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Homework Help: The approx. uncertainty in r

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]V=\pi r^{2} h[/tex]

    The uncertainties in V and h are shown below
    V 7%
    h 3%

    The approximate uncertainty in r is:
    A. 10% B. 5% C. 4% D. 2%

    2. The attempt at a solution
    According to the key the correct answer is B. This answer can be calculated by:
    [tex]r^{2}=\frac{V}{\pi h}[/tex]
    which gives 3%+7% = 10% uncertainty
    and subsequently
    [tex]r=\sqrt{\frac{V}{\pi h}}[/tex]
    giving 10% * 1/2 = 5%

    if the value of 5% uncertainty for r is inserted in the original equation it all doesn't make sense:
    [tex]V=\pi r^{2} h[/tex] should then give an uncertainty for V of 3%+5%+5%=13%

    Can anyone please explain this?
  2. jcsd
  3. May 4, 2010 #2


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    Homework Helper

    I'd say the answer was 2%, but I can't really explain how your book got to that answer.

    [tex]\frac{dV}{V}=2 \frac{dr}{r}+\frac{dh}{h}[/tex]
  4. May 4, 2010 #3
    That was my first thought as well. But the key says no.

    And as proved in my previous post, the 5% uncertainty does make sense, however if used in the original equation (or by simply reversing the second) it doesn't sum up.
  5. Apr 20, 2011 #4
    Bump. Any ideas?
  6. Apr 20, 2011 #5
    you must be clear about what is measured and what is calculated.

    if you measure V and h and are wanting to calculate r, then 5% is the correct error. but then your check does not make sense because there you are assuming r and h are measured and V is calculated.

    If the question asked what error in r (which is measured) would give rise to a 7% error in V (which is calculated) given a 3% error in h (which is measured), then the answer would be 2%.

    hope this helps
  7. Apr 20, 2011 #6
    Yes, that makes sense. Thanks!
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