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Finding possible heights of mercury in two joined manometers

  1. May 21, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-5-22_9-42-0.png



    2. Relevant equations
    ρ=hpg

    3. The attempt at a solution
    I really don't know how to rationalise the answer to this question. I just can't find the correct answer at all. The closest answer I thought was C and that too was also wrong since the pressure of mercury on the right side of manometer would be greater than 8000Pa in the right bulb. And the height difference among the two mercury column would add to the pressure of the left mercury column, making it greater than 13600Pa. So, I really am stuck at this question. Is my reasoning or concept wrong here, since I can't find the answer. By the way, the correct answer is D
     

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  3. May 21, 2015 #2

    TSny

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    Can you express the pressure at point A in terms of the pressure in bulb X and the height h1? See figure below.
     

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  4. May 21, 2015 #3
    Is it Ah1= pressure in bulb X? Since point A is higher than point B
     
  5. May 21, 2015 #4

    TSny

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    What does "A" represent on the left hand side of this equation?

    You should have studied a very fundamental formula that you can use to find the pressure difference between two points in a fluid at rest.
     
  6. May 21, 2015 #5

    TSny

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    In your first post you wrote the equation ρ = hpg? Can you describe what each symbol stands for?
     
  7. May 21, 2015 #6
    Oh, sorry I don't think A represents anything, so is it just h1=pressure in bulb X?
     
  8. May 21, 2015 #7

    TSny

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    No, h is a height. Height does not even have the same dimensions as pressure. So, writing h = pressure can't be correct.
     
  9. May 21, 2015 #8
    Oops,wrote the formula wrongly. Should be pressure=hρg, with h representing height, ρ for density and g for gravity
     
  10. May 21, 2015 #9
    So is h1ρg=pressure in bulb X correct?
     
  11. May 21, 2015 #10

    TSny

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    No, you need to think in terms of pressure differences. If h is the difference in height between two points a and b in a fluid at rest, then the pressure difference is Pb - Pa = ρgh where h is how much deeper b is compared to a.

    How would this apply to the two points A and C shown below?
     

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  12. May 21, 2015 #11
    Is the pressure difference between point A and C is h1ρg?
    Is the difference between point C and the other lowest point at bottom of bulb Y the pressure balancing pressure in bulb Y?
     
  13. May 21, 2015 #12

    TSny

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    Yes. Make sure you understand which point is at the higher pressure. Can you write an equation that relates PA, PB, and h1? [Edit: I meant to refer to points A and C, not A and B]
    Sorry, I don't understand what you are saying here. Can you write another equation that relates the pressures at points B and D in the figure below?
     

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    Last edited: May 21, 2015
  14. May 21, 2015 #13
    Is it PA-PB=h1ρg?
     
  15. May 21, 2015 #14

    TSny

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    Oops, that was my fault. I meant to ask if you can write an equation that relates the pressures at points A and C (not A and B). Points A and C can be considered as two points in the same mercury fluid on the left.
     

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  16. May 21, 2015 #15
    Is it PA+h1ρg= PC?
     
  17. May 21, 2015 #16

    TSny

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    Yes. Do you see that the pressure PA is also the pressure, P, of the air in the middle horizontal section of the apparatus? Also, PC is equal to the pressure in bulb X. So you can rewrite your equation by replacing PA by P and PC by PX.

    Now repeat this process for points B and D.
     
  18. May 21, 2015 #17
    Wow, I've got it. Thanks so much for pointing out the approach:wink:
     
  19. May 21, 2015 #18

    TSny

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    Great! Good work!
     
  20. May 22, 2015 #19
    Since 16000Pa = 12cmHG and 8000Pa = 6cmHg, what pressure P in the intermediate loop would give h1 = 18cm and h2 = 12cm?
     
  21. May 22, 2015 #20

    TSny

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    Good point. I didn't bother to find P. To make it work I guess we could take the given pressures to be gauge pressures rather than absolute pressures.
     
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