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Fluids question: U Shaped Tube.

  • Thread starter Eilrymist
  • Start date
  • #1
2
0

Homework Statement



(1.1.10) The diagram below shows a U-shaped tube. The fluids in the tube are glycerine (blue
shading) and mercury (gray shading). The height of the glycerine column on the left side of the
tube is h1 = 5.00 cm. The density of the mercury is 13.53 g/cm3
.
(a) What is h2?
(b) Oil with a density 0.800 g/cm3
is added to the right side of the tube, so that the right side of
the tube has an oil column sitting on top of the mercury. What must be the depth of this oil
column so that the top of the glycerine column on the left and the top of the oil column on the
right are at the same height? Keep four significant digits in any intermediate steps.

Homework Equations



P = ρgh + Patm
ρ(glycerine) = 1.261 g/cm3
F = (Pressure)(area)

The Attempt at a Solution



I have solved part a. I am having trouble with Part b, where there is a layer of oil on the other side. I have read the text book, and want to use the relationship between the pressures being the same at a certain depth, like what I did in part A. What I am unsure of is how to deal with the two layers when they are in a tube because there is no "bottom" to the tube. This is my attempt at solving it:

P (a) = P (b) <- a and b are located where the bottom of the mercury layer is.

P (oil) = Patm + ρ(oil)gh2
P(a) = P(oil) + ρ(mercury) g (h1 - h2)

P(b) = Patm + ρ(glycerine) g h1

I set these equal to each other, but I believe this is incorrect and I am missing some concept here.

My answer: 7.7 cm
Professors answer: 4.82 cm
 
Last edited:

Answers and Replies

  • #2
6,054
390
Can you post the diagram?
 
  • #3
2
0
Can you post the diagram?
thank you for your response. I would post it, but I did solve the question. It was me neglecting the units. for future questions i will make sure to post the diagrams
 

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