Determining the Density of a sphere.

  1. 1. The problem statement, all variables and given/known data

    The radius of a sphere is measured to be (6.90 0.25 ) cm, and its mass is measured to be (1.75 0.05) kg. The sphere is solid. Determine its density in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures.)

    2. Relevant equations

    Volume = (4/3) (pi)(radius)^3

    = (4/3) (3.1416)(6.9/100)^3 = 1.376 X 10^-3 m^3

    Density = mass / vol = 1.75 / 1.376 X 10^-3 = 1271.8 kg / m^3

    = 1.27 X 10^3 kg /m^3


    3. The attempt at a solution

    % error in radius = (0.25/6.9)X100 = 3.6%
    therefore % error in volume = 3 X 3.6% = 10.8%
    % error in mass = (0.05/1.75) X 100 = 2.86%

    So % error in the density = 10.8 + 2.86 = 13.66%

    Therefore undertainty in density = 13.66% of 127 X 10^3 kg/m^3

    = 173 kg/m^3

    So density = (1.27±0.173) X 10^3 kg/m^3

    According to webassign, my answer is incorrect. I can't figure out what I did wrong. Any help would be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. My first time using this board, and no response for over 4 hours. Yet others who post after me are getting help asap. Thanks for the help guys!
     
  4. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to the forums teamsoulglo,

    Patients is a virtue, and besides, I'd be impressed if on average questions were answered quicker on any other boards. Now, to your question, you've done everything right up until when you calculate the uncertainty in the density. Suppose we have two quantities with their appropriate errors;

    [tex](A\pm\delta A)\hspace{1cm}(B\pm\delta B)[/tex]

    Now, suppose we want to find X = A/B, then the uncertainty in this case is;

    [tex]\frac{\delta X}{X} = \sqrt{\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2}[/tex]

    Also note that when you divide a value by a constant, you must also divide the error by that same constant. Hope this helps.
     
  5. I was a little unclear why the radius uncertainty is multiplied my three. Is there a relation between the exponent and the amt its multiplied by?? Truly wondering.
     
  6. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Indeed there is (as far as I know). If we again have some quantity [itex](A\pm\delta A)[/itex] and we wish to find [itex]Z = A^n[/itex] then the error in this case is;

    [tex]\frac{\delta Z}{Z} = n\cdot\frac{\delta A}{A}[/tex]
     
  7. Thanks hoot. Learn something every day on this forum.
     
  8. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    No problem :smile:
    Oh how true that is, more often that not its me doing the learning!
     
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