9702 s08 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2008 question paper

9702 PHYSICS

9702/01 Paper 1 (Multiple Choice), maximum raw mark 40

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme Syllabus Paper

GCE A/AS LEVEL – May/June 2008 9702 01

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Question Number

Key Question Number

Key

1 C 21 A

2 D 22 B

3 C 23 A

4 C 24 C

5 D 25 B

6 C 26 D

7 B 27 D

8 C 28 B

9 D 29 A

10 D 30 A

11 A 31 C

12 A 32 B

13 A 33 C

14 A 34 C

15 D 35 C

16 A 36 D

17 C 37 A

18 B 38 B

19 C 39 C

20 B 40 B




9702 PHYSICS

9702/02 Paper 2 (AS Structured Questions), maximum raw mark 60

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.

All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated.





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1 (a) allow anything in range 20 Hz → 20 kHz B1 [1]

(b) allow anything in range 10 nm → 400 nm B1 [1]

(c) allow anything in range 10 g → 100 g B1 [1]

(d) allow anything in range 0.1 kg m–3 → 10 kg m–3 B1 [1] 2 (a) (i) k is the reciprocal of the gradient of the graph C1 k = 32 / (4 × 10–2) = 800 N m–1 A1 [2] (ii) either energy = average force × extension or ½kx2 or area under graph line C1 energy = ½ × 800 × (3.5 × 10–2)2 or ½ × 28 × 3.5 × 10–2 M1 energy = 0.49 J A0 [2] (b) (i) momentum before cutting thread = momentum after C1 0 = 2400 × V – 800 × v M1 v / V = 3.0 A0 [2] (ii) energy stored in spring = kinetic energy of trolleys C1

0.49 = ½ × 2.4 × (3

1 v)2 + ½ × 0.8 × v2 C1

v = 0.96 m s–1 A1 [3] (if only one trolley considered, or masses combined, allow max 1 mark) 3 (a) (i) v2 = 2as 1.22 = 2 × a × 1.9 M1 a = 0.38 m s–2 A1 [2] (ii) F = ma = 42 × 0.38 M1 = 16 N A0 [1] (b) power = Fv C1 = 16 × 1.2 = 19 W A1 [2] (c) (i) component = 42 × 9.8 × sin2.8 C1 = 20.1 N A1 [2] (ii) accelerating force = 20.1 – 16 = 4.1 N C1 acceleration of trolley = 4.1 / 42 = 0.098 m s–2 C1 s = ½at2 3.5 = ½ × 0.098 × t2 C1 t = 8.5 s A1 [4]



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(d) either allows plenty of time to stop runaway trolley or speed of trolley increases gradually or trolley will travel faster B1 [1] (answer must be unambiguous when read in conjunction with question) 4 (a) (i) 1. stress = force / (cross-sectional) area B1 [1] 2. strain = extension / original length B1 [1] 3. Young modulus = stress / strain B1 [1] (ratios must be clear in each answer) (ii) either fluids cannot be deformed in one direction / cannot be stretched or fluids can only have volume change or no fixed shape B1 [1]

(b) either unless ∆p is very large or 2.2 × 109 is a large number M1

∆V is very small or ∆V/V is very small, (so ‘incompressible’) A1 [2]

(c) ∆p = hρg 1.01 × 105 = h × 1.08 × 103 × 9.81 C1 h = 9.53 m C1

∆h / h = 0.47 / 10 or 0.47 / 9.53 error = 4.7% or 4.9% or 5% A1 [3] 5 (a) (i) frequency: number of oscillations per unit time M1 of the source / of a point on the wave A1 [2] (ii) speed: speed at which energy is transferred / speed of wavefront B1 [1] (b) (i) does not transfer energy (along the wave) B1 [1] (ii) position (along wave) where amplitude of vibration is a maximum B1 [1] (iii) all three positions marked B1 [1] (c) wavelength = 2 × 17.8 = 35.6 cm C1

v = fλ C1 v = 125 × 0.356 = 44.5 m s–1 C1 44.52 = 4.00 / m C1 m = 2.0 × 10–3 kg m–1 A1 [5]



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6 (a) either P = VI and V = IR or P = V2 / R C1

resistance = 38.4 Ω A1 [2] (b) zero B1 1.5 kW B1 3.0 kW B1 0.75 kW B1 2.25 kW B1 [5]

7 (a) α-particle: either helium nucleus or contains 2 protons + 2 neutrons

or He4

2 B1

β-particle: either electron or e0

1− B1

α speed < β speed (1)

α discrete values of speed/energy, β continuous spectrum (1)

either α ionising power >> β ionising power

or α range << β range (1)

α positive, β negative (only if first two B marks not scored) (1)

α mass > β mass (only if first two B marks not scored) (1) (any two sensible pairs of statements relevant to differences,

– do not allow statements relevant to only α or β, 1 each, max 2) B2 [4]

(b) (i) U236

92 → Th

232

90 M1

+ He4

2 A1 [2]

(ii) 1. correct position for U at Z = 92, N = 145 B1 2. correct position for Np relative to U i.e. Z + 1 and N – 1 B1 [2]




9702 PHYSICS

9702/31 Paper 31 (Advanced Practical Skills 1), maximum raw mark 40







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1 Manipulation, measurement and observation Successful collection of data (b) Apparatus setup without help from supervisor. [1]

(b) Value of 90° ≤ θ ≤ 180°. [1]

(c) Six sets of values for θ and n scores 4 marks, five sets scores 3 marks, etc. [4]

Wrong trend –1 (θ increases, n increases; On graph: negative slope). (c) Repeat readings. [1] Range and distribution of values (c) Need 0/1/2 and 10/11. [1] Presentation of data and observations Table: layout (c) Column headings (n (no unit), θ / °, θ/2 / °, cos(θ/2) (no unit), cos(θ/2(˚)) ). [1] Each column heading must contain a quantity and a unit where appropriate. Ignore units in the body of the table. There must be some distinguishing mark between the quantity and the unit (i.e. solidus is expected, but accept, for example, θ (°)). Allow θ in degrees/θ in °. Table: raw data (c) Consistency of presentation of raw readings of θ. If no θ column –1. [1] Expect integer values. Allow to the nearest degree (e.g. 23, 23.0, 23.5). All values of θ must be given to the same number of decimal places. Table: calculated quantities (c) Significant figures. If no θ column –1. [1] Apply to cos(θ/2). If θ is given to 2 sf, then accept cos(θ/2) to 2 or 3 sf. If θ is given to 3 sf, then accept cos(θ/2) to 3 or 4 sf. If θ is given to 4 sf, then accept cos(θ/2) to 4 or 5 sf. (c) Values of cos(θ/2) correct. Use average if present. [1] Underline and check a value. If incorrect, write in the correct value. Ignore small rounding errors. Graph: layout (Graph) Axes. Allow inverted axis. Wrong axis –1. [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points must occupy at least half the graph grid

in both x (4 ) and y (6 ) directions. Scales must be labelled with the quantity that is being plotted. Ignore units.



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Graph: plotting of points (Graph) All observations must be plotted. [1]

Work to an accuracy of plot ≤ 0.5 small square.

If any plot diameter ≥ 0.5 small square –1. Graph: trend line (Graph) Line of best fit. Allow line from 5 trend plots. [1] Judge by scatter of points about the candidate's line. There must be a fair scatter of points either side of the line. If line thicker than 0.5 small square –1. Quality of data

(Graph) Judge by scatter of points (± 0.4 object) about the examiner’s line. [1] All plots from table are needed (minimum 6) for this mark to be scored. If –ve trend or wrong axis on graph, no mark. Analysis, conclusions and evaluation Interpretation of graph (e) Gradient [1]

The hypotenuse of the ∆ must be equal to or greater than half the length of the drawn line. Read-offs must be accurate to half a small square. Check for ∆y/∆x (i.e. do not allow ∆x/∆y). (e) y-intercept from graph or substitute correct read-offs into y = mx + c (Close to 0). [1] Check false origin. Correct substitution needed and no algebraic error (e.g. y/mc = c). Allow

ecf from gradient. Drawing conclusions (f) Value for T. Allow 1 SF. Valid values: 2, 2.0, 1.96, 1.962 N. [1] (g) Value for m. Use of gradient = mg/2T. Not substitution method. [1] Unit consistent with value. In range 0.010 – 0.050 kg (10 – 50 g). 2 or 3 SF. If no unit is given then this mark cannot be scored. [Total: 20]



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2 Manipulation, measurement and observation Successful collection of data (b) Set up apparatus to get V0. Minor help –1, e.g. incorrect connections of LED. Major help –2, e.g. set up circuit. [2]

(b) Voltmeter reading, V0. Sensible value with unit. V0 ≤ 4.00 V ± 0.01 V. 2/3 d.p. [1]

(c) (i) Evidence of repeats. Consistent unit. Reading ± 0.01 mm or 0.001 mm.

Range 0.5 mm ≤ 16t ≤ 5 mm. [1] (d) (i) Measurement of voltage V. If (d)(i) and/or (d)(ii) negative –1. [1] (d) (ii) Measurement of voltage V. [1] Quality of data

(d) (ii) V0 < V(d)(i) < V(d)(ii) [1] Presentation of data and observations Display of calculation and reasoning (c) (iii) Calculation of one thickness t. 16t / 16. Check calculation ((c)(i)/16). Allow ecf (c)(i). [1] (c) (iv) Justify the number of significant figures in t, related to no. of SF in 16t / raw data. [1] (Same number of significant figures in 16t or one more.) Decimal place arguments scores zero. (e) Calculation to check proportionality. Evidence for (V-V0 ) required. Calculate correct ratio (V-V0)/n in both cases. If n = 16, -1. [1] Analysis, conclusions and evaluation Drawing conclusions (e) Conclusion. [1] Sensible comments relating to calculations and suggested relation. Incorrect ideas score zero. Accept reference back to (c)(ii). Estimating uncertainties

(c) (ii) Percentage uncertainty in 16t. Consistent units. ∆16t = ± 0.01 mm or 0.001 mm. [1] If repeated readings have been done then the uncertainty could be half the range. Correct ratio idea required (0.01 or 0.001/16t x 100 %). Ecf from (c)(i).



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Identifying limitations and suggesting improvements (f) (i) & (ii) Identify limitations and improvements by underlining relevant point and

annotating the tick using the following letters in the grid.

Problem (P) Solution (S)

A Two readings not enough (to draw a conclusion).

Take many readings AND plot a graph/find many values of k.

B Alignment of cylinders/

alignment of LED/LDR.

Guide used; ruler/line on desk./

Adjust LED/LDR to get max voltage/method of fixing LED/LDR in cylinder.

C Stray light coming in/not light tight/cylinders not sealed so let light enter tube/external light hits LDR.

Dark room/black cloth over head/lights off and blinds down/black box/black tape.

D Difficult to hold all together/voltage meter fluctuates.

Method of fixing; clamp/plasticine/tape.

E Separation between LED and LDR changes (as paper added).

Pre-slots in tube.

Max 4 Max 4 X – Other valid limitation or improvement. Do not allow ‘varying thickness of paper, zero error on micrometer’. Do not allow ‘repeated readings, parallax error’. Do not allow ‘use a computer to improve the experiment’. Ignore separation of layer affects light getting through and squashing of paper for micrometer

reading. [Total: 20]




9702 PHYSICS

9702/32 Paper 32 (Advanced Practical Skills 2), maximum raw mark 40







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1 Manipulation, measurement and observation Successful collection of data (b) Value of length 0.470m to 0.490m (to nearest cm or mm). [1] (c) 10T (or more) has been measured (could be evidence in table of results). [1] (c) Repeat readings. At least two readings of 10T or T (could be in table). [1] (d) Results Write the number of readings as a ringed total next to the table of results. Six sets of values for T and l scores 3 marks, five sets scores 2 marks. If raw data shows reverse trend then –1. [3] (d) Apparatus set up without help from Supervisor. [1] Range and distribution of values (d) Range of results (including the value in (b)). Must include 48cm and 18cm (nominal values), with no interval greater than 7cm. [1] Presentation of data and observations Table: layout (d) Column headings. Each column heading must contain a quantity and a unit where appropriate. Ignore units in the body of the table. There must be some distinguishing mark between the quantity and the unit (solidus is expected, but accept, for example, T (s)). [1] Table: raw data (d) Consistency of presentation of raw readings. All values of 10T (or T) must be given to the same number of decimal places. If these are to the nearest second then –1. Allow trailing zeros. [1] Table: calculated quantities (d) Significant figures. Apply toT ². Take trailing zeros into account. If 10T is given to 2 sf, then accept T ² to 2 or 3 sf. If 10T is given to 3 sf, then accept T ² to 3 or 4 sf. If 10T is given to 4 sf, then accept T ² to 4 or 5 sf. [1] (d) Values of T ² correct. Check a value (from candidate’s T). If incorrect, write in the correct value. [1]



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Graph: layout

(Graph) Axes. Sensible scales must be used (not 3:10 etc.), with labels at least every three large

squares. Scales must be such that the plotted points occupy at least half the graph grid in both x

and y directions. Scales must be labelled with the quantity which is being plotted. Ignore units. Indicate false origin with FO. Allow reversed axes, but if wrong graph plotted then –1. [1]

Graph: plotting of points

(Graph) All observations must be plotted. Count and circle the number of plots. Ring and check a suspect plot. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Don’t allow blobs (i.e. large dots with diameter ≥ half a small square). [1]

Graph: trend line

(Graph) Line of best fit. Allow 5 trend plots. Judge by scatter of points about the candidate's line. Indicate best line if candidate's line is not the best line. Don’t allow a line thicker than half a small square. [1]

Quality of data

(Graph) Judge by scatter of points. Allow 2cm (scaled) in the l direction either side of any line that could be drawn. All plots from table are needed for this mark to be scored. Do not award this mark if the trend is wrong or if wrong graph is drawn. [1]

Analysis, conclusions and evaluation

Interpretation of graph

(f) Gradient.

The hypotenuse of the ∆ must be ≥ half the length of the drawn line. Read-offs must be accurate to half a small square. Check for ∆y/∆x (do not allow ∆x/∆y). [1]

(f) The y-intercept value must be read to the nearest half square. Check for false origin. The value can be calculated using ratios or y = mx + c. [1]

Drawing conclusions

(g) Value for M. Check substitution into “gradient = 4π2m/g(m+M)” is correct. Allow 10 – 70g. Unit required. [1]

(g) Value for z. Must equal the y-intercept. Unit required (s2). 2 or 3 s.f. [1]

[Total: 20]



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2 Manipulation, measurement and observation Successful collection of data (b) (i) First measurement of circumference to nearest mm. Unit must be given. [1] (c) Measurement of t1. [1] (c) Repeated measurements for t1 (or t2). [1] (d) Second measurement of c. [1] (d) Second measurement of circumference < first measurement. [1] (d) Measurement of t2. [1] Quality of data (d) t decreases when c decreases. [1] Presentation of data and observations Display of calculation and reasoning (b) (iii) Value of first radius calculated correctly. Consistent unit must be given.

Check correct use of c = 2πr. [1] (d) Value of second radius, with same s.f. (or one more than) c2. [1] (e) Correct calculation to check proportionality. Possibilities include: two calculations of t2/r or ratio of t2 values and ratio of r values both calculated. [1] Analysis, conclusions and evaluation Drawing conclusions (e) Sensible comments relating to calculations and suggested relationship. The only way this mark can be scored without the first (e) mark is if the results show the

wrong trend and it is argued that this disproves the suggested relationship (but don’t credit ‘results show inverse proportionality’). [1]



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Estimating uncertainties (b) (ii) Percentage uncertainty in c. ∆c must be 0.2–0.5cm (or half the range if repeated readings).

Correct ratio idea required (∆c/circumference) × 100%. [1] Identifying limitations (f) (i) Underline and tick relevant point (one from each section): A - two sets of readings are not enough (to draw a conclusion) B - difficult to make accurate cylinder shape - cylinder radius/circumference varies C - cylinder doesn’t roll straight D - human reaction error (in timing) - measured time is very short not ‘difficult to release cylinder and start stopwatch together’ E - difficult to judge end point [4 max] Suggesting improvements (f) (ii) Underline and tick relevant point (one from each section): A - test many cylinders and plot a graph - test many cylinders and find many values of k B - method of making more accurate cylinder D - time over longer distance - use shallower angle ramp - use light gates/pressure switches to measure time - use freeze-frame photography to measure time - use motion sensor to measure time not just ‘use video’ not ‘repeated readings’ not just ‘use computer/data logger’ E - mark distance with lines on ramp (to eliminate parallax) not ‘use a pointer’ X - new method (e.g. vernier calipers) to measure average diameter [4 max] [Total: 20]




9702 PHYSICS

9702/04 Paper 4 (A2 Structured Questions), maximum raw mark 100







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Section A 1 (a) (i) angle (subtended) at centre of circle B1 by an arc equal in length to the radius (of the circle) B1 [2] (ii) angle swept out per unit time / rate of change of angle M1 by the string A1 [2] (b) friction provides / equals the centripetal force B1

0.72 W = mdω2 C1

0.72 mg = m × 0.35ω2

ω = 4.49 (rad s–1) C1

n = (ω /2π) × 60 B1 = 43 min–1 (allow 42) A1 [5] (c) either centripetal force increases as r increases or centripetal force larger at edge M1 so flies off at edge first A1 [2]

(F = mrω2 so edge first – treat as special case and allow one mark) 2 (a) molecule(s) rebound from wall of vessel / hits walls B1 change in momentum gives rise to impulse / force B1 either (many impulses) averaged to give constant force / pressure or the molecules are in random motion B1 [3]

(b) (i) p = 3

1 ρ<c2> C1

1.02 × 105 = 3

1 × 0.900 × <c2>

<c2> = 3.4 × 105 C1 cRMS = 580 m s–1 A1 [3]

(ii) either <c2> ∝ T or <c2> = 2 × 3.4 ×105 C1 cRMS = 830 m s–1 (allow 820) A1 [2] (c) cRMS depends on temperature (alone) B1 so no effect B1 [2]



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3 (a) (i) amplitude = 0.5 cm A1 [1] (ii) period = 0.8 s A1 [1]

(b) (i) ω = 2π / T C1 = 7.85 rad s–1

correct use of v = ω √(x0

2 – x2) B1

= 7.85 × √(0.5 × 10–22 – 0.2 × 10–22) = 3.6 cm s–1 A1 [3] (if tangent drawn or clearly implied (B1)

3.6 ± 0.3 cm s–1 (A2)

but allow 1 mark for > ±0.3 but Ğ ±0.6 cm s–1) (ii) d = 15.8 cm A1 [1] (c) (i) (continuous) loss of energy / reduction in amplitude (from the oscillating system) B1 caused by force acting in opposite direction to the motion / friction / viscous forces B1 [2] (ii) same period / small increase in period B1 line displacement always less than that on Fig.3.2 (ignore first T/4) M1 peak progressively smaller A1 [3] 4 (a) work done moving unit positive charge M1 from infinity to the point A1 [2] (b) (i) x = 18 cm A1 [1] (ii) VA + VB = 0 C1

(3.6 × 10–9) / (4πε0 × 18 × 10–2) + q / (4πε0 × 12 × 10–2) = 0 C1

q = –2.4 × 10–9 C A1 [3]

(use of VA = VB giving 2.4 × 10 –9 C scores one mark) (c) field strength = (–) gradient of graph B1

force = charge × gradient / field strength or force ∝ gradient B1 force largest at x = 27 cm B1 [3] 5 (a) at t = 1.0 s, V = 2.5 V C1 energy = ½CV 2 C1

0.13 = ½ × C × (8.02 – 2.52) M1

C = 4500 µF A0 [3] (b) use of two capacitors in series in all branches of combination M1 connected into correct parallel arrangement A1 [2]



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6 (a) parallel (to the field) B1 [1]

(b) (i) torque = F × d

2.1 × 10–3 = F × 2.8 × 10–2 C1 F = 0.075 N A1 [2] (use of 4.5 cm scores no marks) (ii) zero A1 [1]

(c) F = BILN(sinθ) C1

0.075 = B × 0.170 × 4.5 × 10–2 × 140 M1

B = 7.0 × 10–2 T = 70 mT A0 [2] (d) (i) (induced) e.m.f. is proportional to / equal to rate of change of M1 (magnetic) flux (linkage) A1 [2] (ii) change in flux linkage = BAN

= 0.070 × 4.5 × 10–2 × 2.8 × 10–2 × 140 C1 = 0.0123 Wb turns induced e.m.f = 0.0123 / 0.14 C1 = 88 mV A1 [3] (Note: This is a simplified treatment. A full treatment would involve the

averaging of B cosθ leading to a √2 factor) 7 (a) charge is quantised / discrete quantities B1 [1] (b) (i) parallel so that the electric field is uniform / constant B1 horizontal so that either oil drop will not drift sideways or field is vertical or electric force is equal to weight B1 [2] (ii) qE = mg C1

q × 850 / (5.4 × 10–3) = 7.7 × 10–15 × 9.8 C1

q = 4.8 × 10–19 C and is negative A1 [3]

(c) charge changes by 1.6 × 10–19 C between droplets / integral multiples M1

so charge on electron is 1.6 × 10–19 C A0 [1] 8 (a) since momentum before combining is zero B1 momenta must be equal and opposite after B1 equal momenta so photon energies equal B1 [3] (b) E = mc2 C1

= 9.1 × 10–31 × (3.0 × 108)2

= 8.19 × 10–14 (J) C1

= (8.19 × 10–14) / (1.6 × 10–13) = 0.51 MeV A1 [3]



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Section B 9 (a) blocks labelled sensing device / sensor / transducer B1 processor / processing unit / signal conditioning B1 [2] (b) (i) two LEDs with opposite polarities (ignore any series resistors) M1 correctly identified as red and green A1 [2] (ii) correct polarity for diode to conduct identified M1 hence red LED conducts when input (+)ve or vice versa A0 [1] 10 large / strong (constant) magnetic field B1 nuclei rotate about direction of field / precess (1) radio frequency / r.f. pulse B1 causes resonance in nuclei , nuclei absorb energy (1) (pulse) is at the Larmor frequency (1) on relaxation / nuclei de-excite emit (pulse of) r.f. B1 detected and processed B1 non-uniform field (superimposed) B1 allows for position of nuclei to be determined B1 and for location of detection to be changed (1) (B6 plus any two extra details, 1 each, max 2) B2 [8] 11 (a) (i) frequency of carrier wave varies M1 in synchrony with displacement of information signal A1 [2] (ii) 1. zero (accept constant) B1 [1] 2. upper limit 530 kHz B1 lower limit 470 kHz B1

changes upper limit → lower limit → upper limit at 8000 s–1 B1 [3] (b) e.g. more radio stations required / shorter range more complex electronics larger bandwidth required (any two sensible suggestions, 1 each) B2 [2] 12 (a) (i) picking up of signal in one cable M1 from a second (nearby) cable A1 [2] (ii) random (unwanted) signal / power B1 that masks / added to / interferes with / distorts transmitted signal B1 [2] (allow this mark in (i) or (ii)) (b) if P is power at receiver,

30 = 10lg(P / (6.5 × 10–6) C1

P = 6.5 × 10–3 W C1

loss along cable = 10lg(26 × 10–3 / 6.5 × 10-3) C1 = 6.0 dB C1 length = 6.0 / 0.2 = 30 km A1 [5]




9702 PHYSICS

9702/05 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30







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1 Planning (15 marks) Defining the problem (3 marks) P1 A is the independent variable or vary A. [1] P2 R is the dependent variable or determine R for different A. [1] P3 Keep the temperature (of glass) constant. Do not allow “controlled variable”. [1] Methods of data collection (5 marks) M1 Basic circuit diagram. [1]

Ammeter and voltmeter with power supply, or ohmmeter without power supply, or bridge methods. M2 Correct orientation of glass between electrodes – largest cross-sectional area. [1] M3 A distance (thickness) measured using a micrometer/vernier scale/vernier callipers. [1] M4 Method of determining area perpendicular to current flow. [1]

Distances measured and multiplied together. This mark may only be scored if it is clear that the correct dimensions are being used. M5 Method of determining resistance. [1]

Ohmmeter. R = V/I justified. Description of balancing bridge with correct equation. Method of analysis (2 marks)

A1 A2

R against 1/A ρ = gradient/l

R against l /A ρ = gradient

1/A against R or 1/R against A ρ = 1/(gradient × l)

l /A against R or l /R against A ρ = 1/gradient

lg R against lg A ρ = 10l × y-intercept

[2] Safety considerations (1 mark) S1 Relevant safety precaution related to: [1]

EHT power supply (>100 V) – switch off before changing circuit/use of rubber gloves; or handling glass – wear (thick) gloves. Additional detail (4 marks) D1/2/3/4 Relevant points might include [4]

Calculation of typical resistance of glass using value of resistivity given. Range of ammeter or ohmmeter with reasoning. Use of EHT or power supply >1000 V or microammeter/galvanometer. Take many readings of thickness and average. Good contact between circuit and glass e.g. metal plates, foil, conducting putty. Metal plates/foil/conducting putty to cover all of the cross-sectional area in use. Method of securing good contact between circuit and glass, e.g. g clamps, weights. Clean/dry the glass. [Total: 15]



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2 Analysis, conclusions and evaluation (15 marks)

(a) gradient = –ρη y-intercept = ln R0 or loge R0 [1]

(b)

ln (R / s–1) Errors in ln R

6.620 or 6.62 ± 0.02 – 0.03

6.363 or 6.36 ± 0.03

6.064 or 6.06 ± 0.05

5.799 or 5.80 ± 0.06

5.521 or 5.52 ± 0.08

5.247 or 5.25 ± 0.10 – 0.11

Column heading for ln (R); ln R calculated. All correct for one mark. [1] 2 or 3 dp scores one mark as shown. [1] Errors in ln R: Ignore no. of sf in errors. [1]

(c) (i) Points plotted correctly. All six within half a small square required for this mark. [1] Error bars in ln R plotted correctly. Check x = 0.0300 m. Allow ecf. [1]

(ii) Line of best fit. Must be within tolerances. [1] Worst acceptable straight line. Line should be clearly labelled. [1]

(iii) Gradient of best-fit line. Must be negative. [1]

Award this mark if in range –54.0 to –56.0. If out of range, check the read offs. Work to half a small square. The triangle used should be greater than half the length of the drawn line.

Error in gradient [1]

Method of determining absolute error. Expect to see difference between gradient from best-fit line and the worst acceptable line correctly evaluated.

(d) Value of η [1]

Award mark for method η = candidate’s gradient/11300 evaluated. Error in η [1]

Method of determining absolute error. Unit of η: m2 kg–1. [1]

(e) Value of x in the range 0.0410 m to 0.0425 m [1]

If x in the range 0.0410 m to 0.0425 m, then method of determining error in x. [1]

[Total: 15]

9702 s08 ms_all

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