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underwater bucket Kaplan fl exam

Discussion in 'MCAT Study Question Q&A' started by chiddler, May 5, 2012.

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  1. chiddler

    chiddler

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    SDN Members don't see this ad. (About Ads)
    Take a bucket. Flip it upside down. Dunk it in water so that air remains inside the bucket. As you submerge it more and more, pressure pushes on the air reducing its volume.

    Kaplan is fancy and they call it a "bell". But screw them. it's a bucket.

    [​IMG]

    So how do I find that pressure? I am not looking for a test taking strategy to answer this question (please don't suggest PoE). I am looking for a conceptual understanding of this question.

    Thanks very much.

    edit: Ok I figured out. The pressure of the water and the pressure of the air must be equal. Thus the pressure at the contact point of the water and air must be the pressure of the air = rho*g*h of the water depth.
    Last edited: May 5, 2012
  2. chaser0

    chaser0

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    It is a disguised manometer question, basically.

    So treat it the same way. So the answer would be B.
  3. MedPR

    MedPR

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    I can say with 100% certainty that it is either A or B. 80% sure it's B since dh is h2-h1 and not h2+h1.
  4. Morsetlis

    Morsetlis SGU MS-3

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    h2 + h1 is a nonsensical value.
    h2 - h1 is the distance downward from sea level.
  5. FishyTheFish

    FishyTheFish

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    The answer is A.

    Logically, we know that as D increases, pressure will increase and that as L increases pressure will also increase.

    Thus we must find the equation that satisfies our logic.

    Expand out A and you get Pressure = P_o + pgD + pgL
    As D increases, Pressure increases. As L increases, Pressure increases. Thus equation A satisfies our logic and is the right answer. Do you have the solution to confirm?
  6. Dasypus

    Dasypus

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    Sorry fishy, that logic doesn't swim. The answer is B, because the pressure inside the bell must be at equilibrium with the water level (otherwise the water would rise or sink until equilibrium was reached). The gauge pressure at any height is ρgh, and the height is D-L.
  7. MedPR

    MedPR

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    Yea I'm leaning more towards B than A.
  8. chiddler

    chiddler

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    answer is B.
  9. MedPR

    MedPR

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    Even not knowing the concept, you can answer this question with equation recognition. P=P+pgh.
  10. FishyTheFish

    FishyTheFish

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    I guess my logic doesn't work because D and L effect each other?
  11. chiddler

    chiddler

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    which is why i explicitly said that i was looking for a conceptual understanding of this question. i thought B was a pretty obvious choice.
  12. MedPR

    MedPR

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    Damn, I missed that. Ok so at depth D, you have all the water above it pushing down on the trapped air but since the bell is open on the bottom you have water of height L pushing up (like a buoyant force) on the air. So written out the long way, the equation is actually P=P0+pgD-pgL because the upward pressure by the water in the L height region is negating some of the downward pushing pressure by the D height region.
  13. chiddler

    chiddler

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    you have the up down mixed up!

    the way I see it, the contact point between the two mediums indicates pressures of both. Because if the pressure wasn't equal then fluid would move around until it is equal.

    So. Atmospheric pressure + pressure due to water depth. Patm + rho*g*h. What is h? It is D-L.

    Pressure = Patm + rho*g*(D-L)
  14. MedPR

    MedPR

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    Works for me :D

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