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Homework Help: SigFigs in Volume and Uncertainty?

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data
    A car engine moves a piston with a circular cross section of 7.500 ± 0.005cm diameter a distance of
    3.250 ± 0.001cm to compress the gas in the cylinder.
    (a) By what amount is the gas decreased in volume in cubic centimeters?
    (b) Find the uncertainty in this volume.

    2. Relevant equations
    Area = πr2
    ΔVolume = Area × ΔDistance
    %unc = (ΔA/A) × 100%

    3. The attempt at a solution
    The radius of the cross section of the piston is 3.75cm so the area comes out to 44.18 cm2
    The change in volume then comes out to 143.6cm3
    -This value is measured to 4 significant figures because both the diameter and the distance were given to 4 significant figures.​
    The percent uncertainty in the diameter comes out to 0.0267%
    The percent uncertainty in the distance comes out to 0.0308%
    The percent uncertainty in the change in volume should then be these values added together, giving 0.0575%
    -With the correct number of significant figures this should be 0.06% because the original uncertainties (0.005 and 0.001) both had only 1 significant figure.​
    Then, to get the uncertainty for the change in volume you use %unc = (ΔA/A) × 100% and solve for ΔA
    -For this I got 0.08618 which comes to ± 0.09cm3

    My question is: If the change in volume (143.6cm3) has the correct number of significant figures but is measured only to the nearest tenth of a centimeter, then how can the uncertainty (±0.09cm3) be in hundredths of a centimeter?
  2. jcsd
  3. Apr 25, 2016 #2

    Simon Bridge

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    The "sigfig" rule is only a rule of thumb and should not be used if you have the option.
    If you have actual uncertainty values, as is the case here, those are what you use.
  4. Apr 26, 2016 #3


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    I agree, but would express it a little differently. The sig fig system is a convention for implying the accuracy. When it is given explicitly, the convention does not apply.
  5. Apr 26, 2016 #4
    Thanks, so just to be clear that would mean that I should express the change in volume and the uncertainty in the change in volume to the nearest thousandth of a centimeter because they were measured to that degree of accuracy in the question, right?
  6. Apr 26, 2016 #5


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    Before we worry about such niceties, let's get the answer basically right. I can see one definite problem with your answer and another possible one.
    You added an uncertainty percentage in one distance (diameter) to an uncertainty percentage in another distance (length) to get an uncertainty percentage in volume. Does anything strike you as rather doubtful in that?
    Secondly, there are two approaches to adding up independent uncertainties. The approach you have used, simply adding them, finds the worst case result; the other finds a more likely range by using a root-sum-square rule. Which have you been taught to use?

    Edit: one more problem maybe... how did you calculate the 0.0267%?
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