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Help with percentage uncertainty question?

abitpissy

It's only one mark I cannot answer it :redface:

.
Someone please explain how you get then answer.

Is the percentage uncertainty of a value= Precision/ value x100.

(edited 7 years ago)

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Reply 1

lerjj

Original post by Questioness

It's only one mark I cannot answer it :redface:

.
Someone please explain how you get then answer.

Is the percentage uncertainty of a value= Precision/ value x100.

It's Uncertainty / Value, but in this case you could claim that precision would also mean uncertainty (not always though!)

So, if I tell you that I've measured something as 3.5cm, what is the implied uncertainty in my measurement?

Spoiler

Reply 2

abitpissy

Original post by lerjj

Spoiler

Your using a mm scale? So it's +\- 0.1 cm ?

(edited 7 years ago)

Reply 3

lerjj

Original post by Questioness

Your using a mm scale? So it's +\- 0.1 cm ?

Yep. So if the diffraction grating claims to have 3.5x10^3 lines/mm, then what is the implied uncertainty in that? For % uncertainty you just divide by the value (and multiply by 100%)

Reply 4

abitpissy

Original post by lerjj

Yep. So if the diffraction grating claims to have 3.5x10^3 lines/mm, then what is the implied uncertainty in that? For % uncertainty you just divide by the value (and multiply by 100%)

Do you mean per meter?

Reply 5

lerjj

Original post by Questioness

Do you mean per meter?

Apparently, yes. 3.5x10^3 lines/m is not very many, but I guess it'll still work.

Also, it's a % uncertainty, so the units really don't matter here. Neither does the x10^3 bit either actually. Any measurement given to you as 3.5 x10^p mkgsA... etc will have the same % uncertainty in it.

(edited 7 years ago)

Reply 6

abitpissy

Original post by lerjj

My answer is wrong?

Reply 7

lerjj

Original post by Questioness

My answer is wrong?

I think I may have confused you :redface:

My point with the 3.5cm reading is that if someone gives you a number to 1d.p. then that implies an uncertainty.

In this case, they've said the number of lines is 3.5 x10^3/m

The implied uncertainty in that is 0.1 x10^3/m. As before, the end bit doesn't really matter, your percentage uncertainty =0.1/3.5 = 3%

Reply 8

abitpissy

Original post by lerjj

I think I may have confused you :redface:

Ahah right! Thank you. Makes much more sense now. :tongue:

Might be a one marker but I've had some trouble with this for some time now :colondollar:

Reply 9

Someboady

I don't follow completely, can someone please re-explain this? :/ Also I need to know how to generalise this to any question involving pretty much anything.
Thanks in advance!

Original post by Questioness

Ahah right! Thank you. Makes much more sense now. :tongue:

Might be a one marker but I've had some trouble with this for some time now :colondollar:

Original post by lerjj

I think I may have confused you :redface:

Reply 10

lerjj

Original post by Someboady

I don't follow completely, can someone please re-explain this? :/ Also I need to know how to generalise this to any question involving pretty much anything.
Thanks in advance!

Let's say I measure something as 3.5cm with a ruler. Why didn't I measure it as 3cm or as 3.54cm? Because the the former isn't as precise as I can be and because the latter is beyond the capabilities of my instrument to measure.

Therefore, if I tell you I measured something as 3.5cm, you should assume that I have some level of competence and that the uncertainty in that is 1mm.

Likewise, if somebody tells you e.g.

g=9.8 ms^{-2}

the implied uncertainty is again 0.1m/s^2.

Tl;dr The uncertainty in a measurement is reflected in the number of significant figures you give it to. The percentage uncertainty can therefore be calculated.

Reply 11

console.log

AQA actually specifically talk about this question in their practical handbook, and I thought it might help:

http://filestore.aqa.org.uk/resources/physics/AQA-7407-7408-PHBK.PDF - page 55

(edited 6 years ago)