How to Determine Pressure with Different Density Liquids in a Manometer

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To determine the pressure required to balance the heights of oil and mercury in a manometer, the pressure difference must equal the weight difference between the two liquids. The oil's relative density of 0.920 and mercury's relative density of 13.55 are crucial for calculations. The formula P = ρgh can be used, where ρ is the density derived from the relative density (specific gravity) of the liquids. The atmospheric pressure can be ignored since the focus is on the pressure difference. Understanding these concepts will help clarify the solution to the problem.
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Homework Statement



Determine the pressure required to make the top of the oil, with height H0=443 mm even with the hight of mercury. The oil has relative density.920 and mercury has relative density13.55. To which side is the pressure applied?

Homework Equations


Code:
|      o----
|      o
|      o        ] 443mm
|      o
o      o----
o      o
oooooooo
o are mercury and | is oil

The Attempt at a Solution



my question is how the helllllll do i know what its relative too... and how would i go about balencing them
 
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There is a pressure being applied to one tube of the manometer. The pressure has to suffice to balance the difference in the weights of the contents of the two tubes. Think about force balance at the bottom of the manometer.
 
but i don't know the force being applied at the top am i crazy or are they too many unknowns?
 
The only unknown is the answer
 
Pressure is force per unit area. The mg force it needs to balance is g*density*volume=g*density*length*area. You can put in symbolic variables for area and length. You will find in the end you don't need to know them because they will cancel.
 
hmmm maybe this might help better than my sketchy diagram(see attachment)...

i still don't know how to calculate

P="rho"*g*h
but how do i calculate "rho" if i only have the relative density
 

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Relative density is referring to a multiple of the density of water. It's not all that 'relative'. It's usually called 'specific gravity'.
 
Good question! I guess they mean "specific gravity" or the question cannot be answered. I'm sorry -- I translated "relative density" into "specific gravity" when reading the question.
 
ok so pgh-Patm = P?

please help me its 3am here... i just need someone to actually explain how to get to the answer, because my teacher is a dick. and i hate university with a passion
 
  • #10
Weight/area is rho*g*h, if that's what you mean. And don't worry about Patm - you are just worried about a pressure difference. And the pressure difference has to support the weight difference between the two tubes.
 
  • #11
Damatrix02 said:
... my teacher is a dick ...
Careful -- look who's helping you!
 
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