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Fluid question

  1. May 17, 2015 #1
    1. The problem statement, all variables and given/known data

    upload_2015-5-17_3-55-47.png
    2. Relevant equations
    P1= Patm + ρwatervg
    P2= Patm + ρmer.vg
    3. The attempt at a solution

    so both sides of the tubes are open to atmosphere pressure and are in equilibrium.
    I can get P1 =P2

    the 1 atm from each side cancels out.
    g (gravity) cancels out

    I'm left with volume (or distance in this case) of water times density of water = distance of mercury times density of mercury:

    (1000 kg/m3 ) (0.12m) = distance of mercury ( 13600 kg/m3)

    solving for distance of mercury gives 0.088m or 0.88 cm

    is this correct?
     
  2. jcsd
  3. May 17, 2015 #2

    Orodruin

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    Yes. As long as you interpret the question as how hight the mercury on the other side rises above the mercury water interface. If it is in comparison to the original level the answer will be different.
     
  4. May 17, 2015 #3
    how would it be in comparison to the original level? what would the answer then be?
     
  5. May 17, 2015 #4

    Orodruin

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    What do you think? If one end moves down 1 cm, how does the other end move?
     
  6. May 17, 2015 #5
    up 1 cm.

    so if the side with water went down 12 cm, the other side should have went up 12 cm, but that isn't what I got as the answer, I'm wrong here
     
  7. May 17, 2015 #6

    Orodruin

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    And what is the resulting difference in surface levels?
     
  8. May 17, 2015 #7

    SteamKing

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    Your calculation should read 0.0088 m or 0.88 cm. Watch those decimals and zeroes.
     
  9. May 17, 2015 #8
    if the water pushed mercury down 12 cm, and the other side moved up 0.88 cm, then the difference should be 12- 0.88 = 11.12 cm?
     
  10. May 17, 2015 #9

    Orodruin

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    No, we are talking only about the mercury surfaces now. If one mercury surface goes down by 1 cm, you have said that the other rises by 1 cm - so what is the separation between the two? You have also stated that the separation between the two should be 0.88 cm. What does this mean for how much the mercury rose on the side you did not pour water?
     
  11. May 17, 2015 #10
    hmm, if I pour 1 cm of mercury into a tube full of mercury, when it is in equilibrium, i essentially raised both sides of the tubes by 0.5 cm.

    so If the water is 12 cm deep, that side that water was added, should increase by 12 / 2 = 6 cm

    so the other side should increase by 0.88 / 2 = 0.44 cm
     
  12. May 17, 2015 #11

    Orodruin

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    We are not talking about pouring more mercury in, we are talking about pushing one of the mercury surfaces down.
    This is wrong.
    This is correct, but check your reasoning.
     
  13. May 17, 2015 #12
    so the separation between the 2 when one side is pushed down 1 cm, is 2 cm
     
  14. May 17, 2015 #13

    Orodruin

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    Correct, and since the separation was 0.88 cm, one side is pushed down 0.44 cm and the other up by 0.44 cm. The water level rises 12 cm above the surface that is pushed down and therefore is 11.56 cm above the original location of the mercury surface, i.e., 11.12 cm above the mercury surface on the other side.
     
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