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What contributes the most to the error in the density of a cylinder?

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    If the three quantities; mass, diameter, and height for a cylinder are measured with the same percentage error, which would contribute the most to the error in the density?


    2. Relevant equations

    ρ=M/V
    ρ=4*M/(pi*d^2*h)

    3. The attempt at a solution

    so, I tried to answer it like this:
    since the equation for the cylinder density is ρ=2*M/(pi*d^2*h)
    since the mass and the height aren't squared the error would remain the same, however the diameter is squared which means the error is squared as well.. so it's the diameter!!!!

    BUT my friend says since ρ(density)=M/V , then mass would contribute the most not diameter ;O

    HELP PLEASE , I dont know which answer to put in our report D:?!

    ::
     
  2. jcsd
  3. Oct 2, 2012 #2
    You are right, it's the diameter.
     
  4. Oct 2, 2012 #3
    REALLY?!
    THANKS ALOT!!
    YOU HAVE NO IDEA HOW HARD WAS IT FOR ME TO GET AN ANSWER FOR THIS QUESTION!!!!!

    ::

    Umm, well ..is it the diameter because of the reason i stated =o?
     
  5. Oct 2, 2012 #4
    Yes, but the error isn't exactly squared.
     
  6. Oct 2, 2012 #5

    I like Serena

    User Avatar
    Homework Helper

    Yes.


    In more detail, the error percentage is ##{\Delta \rho \over \rho} \times 100\%##.

    The contribution of an error due to some variable x is:
    $$\left({\Delta \rho \over \rho}\right)_{\text{due to x}} = {{\partial \rho \over \partial x} \Delta x \over \rho}$$

    Do you know how to calculate that for M, d, and h?
     
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