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Buoyant Force and density

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of copper having a density of 8.90 g/cm^3 has an apparent mass 120g in water and 116g when submerged in a liquid. What is the density of the liquid?



    2. Relevant equations




    3. The attempt at a solution
    Don't know if there's a better method, but here is the way I did it.

    Let the copper block's mass be M and its volume V.
    Mass = density x volume, so M = 8.90V.

    Let the liquid's density be D.

    In water apparent loss of mass = M-120
    In liquid apparent loss of mass = M-116

    Apparent loss of weight (mass) = weight(mass) of fluid displaced.
    For water: M-120 = water's density x volume = 1 x V = V (equation 1)
    For liquid: M-116 = liquid's density x volume = DV (equation 2)

    Replace M by 8.9V in equation 1:
    8.9V - 120 = V
    7.9V = 120
    V = 120/7.9 = 15.19cm³

    M = 8.9V = 8.9 x 15.19 = 135.2g

    From equation 2:
    135.2 - 116 = Dx15.19
    D = 19.2/15.1 9 = 1.26 g/cm³
     
  2. jcsd
  3. Jun 26, 2012 #2

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    Hello jgridlock,

    Welcome to Physics Forums!

    It wouldn't hurt to list some relevant equations in this section.
    'Looks correct to me. :approve:
     
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