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Calculating % uncertainties

  1. Mar 6, 2016 #1
    A manufacturer needs to determine the volume of metal used to produce a metal pipe of the following dimensions:
    Length, L 40 + 1 m
    External diameter, D 12.0 + 0.2 cm
    Internal diameter, d 10.0 + 0.2 cm
    Estimate the greatest possible percentage error of the volume.

    the correct ans is 22.5%.

    but no matter how many times i've tried, my ans is always 34%. Help pls thank you! :)
     
  2. jcsd
  3. Mar 6, 2016 #2
    Could you show us your working?
     
  4. Mar 6, 2016 #3
    V=(pi)(R^2)(L) - (pi)(r^2)(L)

    delta(R)=(1/2)(0.2)=0.1cm
    delta(r)=(1/2)(0.2)=0.1cm

    delta(V) = delta(pi*R^2*L) + delta(pi*r^2*L)

    Let x=(pi)(R^2)(L)

    delta(x)/x = 2(0.1/6) + 1/40= 7/120

    delta(x)=7/120 * (pi*R^2*L)=7/120 * 14 400 pi = 840pi

    Let y=pi*r^2*L

    delta(y)/y = 2(0.1/5) + 1/40=13/200

    delta(y)=13/200 * pi*r^2*L=13/200 * 10 000pi= 650pi

    delta(V) = delta(x) + delta(y)
    =1490pi
    delta(V)/V *100% = 1490pi/4400pi *100 = 34%

    fyi 4400pi is the value of V
    thank you
     
  5. Mar 6, 2016 #4

    Drakkith

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    Staff: Mentor

    Hi harvey,

    I've moved your thread to the homework forums, but in the future please post all homework questions in the appropriate homework forum and use the template provided.
     
  6. Mar 6, 2016 #5

    haruspex

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    You have counted the 1/40 twice over. There may be other errors.
    Safest way is just to compute the maximum possible volume.
     
  7. Mar 6, 2016 #6
    i'm sorry, what do you mean by counting 1/40 twice? is my method not the correct approach anyway?
     
  8. Mar 6, 2016 #7

    haruspex

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    You counted it here:
    and here:
    Your problem starts with this line:
    Let's back up one step:
    ##\Delta V = \Delta (\pi*R^2*L-\pi*r^2*L)##
    You cannot split that as ##\Delta (\pi*R^2*L)+\Delta(\pi*r^2*L)## because the two terms are not independent.
    If the actual length is L-ΔL on the inner radius it will be L-ΔL on the outer radius. Your analysis exaggerates the range by allowing a volume like
    ##\pi((R-\Delta R)^2(L-\Delta L)-(r+\Delta r)^2(L+\Delta L))##
     
  9. Mar 7, 2016 #8
    however, it is true that if R=A+B,
    then delta(R) = delta(A) + delta(B).
    i don't get what you mean by independent though. i'm sorry but i just want to fully understand this, thank you!
     
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