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The following question is taken from Physics by Halliday, Resnick and Krane, 5th ed. Vol. 1.

A student is calculating the thickness of a single sheet of paper. She measures the thickness of a stack of $80$ sheets with Vernier calipers, and finds the thickness to be $l = 1.27~\mathrm{cm}$. To calculate the thickness of a single sheet she divides. Which of the following answers has the correct number of significant digits?

(A) $0.15875~\mathrm{mm}$

(B) $0.159~\mathrm{mm}$

(C) $0.16~\mathrm{mm}$

(D) $0.2~\mathrm{mm}$

Now, my question is a general question regarding these type of measurement where measure some quantity and divide it because we do not have precise enough instruments. For example, we can "measure" the time period of one oscillation of a pendulum by measuring the time taken for ten oscillations.

Now, my question is regarding the value and the error in the thickness of sheet of paper or time required for one oscillation. If $L$ is the thickness of one one sheet, then assuming that every sheet is equally thick, $L = l/80~\mathrm{cm} = 0.015875~\mathrm{cm}$. And $\Delta L = \Delta l / |80|$. Assuming $\Delta l = 0.01~\mathrm{cm}$, we get $\Delta L = 0.000125~\mathrm{cm}$. So, the actual thickness is $L = 0.0159 \pm 0.0001~\mathrm{cm}$. Am I right? This would correspond to option (B).

But when performing an experiment to "measure" the time period of one oscillation of a pendulum by measuring the time taken for ten oscillations, my professor told that the time period of one oscillation cannot be reported such that the uncertainty in the value is smaller than the least count of the instrument used to measure the time period of ten oscillations. For example, if the least count of a stopwatch is $1~\mathrm{s}$, and the time taken for ten oscillations is $8~\mathrm{s}$, then what should be the time required for one oscillation? Should it be $0.8 \pm 0.1~\mathrm{s}$? But if this is true, then as the least count of the stopwatch is greater than the value itself, this value does not convey any information according to my professor's saying.

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I think you must have misunderstood your professor. If the time for ten oscillations is $8 \pm 1$ seconds then the time for one oscillation is:

$$ \tau = \frac{8 \pm 1}{10} = \frac{8}{10} \pm \frac {1}{10} $$

So it is $\tau = 0.8 \pm 0.1$ seconds.

If your professor says that:

since the least count of the stopwatch is 1 s, the value of the time period of one oscillation cannot be certain than 1 s

then they are wrong.

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  • $\begingroup$ Yes, I made a mistake in my question and now I have edited it. But according to my professor, since the least count of the stopwatch is $1~\mathrm{s}$, the value of the time period of one oscillation cannot be certain than $1~\mathrm{s}$. So, according to him, the time period cannot be reported as $0.8~\mathrm{s}$. $\endgroup$ – Apoorv Potnis Oct 17 '18 at 9:32
  • $\begingroup$ @ApoorvPotnis then your professor is wrong $\endgroup$ – John Rennie Oct 17 '18 at 9:47

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