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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2013 series 9702 PHYSICS 9702/51…

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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2013 series 9702 PHYSICS 9702/51 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2013 9702 51 1 Planning (15 marks) Defining the problem (3 marks) P f is the independent variable or vary f. [1] P h is the dependent variable or measure h. [1] P Keep current in coil constant. [1] Methods of data collection (5 marks) M Labelled diagram of apparatus: including working circuit with a.c. supply, coil, stand and ring. [1] M Signal generator (or variable frequency power generator). [1] M Measure h with a rule/calipers. [1] M Measure f using oscilloscope/read off signal generator. [1] M Measure h from opposite sides of the ring (and determine mean)/wait for ring to stabilise. [1] Method of analysis (2 marks) A Plot a graph of lg h against lg f. (Allow ln – ln graphs.) [1] A k = 10y-intercept and n = gradient. [1] Safety considerations (1 mark) S Reasoned method to prevent the coil overheating – switch off when not in use. Or reasoned method to prevent injury from hot coil – do not touch hot coil, use gloves. Do not allow “small currents”. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use iron/steel retort stand 2 Method to keep current constant 3 Method to determine period from oscilloscope using time-base 4 When using oscilloscope, f = 1/T 5 Use coil of many turns/large current to have measurable heights 6 Logarithmic equation e.g. lg h = n lg f + lg k 7 Relationship is valid if a straight line is produced Do not allow vague computer methods. [Total: 15] © Cambridge International Examinations 2013 Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2013 9702 51 2 Analysis, conclusions and evaluation (15 marks) Mark Expected Answer Additional Guidance (a) A1 Gradient = v/4 y-intercept must be negative. y-intercept = –k (b) T1 1/f / 10–3 s Column heading. Allow equivalent unit. e.g. f –1/ 10–3 Hz–1 or 1/f (10–3 s) T2 A mixture of 3 s.f. and 4 s.f. is allowed. 3.13 or 3.125 2.94 or 2.941 2.65 or 2.646 2.34 or 2.336 2.08 or 2.083 1.95 or 1.953 (c) (i) G1 Six points plotted correctly Must be less than half a small square. Ecf allowed from table. Penalise “blobs”. U1 All error bars in d plotted Must be within half a small square. Each error correctly bars should be five small squares (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (2.06, 16) and (2.10, 16) and upper end of line should pass between (3.12, 24.5) and (3.16, 24.5). G3 Worst acceptable straight Line should be clearly labelled or dashed. Should line. pass from top of top error bar to bottom of bottom Steepest or shallowest error bar or bottom of top error bar to top of possible line that passes bottom error bar. Mark scored only if error bars through all the error bars. are plotted. (iii) C1 Gradient of best fit line The triangle used should be at least half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT. U2 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient. (iv) C2 y-intercept – must be Check substitution in y = mx + c. negative Allow ecf from (c)(iii). FOX does not score. Should be about –1 (–0.35 to –1.33). U3 Uncertainty in y-intercept Check substitution in y = mx + c. Difference in worst y-intercept and y-intercept. Allow ecf from (c)(iv) but FOX does not score. © Cambridge International Examinations 2013 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2013 9702 51 (d) C3 k = – y-intercept cm or m Method required with appropriate unit. v = 4 × gradient cm s–1 or Do not penalise POT. Allow ecf from (c)(iii) and m s–1 (iv). U4 Determines absolute Uncertainty k = uncertainty in y-intercept. uncertainties in k and v Uncertainty in v = 4 × uncertainty in gradient. (e) (i) C4 Between 250 and 260 given Must be in range. Check method. to 2 or 3 s.f. (ii) U5 Percentage uncertainty in f Absolute uncertainties in d and k must be added. ∆(d + k) = ∆d + ∆k % uncertainty in v + % uncertainty in (d + k). [Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U2] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (c) (iv) y-intercept [U3] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) (d) [U4] Uncertainty in k = uncertainty in y-intercept Uncertainty in v = 4 × uncertainty in gradient (e) [U5] max v max gradient max f = = 4(min d + min k ) (min d + min k ) min v min gradient min f = = 4(max d + max k ) 4(max d + max k ) Percentage uncertainty = % uncertainty in v + % uncertainty in (d + k) Percentage uncertainty = % uncertainty in gradient + % uncertainty in (d + k) © Cambridge International Examinations 2013

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