Aurelius120 said:
View attachment 357249
We have to blow against (weight-buoyancy), Right?
Weight of what and buoyancy of what, exactly?
In the initial state, I see an inverted beaker supported by nothing other than the rim of the vertical pipe.
The upward pressure of water on the flat upper surface is exactly equal to the downward pressure from the atmosphere. The upward pressure of the air in the pipe is exactly equal to the upward pressure from the air in the pipe. Zero net upward force from buoyancy on the bulk volume of the beaker.
We do have some buoyancy from the upward pressure of water on the lower rim of the beaker which is not entirely matched by the downward pressure on the upper rim of the beaker.
So yes, we have the weight of the glass (or metal) beaker minus the buoyancy from the volume of the beaker's glass (or metal) in water. A beaker made of ordinary glass or from borosilicate will have a specific gravity of about 2.4 or 2.5. Stainless steel is three times that.
So for a glass beaker, it would be about 60% as hard as blowing against the same beaker resting on top of the vertical pipe with an airtight seal but no water. (##\frac{2.4 - 1.0}{2.4} \approx 0.6)##. For a stainless steel beaker it would be about 87% as hard as blowing against the same beaker resting on top of the vertical pipe with an airtight seal but no water (##\frac{7.5-1.0}{7.5} \approx 0.87)##
The diameter of the pipe will be a critical factor in how much air pressure we will need to do this.
If the seal is leaky, we may be able to introduce air more easily than that. The diameter of the beaker will then become important.