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Relative density of ice in cylinder

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A measuring cylinder contains 60cm3 of oil at 0 degrees celsius .When a piece of ice is dropped into the cylinder,it sank completely in oil and the oil level rose to 90 cm3 mark.When the ice melted the oil level came down to 87cm3 mark.The relative density of ice is
    (1)0.80 (2) 0.85 (3) 0.90 (4)0.95 (5)0.98

    2. Relevant equations



    3. The attempt at a solution
    I guess the volume of ice is 90-60=30cm3
    and the relative density of ice can be given as = real weight of ice(mg) / apparent loss in weight in oil(mg-m'g)?

    mg would be 30*d*g,where d is the density of oil ,but how do I find its apparent weight in oil.I think it has something to do with the 87cm3 but I can't figure out what.

    Hope someone can help.
    Thanks in advance.
     
  2. jcsd
  3. Jul 27, 2009 #2

    Cyosis

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    Homework Helper

    To calculate the relative density you need to have a reference density. For this problem I guess you're expected to calculate the relative density of ice with respect to water.

    Then [itex]RD=\rho_{ice} / \rho_{water}[/itex]. Now try to find expressions for both densities.
     
  4. Jul 27, 2009 #3
    Thanx.
    so [itex]\rho_{ice}[/itex] = m/30
    & [itex]\rho_{water}[/itex]= m/27

    therefore RD = (m/30)*(27/m) = 0.9,which would be answer no.3?
    but I have a slight problem,in some questions i've used the equations
    RD of the material of a body= weight of body/weight of an equal volume of water
    and to find the RD of a liquid=weight of a given volume of a liquid/weight of an equal vol.of water
    or upthrust in liquid/upthrust in water
    I can't undersatand why we can't solve the above problem using these equations
     
  5. Jul 27, 2009 #4

    Cyosis

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    Homework Helper

    Let's use this definition then. The weight of the body of ice is [itex]W_i=m_ig=\rho_i V_i g[/itex].
    The weight of an equal volume of water is [itex]W_w=m_w g=\rho_w V_i g[/itex]. Putting this in your definition of RD yields:

    [tex]
    RD=\frac{\rho_i V_i g}{\rho_w V_i g}=\frac{\rho_i}{\rho_w}
    [/tex]

    So as you can see these two equations, while perhaps looking different, really are just the same.
     
  6. Jul 27, 2009 #5
    Oh yes!They are the same.
    Thanks so much,Cyosis.
     
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