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1. Can anyone please help me with the question 2(d) ? The marking schemes don't help with this. I studied without the help of any teachers and i don't understand this get myself. Please help..
2. (Original post by Zainabr17)
Can anyone please help me with the question 2(d) ? The marking schemes don't help with this. I studied without the help of any teachers and i don't understand this get myself. Please help..
Which exam board and specification?
3. (Original post by 16Characters....)
Which exam board and specification?
http://www.acethem.com/pastpapers/a-...-5-15323.html/

I need to know how the error in resistivity is found. The explaination. Even the marking schemes give the answers but i dont get it.
Thank you so much.
4. Explanation* -_-
5. (Original post by Zainabr17)
http://www.acethem.com/pastpapers/a-...-5-15323.html/

I need to know how the error in resistivity is found. The explaination. Even the marking schemes give the answers but i dont get it.
Thank you so much.
I'll use a different example. Suppose I was trying to measure g, and I had a value of 9.8 with an error of 0.01. Then I could write that g = 9.8 +/- 0.01. If I then multiplied that by 2, I'd have 2g = 19.6 +/- 0.02. So the error is now 0.02, i.e. it's doubled.

In general, if we multiply a value by a constant, the error is also multiplied. Hence you can use this to calculate the error in the resistivity from the error in the gradient.
6. (Original post by 16Characters....)
I'll use a different example. Suppose I was trying to measure g, and I had a value of 9.8 with an error of 0.01. Then I could write that g = 9.8 +/- 0.01. If I then multiplied that by 2, I'd have 2g = 19.6 +/- 0.02. So the error is now 0.02, i.e. it's doubled.

In general, if we multiply a value by a constant, the error is also multiplied. Hence you can use this to calculate the error in the resistivity from the error in the gradient.
Thank you!

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