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Fluid dynamic problem

  1. Feb 15, 2012 #1
    Hi I am solving some problems in fluid dynamics I came to this problem that I don't understand how to solve it.


    A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are ρ, ρ1 and ρ2 respectively, the ratio of the volume of granite in water to the volume in mercury is
    (a) (ρ2 – ρ) /(ρ – ρ1)
    (b) (ρ2 + ρ) /(ρ+ ρ1)
    (c) ρ1 ρ2 /ρ
    (d) ρ1 /ρ2
    (e) ρ2 /ρ1

    figure of problem is here -->
    http://www.apphysicsresources.com/2009/07/ap-physics-b-multiple-choice-questions.html



    I tried to approach it like this. we know that the granite will sink so we know that the pressure that water will put on the granite untill given height will be the same as buoyancy force of mercury,which is same as weight displaced.

    But I couldn't solve it in the end in terms of those variables I don't also understand the derivation in that link.

    Thanks in advance.
     
  2. jcsd
  3. Feb 16, 2012 #2
    sorry to bump this but I really need to understand this problem I hope this isn't against the rules.
     
  4. Feb 16, 2012 #3

    boneh3ad

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    Science Advisor
    Gold Member

    Well start with Archimedes principle: The mass of the fluid displaced equals the mass of the object displacing the fluid.

    The mass of the granite is
    [tex]\rho (V_1 + V_2)[/tex]

    The mass of displaced water is
    [tex]\rho_1 V_1[/tex]

    The mass of the displaced mercury is
    [tex]\rho_2 V_2[/tex]

    So, we have
    [tex]\rho (V_1 + V_2) = \rho_2 V_2 + \rho_1 V_1[/tex]

    [tex]\rho V_1 + \rho V_2 = \rho_2 V_2 + \rho_1 V_1[/tex]

    [tex]\left(\rho - \rho_1\right) V_1 = \left(\rho_2-\rho\right) V_2[/tex]

    [tex]\frac{V_1}{V_2} = \frac{\rho_2-\rho}{\rho - \rho_1}[/tex]
     
  5. Feb 17, 2012 #4
    I see so we just made the mass of granite to be the amount it displaces on water and the amount it displaces on mercury. that makes sense.
    Thanks.
     
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