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Finding pressure P of air held in between two columns of Mercury

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    2. Relevant equations
    P=hρg

    3. The attempt at a solution
    The length of mercury column ...
    (1) on the left side (exposed to atmosphere) would be ( 760 + x/2 )mm
    (2) on the right side (exposed to atmosphere) would be ( 760 + (50-y)/2 )mm
    (3) on the left (exposed to air at pressure P) would be ( 760 -x/2) mm
    (4) on the right (exposed to air at pressure P) would be ( 760- (50-y)/2 ) mm

    By equating (1)=(2) or (3)=(4) ,
    x=50-y
    A and D are out as x and y does not satisfy this equation.

    I chose B as (by assuming from the diagram) x is shorter than y, but I do not know how to calculate the Air pressure P.. Any hints? Thanks in advance
     
  2. jcsd
  3. Nov 2, 2015 #2
    The results you obtained under items 1 and 2 are both correct expressions for the pressure P (although the factor of 2 in the denominator should not be in there. Where did that come from?). The results you obtained under items 3 and 4 make no sense to me.
    So,

    P = 760 + x

    and

    P = 760 + (50-y)

    So which combination of values is consistent with these two equations?
     
  4. Nov 2, 2015 #3
    Oh I see.. B.
    Oh I thought that the column (1) exposed to air would increase by x/2, and column (2) exposed to air pressure would decrease by x/2. But I guess it's wrong.
    Thanks for the help!
     
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