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I'm not sure how to calculate the uncertainty in this physics paper 5

The paper I am talking about is 9702/51/O/N/19

For question 2b it asks to include the absolute uncertainty in T but it doesn't give any uncertainty in the raw data of t so I don't know how to calculate it. Cab someone help?

Reply 1

Roasted Potato

Can you put a link in to the paper and I will have a look!

Reply 2

Fred Mannings

Original post by Roasted Potato

Can you put a link in to the paper and I will have a look!

https://pastpapers.papacambridge.com/view.php?id=Cambridge%20International%20Examinations%20%28CIE%29/AS%20and%20A%20Level/Physics%20%289702%29/2019-Oct-Nov/9702_w19_qp_51.pdf

Reply 3

Roasted Potato

The interval in which the true value lies is called the uncertainty in the measurement. Absolute Uncertainty or ± value The absolute uncertainty in the mean value of measurements is half the range of the measurements. E.g. Suppose the measurements of the diameter of a pin by a Vernier Calliper are as follows: 0.25mm; 0.24mm;0.26mm; 0.23mm;0.27mm; The mean = (0.25 + 0.24 + 0.26 + 0.23 + 0.27)/5 =125/5 = 0.25mm The range = 0.27 - 0.23 = 0.04mm Absolute Uncertainty = ± 0.04/2 = ± 0.02 So, the mean value = mean ± range/2 = 0.25 ± 0.04/2 = 0.25 ± 0.02

Sometimes, the multiple measurements that you take could be the same, leaving you with no variation or range. In other words, absolute uncertainty turns out to be 0! This situation can also arise when you have a single measurement. In such circumstances, the resolution of the device - the smallest measurement possible - comes to our rescue. The resolution is taken as the absolute uncertainty. E.g. If the device in question is a meter rule, the resolution, 1mm, is the absolute uncertainty.

^^^ Copied from this website as I think it has good explantions, example and interactive section if you want to experiment
https://www.vivaxsolutions.com/physics/errors-and-uncertainties.aspx

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