Have manometer given density and mass

AI Thread Summary
When water is added to one arm of a U-shaped manometer filled with Karo syrup, it will create a difference in liquid heights between the two arms due to the differing densities. The initial equilibrium state must be analyzed, and a sketch is recommended to visualize the changes. The force exerted by the added water can be calculated using the formula F=mg, where m is the mass of the water and g is the acceleration due to gravity. The final solution should detail the height difference in the liquid columns after the water is introduced. Understanding these principles is crucial for solving the problem accurately.
p.mcnamara
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Homework Statement

Assume you have a U-shaped, hollow tube (we call it a manometer) with Karo syrup (density 1200 kgm-3) in it such that both arms are half filled. What would happen if you added a given mass of water (1000 kgm-3) into one of the arms of the manometer? (the mass has less volume than half the arm of the manometer)



Homework Equations





The Attempt at a Solution

F=mg=v*p*g F=(1000kgm-3)(9.8m/s2)=9800N
 
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p.mcnamara said:

Homework Statement

Assume you have a U-shaped, hollow tube (we call it a manometer) with Karo syrup (density 1200 kgm-3) in it such that both arms are half filled. What would happen if you added a given mass of water (1000 kgm-3) into one of the arms of the manometer? (the mass has less volume than half the arm of the manometer)



Homework Equations





The Attempt at a Solution

F=mg=v*p*g F=(1000kgm-3)(9.8m/s2)=9800N

Welcome to the PF.

Your work doesn't really go towards the question, IMO. You need to draw a sketch (or post the one in your textbook) of the manometer in equilibrium before and then after adding the water to one side. Your answer should reflect the difference in the heights of the two tops of liquid in the two sides of the manometer.

Please post a sketch of the before/after situations, and then work out the difference between the two heights in the after state.
 

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