch of the balloon itself. This depends on the _volume_ inside the balloon, not the weight. I'm sure you have noticed that filled balloons tend toward a spherical shape if allowed: this is because the sphere has the smallest surface area for a given volume, and the stretchy skin tries to minimize _area_, but the volume is _conserved_.
When you then immerse this balloon in water you are not only _displacing_ the water that is in the tank, raising its level, but you also still have the stored tension in the balloon's skin due to the volume of water inside the balloon, even though its "weight" has been, you believe, neutralized. (It hasn't really, it's just been transferred to the lifted water in the container.) Since the volume in the balloon isn't changed, the skin tension is still there. Now you open the valve to the tube.... and the skin of the balloon pushes the water level up in the tube. The air that now bubbles out is due to the stretched balloon skin.
If you use a limp sack instead of a balloon, what happens? Nothing.

[vineet_kiran]

@TinselKoala,

1)      Stretching  of balloon requires  force which is provided by  weight of the water and also pressure of the water but not volume of water.    If you attach  a lesser length tube to the balloon,  the stretch in the balloon pushes the water up and water spills  out  bringing the balloon to its normal un stretched  condition but when you fix a lengthy tube (vertically),  water will  rise  to some height  corresponding to the force and pressure offered by the stretch of the balloon and stops at that height holding the balloon in stretched condition.

2)      When you immerse the water balloon with lengthy vertical tube attached to it  in water,  the balloon experiences pressure / force from  outside water which allows it to push  more water into the vertical tube.    The deeper the balloon pushed into the water,  it experiences more pressure and hence pushes more water into the tube.  (it works just like  a pressure indicator, indicating pressure at depth).

3)      When  water balloon is  immersed  in water  it loses its weight totally  with respect to external measuring balance (ie.,  if you fix this water balloon to a spring balance  and then immerse it in water,  the spring balance shows zero weight) because density of water in balloon and outside water  is same.  Rising of water level in the  container is immaterial  as it happens  when you immerse any solid material in water.   But this solid material immersed in water  loses its weight with respect to external spring balance depending on upthrust  of water as per the laws of Archimedes.

4)      Because the water balloon loses its weight with respect external force,  it can be easily moved inside the water applying a very little force.   When you  push the balloon deep in water, it displaces more water into the tube which in turn displaces air  present in upper portion of the tube to the exit  and  air comes out in the form of bubbles with huge force.   

5)      The energy required to move the balloon inside the water container is negligible  whereas air bubbles come out of the other container  with huge force.

6)      By  providing a valve or plug on the tube you can repeat the cycle moving balloon up and down with very little input energy and get comparatively high energy output in the form  of  air bubbles emerging out.

Vineet.K.

[TinselKoala]

Vineet, have you actually done this experiment yourself? 

Your description, if I am understanding you correctly, is not the way objects behave in reality.

When you fill a balloon with water it is primarily the volume that stretches the balloon, not the weight of the water. You can prove this to yourself very easily.

Fill a balloon with red-colored water to the extent required for your experiment,  and tie a knot in the end. No Air !!

Put this balloon into a bucket of water. Does the balloon shrink or expand? Its weight has "decreased".... but it still encloses the same volume and its skin still has the same overall tension. If you punch a hole in the submerged balloon,  now it will leak out the red water until the balloon's skin is relaxed again.

It is the volume inside the balloon that determines its tension. When  you have a balloon filled with water hanging in air, some of the tension in the skin is caused by the weight of the water but to get the effect you are describing you obviously have to put enough water in the balloon to stretch it substantially.

And don't forget: for any gravity or buoyancy effect to be useful, it has to be repeated in a cycle. If your first cycle depends on stored energy-- as yours does -- and the energy must be replaced for the cycle to repeat.... then you've got a problem. Who is filling up that balloon in the first place, working against its skin tension? Who lifted the water to do that with? What did it cost?