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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the November 2005 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. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses’.

Transcript of 9702 w05 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the November 2005 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. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses’.

Page 1 Mark Scheme Syllabus Paper

A AND AS LEVEL – NOVEMBER 2005 9702 1

© University of Cambridge International Examinations 2005

Question Number

Key Question Number

Key

1 B 21 C

2 C 22 D

3 D 23 B

4 B 24 B

5 C 25 C

6 C 26 D

7 D 27 C

8 A 28 A

9 D 29 C

10 B 30 A

11 A 31 A

12 D 32 B

13 A 33 D

14 C 34 C

15 A 35 B

16 B 36 B

17 C 37 C

18 C 38 D

19 B 39 B

20 B 40 D

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE A/AS Level

MARK SCHEME for the November 2005 question paper

9702 PHYSICS

9702/02 Paper 2 maximum raw mark 60

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the

Examination. 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. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper

GCE A/AS LEVEL – November 2005 9702 2

© University of Cambridge International Examinations 2005

1 (a) (i) force per unit area (ratio idea essential) B1 (ii) kg m-1 s-2 B1 [2]

(b) ρ has base unit kg m-3 B1 g has base unit m s-2 B1

hρg has base unit m × kg m-3 × m s-2 M1 same as pressure QED A0 [3] 2 (a) point where whole weight of body (allow mass) M1 may be considered to act (do not allow ‘acts’) A1 [2] (b) when CG below pivot, weight acts through the pivot B1 (so) weight has no turning effect about pivot B1 [2] 3 (a) change in velocity/time (taken) B1 [1]

(b) velocity is a vector/velocity has magnitude & direction B1 direction changing so must be accelerating B1 [2]

(c) either 6.1 × cos35 = 4.99 N or scale shown B1 so no resultant vertical force triangle of correct shape B1

6.1 sin35 = 3.5 N resultant = 3.5 ± 0.2 N B1

horizontally horizontal ± 3° B1 [4] allow answer based on centripetal force: resultant is centripetal force (which is horizontal) (B1) resultant is horizontal component of tension (B1)

6.1 sin35 = 3.5 N (B1) horizontally (B1)

4 (a) (i) use of tangent at time t = 0 B1

acceleration = 42 ± 4 cm s-2 A1 [2] (ii) use of area of loop B1

distance = 0.031 ± 0.001 m B2 [3]

allow 1 mark if 0.031 ± 0.002 m) (b) (i) F = ma C1

= 0.93 × 0.42 {allow e.c.f. from (a)(i)} = 0.39 N A1 [2] (ii) force reduces to zero in first 0.3 s B1 then increases again in next 0.3 s M1 in the opposite direction A1 [3]

Page 2 Mark Scheme Syllabus Paper

GCE A/AS LEVEL – November 2005 9702 2

© University of Cambridge International Examinations 2005

5 (a) similarity: e.g. same wavelength/frequency/period, constant phase difference B1

difference: e.g. different amplitude/phase B1 [2] (do not allow a reference to phase for both similarity and difference) (b) constant phase difference so coherent B1 [1]

(c) (i) intensity ∝ amplitude2 C1

I ∝ 32 and IB ∝ 22 leading to M1

IB = 9

4I A0 [2]

(ii) resultant amplitude = 1.0 × 10-4 cm C1

resultant intensity = 9

1I A1 [2]

(d) (i) displacement = 0 B1 [1]

(ii) xA = -2.6 × 10-4 cm and xB = +1.7 × 10-4 cm C1

allow ± 0.5 × 10-4 cm)

resultant displacement = (-) 0.9 × 10-4 cm A1 [2] 6 (a) force must be upwards (on positive charge) M1 so plate Y is positive A1 [2] (b) (i) E = V / d C1

= 630/(0.75 × 10-2)

= 8.4 × 104 N C-1 A1 [2] (ii) qE = mg C1

q = (9.6 × 10-15 × 9.8) / (8.4 × 104) C1

= 1.12 × 10-18 C A1 [3] 7 (a) either V = E R1 / (R1 + R2) or I = E / (R1 + R2) C1

= 3000

1800 × 4.50 V =

3000

1800 × 4.50 M1

= 2.70 V = 2.70 V A0 [2]

(b) (i) for a wire, V = I x (ρL/A) M1

I, ρ and A are constant A1

so V ∝ L A0 [2]

Page 3 Mark Scheme Syllabus Paper

GCE A/AS LEVEL – November 2005 9702 2

© University of Cambridge International Examinations 2005

(ii) 1 2.70 V A1 [1]

2 100

L =

50.4

70.2 C1

L = 60.0 cm A1 [2] (iii) thermistor resistance decreases as temperature rises M1 so QM is shorter A1 [2] 8 (a) product of force and distance M1 moved in the direction of the force A1 [2] (b) (i) falls from rest B1 decreasing acceleration B1 reaches a constant speed B1 [3] (ii) straight line with negative gradient B1 y-axis intercept above maximum EK B1 reasonable gradient (same magnitude as that for EK initially) B1 [3]

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE A Level

MARK SCHEME for the November 2005 question paper

9702 PHYSICS

9702/03 Paper 3 (Practical Paper) maximum raw mark 25

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the

Examination. 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. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 3

© University of Cambridge International Examinations 2005

Part (b) (ii) Absolute uncertainty = 1 mm (1 mark). [2]

Percentage uncertainty in first value of d (i.e. ratio correct) (1 mark).

(iii)Measure each end of rule to see if distance above bench is the same (1 mark) [2]

Use of set square to ensure that half-metre rule is 90O to bench (1 mark)

Part (c) Readings [3]

6 sets of readings scores 1 mark.

Repeated readings scores 1 mark.

Reasonable interval between values of d (> 5 cm) scores 1 mark.

Help given by Supervisor, then -1. Excessive help then -2.

Quality of results [2]

Judge by scatter of points about the line of best fit.

6 trend points with little scatter scores 2 marks.

6 trend points with ‘a fair amount of scatter’ scores 1 mark.

5 trend points with little scatter scores 1 mark.

4 trend points (or less) scores zero.

Considerable scatter scores zero.

Column headings [2]

Apply to F and d. One mark each.

The headings must contain a quantity and a unit.

Consistency [2]

Apply to F and d only. One mark each.

Values of d must be given to the nearest millimetre.

Part (d) Axes [1]

Scales must be such that the plotted points occupy at least half the graph

grid in both the x and y directions.

Scales must be labelled. Do not allow awkward scales.

Plotting of points [1]

Check a suspect plot. Circle and tick if correct.

If incorrect, show correct position with arrow, and -1.

Work to half a small square.

Line of best fit [1]

There must be a reasonable balance of points about the line of best fit.

Page 2 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 3

© University of Cambridge International Examinations 2005

Part (e) Determination of gradient [2]

∆ used must be greater than half the length of the drawn line;

read-offs correct; ratio correct (one mark).

Gradient given as negative (one mark).

y-intercept [1]

The value may be read directly or calculated using y = mx + c and a

point on the line.

Part (f) Gradient equated with –W/L [1]

Value of W [1]

Significant figures in W [1]

Accept 2 or 3 sf only.

Intercept equated with Wmg

+

2 [1]

Value of m [1]

Units of m and W both correct [1]

25 marks in total.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the November 2005 question paper

9702 PHYSICS

9702/04 Core maximum raw mark 60

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the

Examination. 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. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper

A LEVEL – NOVEMBER 2005 9702 4

© University of Cambridge International Examinations 2005

1 (a) GM / R2 = Rω2 …………….…………………...…..………………….. C1

ω = 2π / (24 × 3600) ………………………………..……..…………… C1

6.67 × 10–11 × 6.0 × 1024 = R3 × ω2

R3 = 7.57 × 1022 ………………………………………………………… M1

R = 4.23 × 107 m ……………………………………………………….. A0 [3]

(b) (i) ∆Φ = GM/Re – GM/Ro …...………………………………………….….. C1

= (6.67 × 10–11 × 6.0 × 1024) ( 1 / 6.4 × 106 – 1 / 4.2 × 107)

= 5.31 × 107 J kg–1 …………………………………………………. C1

∆EP = 5.31 × 107 × 650 …………………………………………………. C1

= 3.45 × 1010 J …………………………………………………….. A1 [4]

(c) e.g. satellite will already have some speed in the correct direction … B1 [1]

2 (a) obeys the law pV = constant × T ………………………..……………….. M1 at all values of p, V and T ………………………………………………. A1 [2]

(b) n = (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………... C1 = 3.73 mol ………………………………………………………………. A1 [2]

(c) at new pressure, nn = 3.73 × 9.2

4.3 ×

300

290

= 4.23 mol ….………………………………………. C1 change = 0.50 mol ……………………………………………………….… C1 number of strokes = 0.50 / 0.012 = 42 (must round up for mark) ……. A1 [3]

3 (a) correct statement, words or symbols …..…………………………...….. B1 [1]

(b) (i) w = p∆V ………………………………………………………………….. C1

= 1.03 × 105 × (2.96 × 10–2 – 1.87 × 10–5) = (–) 3050 J …………………..……………….…………..…………… A1 [2]

(ii) q = 4.05 × 104 J …………………………………………………………. B1 [1]

(iii) ∆U = 4.05 × 104 – 3050 = 37500 J …no e.c.f. from (a)………………… A1 [1] penalise 2 sig.fig. once only

(c) number of molecules = NA ………………………………………………. C1

energy = 37500 / (6.02 × 1023)

= 6.2 × 10–20 J (accept 1 sig.fig.) …………………………..…. A1 [2]

4 (a) (i) ω = 2πf ………………………………………………………....………….. C1

= 2π × 1400 = 8800 rad s–1 ………………………………………………………….. A1 [2]

(ii) a0 = (–)ω2x0 ……………………………..………………………………… C1

= (8800)2 × 0.080 × 10–3 = 6200 m s–2 …………………….……………………………………. A1 [2]

(b) straight line through origin with negative gradient …….…………….... M1 end points of line correctly labelled …………………………………….. A1 [2]

(c) (i) zero displacement ………………………………………………………… B1 [1]

(ii) v = ωx0 ……………………………………………………………………. C1

= 8800 × 0.080 × 10–3 = 0.70 m s–1 ……………………………………………………………. A1 [2]

Page 2 Mark Scheme Syllabus Paper

A LEVEL – NOVEMBER 2005 9702 4

© University of Cambridge International Examinations 2005

5 (a) ½mv2 = qV ……(or some verbal explanation) …..……………..…… B1

½ × 9.11 × 10-31 × v2 = 1.6 × 10-19 × 1.2 × 104 ………………………… B1

v = 6.49 × 107 m s–1 ………….………………………………………… A0 [2] (b) (i) within field: circular arc ………………..………………..…………….. B1 in ‘downward’ direction ……………..………………….. B1 beyond field: straight, with no ‘kink’ on leaving field ………………… B1 [3] (ii) 1. v is smaller …………………………………………………………………. M1 deflection is larger ………………………………………………………… A1 [2] 2. (magnetic) force is larger ………………………………………………… M1 deflection is larger ……………………………………………………….. A1 [2] 6 (a) (numerically equal to) force per unit length …………………….…….… M1 on straight conductor carrying unit current ……………………………. A1 normal to the field ………………………………………………………… A1 [3]

(b) flux through coil = BA sinθ ……………………………………………….. B1

flux linkage = BAN sinθ ………………..………………………………… B1 [2] (c) (i) (induced) e.m.f. proportional to ………………………………..…..…..… M1 rate of change of flux (linkage) …………………….……………………. A1 [2] (ii) graph: two square sections in correct positions, zero elsewhere ….. B1 pulses in opposite directions …………………………………… B1 amplitude of second about twice amplitude of first ………….. B1 [3] 7 (a) (i) energy required to separate the nucleons in a nucleus ..………..…….. M1 nucleons separated to infinity / completely …………..……………….. A1 [2] (ii) S shown at peak …..……………………………………………………… B1 [1] (b) (i) 4 ……...…………………………………………………………………… A1 [1] (ii) 1. idea of energy as product of A and energy per nucleon …………… C1

energy = (8.37 × 142 + 8.72 × 90) – 235 × 7.59 = 1189 +785 – 178 = 190 MeV ………(–1 for each a.e.) ………………………… A2 [3] 2. energy = mc2 ……………………………………………………………. C1

1 MeV = 1.6 × 10–13 J …………………………………………………. C1

energy = (190 × 1.6 × 10–13) / (3.0 × 108)2

= 3.4 × 10–28 kg …………………………………………..…… A1 [3]

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced/Advanced Subsidiary Level

MARK SCHEME for the November 2005 question paper

PHYSICS

9702/05 Paper 5

maximum raw mark 30

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. 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. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper

GCE A/A LEVEL – November 2005 9702 05

© University of Cambridge International Examinations 2005

1 (b) (i) Repeated readings of raw times [1]

(ii) Absolute uncertainty in t (one mark) [2] Working to get the number of oscillations for 1% uncertainty (one mark).

(c) Readings [3] 6 sets scores one mark. Allow more than 6 sets without penalty. lg (T/s) and lg n correct; one mark each. Values must be checked. Minor help from the Supervisor, -1. Major help, then -2. If help has been given then write SR at the top of the front page of the script, and give a brief explanation of the type of help that has been given by the

table of results.

Most raw times > 10 s [1]

Quality of results [1] Judge by scatter of points about the line of best fit.

Column headings [1]

Each column heading must contain a quantity and a unit. There must be some distinguishing feature between the quantity and the unit.

Consistency of raw readings [1] All the raw readings of t should be given to the same number of d.p.

(d) (i) Axes [1] The axes must be labelled with the quantities plotted. Ignore units on the axes. The plotted points must occupy at least half the graph grid in both the x and y. directions (i.e. 4 large squares in the x-direction and 6 large squares in the y-direction). Do not allow more than 3 large squares between the labels on an axis. Do not allow awkward scales (e.g. 3:10, 6:10 etc.).

Plotting of points [1] All the observations must be plotted. Count the number of plots and ring this total on the grid. Do not allow plots in the margin area.

Check one suspect plot. Circle this plot. Tick if correct. If incorrect, mark the Correct position with a small cross and use an arrow to indicate where the plot should have been, and -1. Allow errors up to and including half a small square.

Page 2 Mark Scheme Syllabus Paper

GCE A/A LEVEL – November 2005 9702 05

© University of Cambridge International Examinations 2005

(ii) Line of best fit [1]

Only a drawn straight line through a linear trend is allowable for this mark. This mark can only be awarded for 5 or more plots on the grid. There must be a reasonable balance of points about the drawn line. Do not allow a line of thickness greater than half a small square.

(iii) Gradient [1] Ignore any units given with the value. Hypotenuse of ∆ must be > half the length of line drawn. Check the read-offs. Work to half a small square. ∆x/∆y gets zero.

Values taken from the table that lie on the line to within half a small square are acceptable.

(iii) y-intercept [1]

(e) q = gradient [1]

(e) p = 10y-intercept [1]

(f) Value of k [1]

Must be in range 20 to 30 N m-1

(f) Unit of k (e.g. N m-1 or kg s-2) [1]

(f) sf in k (accept 2 or 3 sf only) [1]

20 marks in total

Page 3 Mark Scheme Syllabus Paper

GCE A/A LEVEL – November 2005 9702 05

© University of Cambridge International Examinations 2005

2 A1 Procedure OK (i.e. measure pressure and electrical power; change pressure [1] and measure new power and repeat). This mark can be scored even if the method is unworkable. A2 Diagram of workable arrangement [2]

e.g. wire connected to joulemeter and power supply. A stopwatch must be used if this method is employed. Allow alternative circuit using ammeter and voltmeter. The wire must be shown inside a closed container with some means of changing and measuring the pressure.

One mark for the electrical arrangement; one mark for the mechanical arrangement.

A3 Measurement of pressure [1] (e.g. Bourdon gauge/pressure gauge/manometer)

B1 Change setting on the power supply to keep the wire at the same [1] temperature when the pressure is changed.

B2 Power supplied to wire = V x I (or reading on joulemeter ÷ time) [1]

B3 Use of thermocouple thermometer to monitor temperature whilst pressure [1] is changed or adjust V or I to keep R (and hence T) constant

C Any one safety precaution [1] e.g. safety screens/goggles/wire mesh surrounding vacuum chamber

D1/2 Any further good design features [2] Some of these might be: Thermocouple thermometer shown attached to the wire to monitor temperature. Light spot galvanometer connected to thermocouple. Use of a needle valve to control pressure. Vacuum grease the connecting wires to the heater wire. Allow time between readings for experiment to stabilise. Do not allow the wire to become too hot or the thermocouple may melt.

10 marks in total

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the November 2005 question paper

9702 PHYSICS

9702/06 Options maximum raw mark 40

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the

Examination. 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. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 6

© University of Cambridge International Examinations 2005

Option A - Astrophysics and Cosmology 1 (a) (i) (mean) distance between Earth and Sun B1 [1] (ii) distance at which 1 AU subtends an angle M1 of one arc-second A1 [2]

(b) arc = rθ C1

1.5 × 1011 = r × 2π / (360 × 60 × 60) M1

1.0 pc = 3.09 × 1016 m A1 [3] 2 (a) e.g. 3 K microwave background radiation redshift of light from galaxies any two sensible suggestions, 1 each, max 2 B2 [2] (b) If Universe is static and infinite B1 then every line of sight would end on a star M1 so night would be as bright as day A1 [3] (c) depends on (mean) density of matter in the Universe B1 greater than a certain value, Universe will expand and then contract B1 below this certain value, Universe will expand indefinitely B1 [3] 3 (a) e.g. absorption of IR by water vapour in atmosphere much stray IR at Earth’s surface any two sensible suggestions, 1 each, max 2 B2 [2] (b) e.g. distant galaxies B1 moving so fast that they are red-shifted into IR B1 e.g. cool objects (brown dwarfs) B1 give off IR but not visible light B1 [4] allow any two sensible suggestions (2) + reasoning (1 + 1)

Page 2 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 6

© University of Cambridge International Examinations 2005

Option F - The Physics of Fluids 4 (a) e.g. incompressible fluid / constant density horizontal flow non-viscous streamline any three, 1 each, max 3 B3 [3]

(b) force = A∆p C1

= 25 × ½ × 1.2 × (852 – 752) C1

= 2.4 × 104 N A1 [3] 5 (a) (i) centre of mass of displaced fluid B1 [1] (ii) B shown at centre of submerged section B1 [1] (iii) upthrust acts upwards through B B1 weight acts downwards through C B1 these two forces provide a restoring couple B1 [3] (b) (i) becomes less B1 [1] (ii) decrease B1 [1] (iii) increases B1 [1] (c) C and B coincide M1 no longer providing a restoring couple A1 [2] 6 (a) non-steady / haphazard flow of fluid B1 [1] (b) turbulence represents (continuous) transfer of kinetic energy B1 this transfer of energy per unit time represents power B1

power = FD × speed so more power means larger FD B1 [3]

Page 3 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 6

© University of Cambridge International Examinations 2005

Option M - Medical Physics 7 (a) electrons accelerated / high speed electrons B1 bombard metal target B1

electrons decelerated greatly → e.m. radiation B1 wide range of decelerations gives continuous spectrum B1 electrons in target atoms excited B1 de-excitation of these electrons gives line spectrum B1 [6] (b) (i) sharpness: ease with which edges of structures can be seen B1 [1] (ii) contrast: difference in blackening between structures B1 [1] 8 (a) short sight (myopia) B1 [1] (b) (i) concave lens drawn B1 [1] (ii) rays diverge after passing through the concave lens B1 rays converge on the retina B1 [2] 9 (a) (i) intensity: energy per unit area per unit time (normal to area) B1 loudness: subjective response (of a person) to (a given) intensity B1 (ii) ability to distinguish between two different intensities of sound B1 [3] (b) intensity level = 10 lg(I / I0)

89 = 10 lg I / (1.0 × 10–12) C1

I89 = 7.94 × 10–4 W m–2 C1

92= 10 lg I / (1.0 × 10–12)

I92 = 1.58 × 10–3 W m–2 C1 ratio = I89 / (I92 – I89) C1 = 1.0 A1 [5]

Page 4 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 6

© University of Cambridge International Examinations 2005

Option P - Environmental Physics 10 (a) diagram: closed box with glass top B1 metal base and water tubing B1 blackened interior B1 [3] (b) largest area normal to sunlight B1 [1]

(c) power = flow rate × c × ∆θ C1

800 × 0.35 × 1.4 = flow rate × 4200 × 15 C1

flow rate = 6.2 × 10–3 kg s–1 A1 [3] 11 (a) (i) change in pressure and volume (and temperature) M1 without any (thermal) energy entering or leaving the system A1 [2] (ii) the change takes place rapidly B1 no time for energy to flow in/out of the gas B1 [2] (b) (i) correct direction shown (clockwise) B1 [1] (ii) correct section marked (vertical section on left of diagram) B1 [1] 12 (a) the lead compounds are released as air pollution B1 any further comment e.g. cause mental disorders, enter food chain via plants B1 [2] (b) e.g. noise, visual any two sensible suggestions, 1 each, max 2 B2 [2] (c) (i) available without using (fossil) fuels B1 [1] (ii) e.g. do not produce air pollution, no mining/transportation any two sensible suggestions, 1 each, max 2 B2 [2]

Page 5 Mark Scheme Syllabus Paper

GCE A LEVEL – November 2005 9702 6

© University of Cambridge International Examinations 2005

Option T - Telecommunications 13 (a) signal is in the form of a series of pulses of light/IR B1 pulses pass along a glass fibre B1 as a result of total internal reflection B1 [3] (b) technological: e.g. greater bandwidth, less noise, less power loss per unit length any two sensible suggestions, 1 each, max 2 B2 social: e.g. increased security, cheaper, less bulky any two sensible suggestions, 1 each, max 2 B2 [4] 14 (a) (i) thermal energy (in the cable) / resistance B1 [1] (ii) loss = 10 lg(0.55 / 0.60) C1 = (-) 0.38 dB C1

loss per unit length = 0.38 / 75 × 10–3 C1 = 5.0 dB km–1 A1 [4] (b) (i) unwanted (random) signal power B1 [1] (ii) e.g. molecular/lattice vibrations, pick-up of e.m. signals any two sensible suggestions, 1 each, max 2 B2 [2] 15 digital more reliable than analogue

fewer people employed in telephone industry greater multiplexing means reduced cost per call reduced costs means available to more people huge expansion international calls huge expansion of non-voice communications development/expansion of internet introduction of multichannel cable TV companies any five sensible statements, 1 each, max 5 B5 [5]