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Fluids - maximum error - diameters

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data:
    The expressions you obtained for acceleration and drainage time (equations 3 and 4 in the manual) both contain the following factor or something equivalent: (D/d)^4-1
    In the lab, we will be dealing with a bottle with a diameter around 9 cm. A typical diameter of the drainage hole is 8 mm. With these numbers, what is the maximum error that we introduce in the expression above if we use the approximation:
    (D/d)^4-1 ~ (D/d)^4
    This is a small error compared to the uncertainties in the lab (the motion detector alone has a resolution of 1 mm. Since we will be measuring distances of the order of 20 cm, this is already a 0.5% uncertainty.)

    2. Relevant equations
    This question makes NO sense to me. I don't get the point of the approximation equation.

    3. The attempt at a solution
    Wouldn't the maximum error just be 9cm/8mm or 9/.08 = 112.5 OR would it be .08/9 =
    .008 or 2/225? I've tried both of these answers but the program didn't accept either. What am I doing wrong? How am I supposed to use the approximation equation listed in the question??? Thank you sooooo much, this is due tomorrow night so I'm allowed to enter the lab/participate on Monday.
  2. jcsd
  3. Nov 3, 2012 #2
    Look, you just need to think about these things. Think about what the meaning of "error" is. Think about what the "true" value is, what the "approximation" is, and what you can do to arrive at the difference between the two.
  4. Nov 3, 2012 #3
    Error: The difference between the result of the measurement and the true value of what you're measuring.
    True value: What the actual value should be.
    Approximation: What you approximated or rounded the measurement to.
    Well, I believe from the above problem that the approximation would be within 8 mm or .08 cm of the true value. Is that correct?
    I'm still confused on what the question is asking. Why do they say something about (D/d)^4 -1 = (D/d)^4 ; I don't understand how those equations are equal. And I don't understand how I'm supposed to use them to answer the question.
  5. Nov 4, 2012 #4


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    What is it that is being approximated? (Hint: it is not a distance.)
  6. Nov 4, 2012 #5
    Acceleration and drainage time?
  7. Nov 4, 2012 #6


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    "the maximum error that we introduce in the expression above" where the expression above was "(D/d)^4-1". OK, it did earlier refer to "The expressions you obtained for acceleration and drainage time", but those expressions did not appear "above".
    Anyway, the question says "(D/d)^4-1" appears as a factor in the expressions for acceleration and drainage times, so assuming they want the percentage error, it's the same answer in all three cases. So just take it as asking for the percentage error that arises from using (D/d)^4 in place of (D/d)^4-1.
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