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Specific Gravity

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A 8.3 kg block of some unknown material is suspended from a spring scale and submerged in water. The spring scale reads 46.9 N. What is the specific gravity of the block?




    2. Relevant equations

    SG = density of object/density of reference material (water = 1000 kg/m^3)

    density = mass/volume

    3. The attempt at a solution

    How do I find the density of the object if I do not know its volume?
     
  2. jcsd
  3. Nov 13, 2009 #2

    rl.bhat

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    Find the weight of the block mg in air.
    Weight in water is given. Find loss of weight. That is equal to weight of the displaced liquid. From that you can find the volume of the block.
     
  4. Nov 13, 2009 #3
    so......

    weight in air = 8.3 * 9.81 = 81.42 N

    differnce = 81.42 - 46.9 = 35.523 N (weight of displaced liquid)

    Density = m/v so

    1000 = 35.523/v ....v = .035523 (volume of block)

    density of block = m/v = 8.3/.035523 = 233.65

    SG ( of block) = 233.65/1000 = .233

    I did this and I get the wrong answer
     
  5. Nov 13, 2009 #4

    rl.bhat

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    35.523 N is the weight of the displaced water. Find the mass of the displaced water. And then find the density of the block.
     
  6. Nov 13, 2009 #5
    ok...

    mass of displaced water is 3.21 kg

    1000 = 3.21/v....v= 0.00321

    Density of block = 8.3/0.00321 = 2585.669

    SG = 2585.669/1000 = 2.585

    Still getting wrong answer
     
  7. Nov 13, 2009 #6
    can someone just show me the step by step solution please.....That way I can know exactly know what I am doing wrong instead of doing the wrong steps over and over....TIA
     
  8. Nov 13, 2009 #7

    rl.bhat

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    35.523n/9.8 =---?
     
  9. Nov 13, 2009 #8
    SG = 81.42/(81.42 - 46.9) = 2.36

    Thanks for your help though
     
  10. Nov 13, 2009 #9
    Now for the density of block....

    D = 8.3/.0035 = 2.37 * 10^3 kg/m^3
     
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