Physics (AS & A Level) · May / June 2016 · Paper 52 · Question 2
9702
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The question as text
A student is investigating how the resistance of a wire depends on the diameter of the wire. The circuit is set up as shown in Fig. 2.1. 1 ohmmeter wire Fig. 2.1 The resistance R of the wire is measured using an ohmmeter. The experiment is repeated for wires of the same material and same length L but different diameter d. It is suggested that R and d are related by the equation R = 4ρL πd 2 where ρ is a constant. 1
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2(a)[1]
ρ π 4 L
2(b)[3]
T1 1 d / 106 m–2 T2 1.2 or 1.21 3.2 or 3.19 4.7 or 4.73 6.9 or 6.93 9.8 or 9.77 14 or 13.7 All values to 2 s.f. or 3 s.f. Allow a mixture of significant figures. Must be values in table. U1 From ± 0.03 to ± 1 Allow more than one significant figure. Allow zero for first uncertainty and up to 1.2 for largest uncertainty.
2(c)(i)[2]
G1 Six points plotted correctly Must be within half a small square. Do not allow “blobs”.
ECF allowed from table. U2 Error bars in 1 d plotted correctly All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
2(c)(ii)[2]
G2 Line of best fit Lower end of line must pass between (2.6, 4.0) and (3.0, 4.0) and upper end of line must pass between (12.4, 18.0) and (13.0, 18.0). G3 Worst acceptable straight line. Steepest or shallowest possible line that passes through all the error bars. Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Must be steepest/shallowest line. Mark scored only if error bars are plotted.
2(c)(iii)[2]
Gradient of the line of best fit; the triangle used should be at least half the length of the drawn line. (1)
Absolute uncertainty: gradient of line of best fit − gradient of worst acceptable line, or ½(steepest worst line gradient − shallowest worst line gradient). (1)
(Should be about 1.4–1.5 × 10⁻⁶.)
2(d)(i)[2]
π× = × gradient 0.7854 gradient 4L Must use gradient value. Do not penalise POT (Should be about 1 × 10–6.) Ω m Correct unit and correct power of ten.
2(d)(ii)[1]
Percentage uncertainty in ρ: percentage uncertainty in gradient + 1%, i.e. (Δgradient/gradient + 0.01/1.00) × 100. (1)
2(e)[2]
R in the range 25.5 to 28.4 Ω, given to 2 or 3 s.f. Allow 26, 27 or 28. (1)
Absolute uncertainty in R, using percentage uncertainty = (Δρ/ρ + 2 × 0.01/0.23) × 100. Percentage uncertainty must be greater than 8.6%. (1)
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