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’.

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

GCE A/AS Level
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.
Examination.
schemes.
Level syllabuses.

GCE A/AS LEVEL – November 2005 9702 2
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]

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
4
I A0 [2]
(ii) resultant amplitude = 1.0 × 10-4
cm C1
resultant intensity =
9
1
I 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]

(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]

GCE A Level
9702 PHYSICS
9702/03 Paper 3 (Practical Paper) maximum raw mark 25
Examination.
demonstrated.
Examination.
schemes.
Level syllabuses.

GCE A LEVEL – November 2005 9702 3
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.

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 W
mg
+
2
[1]
Value of m [1]
Units of m and W both correct [1]
25 marks in total.

GCE Advanced Level
9702 PHYSICS
9702/04 Core maximum raw mark 60
Examination.
demonstrated.
Examination.
schemes.
Level syllabuses.

A LEVEL – NOVEMBER 2005 9702 4
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 = (–)ω2
x0 ……………………………..………………………………… 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]

A LEVEL – NOVEMBER 2005 9702 4
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]

GCE Advanced/Advanced Subsidiary Level
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.
demonstrated.
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.

GCE A/A LEVEL – November 2005 9702 05
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.

(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

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

GCE Advanced Level
9702 PHYSICS
9702/06 Options maximum raw mark 40
Examination.
demonstrated.
Examination.
schemes.
Level syllabuses.

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
(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)

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]

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]

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
(c) (i) available without using (fossil) fuels B1 [1]
(ii) e.g. do not produce air pollution, no mining/transportation

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
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
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]

9702 w05 ms_all

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