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Calculating the Uncertainty of a Pool's Volume 
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#1
Jan1809, 04:36 PM

P: 400

1. The problem statement, all variables and given/known data
Molly decides to find the volume of a swimming pool. She measures its length to 3 significant figures, its width to four significant figures, and its height to 2 signficant figures. A) How many sig figs are in the volume of the pool? B) How many sig figs should there be in the uncertainty of the volume of the pool? 2. Relevant equations Volume = length x width x height 3. The attempt at a solution A) The answer is 2, because it is the least amount of sig figs. B) I don't know what an uncertainty is, or how to calculate it  maybe someone can explain it or give me a hint and I can try the problem again? Edit: Maybe there's nothing to calculate  I just remembered there's a rule that says only 1 or 2 sig figs should be present in any error calculation. 


#2
Jan1809, 06:19 PM

P: 33

In the length: 3 lots of ten meters 0 lots of single meters 0 lots of ten centimeters Take the last measurement, if we are only accurate to the nearest 10cm then we could be anywhere between +5 and 5 of a single ten centremetre graduation. In terms of meters that makes it +/0.05. Where the '+/' you should read 'plus minus' in reference to plus this amount to the value, or take this amount from the value. This is because in rounding you round up on a .5 and round down on a .499999999... but as .4999999 is basically .5 we might as well call it .5 Thus we actually should write the measurements; (30.0 +/ 0.05)m by (20.00 +/ 0.005)m As you have correctly deduced that if the length can only be accurate to the nearest 10cm then there is little point knowing the width to the nearest 1cm because the first measurement wasn't that accurate. So how many significant figures should we have in the uncertainty. Well witting the measurements with their margin of error should give you some idea. Hint: Uncertainties are/never should be accurate... Highlight the below if still stuck: Well in the maximum case the length could be 30.5 and in the minimum case 29.5, the width 20.005 and 19.995 Therefore: Max 610.1525m^{2} Min 589.8525m^{2} Now if we take the max away from the min; = 20.3m^{2} We are out by an entire 20 metres! If we half this for our +/ we get 10.15m^{2} in the strict sense this could be our uncertianty in the overal area messurement, but you might as well round it to 1 significant figure. This is because your lenght messurement is only to 2 significant figures and if I write the numbers like; 610.000000 589.000000 010.000000 You can see that the 10m is going to dominate the 15cm after the decimal point. Hence you should only quote the uncertainty to the first significant figure of your +/ uncertainty. Hence (600 +/ 10)m Note in practice, you don't need to go though the entire calculation, simply quoting a 1 significant figure above the lowest in your product will do, although try it for the volume and then see what result that shows... You want the volume so you might want to do the method again with your third measurement. You can't really be 'wrong' with uncertainties, because they are a judgment of how well you can measure something, and if your unsure, just dock a decimal figure or two from your answer. Haths 


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