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Homework Help: Percent uncertainty in the volume of a spherical beach ball

  1. Aug 19, 2004 #1
    Question is : What is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 3.86 ± 0.08 m ?

    The answer is 6 % , but Im not getting that, Im pretty sure im on the right path, I found the volume of the ball using V = (4 x 3.14 x r^3) / 3

    Btw, Im pretty sure you need to take into account the ± .08m for 3.86 when you find radius. I got 3.94 and 3.78 Radius's. But I still dont see how I can get 6 %. Heres the percent uncertainty formula. (uncertainty)/(value) x 100

    Maybe Im on the wrong track, please help me through this problem, I am new to Physics and a little rusty on my math, so any pointers are greatly appreciated.

  2. jcsd
  3. Aug 19, 2004 #2


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    If you get in trouble with these questions, one idiot p-proof method is to work out the volume: with a) the measured value, b) the maximum value, c) the minimum value, then by comapring the volumes the uncertainty is obvious.
  4. Aug 19, 2004 #3
    a) V=240
    b) V=256
    c) V=226

    Please point out the obvious.

    (I know the answer, I just want some help how to find it.)
  5. Aug 19, 2004 #4


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    (max. - min.)/2 then just divide and mutiply by 100 to get the answer in percent.
    Last edited: Aug 19, 2004
  6. Aug 19, 2004 #5
    Here is how you do it:

    [tex] uncertanty = \frac{V_{u}- V_{e}}{V_{e}} * 100 [/tex]

    [tex] uncertanty = \frac{256 - 240}{240} * 100 [/tex]

    [tex] uncertanty = 6.66 [/tex]

    Does that answer your question.
    Last edited: Aug 19, 2004
  7. Aug 19, 2004 #6
    there we go, sorry for all of the posts, I had a big error, the answer is 3 posts up.
  8. Jan 8, 2009 #7
    [tex]V=\frac{4}{3} \pi r^3[/tex]

    [tex]dV=\frac{4}{3}(3r^2) \pi[/tex]

    [tex]dV=\frac{4}{3}(44.6988) \pi \times (0.08)=14.97971165[/tex]

    [tex]\frac{14.97971165}{256} \times 100 = 5.8[/tex]
  9. Jan 8, 2009 #8


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    Staff: Mentor

    Do you realize this thread is more than 4 years old...?
  10. Jan 8, 2009 #9
    Russ - "what happens online, STAYS online..." apparently forever. That's the beauty & the curse of the medium.
  11. Jan 9, 2009 #10
    woah, I am completely sorry.
    I was just searching in Google and found this. I didn't realize that it would be that old.
    BTW, people who were searching like me now know the answer.
  12. Jan 11, 2009 #11
    I agree with that!
  13. Feb 8, 2010 #12
    6 years later and still useful.
  14. Sep 14, 2010 #13
    Almost 7 years later and we just did this problem in class! Still useful.
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