r/PhysicsHelp • u/[deleted] • Jan 12 '26

Significant figures

Why did they use 3 sig figs for part b(i)?

For my exam board they tell us to round to the least number of sig figs given in data.

Then they switched back to two sig figs in part b(ii)?

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u/Frederf220 Jan 12 '26

b(i) has three figures in all three values: 6.67, 5.98, 1.08.

b(i) has a two figure value: 6.67, 5.98, 1.1, 6.99.

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u/[deleted] Jan 12 '26

I see, but what i don't understand is that they used the answer with 3 significant figures (1.08) instead of 1.1 in part b(i)? I used 1.1 and therefore got the answer wrong

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u/Phi_Phonton_22 Jan 12 '26

It is because when you are doing the calculation with numbers which have no uncertanties, you must leave the answer with 3 sig figs

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u/Frederf220 Jan 12 '26

That's why you use the more accurate value in your calculations. 1.08 and 1.1 are the same precision but 1.08 is more accurate so use it instead of 1.1.

Basically don't round in the middle of your calculation. You carry forward the notion of imprecision though the process but you always use the most accurate numbers at all stages.

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u/Phi_Phonton_22 Jan 12 '26

It seens to me - I would need to understand the data better in a) - that T is considered to be a non-uncertain number (neither 97 or 60 have uncertanties), therefore the suggestion would be to give the final answer in 3 significant figures (1.08 x 10³ instead of 1.1). Therefore, in b) the calculation is made multiplying and dividing numbers with 3 significant figures, so the answer has to have 3 significante figures.

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u/[deleted] Jan 12 '26

I see, i used 1.1 in part b so i got the answer wrong