Calculate Percent Error: Measuring 75cm with Meter Stick

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The discussion revolves around calculating the percent error when measuring a distance of approximately 75 cm with a meter stick. Participants estimate the uncertainty in measurement, suggesting values like ±0.2 cm or ±1 mm, leading to different percent error calculations. One participant calculates a percent error of 0.27% based on ±0.2 cm, while another suggests that if the measurement could be refined to ±1 mm, the error would be 0.13%. There is also confusion regarding the conversion of fractional errors to percentages, with clarification that multiplying by 100 is necessary for percentage representation. Overall, the conversation highlights the nuances in estimating measurement errors and the importance of understanding significant figures.
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Homework Statement


Estimate the percent error in measuring a distance of about 75cm with a meter stick.



Homework Equations





The Attempt at a Solution


I'm assuming one could accurately measure this value to a couple mm. so it would be 75+-0.2cm. Now the book gives an answer of 0.13% i just have no clue how they get there. Thanks in advance.
 
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If the meter stick has mm's as the smallest division, you would be rounding at most .5 mm on each end. Divide by 75 and multiply by 100 to convert to per cent.
 
sergey90 said:

Homework Statement


Estimate the percent error in measuring a distance of about 75cm with a meter stick.



Homework Equations





The Attempt at a Solution


I'm assuming one could accurately measure this value to a couple mm. so it would be 75+-0.2cm. Now the book gives an answer of 0.13% i just have no clue how they get there. Thanks in advance.

Your uncertainty of 0.2 cm means 0.2 in 75, which is 2 in 750 which is 1 in 375

1/375 * 100 = 0.26666 %

If you had thought you could get it to within 1 mm, the uncertainly would have been 0.13 %
 
I would say the error on any measuring instrument is +/- 1 scale division or +/1 1 in the last digit of a digital meter.
If the metre stick has mm divisions then I would say the error is +/- 1mm which is +/-1 in 750 (similar to PeterO with his +/-)
The question refers to a distance of ABOUT 75cm so it seems to me they want a rough estimate of the error rather than an answer to several significant figures.
 
thanks for the help guys. One last clarification on the matter: in previous problems of fractional errors if for example we have a value of 9.69+-0.07cm^2 the book just divides 0.07 by 9.69 and says that's my fractional error % without multiplying by 100. Why so?
 
sergey90 said:
thanks for the help guys. One last clarification on the matter: in previous problems of fractional errors if for example we have a value of 9.69+-0.07cm^2 the book just divides 0.07 by 9.69 and says that's my fractional error % without multiplying by 100. Why so?

Perhaps it was given as a fractional error, to convert a fraction to a percentage, you multiply by 100.

It is unusual to fractional error % - usually either fractional error or percentage error.
 
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