## 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.

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.

#### Attachments

• example.pdf
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verty
Homework Helper
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.

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.

verty
Homework Helper
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?

verty
Homework Helper
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.

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.

olivermsun
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.

Pl recommend some book to be helpful in this case.

verty
Homework Helper
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.

Many thanks for the source you provided and the further explanation you added. The things seem settled now.
High regards.
Zahid

olivermsun