Physics Help Forum uncertainty and relative error
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 Apr 3rd 2017, 02:04 AM #1 Junior Member   Join Date: Apr 2017 Location: Australia Posts: 1 uncertainty and relative error 1. The problem statement, all variables and given/known data You have an object composed of an unknown material that is in the form of a cube. You use a metre rule to measure the dimensions of the object, which are found to be 2.0 cm, 2.0 cm and 2.0 cm. The rule has a resolution of 1 mm and there is no other information available. You weigh the object using the electronic balance in the Physics Lab. The smallest scale on the balance is 0.1 grams, and the manufacturer’s specifications indicate that the “accuracy” of the balance is “1% reading + 2 digits”. It is found that the “mass” of the object is 75.1 grams. What is the uncertainty and relative error in the volume of the object? What is the uncertainty and relative error in the mass of the object? What is its density of the object, with uncertainty? What material do you think it is?
 Apr 3rd 2017, 02:44 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 508 Do you have and understand definitions for relative error and uncertainty, before you try to calculate examples?
 Apr 3rd 2017, 05:00 AM #3 Senior Member   Join Date: Aug 2010 Posts: 173 There is an engineer's "rule of thumb" that "when quantities are added (or subtracted) the errors add, when quantities are multiplied (or divided) the relative errors add". To see this, think of the "errors" as "differentials": if A= x+y+z then dA= dx+ dy+ dz. If A= xyz then dA= yzdx+ xzdy+ xydz so that dA/A= yzdx/xyz+ xzdy/xyz+ xydx/xyz= dx/x+ dy/y+ dz/z. Here the error in each length measurement is 0.1 cm while the measurements themselves are 2.0 cm. The relative error is $\displaystyle \frac{0.1}{2.0}= 0.05$. The three length measurements are multiplied so the relative error in the volume is 3(0.05)= 0.15. Since the calculated volume is 8.0 cubic cm the error (I think your "uncertainty") is (0.15)(8.0)= 1.2 cm.

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