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A vessel filled with two liquid how to determine boant force?

  1. Dec 30, 2011 #1
    A vessel filled with two liquid how to determine boyant force?

    A vessel contains oil of density(800kgm-3) over mercury (density 13600kgm-3) A homogeneous Sphere is floats with half of its volume immersed in mercury and other half in oil. What is density of the material of the sphere?

    How to apply Archimedes principle of floating? can you please help.
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 30, 2011 #2

    ehild

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    What does Archimedes' Principle say?


    ehild
     
  4. Dec 31, 2011 #3
    It says that buoyant force is equal to weight of liquid displaced.
     
  5. Dec 31, 2011 #4

    ehild

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    There are two kinds of liquid displaced. What is the volume and the weight of each when the volume of the sphere is V?

    ehild
     
  6. Dec 31, 2011 #5
    But lighter liquid seems to push the sphere downward rather than pushing it upward. and lower liquid seems to push with greater power.
    even after this. can i apply that principle?
     
  7. Dec 31, 2011 #6

    ehild

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    You get upward force at the bottom of the immersed volume, and an upward force at the top.
    Pascal's law states that the force a liquid exerts on a surface is normal to the surface and F=PA.
    As the pressure is greater at the bottom of the immersed object then at the top, the liquid pushes the bottom surface upward by a greater force than it pushes the top surface downward. The resultant force is always upward.


    ehild
     
  8. Dec 31, 2011 #7
    thnank's sir(are u male) for helping me,
     
  9. Dec 31, 2011 #8

    ehild

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    Can you proceed from here?

    ehild
     
  10. Dec 31, 2011 #9
    Actually i have already(before posting) done this question by this method. But I did not understand that is it correct or not. that's why i post it here.
     
  11. Dec 31, 2011 #10

    ehild

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    Do you understand it now?

    Take a block instead of the sphere. The area of the top and bottom sides is A. The pressure at depth d in the oil is
    Ptoilgd +P0, so the downward force on the top surface of the block is Ft= A(gρoilgd +P0).

    At the bottom of the block, the pressure is that of a column of oil of height d+h/2 and a column of mercury of height h/2. Pboilg(d+h/2)+ ρHgg(h/2)+P0, so the force Fb =A(ρoilg(d+h/2)+ ρHgg(h/2)+P0)
    The resultant is the buoyant force: BF=Ag( ρoil(d+h/2)+ ρHg(h/2)-ρoild)=Ag(ρoil(h/2)+ ρHg(h/2))=g(V/2)(ρoil+ ρHg).

    ehild
     

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