0
$\begingroup$

Consider we want to prepare a bigger measuring cylinder (in which the scale division would be 10 liters i.e. 0,10,20,30, 40)

We want to do it with a smaller measuring cylinder in which the scale division of the cylinder is 1 liter and the maximum volume of the smaller measuring cylinder is 2 liters.

Now how could we calculate the uncertainty of volume measuring using the big cylinder? For example, what would be the uncertainty of a 15-liter liquid which is measured by the bigger measuring cylinder?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Suppose you volume of $2$ litres has an uncertainty of $\pm 0.1$ litre which you may judge to be too large.
(You will have to decide on the possible error after inspecting your $2$ litre container.

The problem is that you are going to get a cumulative error.

Fill the container up to the $2$ litre mark and empty it into the larger container and mark the water level.
Repeat until there are 20 calibration marks ($20$ litres) on the larger container.
You can then repeat the process and that will give you some idea of the variation in your scale readings. If the larger container is of uniform cross-section then you can interpolate where the $1$ litre interval marks are otherwise you need add $1$ litre to the empty larger container then $2$ litres, 2 litres etc.

The $40$ litre mark was the result of the addition of $2\pm 0.1$ litres $20$ times so the maximum error is about $40 \pm(20 \times 0.1) = 40\pm 2$ litres.
A better estimate of the error is that it is $\pm \sqrt{20 \times 0.1^2}\approx \pm 0.45$ ie $40.0 \pm 0.45$ litres.

For smaller volumes eg $15$ litres you reason that it is $7 \times 2 + 1 \times 1$ litres and work the maximum error $\pm 8 \times 0.1 = \pm 0.8$ litres or the better estimator $\pm \sqrt{8 \times 0.1^2}\approx \pm 0.3$ litres.

Improve this answer
$\endgroup$
4
  • $\begingroup$ But reading the scale of the big container also has uncertainty (in addition to the cumulative uncertainty), so how can we include the big container scale reading uncertainty if we marked and divided the scale of the big container for example 10 to 10 liters? $\endgroup$
    – MENG
    Commented Jun 13, 2022 at 9:16
  • $\begingroup$ We could just estimate that the volume we have in the container is 15 (assume we read the scale between 10 and 20), so the actual value might not be 15, it might be 15.5 $\endgroup$
    – MENG
    Commented Jun 13, 2022 at 9:21
  • $\begingroup$ I think we need to add up the above number (0.3) with (uncertainty of big container) $\times$ (uncertainty of small volume container) $\endgroup$
    – MENG
    Commented Jun 13, 2022 at 12:03
  • 1
    $\begingroup$ @MENG I have tried to give an indication of how one might estimate the error. It might help to improve on the process but more importantly show where one could improve the accuracy of the process. $\endgroup$
    – Farcher
    Commented Jun 13, 2022 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.