CAMBRIDGE EXAM MARK SCHEME

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced/Advanced Subsidiary Level
MARK SCHEME for the May/June 2006 question paper
9702 PHYSICS
9702/01 Paper 1 – Multiple Choice
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.
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 May/June 2006 question papers for most IGCSE and GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme Syllabus Paper
GCE A/AS LEVEL – May/June 2006 9702 01
© University of Cambridge International Examinations 2006
Question
Number
Key
Question
Number
Key
1 B 21 C
2 D 22 B
3 D 23 A
4 D 24 B
5 C 25 B
6 C 26 C
7 A 27 B
8 C 28 D
9 D 29 B
10 B 30 A
11 B 31 C
12 A 32 B
13 D 33 B
14 B 34 A
15 A 35 B
16 D 36 C
17 D 37 B
18 A 38 A
19 C 39 D
20 A 40 B

GCE Advanced Level
9702 PHYSICS
9702/02 Paper 2
Maximum raw mark 60
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.

GCE A Level – May/June 2006 9702 02
1 (a) kg m s–2
B1 [1]
(b) kg m–1
s–1
B1 [1]
(c) (i) v2
= 2gs
= 2 × 9.8 × 4.5 C1
v = 9.4 m s–1
A1 [2]
(ii) either
F (= 3.2 × 10–4
× 1.2 × 10–2
× 9.4) = 3.6 × 10–5
N M1
weight of sphere (= mg = 15 × 10–3
× 9.8) = 0.15 N M1
3.6 × 10–5
<< 0.15, so justified A1 [3]
or
mg = crvT (M1)
terminal speed = 3.8 × 104
m s–1
(M1)
9.4 << 3.8 × 104
, so justified (A1)
2 (a) (i) point at which whole weight of body M1
may be considered to act A1 [2]
(ii) sum of forces in any direction is zero B1
sum of moments about any point is zero B1 [2]
(b) either:
T and W have zero moment about P M1
so F must have zero moment, i.e. pass through P A1 [2]
or:
if all pass through P, distance from P is zero for all forces (M1)
so sum of moments about P is zero (A1)
(c) (i) Fcosα = Tcosβ B1 [1]
(ii) W = Fsinα + Tsinβ B1 [1]
(iii) 2W = 3Tsinβ B1 [1]
3 (a) sum of (random) kinetic and potential energies M1
of the atoms/molecules of the substance A1 [2]
(b) (i) potential energy unchanged as atoms remain in same positions M1
allow ‘reduced because atoms slightly closer together’
vibrational kinetic energy reduced because temperature lower M1
so internal energy less A1 [3]
(ii) potential energy increases because separation increases M1
kinetic energy unchanged because temperature unchanged M1
so internal energy increases A1 [3]
4 (a) mass per unit volume (ratio idea must be clear, not units) B1 [1]
(b) (i) pressure is same at the surface of mercury
because at same horizontal level B1 [1]
(ii) hρg is same for both B1
53 × 10–2
× 1.0 × 103
× g = 71 × 10–2
× ρ × g C1
ρ = 7.5 × 102
kg m–3
A1 [3]

GCE A Level – May/June 2006 9702 02
5 (a) no hysteresis loop/no permanent deformation M1
(do not allow ‘force proportional to extension’)
so elastic change A0 [1]
(b) work done = area under graph line OR average force × distance B1
= ½Fx ½(F2 + F1)(x2 – x1) A1
F = kx, so work done = = ½kx2
½k(x2 + x1)(x2 – x1) A1
work done = ½k(x2
2
– x1
2
) A0 [3]
(c) gain in energy of trolley = ½k(0.0602
– 0.0452
) + ½k(0.0302
– 0.0452
) C1
= 0.36 J C1
kinetic energy = ½ × 0.85 × v2
= 0.36 C1
v = 0.92 m s–1
A1 [4]
6 (a) (i) correct shape drawn B1 [1]
(ii) two nodes marked correctly B1 [1]
(b) ½λ = 0.324 m C1
v = fλ C1
= 512 × 2 × 0.324
= 332 m s–1
A1 [3]
(c) ¼λ = 16.2 cm C1
either antinode is 0.5 cm above top of tube
or antinode is 16.2 cm above water surface A1 [2]
7 (a) lamp C M1
lamp is shorted A1 [2]
(b) shorted lamp A would cause damage to the supply/lamps
/blow fuse in supply B1 [1]
(c) 15 Ω B1 [1]
(d) (i) V = I R C1
R = 30 Ω A1 [2]
(ii) P = VI or I2
R or V2
/ R C1
P = 1.2 W A1 [2]
(e) filament is cold when measuring with ohm-meter in (b) B1
resistance of filament rises as temperature rises B1 [2]
8 (a) nucleus emits M1
α- or β- particles and/or γ-rays A1 [2]
(b) decay unaffected by environmental changes M1
such as temperature, pressure etc. (one e.g. is sufficient) A1 [2]
(c) constant probability of decay (per unit time) of a nucleus B1
cannot predict which particular nucleus will decay next B1 [2]

GCE Advanced Level and GCE Advanced Subsidiary Level
9702 PHYSICS
9702/03 Paper 3
Maximum raw mark 25
demonstrated.
Examination.
CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.

GCE A/AS Level – May/June 2006 9702 03
(c) (i) First value of θ (less than 15°) [1]
(iv) Absolute uncertainty = 1 or 2 mm (1 mark). [2]
Percentage uncertainty in first value of d (i.e. ratio correct) (1 mark).
(v) Source of error in d: [1]
meniscus effects, shape of bottle leads to parallax problems
(vi) Source of error in θ : [1]
parallax effects, difficult to keep head still, difficult to move head with rule,
difficult to judge point of toppling
(d) Readings [6]
9 sets of readings scores 6 marks; 8 sets, 5 marks; 7 sets, 4 marks etc.
Less than 4 sets scores zero.
Help given by Supervisor, then –1. Excessive help then –2.
Repeated readings [1]
Reasonable interval between values of d (Y 1.0 cm) [1]
Quality of results [2]
Judge by scatter of points about the curve (one mark)
Curve for small d values steeper than curve for large d values (one mark)
Column headings [1]
Apply to d.
The heading must contain a quantity and a unit.
Consistency [1]
Apply to d.
Values of d must be given to the nearest millimetre.
(e) 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.
Curve of best fit [1]
There must be a reasonable balance of points about the curve.
(f) Determination of max value [1]
(g) Second curve; all values of θ larger than before (1 mark) [2]
Similar shape (1 mark)
(h) Suggestions for improvements (one mark each) [2]
[Total marks: 25]

9702 PHYSICS
9702/04 Paper 4
Maximum raw mark 60
demonstrated.
Examination.
CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.

1 (a) centripetal force is provided by gravitational force B1
mv2
/ r = GMm / r2
B1
hence v = √(GM / r) A0 [2]
(b) (i) EK (= ½mv2
) = GMm / 2r B1 [1]
(ii) EP = - GMm / r B1 [1]
(iii) ET = - GMm / r + GMm / 2r C1
= - GMm / 2r. A1 [2]
(c) (i) if ET decreases then - GMm / 2r becomes more negative
or GMm / 2r becomes larger M1
so r decreases A1 [2]
(ii) EK = GMm / 2r and r decreases M1
so (EK and) v increases A1 [2]
2 (a) e.g. fixed mass/ amount of gas
ideal gas
(any two, 1 each) B2 [2]
(b) (i) n = pV / RT C1
= (2.5 × 107
× 4.00 × 104
x 10-6
) / (8.31 × 290) C1
= 415 mol A1 [3]
(ii) volume of gas at 1.85 × 105
Pa = (2.5 × 107
× 4.00 × 104
) / (1.85 × 105
)
= 5.41 × 106
cm3
C1
so, 5.41 × 106
= 4.00 × 104
+ 7.24 × 103
N C1
N = 741 A1 [3]
(answer 740 or fails to allow for gas in cylinder, max 2/3)
3 (a) gradient of graph is (a measure of) the sensitivity M1
the gradient varies with temperature A1 [2]
(b) 2040 ± 20 Ω corresponds to 15.0 ± 0.2 °C C1
T / K = T / °C + 273.15 (allow 273.2) C1
temperature is 288.2 K A1 [3]
4 (a) (i) 1.0 B1 [1]
(ii) 40 Hz B1 [1]
(b) (i) speed = 2πfa C1
= 2π × 40 × 42 × 10-3
= 10.6 m s-1
A1 [2]
(ii) acceleration = 4π2
f2
a C1
= (80π)2
× 42 × 10-3
= 2650 m s-2
A1 [2]
(c) (i) S marked correctly (on ‘horizontal line through centre of wheel) B1
(ii) A marked correctly (on ‘vertical line’ through centre of wheel) B1 [2]

5 (a) (i) force per unit positive charge (ratio idea essential) B1 [1]
(ii) E = Q / 4πε0r2
M1
ε0 being the permittivity of free space A1 [2]
(b) (i) 2.0 × 106
= Q / (4π × 8.85 × 10-12
× 0.352
) C1
Q = 2.7 × 10-5
C A1 [2]
(ii) V = (2.7 × 10-5
) / (4π × 8.85 × 10-12
× 0.35) C1
= 7.0 × 105
V A1 [2]
(c) electrons are stripped off the atoms B1
electrons and positive ions move in opposite directions,
(giving rise to a current) B1 [2]
6 (a) (i) arrow B in correct direction (down the page) B1
(ii) arrow F in correct direction (towards Y) B1 [2]
(b) (i) When two bodies interact, force on one body is equal but opposite in
direction to force on the other body. B1 [1]
(ii) direction opposite to that in (a)(ii) B1 [1]
(c) suggested reasonable values of I and d B1
mention of expression F = BIL B1
force between wires is small M1
compared to weight of wire A1 [4]
7 (a) ‘uniform’ distribution B1 [1]
(b) concentric rings B1 [1]
(c) higher speed, more momentum M1
λ = h / p M1
so λ decreases and ring diameter decreases A1 [3]
8 (a) arrow labelled E pointing down the page B1 [1]
(b) (i) Bqv = qE M1
forces are independent of mass and charge ‘cancels’ M1
so no deviation A1 [3]
(ii) magnetic force > electric force M1
so deflects M1
‘downwards’ A1 [3]

9702 PHYSICS
9702/05 Paper 5
Maximum mark 30
demonstrated.
Examination.

GCE A LEVEL – May/June 2006 9702 05
1 (b) (ii) Expect to see x read to nearest mm and then:
Use connector with sharp edge [1]
Position the wire on top of the scale on the rule to reduce parallax error
(c) Readings [3]
Write the number of readings as a ringed total by the results table.
6 sets of readings scores 1 mark.
Check a value for lg(x/m) and a value for lg(IA). Underline checked values.
Ignore small rounding errors. Tick if correct; 1 mark each.
If incorrect then write in correct value.
Excessive alterations in the table then –1.
If minor help is given with the circuit, then –1. If excessive help is given then –2.
Please indicate when help has been given to a candidate by writing SR at the top of the
front page of the candidate's script. Also, please indicate the type of help that has been
given by writing a brief comment by the table of results.
Column headings. Allow lg(x/m) and lg(I/A). Allow POT errors. [1]
There must be some distinguishing mark between the quantity and its unit.
Please or underline each correct column heading to show that it has been seen.
Consistency of raw readings in the table of results [1]
Apply to x and I only.
Expect to see I to either 0.01 A or 0.1 A.
Expect to see all the values of x given to the nearest millimetre.
Indicate using a vertical line down each column of raw readings to show it has been seen
and a C if correct.
(d) (i) Axes [1]
Each axis must be labelled with a quantity. Allow lg(x), lgx, lg(x/m) but not lgx/m.
Scales must be such that the plotted points occupy more than half the graph grid in
both the x and y directions. Do not allow more than 3 large squares between scale
markings. Do not allow awkward scales (e.g. 3:10, 6:10, 7:10, etc.).
Plotting of points [1]
Count the number of plots on the grid and write this value by the line and ring it.
Do not allow plots in the margin area.
The number of plots must correspond to the number of observations (at least 6).
Do not award this mark if the number of plots is less than the number of observations.
Check one suspect plot. Circle this plot. Tick if correct. If incorrect then mark the
correct position with a small cross and use an arrow to indicate where the plot should
have been. Allow errors less than half a small square.

(ii) Line of best fit [1]
There must be a reasonable balance of points about the line of best fit.
If one of the plots is a long way from the trend of the other plots then allow this
plot to be ignored when the line is drawn.
One mark can be awarded if the line of best fit is ‘reasonable’, but not quite right.
(iii) Measurement of gradient [1]
The hypotenuse of the triangle must be greater than half the length of the drawn line.
If read-offs are inaccurate by half a small square, or more, then zero.
Please indicate the vertices of the triangle used by labelling with ∆.
One mark for the read-offs. The value must be negative. Value should be around -1.0.
∆x/∆y scores zero. Ignore units.
y-intercept. Expect –0.8. Ignore units. [1]
Check the read-off. Value must be within half a square.
Accept correct substitution from a point on the line into y = mx + c. Allow ecf from
(d)(iii) gradient for m.
(e) lg x = n lg I + lg k [1]
This can be implied from the working.
Value for n (from gradient). Allow ecf from (d)(iii) gradient. n = –1.0. Ignore unit. [1]
Value for k (from 10y-intercept
). Allow ecf from (d)(iii) y-intercept. k = 0.158. Ignore unit. [1]
Working must be checked.
(f) (i) Value of diameter of wire (± 0.02 mm of SV). (36 swg, diameter = 0.18 mm) [1]
(ii) Cross-sectional area correct. (36 swg, Area = 2.5 x 10 –8
m2
) [1]
(iii) Percentage uncertainty in area [2]
One mark for percentage uncertainty in r. One mark for % uncertainty x 2.
(g) Correct value of ρ. No POT error allowed. Expect to see ρ = 4.7 x 10–7
Ωm [1]
Check the substitution and consistency of units.
(h) k = 4 x previous value. Allow ecf (g). Expect to see k = 1.9 x 10–6
Ωm [1]
[Total: 20 marks]

Question 2
A1 Diagram of arrangement (light source/mesh/screen or collimator/mesh/telescope) [1]
A2 Fringes or dots shown on screen. May be shown on diagram. [1]
A3 Some sensible discussion of coherence [1]
A4 Use of laser or single slit (and lens) in collimator [1]
B1 Measurements: distance from mesh to screen and separation between fringes [1]
OR measure an angle from the spectrometer table
B2 n = 1; find separation between central fringe and first bright fringe [1]
OR measure angle between central bright beam and first order beam using scale on table
B3 Use of θλ sin=n to find d. n must be clearly identified [1]
C Any safety precaution [1]
e.g. use goggles/do not look directly into laser beam/cover over sodium lamp
do not touch the bulb
D Any good/further detail [2]
Examples of creditworthy points might be:
Take readings with mesh in different positions to average d
Sketch/suggestion of two-dimensional array of dots on screen
Laser + mesh + screen all at same height
λ = 589 nm for sodium lamp or about λ = 630 nm for He/Ne laser/semiconductor laser
D of the order of 1 m to 4 m (laser method)
Measure 2θ and divide by two to reduce uncertainty in θ
Repeat experiment with 2nd
order (3rd
order etc.) beams
Detail relating to setup/use of spectrometer
Allow other valid points.
[Total: 10 marks]

GCE Advanced Level
9702 PHYSICS
9702/06 Paper 6
Maximum raw mark 40
demonstrated.
Examination.

GCE A – May/June 2006 9702 06
Option A - Astrophysics and Cosmology
1 Planet: almost circular orbits B1
all in nearly the same plane B1
Comet: highly elliptical orbits B1
in many different planes B1 [4]
2 (a) (mean) density M1
of matter in the Universe A1 [2]
(b) (i) symmetrical curve below given line M1
touching given line at ‘present time’ A1 [2]
(ii) H0 not known with any certainty B1
mass of matter in the Universe not known B1
extent of Universe unknown B1 [3]
(allow 1 of the last 2 marks for ρ0 not known)
3 1 light-year = 0.306 pc (allow 0.3 pc) C1
1.3 × 1010
light-years = 3.98 × 103
Mpc C1
v = H0d C1
speed = 60 × 3.98 × 103
= 2.39 × 105
km s-1
ratio = (2.39 × 105
x 103
)/(3 .0 × 108
)
= 0.8 A1 [4]
4 e.g. vast expense (M1)
money could be spent on humanitarian aid (A1)
observations possible that cannot be made on Earth (M1)
since atmosphere limits observations (A1)
technological/scientific developments on Earth (M1)
greater understanding of Universe (M1)
leads to ‘spin off’ benefits for individuals (A1)
Any sensible comments, 1 each to max 5 B5 [5]
Option F - The Physics of Fluids
5 (a) conservation of volume/mass/density or incompressible B1 [1]
(b) conservation of energy B1 [1]
6 (a) air near jet is moving at speed OR water in jet is moving at speed B1
higher speed air has a lower OR high-speed water has lower pressure B1
pressure
(because) air is dragged along by OR air is drawn into water jet B1
water jet
air (outside pump) is not moving OR loss of air reduces pressure B1 [4]
(b) (i) air/water in pump has a higher speed M1
so greater pressure difference A1 [2]

GCE A – May/June 2006 9702 06
(ii) no change in speed of air OR reference to greater ρ in Bernoulli eqn M1
so no change in pressure OR greater pressure difference A1 [2]
difference
(allow any logical argument based on liquid causing more/less drag on air)
7 (a) eddy currents have kinetic energy OR cause extra drag M1
eddy currents caused by
movement of the car OR energy required to overcome drag A1
extra energy (of eddy currents) is derived from car’s fuel A1 [3]
(b) (i) power = force × speed B1
so power = ½CDAρv2
× v and A and ρ are constants B1 [2]
(ii) 84 × 103
= ½ × 0.34 × 1.8 × 1.1 × vmax
3
C1
vmax = 63 m s-1
A1 [2]
(iii) P = ½ × 0.34 × 1.8 × 1.1 × (63 + 9)3
C1
P = 126 kW C1
ratio = 126 / 84 = 1.5 A1 [3]
Option M - Medical Physics
8 (a) alternating voltage B1
applied across (piezo-electric) crystal B1
causes crystal to vibrate B1
crystal dimensions such as to give resonance (in US range) B1 [4]
(b) wavelength at 1 MHz is shorter B1
so greater detail is possible B1 [2]
9 e.g. used as a scalpel (1)
further detail: causes (explosive) vaporisation of intracellular water (1)
CO2 laser (1)
IR radiation strongly absorbed by water (1)
laser beam focused to give high power density (1)
no/very little bleeding (1)
accurate guidance (1)
e.g. repair of retina (1)
further detail: focused laser beam onto retina (1)
melts tissue and forms a weld (1)
(pulsed) ruby or argon laser (1)
any two examples: named (1) plus further detail (2) B6 [6]
(allow up to two marks for each diagnostic technique)
10 (a) minimum intensity (of sound) detected M1
where intensity = (sound) power per unit area at a stated frequency A1
value is 1 × 10-12
W m-2
B1
at 3 kHz (allow 2 kHz → 3 kHz) B1 [4]

GCE A – May/June 2006 9702 06
(b) (i) intensity = (0.14 × 10-6
)/(54 × 10-6
) = 2.6 × 10-3
W m-2
C1
IL = 10 lg (2.6 × 10-3
)/(1 × 10-12
) C1
= 94 dB A1 [3]
(ii) comment e.g. would be perceived as being loud
could cause tinnitus over a short period of time
could cause deafness over a long period of time
higher level than is acceptable in the workplace
any appropriate comment, 1 mark B1 [1]
Option P - Environmental Physics
11 (a) at times of low usage of electrical power B1
water pumped from low-level to high-level reservoir B1
at times of high/sudden demand for electrical power B1
water released to pass through turbines B1 [4]
(b) electrical energy generated = 78 × 106
× 4.0 × 3600 = 1.12 × 1012
J C1
energy to be stored = (1.12 × 1012
)/0.75 = 1.5 × 1012
J C1
1.5 × 1012
= ρVgh C1
= 1.0 × 103
× V × 9.8 × 95
V = 1.6 × 106
m3
A1 [4]
12 (a) law: it is impossible to convert all of a given amount of thermal energy into work B1
(that is) W QH B1
(QH – W ) is energy rejected at temperature TL B1 [3]
(b) W/QH = 1 – TL/TH B1 [1]
(c) efficiency = 1 – 313/393 C1
= 0.20 A1 [2]
13 (a) (i) e.g. industry setting up
people preparing to go to work
starting to cook breakfast
(allow any two sensible suggestions, 1 each) B2 [2]
(ii) e.g. change in temperature with use of heaters/air conditioning
holiday or workday with more power used by industry when not on holiday
(allow any two sensible suggestions, 1 each) B2 [2]

GCE A – May/June 2006 9702 06
(b) (i) sudden increase in demand (as appliances are used) B1
(ii) increased demand in the afternoon B1 [2]
(allow any two sensible suggestions in (i) and (ii))
Option T - Telecommunications
14 (a) (instantaneous) displacement of information signal M1
determines the frequency of the carrier wave A1 [2]
(b) (i) 12 V B1 [1]
(ii) 650 kHz B1 [1]
(iii) 550 kHz B1 [1]
(iv) 3000 B1 [1]
15 (a) analogue-to-digital converter (do not allow ADC) B1 [1]
(b) controls the time at which samples are taken B1 [1]
(c) enables higher frequency components in signal to be ‘detected’ B1 [1]
16 (a) electromagnetic shielding for the inner conductor B1
the braid is earthed B1 [2]
(b) increased bandwidth means more information can be carried B1
so more calls can be transmitted simultaneously B1
fewer links are required B1 [3]
17 (a) e.g. cross-talk/cross-linking
interference/picking up atmospherics/picking up man-made radiation
white noise associated with vibrating atoms
(any two, 1 each) B2 [2]
(b) (i) number of dB = 10 lg (P2/P1)
35 = 10 lg (P/{7.6 × 10-6
}) C1
P = 0.024 W A1 [2]
(ii) number of dB = 10 lg (2.6/0.024) = 20.3 C1
length = 20.3/5.8 = 3.5 km A1 [2]

CAMBRIDGE EXAM MARK SCHEME

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