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Oct/Nov 2004, Paper2, Question 1, Part B) (ii)
I got the percentage uncertainty of diameter of wire as (0.02 / 0.50) * 100 = 4 %
Now when I try to get the uncertainty of the area of cross-section of the wire, the answer doesn't come up as 8%.
This is how I do it:
Cross-sectional area = pi * (r^2)
Diameter = 0.50 mm / 2 = 0.25 mm
pi (0.25^2) = 0.196 ~= 0.2 mm^2
% uncertainty of cross-sectional area = (uncertainty of area / area)* 100
% uncertainty of cross-sectional area = (0.02 / 0.2) * 100 = 10 % ????
Do I have to separately calculate the uncertainty of cross-sectional area? If yes, how?
QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_2.pdf
Thanks in advance.
All that working for the second part is completely unnecessary, its just for 1 mark! First of all you should know that the percentage uncertainty in r is the same as in the diameter (4%) as dividing or multiplying by a constant has no effect on uncertainty. Now Area = pie x (r^2), and pie is a constant so again that will have no effect on uncertainty. so we only need the r^2. Now when there are indices involved you have to multiply the uncertainty by the index (for example, uncertainty in r = 4%, so uncertainty in (r^2) is 4% x 2 = 8%, so if it was (r^3) like maybe volume or something, the uncertainty would be 4% x 3 = 12%) so in this case all we have to do is 4 x 2 which gives us 8%, that's it!