Report

Use the instrument precision. Instrument precision is defined as the ‘smallest non-zero reading measured by the instrument’, and refers to the actual graduations on the scale of the instrument. A metre ruler with mm graduations would have a precision of + 1 mm. For single readings, or multiple identical readings, taken by this type of ruler the estimate of uncertainty would be + 1mm Calculate the mean value Estimate the uncertainty using: Uncertainty = 0.5 X spread e.g. 5 readings of length: 0.15mm, 0.12mm, 0.16mm, 0.13mm, 0.14mm Mean = 0.14mm Uncertainty = 0.5 x (0.16-0.12) = 0.02mm Length = 0.14 ± 0.02mm Calculate the % uncertainty in the area of a metal wire given that diameter is 0.16 ± 0.01 mm % uncertainty in diameter = (0.01/0.16) × 100 = ± 6.3 % area is proportional to diameter squared % uncertainty in area = 6.3 + 6.3 = ± 12.6 % Calculate % uncertainty in the speed measurement given the following uncertainties in distance and time measurements % uncertainty in distance = ± 0.5 % % uncertainty in time = ± 2.5 % % uncertainty in speed = 0.5 + 2.5 = ± 3.0 % Use the format: variable symbol – solidus – unit You should not name the variable in full. e.g. eg ‘V/mV’ is correct, ‘output pd of solar cell in millivolts’ is not Unless directed otherwise, put all data in one table, from raw data on the left, to processed values for plotting on the right. 1 mark for each of the following points: Labelling of axes Suitable Scales Line of best Fit 1 mark for each of the following points: Plotting of points Line of best fit Reading data for calculation of gradient, and suitable size ‘triangle’ Gradient value, with appropriate significant figures Marking the origin correctly on a graph, eg PHAB3X Sec A Part 1 Q2(b) 100 100 80 80 60 60 0 0 0 1 2 3 Unacceptable: the marking of the origin as above produces a non-linear scale which will always be penalised. 0 1 2 3 Solution: use of the broken scale convention resolves the problem but watch out if a gradient calculation is then required. 100 100 80 80 60 60 40 0 1 2 3 Unacceptable: leaving an origin unmarked on either axis will not be accepted; the scale will still be treated as non-linear since the origin is now ambiguous. 0 1 2 3 Solution: use of a false origin is acceptable but candidates should be careful if they are then asked to calculate the gradient. Finding an intercept which cannot be read directly, eg PHAB3X Section B Q1(a) white rectangle represents edge of grid printed in answer booklet While the intercept on the horizontal axis can be read directly, the vertical intercept can not. Candidates gain no credit for extending the line off the grid into the margin. When answering Section B, candidates should not be given the opportunity to re-plot graphs. The use of algebra is expected if the intercept cannot be read directly. 0 0 Significant Figures in a Previous Paper: foruse usein inanswering answering part for part(a) (a) L/cm R/W R/W 6.6 6.6 10.6 10.6 13.8 13.8 17.8 17.8 21.4 21.4 2.9 2.9 7.6 7.6 13.0 13.0 21.6 21.6 30.4 30.4 0.067 0.0666 [0.067] 0.068 0.0676 [0.068] 0.068 0.0683 0.068 0.0682 0.066 0.0664 To test the theory that R = kL2 candidates were expected to evaluate R/L2 for every row of the table. Many candidates forfeited marks because they truncated their results to 2 significant figures; in at least 3 rows of the table, 3 sf was justified. Tables Independent variable must be in left hand column of table Quote to max possible significant figures (e.g. 1.10 V if instrument precision is 0.01V) Graphs Points on a graph should cover at least half the grid horizontally and vertically. All points must be plotted Marks are forfeited if plotted > 1mm from correct position Marks can be forfeited if > 2mm from trend line Use the convention variable symbol – solidus – unit for table headings and graph axes; a bracket is essential if the log of a variable is involved, eg ln(variable/unit) When compressing a graph scale use the broken scale convention if marking the origin (0, 0) otherwise mark a false origin AS and A2 candidates should be able to calculate the intercept on a graph if this cannot be read directly The result of a calculation should be to the same number of significant figures as the least accurate data used in the calculation