Adjustment of Significant Figures

In summary, the volume of a cylinder with a diameter of 1.22 cm and a length of 5.35 cm, measured with a vernier caliper of least count 0.01 cm, is calculated to be 6.2509 cm3 with an uncertainty of 0.11 cm3. However, after considering the rule for significant figures and the best estimate of the uncertainty, the final answer is revised to 6.25 ± 0.12 cm3. It is important to clearly describe the method used for determining uncertainties and to follow the conventions set by organizations such as NIST and ISO.
  • #1

Homework Statement


The diameter and length of solid cylinder measured with a vernier calipers of least count 0.01 cm are 1.22 cm and 5.35 cm respectively. Calculate the volume of the cylinder and the uncertainty involved within it.

Homework Equations


V= 1/4 πd2 l

The Attempt at a Solution


After substituting values in above formula V= 6.2509 cm3. The uncertainty involved is 0.11 cm3. Please see the pdf file attached for full solution as given in my book. My observation is that the answer should be 6.25± 0.11 cm3. I suggest there should be 3 significant figures in the final answer (volume) because as per rule least significant digits in the final result have to adjusted according to the measurement with least significant figures (3 in both given measurements). Moreover in the absolute uncertainty the decimal places have to be accommodated according to the decimal places in the final result. Please guide me on this.
 

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  • #2
I agree with you that the answer given in the book is not correct. By my calculations, the volume could be as much as 6.37, so 6.2 +- 0.1 is not good enough. The answer I get is 6.255 +- 0.115. This covers the range that the volume could be.
 
  • #3
Thanks for the reply. But you have shown four significant figures in the final result 6.225 which should be three only as demanded by the given measurements and the least count/absolute uncertainty has to be 0.11 not 0.115 as you wrote. Please reply.
 
  • #4
Zahid Iftikhar said:
Thanks for the reply. But you have shown four significant figures in the final result 6.225 which should be three only as demanded by the given measurements and the least count/absolute uncertainty has to be 0.11 not 0.115 as you wrote. Please reply.

Then I think it should be 6.25 +- 0.12. What results do you get with pi = 3.14159?
 
  • #5
See, if I use pi = 3.14159 I get 6.254 +- 0.114, but if you write that as 6.25 +- ?, the ? should be 1.2 because it can be as high as 6.368. Does that make sense? This is my belief, that it should cover the gap.
 
  • #6
Thank you very much indeed for the time.
My point is still not taken. I am referring to the rule of adjustment of significant figures in the final answer. The rule says, number of significant figures in the final answer, when measurements are multiplied or divided , could not be more than least number of significant figures in either of the measurements. In the current case 1.22 cm and 5.35 cm both have coincidentally, 3 significant figures, so 6.254---- cm has to be rounded off till there are 3 significant figures left. In this case it will be 6.25 cm, not 6.254 cm (for it has 4 significant figures). Once we have decided 6.25 cm as the final answer with correct number of significant figures, what remains is its synchronization with the absolute uncertainty. 6.25 cm allows only two decimal places to be retained in the absolute uncertainty. Hence 0.1125 cm has to be rounded off as 0.11 cm and final answer may be written as 6.25± 0.11 cm3. Please reflect.
 
  • #7
The method used in the book answer is to carry the errors as percent (proportional) errors because you are multiplying two numbers with uncertainty. This is a more accurate way to deal with the error than just counting significant digits. However, I think that the absolute error due to least count of 0.01 cm might be better estimated as ±0.005 cm, which changes the answer.
 
  • #8
Pl recommend some book to be helpful in this case.
 
  • #9
This is the best resource I can find. It says that the uncertainty is the scientist's "best estimate" of the range of values. It also says that uncertainties should have one significant figure, but some scientists use 2 significant figures if the uncertainty starts with 1.

Now in the case above, we have 3 significant figures and the result of the calculation is 6.254 +- 0.114. (Do you agree? Use pi = 3.14159 because pi is exact so we can do that.) Now we can use 2 figures for the uncertainty but it should be a best estimate. I think a best estimate is 6.25 +- 0.12 because we know the range is from 6.24 to 6.37.
 
  • #10
Many thanks for the source you provided and the further explanation you added. The things seem settled now.
High regards.
Zahid
 
  • #11
verty said:
This is the best resource I can find. It says that the uncertainty is the scientist's "best estimate" of the range of values. It also says that uncertainties should have one significant figure, but some scientists use 2 significant figures if the uncertainty starts with 1.
Yeah, just be careful that this is not the convention set by, e.g., NIST and ISO. There the uncertainty is reported to whatever significant digits are applicable, based on the method (such as combined standard uncertainty, confidence intervals, etc.), Also, the methods used should be clearly described.
 
  • #12
Thanks Olivermsun for the help.
Regards
Zahid
 

1. What are significant figures and why are they important?

Significant figures are the digits in a number that represent the precision of a measurement. They are important because they indicate the level of accuracy of a measurement and help ensure that calculations are carried out with the appropriate level of precision.

2. How do I determine the number of significant figures in a measurement?

The rules for determining significant figures are as follows:1. All non-zero digits are significant.2. Leading zeros are not significant.3. Trailing zeros are significant if there is a decimal point in the number.4. Trailing zeros in a whole number with no decimal point are not significant.

3. Can I round off a number to a certain number of significant figures?

Yes, you can round off a number to a specific number of significant figures by following these steps:1. Identify the digit in the last significant figure.2. If the digit is 5 or higher, round up the last significant figure.3. If the digit is less than 5, leave the last significant figure as is.4. Remove all the non-significant figures after the desired number of significant figures.

4. How do I perform calculations with significant figures?

The general rule for calculations with significant figures is to round the final answer to the same number of significant figures as the measurement with the least number of significant figures. For addition and subtraction, round to the least number of decimal places. For multiplication and division, round to the least number of significant figures.

5. Why do we use scientific notation when dealing with very large or very small numbers?

We use scientific notation to express very large or very small numbers in a more compact and manageable form. It also allows us to easily compare and perform calculations with these numbers. In scientific notation, a number is expressed as a coefficient multiplied by 10 to the power of an exponent.

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