Fluids and pressures

  • #1
187
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Mercury is poured into a u-tube. The left arm has cross sectional area A1 = 10 cm^2, the right has area A2, 5cm^2.
100 g of water are then poured into the right arm.
determine the length of the water in the right arm and given the density of Hg = 13.6 g/cm^3, what distance h, does the mercury rise in the left arm.

i have a diagram here and have calculated the length of the water in the right arm to be 0.2 m. i dont know now how to go about calculating the height that the mercury rises though. any help?
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
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Remember the total volume of mercury has not changed, therefore the volume displaced on the right must be the same volume that rose a distance h.
 
  • #3
lightgrav
Homework Helper
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At the bottom of the water column, it touches the mercury.
On the other side of the U-tube "balance", at that height,
there's an equal *Pressure* caused by mercury above it.
 
  • #4
187
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ok so i know at that depth on the right side the pressure = external pressure + densityofwater*g*0.2
so i equate this and get [tex]\rho_{Hg}gh=\rho_{water}g(0.2)[/tex]
but the h here is not the h i require, it is height i require plus some other height...
where that other height is the length displaced on the right i think??
 
  • #5
lightgrav
Homework Helper
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If you would replace the water height
with half of the new excess mercury height,
the two sides would be the same height.
just like they started.
(see cyclovenom's post, above)
 
  • #6
187
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quite a tricky little problem, but thanks for all your help i got the correct answer now
 

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