9702 s13 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9702 PHYSICS
9702/11 Paper 1 (Multiple Choice), maximum raw mark 40
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Mark Scheme Syllabus Paper
GCE A LEVEL – May/June 2013 9702 11
© Cambridge International Examinations 2013
Question
Number
Key
Question
Number
Key
1 D 21 D
2 C 22 A
3 C 23 A
4 D 24 C
5 D 25 A
6 D 26 C
7 B 27 A
8 D 28 C
9 A 29 B
10 D 30 B
11 D 31 D
12 A 32 A
13 C 33 A
14 B 34 A
15 A 35 B
16 C 36 D
17 B 37 D
18 B 38 C
19 B 39 A
20 B 40 B

9702 PHYSICS

Question
Number
Key
Question
Number
Key
1 B 21 C
2 B 22 B
3 A 23 B
4 D 24 D
5 C 25 D
6 B 26 B
7 D 27 C
8 B 28 B
9 D 29 C
10 B 30 D
11 D 31 A
12 A 32 C
13 B 33 A
14 D 34 A
15 B 35 D
16 B 36 D
17 C 37 C
18 A 38 C
19 C 39 B
20 D 40 B

9702 PHYSICS

Question
Number
Key
Question
Number
Key
1 B 21 D
2 A 22 A
3 C 23 C
4 A 24 C
5 D 25 C
6 B 26 B
7 B 27 D
8 C 28 D
9 B 29 C
10 B 30 D
11 A 31 D
12 B 32 C
13 D 33 C
14 A 34 B
15 A 35 C
16 A 36 B
17 C 37 C
18 A 38 C
19 C 39 D
20 A 40 B

9702 PHYSICS
9702/21 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,
which would have considered the acceptability of alternative answers.

GCE AS/A LEVEL – May/June 2013 9702 21
1 (a) the wire returns to its original length (not ‘shape’) M1
when the load is removed A1 [2]
(b) energy: N m / kg m2
s–2
and volume m3
C1
energy / volume: kgm2
s–2
/ m3
M1
energy / volume: kgm–1
s–2
A0 [2]
(c) ε has no units B1
E: kg m s–2
m–2
M1
units of RHS: kgm–1
s–2
= LHS units / satisfactory conclusion to show C has
no units A1 [3]
2 (a) mass is the property of a body resisting changes in motion / quantity of
matter in a body / measure of inertia to changes in motion B1
weight is the force due to the gravitational field/force due to gravity
or gravitational force B1 [2]
Allow 1/2 for ‘mass is scalar weight is vector’
(b) (i) arrow vertically down through O B1
tension forces in correct direction on rope B1 [2]
(ii) 1. weight = mg = 4.9 × 9.81 (= 48.07) C1
69 sin θ = mg C1
θ = 44.(1)° scale drawing allow ± 2° A1 [3]
use of cos or tan 1/3 only
2. T = 69 cos θ C1
= 49.6 / 50N scale drawing 50 ±2 (2/2) 50 ±4 (1/2) A1 [2]
correct answers obtained using scale diagram or triangle of forces will score
full marks
cos in 1. then sin in 2. (2/2)
3 (a) loss in potential energy due to decrease in height (as P.E. = mgh) (B1)
gain in kinetic energy due to increase in speed (as K.E. = ½ mv2
) (B1)
special case ‘as PE decreases KE increases’ (1/2)
increase in thermal energy due to work done against air resistance (B1)
loss in P.E. equals gain in K.E. and thermal energy (B1)
max. 3 [3]

(b) (i) kinetic energy = ½ mv2
C1
= ½ × 0.150 × (25)2
C1
= 46.875 = 47J A1 [3]
(ii) 1. potential energy (= mgh) = 0.150 × 9.81 × 21 C1
loss = KE – mgh = 46.875 – (30.9) C1
= 15.97 = 16J A1 [3]
2. work done = 16J
work done = force × distance C1
F = 16 / 21 = 0.76N A1 [2]
4 (a) pressure = force / area (normal to force) A1 [1]
(b) molecules/atoms/particles in (constant) random/haphazard motion B1
molecules have a change in momentum when they collide with the walls M1
(force exerted on molecules) therefore force on the walls A1
reference to average force from many molecules/many collisions A1 [4]
(c) elastic collision when kinetic energy conserved B1
temperature constant for gas B1 [2]
5 (a) waves overlap / meet / superpose (B1)
coherence / constant phase difference (not constant λ or frequency) (B1)
path difference = 0, λ, 2λ or phase difference = 0, 2π, 4π (B1)
same direction of polarisation/unpolarised (B1)
max. 3 [3]
(b) λ = v / f C1
f = 12 × 109
Hz C1
λ = 3 × 108
/ 12 × 109
(any subject) M1
= 0.025m A0 [3]
(c) maximum at P B1
several minima or maxima between O and P B1
5 maxima / 6 minima between O and P
or 7 maxima / 6 minima including O and P B1 [3]
(d) slits made narrower B1
slits put closer together B1 [2]
(not just ‘make slits smaller’)
Allow tilting the slits M1 and explanation of axes of rotation A1

6 (a) (i) chemical to electrical B1 [1]
(ii) electrical to thermal / heat or heat and light B1 [1]
(b) (i) (PB =) EI or I2
(R1 + R2) A1 [1]
(ii) (PR =) I2
R1 A1 [1]
(c) R = ρl / A or clear from the following equation B1
ratio = I2
R1 / I2
R2 =
( ) ( ) 212
2
ofresistance8hasor
22
RR
d
d
×
π/
π/
l
l
ρ
ρ
C1
= 8 or 8:1 A1 [3]
(d) P = V2
/ R or E2
/ R C1
(V or E the same) hence ratio is 1/8 or 1:8 = 0.125 (allow ecf from (c)) A1 [2]
7 (a) the majority/most went straight through
or were deviated by small angles B1
a very small proportion/a few were deviated by large angles B1
small angles described as < 10° and large angles described as >90° B1 [3]
(b) most of the atom is empty space/nucleus very small compared with atom B1
mass and charge concentrated in (very small) nucleus B1
correct links made with statements in (a) B1 [3]

9702 PHYSICS

1 (a) power = energy / time C1
= (force × distance / time) = kg m2
s–2
/ s C1
= kg m2
s–3
A1 [3]
(b) (i) units of L2
: m2
and units of ρ: kgm–3
and units of v3
: m3
s–3
C1
(C = P / L2
ρv3
) hence units of C: kg m2
s–3
m–2
kg–1
m3
m–3
s3
or any correct statement of component units M1
argument /discussion / cancelling leading to C having no units A1 [3]
(ii) power available from wind = 3.5 × 105
× 100 / 55 (= 6.36 × 105
) C1
v3
= 3.5 × 105
× 100 / (55 × 0.931 × (25)2
× 1.3) C1
v = 9.4 m s–1
A1 [3]
(iii) not all kinetic energy of wind converted to kinetic energy of blades B1
generator / conversion to electrical energy not 100% efficient / heat
produced in generator / bearings etc B1 [2]
(there must be cause of loss and where located)
2 (a) force = rate of change of momentum A1 [1]
(b) (i) horizontal line on graph from t = 0 to t about 2.0 s ± ½ square, a > 0 M1
horizontal line at 3.5 on graph from 0 to 2s A1
vertical line at t = 2.0s to a = 0 or sharp step without a line B1
horizontal line from t = 2s to t = 4s with a = 0 B1 [4]
(ii) straight line and positive gradient M1
starting at (0,0) A1
finishing at (2,16.8) A1
horizontal line from 16.8 M1
from 2.0 to 4.0 A1 [5]
3 (a) the point where (all) the weight (of the body) M1
is considered / seems to act A1 [2]
(b) (i) vertical component of T (= 30cos40°) = 23N A1 [1]
(ii) the sum of the clockwise moments about a point equals the sum of the
anticlockwise moments (about the same point) B1 [1]
(iii) (moments about A): 23 × 1.2 (27.58) M1
= 8.5 × 0.60 + 1.2 × W M1
working to show W = 19 or answer of 18.73 (N) A1 [3]
(iv) (M = W / g = 18.73 / 9.81 =) 1.9(09) kg A1 [1]

(c) (for equilibrium) resultant force (and moment) = 0 B1
upward force does not equal downward force / horizontal component of T
not balanced by forces shown B1 [2]
4 (a) apparatus: cell with particles e.g. smoke (container must be closed) B1
diagram showing suitable arrangement with light illumination and microscope B1 [2]
(b) specks / flashes of light M1
in random motion A1 [2]
(c) cannot see what is causing smoke to move hence molecules smaller than
smoke particles (B1)
continuous motion of smoke particles implies continuous motion of molecules (B1)
random motion of particles implies random motion of molecules (B1)
max. 2 [2]
5 (a) (i) v = fλ C1
λ = 40 / 50 = 0.8(0) m A1 [2]
(ii) waves (travel along string and) reflect at Q / wall / fixed end B1
incident and reflected waves interfere / superpose B1 [2]
(b) (i) nodes labelled at P, Q and the two points at zero displacement B1
antinodes labelled at the three points of maximum displacement B1 [2]
(ii) (1.5λ for PQ hence PQ = 0.8 × 1.5) = 1.2m A1 [1]
(iii) T = 1 / f = 1/50 = 20ms C1
5ms is ¼ of cycle A1
horizontal line through PQ drawn on Fig. 5.2 B1 [3]
6 (a) charge = current × time B1 [1]
(b) (i) P = V2
/ R C1
= (240)2
/ 18 = 3200W A1 [2]
(ii) I = V / R = 240 / 18 = 13.3A A1 [1]
(iii) charge = It = 13.3 × 2.6 × 106
C1
= 3.47 × 107
C A1 [2]
(iv) number of electrons = 3.47 × 107
/ 1.6 × 10–19
(= 2.17 × 1026
) C1
number of electrons per second = 2.17 × 1026
/ 2.6 × 106
= 8.35 × 1019
A1 [2]

7 (a) (i) W = 206 and X = 82 A1
Y = 4 and Z = 2 A1 [2]
(ii) mass-energy is conserved B1
mass on rhs is less because energy is released B1 [2]
(b) not affected by external conditions/factors/environment B1 [1]
or two examples temperature and pressure

9702 PHYSICS

1 (a) force: kg m s–2
A1 [1]
(b) (i) I2
: A2
l: m x: m C1
K: kg m s–2
A–2
A1 [2]
(ii) curve of the correct shape (for inverse proportionality) M1
clearly approaching each axis but never touching the axis A1 [2]
(iii) curving upwards and through origin A1 [1]
2 (a) (i) 1. distance of path / along line AB B1 [1]
2. shortest distance between AB / distance in straight line between AB
or displacement from A to B B1 [1]
(ii) acceleration = rate of change of velocity A1 [1]
(b) (i) distance = area under line or (v/2)t or s = (8.8)2
/ (2 × 9.81) C1
= 8.8 / 2 × 0.90 = 3.96m or s = 3.95m = 4(.0)m A1 [2]
(ii) acceleration = (– 4.4 – 8.8) / 0.50 C1
= (–) 26(.4)ms–2
A1 [2]
(c) (i) the accelerations are constant as straight lines B1
the accelerations are the same as same gradient or
no air resistance as acceleration is constant or
change of speed in opposite directions (one speeds up one slows down) B1 [2]
(ii) area under the lines represents height
or KE at trampoline equals PE at maximum height B1
second area is smaller / velocity after rebound smaller hence KE less B1
hence less height means loss in potential energy A0 [2]
3 (a) (i) the total momentum of a system (of interacting bodies) remains constant M1
provided there are no resultant external forces / isolated system A1 [2]
(ii) elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy B1 [1]
[allow elastic: relative speed of approach equals relative speed of separation]

(b) (i) initial mom: 4.2 × 3.6 – 1.2 × 1.5 (= 15.12 – 1.8 = 13.3) C1
final mom: 4.2 × v + 1.5 × 3 C1
v = (13.3 – 4.5) / 4.2 = 2.1ms–1
A1 [3]
(ii) initial kinetic energy = ½ mA(vA)2
+ ½ mB(vB)2
= 27.21 + 1.08 = 28(.28) M1
final kinetic energy = 9.26 + 6.75 = 16 M1
initial KE is not the same as final KE hence inelastic A1 [3]
provided final KE less than initial KE
[allow in terms of relative speeds of approach and separation]
4 (a) (i) stress = force / cross-sectional area B1 [1]
(ii) strain = extension / original length B1 [1]
(b) (i) E = stress / strain C1
E = 0.17 × 1012
C1
stress = 0.17 × 1012
× 0.095 / 100 C1
= 1.6(2) × 108
Pa A1 [4]
(ii) force = (stress × area) = 1.615 × 108
× 0.18 × 10–6
C1
= 29(.1)N A1 [2]
5 (a) when waves overlap / meet B1
the resultant displacement is the sum of the individual displacements of the waves B1 [2]
(b) (i) 1. phase difference = 180º / (n + ½) 360º (allow in rad) B1 [1]
2. phase difference = 0 / 360º / (n360º) (allow in rad) B1 [1]
(ii) v = fλ C1
λ = 320 / 400 = 0.80m A1 [2]
(iii) path difference = 7 – 5 = 2 (m)
= 2.5λ M1
hence minimum
or maximum if phase change at P is suggested A1 [2]
6 (a) p.d. = work done / energy transformed (from electrical to other forms)
charge B1 [1]
(b) (i) maximum 20V A1 [1]
(ii) minimum = (600 / 1000) × 20 C1
= 12V A1 [2]

(c) (i) use of 1.2kΩ M1
1/1200 + 1/600 = 1/R, R = 400Ω A1 [2]
(ii) total parallel resistance (R2 + LDR) is less than R2 M1
(minimum) p.d. is reduced A1 [2]
7 (a) (i) nucleus contains 92 protons B1
nucleus contains 143 neutrons (missing ‘nucleus’ 1/2) B1
outside / around nucleus 92 electrons (B1)
most of atom is empty space / mass concentrated in nucleus (B1)
total charge is zero (B1)
diameter of atom ~ 10–10
m or size of nucleus ~ 10–15
m (B1)
any two of (B1) marks [4]
(ii) nucleus has same number / 92 protons B1
nuclei have 143 and 146 neutrons (missing ‘nucleus’ 1/2) B1 [2]
(b) (i) Y = 35 A1
Z = 85 A1 [2]
(ii) mass-energy is conserved in the reaction B1
mass on rhs of reaction is less so energy is released
explained in terms of E = mc2
B1 [2]

9702 PHYSICS
9702/31 Paper 3 (Advanced Practical Skills 1),
maximum raw mark 40

1 (a) Value of L in the range 0.790–0.810m. [1]
(c) (ii) Value of d to the nearest mm and d < 0.600m. [1]
(d) Six sets of readings of m and d scores 5 marks, five sets scores 4 marks etc. [5]
Correct trend is d decreases as m increases.
Help from Supervisor –1.
Range of m: mmin = 0g or 10g; mmax > 100g. [1]
Column headings: [1]
Each column heading must contain a quantity and a unit.
The presentation of quantity and unit must conform to accepted scientific convention
e.g. 1/d/m–1
.
Consistency: [1]
All values of d must be given to the nearest mm.
Significant figures: [1]
Significant figures for every row of values of 1/d same as, or one greater than, d as
recorded in table.
Calculation: [1]
Values of 1/d calculated correctly.
(e) (i) Axes: [1]
Sensible scales must be used, no awkward scales (e.g. 3:10).
Scales must be chosen so that the plotted points occupy at least half the graph
grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted.
Diameter of plotted points must be ≤ half a small square (no “blobs”).
Work to an accuracy of half a small square.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
Scatter of points must be less than ± 0.05m–1
of 1/d from a straight line.
(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidate’s line (at least 5 points).
There must be an even distribution of points either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the
candidate. Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be at least half the length of the drawn line.
Both read-offs must be accurate to half a small square in both the x and y directions.
The method of calculation must be correct.
y–intercept: [1]
Either:
Correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Correct read-off of the intercept directly from the graph.
(f) Value of P = candidate’s gradient. Value of Q = candidate’s intercept.
Unit for P (e.g. kg–1
m–1
) and Q (m–1
). [1]
(g) Value of k in range 1.0–2.0. [1]
[Total: 20]
2 (a) (ii) Value of θ with unit. Help from Supervisor –1. [1]
θ in range 72°–92°. [1]
(iii) Absolute uncertainty in θ in range 2º–10º.
If repeated readings have been taken, then the uncertainty can be half the range (but
NOT zero if values are equal). Correct method of calculation to obtain percentage
uncertainty. [1]
(iv) Correct calculation of sin θ. Ignore unit. Do not allow sin θ = O/H ideas as triangle
not a right-angled triangle. [1]
(b) Value of T with unit in range 1.0 ≤ T ≤ 2.0s. [1]
Evidence of repeats here or in (c)(ii). [1]
(c) (ii) Second value of θ. [1]
Second value of T. [1]
Second value of T < first value of T. [1]
(d) (i) Two values of k calculated correctly. [1]
(ii) Justification of s.f. in k linked to significant figures in T (or t) and θ. [1]
(iii) Sensible comment relating to the calculated values of k, testing against a criterion
specified by the candidate. [1]

(e)
(i) Limitations max. 4 (ii) Improvements max. 4 Do not credit
A two readings not enough (to draw a
conclusion)
take more readings and plot a
graph/
calculate more k values and
compare
“repeat readings”
on its own
/few readings/only
one reading
/take more readings
and (calculate)
average k
B end of nail slips in bracket/bracket
moves/is not stable
use something with a sharper
point e.g. cocktail stick/dent in
bracket (to seat head of nail)
valid method to fix bracket e.g.
use blu-tack/glue/use
bigger/heavier bracket/fix
bracket/ clamp to bench
method of fixing nail
C difficult to measure T with reason
e.g. heavily damped/oscillations die
away quickly
‘too few oscillations’
on its own/
T small
D difficult to judge start of/end
of/complete oscillation
use a fixed/fiducial marker
/improved timing method e.g.
video with timer/video and view
frame-by-frame
multiflash photography with
strobe rate
human error/
reaction time
/record time for
more oscillations
marker fixed to rod
/marker placed at
extreme of
oscillation
use light gate
E difficult to read θ/angle/protractor
with reason e.g. difficult to hold
steady in the air
clamp protractor parallax error
use a larger
protractor
F fans/air conditioning affect
oscillations
[Total: 20]

9702 PHYSICS
maximum raw mark 40

1 (b) (ii) Value of t in the range 10.0 s ≤ t ≤ 20.0 s. [1]
Evidence of repeat measurements of t. [1]
(c) Six sets of readings of S and t scores 5 marks, five sets scores 4 marks etc.
If trend wrong or no S or t column –1.
Major help from Supervisor –2 (setting up circuit). Minor help from
Supervisor –1.
[5]
Range:
Values of S must include 22 (kΩ) or 10 (kΩ) and 1.2 (kΩ) or 1.0 (kΩ).
[1]
Column headings:
The presentation of quantity and unit must conform to accepted scientific
convention e.g. 1/t / s–1
1/S / kΩ –1
1/S (kΩ –1
) t / s t (s).
( 1/t(s) 1/S 1/kΩ 1/S (kΩ) –1
are not allowed.)
[1]
Consistency:
All values of raw t must be given to the same precision (either 0.1s or 0.01s).
Significant figures:
Significant figures for every row of values of 1/S must be the same as or one
greater than the s.f. in S as recorded in table.
[1]
[1]
Calculation:
Values of 1/t calculated correctly.
[1]
(d) (i) Axes:
[1]
Plotting of points:
Points must be plotted to an accuracy of half a small square.
Diameter of points must be ≤ half a small square (no “blobs”).
[1]
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded.
Judge by the scatter of all the points about the straight line. Points must be
less than 0.05 kΩ ─1
from a straight line on the 1/S axis.
[1]
(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5
points).
There must be an even distribution of points either side of the line along the full
length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by
the candidate.
Line must not be kinked or thicker than half a small square.
[1]

(iii) Gradient:
The hypotenuse of the triangle must be at least half the length of the drawn
line.
Both read-offs must be accurate to half a small square in both the x and y
directions. The method of calculation must be correct.
[1]
y-intercept:
Either:
Correct read-off from a point on the line substituted into y = mx + c.
Or:
Intercept read off directly from the graph.
[1]
(e) Value of a = candidate’s gradient.
Value of b = candidate’s intercept / candidate’s gradient
= candidate’s intercept / a [1]
Unit for a correct and consistent with value e.g. kΩs–1
, Ωs–1
. [1]
[Total: 20]
2 (a) (i) Measurement of d with unit in range 0.5mm – 2.5mm. [1]
Evidence of repeated readings of d. [1]
(ii) Absolute uncertainty in d in the range 0.2 – 0.5mm.
If repeated readings have been taken, then the absolute uncertainty can be half
the range (but not zero if values are equal).
Correct method of calculation to get percentage uncertainty.
[1]
(c) (i) Measurement of r1 recorded to nearest 0.1cm3
, and in range 1 to 5cm3
. [1]
(ii)
(iii)
Value for n.
Correct calculation of V.
[1]
[1]
(d) Justification of s.f. in V linked to significant figures in (r1 – r2) and in n. [1]
(e) (ii)
(iii)
Second value of d.
Second value of n.
Quality: V larger for larger d.
[1]
[1]
[1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Sensible comment relating to the calculated values of k, testing against a criterion

(g)
(i) Limitations 4 max. (ii) Improvements 4 max. Do not credit
A two readings not enough (to draw a
conclusion)
take many readings and plot a
graph/take many readings and
compare
repeat readings/few
readings/take more
readings and
(calculate) average
k/only one reading
B1 large uncertainty in d because d is
small
improved method to measure d
e.g. measure OD and wall
thickness/image of cross-section
with scale
flexible tube/rigid
tube/wire in
tube/micrometer/d is
small
B2 difficult to measure d with reason
e.g. tube distorts/difficult to insert
both prongs inside tube/difficult to
judge when jaws are in line with
edges of d
use travelling
microscope/measure volume
and calculate d/use a
magnifying glass with scale
C difficult to count bubbles with
reason
e.g. plunger moves
unsteadily/plunger sticks/bubbles
unexpected/bubbles emerge too
quickly
method to improve control
e.g. use a G-clamp or screw to
move plunger/use a narrower
diameter syringe with reason
(e.g.smaller force needed so
better control)
bubbles difficult to
count/lubrication
D difficult to watch the syringe scale
and the bubbles at the same time
description of method to allow
simultaneous measurement
e.g. use video with playback/use
video to allow slow motion/use
video to see bubbles
parallax/use
assistant/change
syringe scale/high
speed camera/slow
motion camera/light
gates
E difficult to keep end of tube at 2 cm
depth
method to keep tube at 2cm
depth
e.g. use of clamp/Blu-Tack/tape
stick tube/attach
tube/water in
tube/comments
about Blu-Tack seal
[Total: 20]

9702 PHYSICS
maximum raw mark 40
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced
Level and Advanced Subsidiary Level components and some Ordinary Level components.

1 (a) (i) Value of raw d in the range 0.15mm ≤ d ≤ 0.44mm. [1]
(b) (v) Value of l in range 0.1m < l < 1m. Value of V in range 0.1V ≤ V ≤ 2.0V. [1]
(d) Six sets of readings of l and V scores 5 marks; five sets scores 4 marks etc. [5]
Major help from Supervisor –2 (setting up apparatus). Minor help from Supervisor –1.
Range of l :∆l ≥ 60cm. [1]
The presentation of quantity and unit must conform to accepted scientific convention.
e.g. 1/l /m–1
, V/l / Vm–1
.
Do not allow 1/l (m), V(V) / l (m).
Consistency: [1]
All values of raw l must be given to the nearest mm.
Significant figures for every row of values of 1/l same as or one greater than l as
recorded in table.
Calculation: [1]
Values of V/l calculated correctly
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph grid in
both x and y directions
Check that the points are plotted correctly.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter
of points must be less than 0.1m–1
from a straight line on the 1/l axis.
Judge by balance of all points on the grid about the candidate’s line (at least 5 points)

(iii) Gradient: [1]
The method of calculation must be correct.
y-intercept: [1]
Either:
Correct read-off from a point on the line and substituted into y = mx + c.
Or:
(f) (i) Value of M = candidate’s gradient. Value of N = –(candidate’s intercept). [1]
(ii) Answer in range ρ: 2.0 ≤ ρ ≤ 20.0 × 10–7
Ωm. Consistent with units. [1]
[Total: 20]
2 (a) (ii) Measurement of raw H in range 10.0cm < H < 20.0cm consistent with unit. [1]
(b) (ii) Measurement of raw h1 to nearest mm with unit. [1]
(iii) Absolute uncertainty in h1 in the range 2–5mm. If repeated readings have been taken,
then the absolute uncertainty can be half the range. Correct method of calculation to get
percentage uncertainty. [1]
(c) (iii) Measurement of h2 less than h1. [1]
Evidence of repeat readings here or in (e). [1]
(d) Correct calculation of F with no units. [1]
(e) Second value of h1. [1]
Second value of h2. [1]
Second value of h2 < first value of h2. [1]
(ii) Justification of s.f. in k linked to significant figures in h1 and (h1 – h2). [1]

(g)
A two readings not enough (to
draw a conclusion)
take many readings and plot a
graph/calculate more k values
and compare
repeat readings
/few readings
/take more readings and
(calculate) average k
/only one reading
B discontinuous movement at
bottom
method of providing
continuous ramp e.g. tape join
alignment
/stick
/fix
C parallax error (or wtte) in h1 or
h2 or heights
ruler and set square with
detail e.g. set square from
ruler to track or ball
ruler perpendicular to bench
/parallax error in height
D difficult to measure h1/h2 with
reason e.g. cannot see
bottom of marble/bottom of
track not at bottom of
marble/thickness of track not
taken into account
measure to top of marble.
/measure diameter of marble
and subtract it from height to
top of marble
H
/clear ramps
E difficult to release marble
without applying a force
description of mechanical
method of releasing marble
e.g. card gate
string method
/use of helpers
F difficult to measure h2 with
reason related to time e.g.
short time interval/doesn’t
stay still at h2 for long
method of improved
measurement of h2 e.g. video
with (clamped) rule/multiflash
photography with (clamped)
rule/trial and improvement
method/position sensore at
top of ramp/grid behind
runway/scale on runway
too fast/ball travelling too
quick, etc.
/high speed camera or slow
motion camera
[Total: 20]

9702 PHYSICS
maximum raw mark 40

1 (b) (i) Value of d in the range 0.480 – 0.500m, with unit. [1]
(c) Six sets of readings of d and F scores 6 marks, five sets scores 5 marks etc.
Incorrect trend or no d data –1.
Minor help from Supervisor –1, major help –2.
[6]
Range: dmax – dmin ≥ 30cm. [1]
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention
e.g. 1/d / m–1
.
[1]
Consistency:
All values of d must be given to the nearest mm and all values of F must be given to
the nearest 0.1N.
[1]
Significant figures:
Significant figures for every row of values of 1/d same as, or one greater than, d as
recorded in table.
[1]
Calculation:
Values of 1/d calculated correctly.
[1]
(d) (i) Axes:
[1]
Plotting of points:
All observations in the table must be plotted on the grid.
Points must be plotted to an accuracy of half a small square.
Diameter of plotted points must be ≤ half a small square (no blobs).
[1]
Quality:
All points in the table must be plotted on the grid (at least 5) for this mark to be
awarded. Judge by the scatter of all the points about a straight line.
Scatter of points must be less than ± 0.001cm–1
from a straight line in the 1/d
direction.
[1]
(ii) Line of best fit:
Judge by balance of all the points on the grid about the candidate’s line (at least 5
points). There must be an even distribution of points either side of the line along
the full length.
One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by
the candidate.
Line must not be kinked or thicker than half a square.
[1]

(iii) Gradient:
Both read-offs must be accurate to half a small square in both the x and y
directions. The method of calculation must be correct.
[1]
y-intercept:
Either:
Correct read-off from a point on the line substituted into y = mx + c or an
equivalent expression, with read-off accurate to half a small square in both
x and y directions.
Or:
Intercept read directly from the graph, with read-off accurate to half a small square
in both x and y directions.
[1]
(e) Value of a = candidate’s gradient. Value of b = candidate’s intercept.
Unit for a correct and consistent with value, e.g. Ncm.
Unit for b is correct and consistent with value, e.g. N.
[1]
[1]
[Total: 20]
2 (a) (i) All values of D to nearest 0.01cm or all to nearest 0.001cm, and in
range 3.0 to 5.0mm.
Evidence of repeat readings of D.
[1]
[1]
(ii) Absolute uncertainty in D in range 0.02 to 0.05cm and correct method of
calculation to obtain percentage uncertainty.
If repeated readings have been taken, then the absolute uncertainty can be half
the range (but not zero if values are equal).
[1]
(c) (i) l in range 19.0 to 21.0cm, with unit, to nearest mm. [1]
(iii) t in range 2.0 to 10.0s and value(s) to nearest 0.1s or 0.01s. [1]
(iv) Correct calculation of v. [1]
(d) Justification for s.f. in v linked to s.f. in D and t. [1]
(e) (ii) Second value of l.
Second value of t.
Second value of t > first value of t.
[1]
[1]
[1]
(ii) Sensible comment relating to the calculated values of k, testing against a criterion
specified by the candidate.
[1]

(g)
A two readings are not enough (to
draw a conclusion)
graph/take more readings and
compare
“repeat readings” on
its own/few readings/
only one
reading/take more
readings and
(calculate) average k
B1 large uncertainty in D because D is
small
measure outside diameter and
wall thickness/measure an
image showing the cross-
section and a scale
use micrometer
B2 tube distorts when measuring D use travelling
microscope/measure volume
and calculate D
C tube not straight so difficult to
make tube vertical/tube not
straight so difficult to measure
length
tape to a straight rod/increase
attached mass
use stiffer tube
D difficult to judge moment (or
operate stopwatch) when level
reaches syringe graduations
use video with timer/view video
frame by frame/collect water for
a timed interval and measure
volume/use light gates and
timer with practical detail/use
different diameter syringe with
reason/use position sensor
above water surface
‘reaction time’ on its
own/human error /
‘light gates’ on its
own/slow motion (or
high speed) camera
E difficult to see water level use coloured water (or
dye) /use clear syringe/view
against black background
F clay stretches (or squashes) tube measure length after attaching
clay
References to parallax error are ignored for this experiment.
[Total: 20]

9702 PHYSICS
maximum raw mark 40

1 (a) (ii) Value of T in range 0.4 ≤ T ≤ 1.4s. [1]
Evidence of repeats. [1]
(b) Six sets of readings of m and t (or T) scores 5 marks, four sets scores 4 marks etc. [5]
Help from Supervisor –1.
Range of m : ∆m ≥ 0.600kg [1]
The presentation of quantity and unit must conform to accepted scientific
convention e.g. 1/T2
/ s–2
.
Do not allow 1/T2
(s)2
Consistency: [1]
All values of raw t must all be given to the nearest 0.1s or 0.01s.
Significant figures for every row of values of 1/T2
same as or one greater than
t (or T) as recorded in table.
Calculation: [1]
Values of 1/T2
calculated correctly
(c) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph grid in
both x and y directions.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of
points must be less than 0.1s–2
of 1/T2
from a straight line.
Judge by balance of all points on the grid about the candidate’s line (at least 5 points)
(iii) Gradient: [1]

y-intercept: [1]
Either:
Correct-read off from a point on the line and substituted into y = mx + c.
Or:
(d) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]
Unit for P (kg–1
s–2
) correct and consistent with value and Q (s–2
) [1]
[Total: 20]
2 (a) (i) Value of L in range 8.0 ≤ L ≤ 12.0cm with consistent unit to the nearest mm. [1]
(ii) Absolute uncertainty 1 ≤ ∆L ≤ 3mm.
If repeated readings have been taken, then the uncertainty can be half the range.
Correct method of calculation to get percentage uncertainty. [1]
(b) (iii) Value of raw N1 an integer. [1]
(c) (iii) Value of N2 ≥ N1. [1]
Evidence of repeats for N1 or N2 either here or in (b)(iii). [1]
(d) Correct calculation of F. [1]
(e) Second value of L. [1]
Second values of N2 and N1. [1]
Second (average) value of N1 > first (average) value of N1. [1]
(ii) Justification of s.f. in k linked to significant figures in L and (N1 – N2) and m. [1]

(g)
A two readings not enough (to
draw a conclusion)
graph/calculate more k values and
compare
repeat
readings/few
readings/only one
reading/take more
readings and
average k
B friction at pulley method of reducing friction of
pulley with location
C wet string added to force/
mass of string not accounted
for
waterproof/nylon/wire
D can only measure to nearest
0.4g/paperclip
use smaller masses e.g. half
paperclips, riders, graph paper
newton meter
E change in N are very small reasoned explanation for changing
length of wire
helpers
parallax errors
F copper wire is not
flat/straight/exit not parallel to
water level
circular wire shape change liquid/wire
[Total: 20]

9702 PHYSICS
9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100

Section A
1 (a) region of space area / volume B1
where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1
either reference to point masses or separation >> ‘size’ of masses A1 [3]
(ii) field strength = GM / x2
or field strength ∝ 1 / x2 C1
ratio = (7.78 × 108
)2
/ (1.5 × 108
)2
C1
= 27 A1 [3]
(c) (i) either centripetal force = mRω2
and ω = 2π / T
or centripetal force = mv2
/ R and v = 2πR /T B1
gravitational force provides the centripetal force B1
either GMm / R2
= mRω2
or GMm / R2
= mv2
/ R M1
M = 4π2
R3
/ GT2
A0 [3]
(allow working to be given in terms of acceleration)
(ii) M = {4π2
× (1.5 × 1011
)3
} / {6.67 × 10–11
× (3.16 × 107
)2
} C1
= 2.0 × 1030
kg A1 [2]
2 (a) obeys the equation pV = constant × T or pV = nRT M1
p, V and T explained A1
at all values of p, V and T/fixed mass/n is constant A1 [3]
(b) (i) 3.4 × 105
× 2.5 × 103
× 10–6
= n × 8.31 × 300 M1
n = 0.34mol A0 [1]
(ii) for total mass/amount of gas
3.9 × 105
× (2.5 + 1.6) × 103
× 10–6
= (0.34 + 0.20) × 8.31 × T C1
T = 360K A1 [2]
(c) when tap opened
gas passed (from cylinder B) to cylinder A B1
work done on gas in cylinder A (and no heating) M1
so internal energy and hence temperature increase A1 [3]

3 (a) (i) 1. amplitude = 1.7cm A1 [1]
2. period = 0.36cm C1
frequency = 1/0.36
frequency = 2.8Hz A1 [2]
(ii) a = (–)ω2
x and ω = 2π/T C1
acceleration = (2π/0.36)2
× 1.7 × 10–2
M1
= 5.2ms–2
A0 [2]
(b) graph: straight line, through origin, with negative gradient M1
from (–1.7 × 10–2
, 5.2) to (1.7 × 10–2
, –5.2) A1 [2]
(if scale not reasonable, do not allow second mark)
(c) either kinetic energy = ½mω2
(x0
2
– x2
)
or potential energy = ½mω2
x2
and potential energy = kinetic energy B1
½mω2
(x0 – x2
) = ½ × ½mω2
x0
2
or ½mω2
x2
= ½ × ½mω2
x0
2
C1
x0
2
= 2x2
x = x0 / √2 = 1.7 / √2
= 1.2cm A1 [3]
4 (a) work done moving unit positive charge M1
from infinity (to the point) A1 [2]
(b) (gain in) kinetic energy = change in potential energy B1
½mv2
= qV leading to v = (2Vq/m)½
B1 [2]
(c) either (2.5 × 105
)2
= 2 × V × 9.58 × 107
C1
V = 330V M1
this is less than 470V and so ‘no’ A1 [3]
or v = (2 × 470 × 9.58 × 107
) (C1)
v = 3.0 × 105
ms–1
(M1)
this is greater than 2.5 × 105
ms–1
and so ‘no’ (A1)
or (2.5 × 105
)2
= 2 × 470 × (q/m) (C1)
(q/m) = 6.6 × 107
Ckg–1
(M1)
this is less than 9.58 × 107
Ckg–1

5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1
(creates) force per unit length of 1Nm–1
A1 [2]
(b) (i) flux density = 4π × 10–7
× 1.5 × 103
× 3.5 C1
= 6.6 × 10–3
T A1 [2]
(ii) flux linkage = 6.6 × 10–3
× 28 × 10–4
× 160 C1
= 3.0 × 10–3
Wb A1 [2]
(c) (i) (induced) e.m.f. proportional to rate of M1
change of (magnetic) flux (linkage) A1 [2]
(ii) e.m.f. = (2 × 3.0 × 10–3
) / 0.80 C1
= 7.4 × 10–3
V A1 [2]
6 (a) (i) to reduce power loss in the core B1
due to eddy currents/induced currents B1 [2]
(ii) either no power loss in transformer
or input power = output power B1 [1]
(b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1
peak voltage across load = √2 × 243
= 340V A1 [2]
or peak voltage across primary coil = 9.0 × √2 (C1)
peak voltage across load = 12.7 × (8100/300)
= 340V (A1)
7 (a) (i) lowest frequency of e.m. radiation M1
giving rise to emission of electrons (from the surface) A1 [2]
(ii) E = hf C1
threshold frequency = (9.0 × 10–19
) / (6.63 × 10–34
)
= 1.4 × 1015
Hz A1 [2]
(b) either 300nm ≡ 10 × 1015
Hz (and 600nm ≡ 5.0 × 1014
Hz)
or 300nm ≡ 6.6 × 10–19
J (and 600nm ≡ 3.3 × 10–19
J)
or zinc λ0 = 340nm, platinum λ0 = 220nm (and sodium λ0 = 520nm) M1
emission from sodium and zinc A1 [2]
(c) each photon has larger energy M1
fewer photons per unit time M1
fewer electrons emitted per unit time A1 [3]

8 (a) two (light) nuclei combine M1
to form a more massive nucleus A1 [2]
(b) (i) ∆m = (2.01410u + 1.00728u) – 3.01605u
= 5.33 × 10–3
u C1
energy = c2
× ∆m C1
= 5.33 × 10–3
× 1.66 × 10–27
× (3.00 × 108
)2
= 8.0 × 10–13
J A1 [3]
(ii) speed/kinetic energy of proton and deuterium must be very large B1
so that the nuclei can overcome electrostatic repulsion B1 [2]
Section B
9 (a) (i) light-dependent resistor/LDR B1 [1]
(ii) strain gauge B1 [1]
(iii) quartz/piezo-electric crystal B1 [1]
(b) (i) resistance of thermistor decreases as temperature increses M1
etiher VOUT = V × R / (R + RT)
or current increases and VOUT = I R A1
VOUT increases A1 [3]
(ii) either change in RT with temperature is non-linear
or VOUT is not proportional to RT/ change in VOUT with RT is non-linear M1
so change is non-linear A1 [2]
10 (a) sharpness: how well the edges (of structures) are defined B1
contrast: difference in (degree of) blackening between structures B1 [2]
(b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid
large penumbra on shadow/large area anode/wide beam
large pixel size
(any two sensible suggestions, 1 each) B2 [2]
(c) (i) I = I0e–µx
C1
ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1
= (4.65 × 10–5
) / (5.00 × 10–4
)
= 0.093 A1 [3]
(ii) either large difference (in intensities)
or ratio much less than 1.0 M1
so good contrast A1 [2]
(answer given in (c)(ii) must be consistent with ratio given in (c)(i))

11 (a) (i) amplitude of the carrier wave varies M1
(in synchrony) with the displacement of the information signal A1 [2]
(ii) e.g. more than one radio station can operate in same region/less interference
enables shorter aerial
increased range/less power required/less attenuation
less distortion
(any two sensible answers, 1 each) B2 [2]
(b) (i) frequency = 909kHz C1
wavelength = (3.0 × 108
) / (909 × 103
)
= 330m A1 [2]
(ii) bandwidth = 18kHz A1 [1]
(iii) frequency = 9000Hz A1 [1]
12 (a) for received signal, 28 = 10lg(P / {0.36 × 10–6
}) C1
P = 2.3 × 10–4
W A1 [2]
(b) loss in fibre = 10lg({9.8 × 10–3
} / {2.27 × 10–4
}) C1
= 16dB A1 [2]
(c) attenuation per unit length = 16 / 85
= 0.19dB km–1
A1 [1]

9702 PHYSICS

Section A
1 (a) equatorial orbit / above equator B1
satellite moves from west to east / same direction as Earth spins B1
period is 24 hours / same period as spinning of Earth B1 [3]
(allow 1 mark for ‘appears to be stationary/overhead’ if none of above marks scored)
(b) gravitational force provides/is the centripetal force B1
GMm/R2
= mRω2
or GMm/R2
= mv2
/R M1
ω = 2π /T or v = 2πR / T or clear substitution M1
clear working to give R3
= (GMT2
/ 4π2
) A1 [4]
(c) R3
= 6.67 × 10–11
× 6.0 × 1024
× (24 × 3600)2
/ 4π2
C1
= 7.57 × 1022
C1
R = 4.2 × 107
m A1 [3]
(missing out 3600 gives 1.8 × 105
m and scores 2/3 marks)
2 (a) (i) 1. pV = nRT
1.80 × 10–3
× 2.60 × 105 = n × 8.31 × 297 C1
n = 0.19 mol A1 [2]
2. ∆q = mc∆T
95.0 = 0.190 × 12.5 × ∆T B1
∆T = 40 K A1 [2]
(allow 2 marks for correct answer with clear logic shown)
(ii) p/T = constant
(2.6 × 105
) / 297 = p / (297 + 40) M1
p = 2.95 × 105
Pa A0 [1]
(b) change in internal energy is 120 J / 25 J B1
internal energy decreases / ∆U is negative / kinetic energy of molecules decreases M1
so temperature lower A1 [3]

3 (a) (i) ω = 2π / T
= 2π / 0.69 C1
= 9.1 rad s–1
A1 [2]
(allow use of f = 1.5 Hz to give ω = 9.4 rad s–1
)
(ii) 1. x = 2.1 cos 9.1t
2.1 and 9.1 numerical values B1
use of cos B1 [2]
2. v0 = 2.1 × 10–2
× 9.1 (allow ecf of value of x0 from (ii)1.)
v0 = 0.19 m s–1
B1
v = v0 sin 9.1t (allow cos 9.1t if sin used in (ii)1.) B1 [2]
(b) energy = either ½ mv0
2
or ½ mω2
x0
2
= either ½ × 0.078 × 0.192
or ½ × 0.078 × 9.12
× (2.1 × 10–2
)2
C1
= 1.4 × 10–3
J A1 [2]
4 (a) (i) V = q / 4πε0R B1 [1]
(ii) (capacitance is) ratio of charge and potential or q/V M1
C = q/V = 4πε0R A0 [1]
(b) (i) C = 4π × 8.85 × 10–12
× 0.45 C1
= 50 pF A1 [2]
(ii) either energy = ½ CV2
or energy = ½ QV and Q = CV C1
energy of spark = ½ × 50 × 10–12
{(9.0 × 105
)2
– (3.6 × 105
)2
} C1
= 17 J A1 [3]
(creates) force per unit length of 1 N m–1
A1 [2]
(b) (i) sketch: concentric circles M1
increasing separation (must show more than 3 circles) A1
correct direction (anticlockwise, looking down) B1 [3]
(ii) B = (4π × 10–7
× 6.3) / (2π × 4.5 × 10–2
) C1
= 2.8 × 10–5
T A1 [2]
(iii) F = BIL (sinθ) C1
= 2.8 × 10–5
× 9.3 × 1
F/L = 2.6 × 10–4
Nm–1
A1 [2]
(c) force per unit length depends on product IXIY / by Newton’s third law / action and
reaction are equal and opposite M1
so same for both A1 [2]

6 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) (i) positive terminal identified (upper connection to load) B1 [1]
(ii) VP = √2 × VRMS C1
ratio = 240 √2 / 9 C1
ratio = 38 A1 [3]
(VP = VRMS / √2 gives ratio = 18.9 and scores 1/3)
(ratio = 240 / 9 = 26.7 scores 1/3)
(ratio = 9 / (240 / √2) = 0.0265 is inverted ratio and scores 1/3)
(c) (i) e.g. (output) p.d. / voltage / current does not fall to zero
e.g. range of (output) p.d. / voltage / current is reduced (any sensible answer) B1 [1]
(ii) sketch: same peak value at start of discharge M1
correct shape between one peak and the next A1 [2]
7 (a) each wavelength is associated with a discrete change in energy M1
discrete energy change / difference implies discrete levels A1 [2]
(b) (i) 1. arrow from –0.54 eV to –0.85 eV, labelled L B1 [1]
2. arrow from –0.54 eV to –3.4 eV , labelled S B1 [1]
(two correct arrows, but only one label – allow 2 marks)
(two correct arrows, but no labels – allow 1 mark)
(ii) E = hc / λ C1
(3.4 – 0.54) × 1.6 × 10–19
= (6.63 × 10–34
× 3.0 × 108
) / λ C1
λ = 4.35 × 10–7
m A1 [3]
(c) –1.50 → –3.4 = 1.9 eV
–0.85 → –3.4 = 2.55 eV (allow 2.6 eV)
–0.54 → –3.4 = 2.86 eV (allow 2.9 eV)
3 correct, 2 marks with –1 mark for each additional energy
2 correct, 1 mark but no marks if any additional energy differences B2 [2]

8 (a) energy is given out / released on formation of the α-particle (or reverse argument) M1
either E = mc2
so mass is less
or reference to mass-energy equivalence A1 [2]
(b) (i) mass change = 18.00567u – 18.00641 u C1
= 7.4 × 10–4
u (sign not required) A1 [2]
(ii) energy = c2
∆m
= (3.0 × 108
)2
× 7.4 × 10–4
× 1.66 × 10–27
C1
= 1.1 × 10–13
J A1 [2]
(allow use of u = 1.67 × 10–27
kg)
(allow method based on 1u equivalent to 930 MeV to 933 MeV)
(iii) either mass of products greater than mass of reactants M1
this mass/energy provided as kinetic energy of the helium-4 nucleus A1
or both nuclei positively charged (M1)
energy required to overcome electrostatic repulsion (A1) [2]

Section B
9 (a) 30 litres → 54 litres (allow ± 4 litres on both limits) A1 [1]
(b) (i) only 0.1 V change in reading for 10 litre consumption (or similar numbers) B1
above about 60 litres gradient is small compared to the gradient at about 40 litres
B1 [2]
(ii) voltmeter reading (nearly) zero when fuel is left C1
voltmeter reads only about 0.1 V when 10 litres of fuel left in tank A1 [2]
(“voltmeter reads zero when about 4 litres of fuel left in tank” scores 2 marks)
10 (a) product of density and speed of sound / wave M1
(density of medium and) speed of sound / wave in medium A1 [2]
(b) if (Z1 – Z2) is small, mostly transmission M1
if (Z1 – Z2) is large, mostly reflection M1
(if ‘mostly’ not stated allow 1/2 marks for these first two marks)
either reflection / transmission also depends on (Z1 + Z2)
or intensity reflection coefficient = (Z1 – Z2)2
/ (Z1 + Z2)2
A1 [3]
(c) e.g. smaller structures can be distinguished B1
because better resolution at shorter wavelength / higher frequency B1 [2]
11 (a) changing voltage changes energy / speed of electrons M1
changing electron energy changes maximum X-ray photon energy A1 [2]
(b) (i) 1. loss of power / energy / intensity B1 [1]
2. intensity changes when beam not parallel C1
decreases when beam is divergent A1 [2]
(ii) ratio = (exp {–2.9 × 2.5}) / (exp {–0.95 × 6.0}) C1
= 0.21 (min. 2 sig. fig.) A1 [2]
(values of both lengths incorrect by factor of 10–2
to give ratio of 0.985 scores 1 mark)

12 (a) takes all the simultaneous digits for one number B1
and ‘sends’ them one after another (along the transmission line) B1 [2]
(b) (i) 0111 A1 [1]
(ii) 0110 A1 [1]
(c) levels shown
t 0 0.2 0.4 0.6 0.8 1.0 1.2
0 8 7 15 6 5 8
(–1 for each error or omission) A2
correct basic shape of graph i.e. series of steps M1
with levels staying constant during correct time intervals A1 [4]
(vertical lines in steps do not need to be shown)
(d) increasing number of bits reduces step height M1
increasing sampling frequency reduces step depth / width M1
reproduction of signal is more exact A1 [3]

9702 PHYSICS

Section A
1 (a) region of space area / volume B1
where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1
either reference to point masses or separation >> ‘size’ of masses A1 [3]
(ii) field strength = GM / x2
or field strength ∝ 1 / x2 C1
ratio = (7.78 × 108
)2
/ (1.5 × 108
)2
C1
= 27 A1 [3]
(c) (i) either centripetal force = mRω2
and ω = 2π / T
or centripetal force = mv2
/ R and v = 2πR /T B1
gravitational force provides the centripetal force B1
either GMm / R2
= mRω2
or GMm / R2
= mv2
/ R M1
M = 4π2
R3
/ GT2
A0 [3]
(allow working to be given in terms of acceleration)
(ii) M = {4π2
× (1.5 × 1011
)3
} / {6.67 × 10–11
× (3.16 × 107
)2
} C1
= 2.0 × 1030
kg A1 [2]
2 (a) obeys the equation pV = constant × T or pV = nRT M1
p, V and T explained A1
at all values of p, V and T/fixed mass/n is constant A1 [3]
(b) (i) 3.4 × 105
× 2.5 × 103
× 10–6
= n × 8.31 × 300 M1
n = 0.34mol A0 [1]
(ii) for total mass/amount of gas
3.9 × 105
× (2.5 + 1.6) × 103
× 10–6
= (0.34 + 0.20) × 8.31 × T C1
T = 360K A1 [2]
(c) when tap opened
gas passed (from cylinder B) to cylinder A B1
work done on gas in cylinder A (and no heating) M1
so internal energy and hence temperature increase A1 [3]

3 (a) (i) 1. amplitude = 1.7cm A1 [1]
2. period = 0.36cm C1
frequency = 1/0.36
frequency = 2.8Hz A1 [2]
(ii) a = (–)ω2
x and ω = 2π/T C1
acceleration = (2π/0.36)2
× 1.7 × 10–2
M1
= 5.2ms–2
A0 [2]
(b) graph: straight line, through origin, with negative gradient M1
from (–1.7 × 10–2
, 5.2) to (1.7 × 10–2
, –5.2) A1 [2]
(if scale not reasonable, do not allow second mark)
(c) either kinetic energy = ½mω2
(x0
2
– x2
)
or potential energy = ½mω2
x2
and potential energy = kinetic energy B1
½mω2
(x0 – x2
) = ½ × ½mω2
x0
2
or ½mω2
x2
= ½ × ½mω2
x0
2
C1
x0
2
= 2x2
x = x0 / √2 = 1.7 / √2
= 1.2cm A1 [3]
4 (a) work done moving unit positive charge M1
from infinity (to the point) A1 [2]
(b) (gain in) kinetic energy = change in potential energy B1
½mv2
= qV leading to v = (2Vq/m)½
B1 [2]
(c) either (2.5 × 105
)2
= 2 × V × 9.58 × 107
C1
V = 330V M1
this is less than 470V and so ‘no’ A1 [3]
or v = (2 × 470 × 9.58 × 107
) (C1)
v = 3.0 × 105
ms–1
(M1)
this is greater than 2.5 × 105
ms–1
or (2.5 × 105
)2
= 2 × 470 × (q/m) (C1)
(q/m) = 6.6 × 107
Ckg–1
(M1)
this is less than 9.58 × 107
Ckg–1

(creates) force per unit length of 1Nm–1
A1 [2]
(b) (i) flux density = 4π × 10–7
× 1.5 × 103
× 3.5 C1
= 6.6 × 10–3
T A1 [2]
(ii) flux linkage = 6.6 × 10–3
× 28 × 10–4
× 160 C1
= 3.0 × 10–3
Wb A1 [2]
(c) (i) (induced) e.m.f. proportional to rate of M1
change of (magnetic) flux (linkage) A1 [2]
(ii) e.m.f. = (2 × 3.0 × 10–3
) / 0.80 C1
= 7.4 × 10–3
V A1 [2]
6 (a) (i) to reduce power loss in the core B1
due to eddy currents/induced currents B1 [2]
(ii) either no power loss in transformer
or input power = output power B1 [1]
(b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1
peak voltage across load = √2 × 243
= 340V A1 [2]
or peak voltage across primary coil = 9.0 × √2 (C1)
peak voltage across load = 12.7 × (8100/300)
= 340V (A1)
7 (a) (i) lowest frequency of e.m. radiation M1
giving rise to emission of electrons (from the surface) A1 [2]
(ii) E = hf C1
threshold frequency = (9.0 × 10–19
) / (6.63 × 10–34
)
= 1.4 × 1015
Hz A1 [2]
(b) either 300nm ≡ 10 × 1015
Hz (and 600nm ≡ 5.0 × 1014
Hz)
or 300nm ≡ 6.6 × 10–19
J (and 600nm ≡ 3.3 × 10–19
J)
or zinc λ0 = 340nm, platinum λ0 = 220nm (and sodium λ0 = 520nm) M1
emission from sodium and zinc A1 [2]
(c) each photon has larger energy M1
fewer photons per unit time M1
fewer electrons emitted per unit time A1 [3]

8 (a) two (light) nuclei combine M1
to form a more massive nucleus A1 [2]
(b) (i) ∆m = (2.01410u + 1.00728u) – 3.01605u
= 5.33 × 10–3
u C1
energy = c2
× ∆m C1
= 5.33 × 10–3
× 1.66 × 10–27
× (3.00 × 108
)2
= 8.0 × 10–13
J A1 [3]
(ii) speed/kinetic energy of proton and deuterium must be very large B1
so that the nuclei can overcome electrostatic repulsion B1 [2]
Section B
9 (a) (i) light-dependent resistor/LDR B1 [1]
(ii) strain gauge B1 [1]
(iii) quartz/piezo-electric crystal B1 [1]
(b) (i) resistance of thermistor decreases as temperature increses M1
etiher VOUT = V × R / (R + RT)
or current increases and VOUT = I R A1
VOUT increases A1 [3]
(ii) either change in RT with temperature is non-linear
or VOUT is not proportional to RT/ change in VOUT with RT is non-linear M1
so change is non-linear A1 [2]
10 (a) sharpness: how well the edges (of structures) are defined B1
contrast: difference in (degree of) blackening between structures B1 [2]
(b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid
large penumbra on shadow/large area anode/wide beam
large pixel size
(any two sensible suggestions, 1 each) B2 [2]
(c) (i) I = I0e–µx
C1
ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1
= (4.65 × 10–5
) / (5.00 × 10–4
)
= 0.093 A1 [3]
(ii) either large difference (in intensities)
or ratio much less than 1.0 M1
so good contrast A1 [2]
(answer given in (c)(ii) must be consistent with ratio given in (c)(i))

11 (a) (i) amplitude of the carrier wave varies M1
(in synchrony) with the displacement of the information signal A1 [2]
(ii) e.g. more than one radio station can operate in same region/less interference
enables shorter aerial
increased range/less power required/less attenuation
less distortion
(any two sensible answers, 1 each) B2 [2]
(b) (i) frequency = 909kHz C1
wavelength = (3.0 × 108
) / (909 × 103
)
= 330m A1 [2]
(ii) bandwidth = 18kHz A1 [1]
(iii) frequency = 9000Hz A1 [1]
12 (a) for received signal, 28 = 10lg(P / {0.36 × 10–6
}) C1
P = 2.3 × 10–4
W A1 [2]
(b) loss in fibre = 10lg({9.8 × 10–3
} / {2.27 × 10–4
}) C1
= 16dB A1 [2]
(c) attenuation per unit length = 16 / 85
= 0.19dB km–1
A1 [1]

9702 PHYSICS
9702/51 Paper 5 (Planning, Analysis and Evaluation),
maximum raw mark 30

1 Planning (15 marks)
Defining the problem (3 marks)
P h is the independent variable or vary h. [1]
P Q is the dependent variable or measure Q (allow t). [1]
P Keep l constant. [1]
Methods of data collection (5 marks)
M Labelled diagram of apparatus: including labelled measuring cylinder/calibrated beaker to
receive water. (Measurement may be credited in the text.) [1]
M Vary position of vertical/larger tube. [1]
M Measure h and l with a rule/caliper. [1]
M Measure d with a travelling microscope or vernier calipers. [1]
M Measure t with stopwatch. [1]
Method of analysis (2 marks)
A Plot a graph of Q against h. [Allow lg Q against lg h] [1]
A
gradient
2 4
×
=
l
gdπρ
η
Must include gradient and η must be subject of formula. [1]
Safety considerations (1 mark)
S Reasoned method to prevent spills, e.g. use tray/sink/cloths on floor. Reasoned method to
prevent injury when adjusting metal/glass tubes by wearing protective gloves. Mark may not
be scored from the diagram. [1]
Additional detail (4 marks)
D Relevant points might include [4]
1 Repeat experiment for same h and average
2 Method to determine the density of water including method to measure mass and volume
and equation
3 Take many readings of d and average
4 Relationship is valid if straight line passing through origin [if lg-lg graph allow straight line with
gradient = 1]
5 Method to check that tube is horizontal
6 Detail on measuring h to the centre of the horizontal tube e.g. add radius of tube
7 Keep temperature of water constant
Do not allow vague computer methods.
[Total: 15]

2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a) A1 Gradient = 1 / EW
y-intercept = 1 / E
(b) T1
V
1
/ V–1 Column heading. Allow equivalent unit.
e.g. V–1
/ V–1
or 1/V / 1/V or 1/V (V–1
)
T2 0.20 or 0.196
0.22 or 0.222
0.25 or 0.250
0.30 or 0.303
0.33 or 0.333
0.37 or 0.370
A mixture of 2 s.f. and 3 s.f. is allowed.
(c) (i) G1 Six points plotted correctly Must be less than half a small square. Ecf allowed
from table. Penalise ‘blobs’.
U1 All error bars in C plotted
correctly
Must be within half a small square. Ecf allowed
from table. Horizontal.
(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line
should pass between (0.75, 0.200) and (0.75,
0.205) and upper end of line should pass between
(3.0, 0.352) and (3.0, 0.358). Allow ecf from points
plotted incorrectly – examiner judgement.
G3 Worst acceptable straight
line.
Steepest or shallowest
possible line that passes
through all the error bars.
Line should be clearly labelled or dashed. Should
pass from top of top error bar to bottom of bottom
error bar or bottom of top error bar to top of bottom
error bar. Mark scored only if error bars are plotted.
(c) (iii) C1 Gradient of best fit line The triangle used should be at least half the length
of the drawn line. Check the read offs. Work to half
a small square. Do not penalise POT.
U2 Uncertainty in gradient Method of determining absolute uncertainty
Difference in worst gradient and gradient.
(c) (iv) C2 y-intercept Expect to see point substituted into y = mx + c
FOX does not score. Do not penalise POT.
Should be about 0.15.
U3 Uncertainty in y-intercept Difference in worst y-intercept and y-intercept.
FOX does not score. Allow ecf from (c)(iv).

(d) (i) C3 E = 1/ y-intercept and V Method required. Do not check calculation.
Allow ecf from (c)(iv).
U4 Absolute uncertainty in E
(d) (ii) C4 Between 2.00 × 10–3
F and
2.40 × 10–3
F and given to 2
or 3 s.f.
Must be in range. Allow use of mF.
(d) (iii) U5 Percentage uncertainty in W %uncertainty in E + %uncertainty in gradient
[Total: 15]
Uncertainties in Question 2
(c) (iii) Gradient [U2]
Uncertainty = gradient of line of best fit – gradient of worst acceptable line
Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(c) (iv) y-intercept [U3]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (i) [U4]
Absolute uncertainty = max E – E = E – min E =
2
minmax EE −
Absolute uncertainty = E
c
c
×
∆
(d) (iii) [U5]
Percentage uncertainty = 100×
W
W∆
∆W = max W – W = W – min W =
2
minmax WW −
max W =
mE minmin
1
×
min W=
mE maxmax
1
×
Percentage uncertainty = 100×





+
E
E
m
m ∆∆
= 100×





+
c
c
m
m ∆∆

9702 PHYSICS
maximum raw mark 30

P f is the independent variable or vary f. [1]
P I0 is the dependent variable or measure I0. [1]
P Keep V0 constant. [1]
M Labelled/ workable circuit diagram of apparatus: including coil connected to power
supply/signal generator. [1]
M Variable frequency ac power supply or signal generator (method of varying frequency). [1]
M Measure current using ammeter or p.d. across resistor. [1]
M Measure V across power supply or across coil e.g. voltmeter or c.r.o. explained. [1]
M Measure f using oscilloscope/read off signal generator. [1]
A Plot a graph of 1/I0
2
against f2
or V0
2
/I0
2
against f2
or equivalent.
Do not allow log-log graph. [1]
A gradient
24
gradient 0
2
2
0
π
=
π
=
VV
L or 2
4
gradient
π
=L [1]
S Reasoned method to prevent overheating of coil or burns from coil. [1]
1 Use lower frequencies to produce larger currents
2 Keep resistance of circuit/coil constant
3 Additional detail on measuring T using timebase
4 f = 1 / period
5 Additional detail on measuring V0 using y-gain
6 Relationship is valid if straight line, provided plotted graph is correct
7 Relationship is valid if straight line not passing through origin, provided plotted graph is
correct (any quoted expression must be correct)
8 Detail on changing r.m.s. to peak
Do not allow vague computer methods. Ignore reference to iron core / other magnetic fields.
[Total: 15]

(a) A1
Gradient =
E
b4
(b) T1 a2
/ 10–4
m2
Column heading. Allow equivalent unit.
Do not award if 10–4
omitted.
T2 44 or 43.6
52 or 51.8
61 or 60.8
71 or 70.6
79 or 79.2
88 or 88.4
U1 1.3 increasing to 1.9 Allow 1 increasing to 2.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Ecf
allowed from table. Penalise ‘blobs’.
U2 All error bars in a2
plotted
correctly
Must be within half a small square. Ecf
allowed from table.
(c) (ii) G2 Line of best fit If points are plotted correctly then lower end
of line should pass between (1000, 43) and
(1000,44.5) and upper end of line should
pass between (2000, 87) and (2000, 89).
Allow ecf from points plotted incorrectly –
examiner judgement.
G3 Worst acceptable straight line.
Steepest or shallowest possible
line that passes through all the
error bars.
Line should be clearly labelled or dashed.
Should pass from top of top error bar to
bottom of bottom error bar or bottom of top
error bar to top of bottom error bar. Mark
scored only if error bars are plotted.
(c) (iii) C1 Gradient of best fit line The triangle used should be at least half the
length of the drawn line. Check the read offs.
Work to half a small square. Do not penalise
POT.

(d) (i) C2 E = 4b/gradient Method required. Do not check calculation.
Should be about 36 000. Do not penalise
POT error.
C3 Vm–1
or NC–1
Penalise POT error.
(d) (ii) U4 Percentage uncertainty in E Must be greater than 2.5%.
(e) C4 V = 540 to 580 V and given to 2
or 3 s.f.
Working must be correct; check for
inconsistent units. Must be in stated range.
U5 Absolute uncertainty in V
[Total: 15]
(d) [U4]
∆
E
E
max E =
m
b
min
max
min E =
m
b
max
min
Percentage uncertainty = 2.5100100 +×




 ∆
=×




 ∆
+
∆
m
m
b
b
m
m
(e) [U5]
Absolute uncertainty = max V – V = V – min V =
2
minmax VV −
max V =
b
aE
min4
maxmax 2
×
=
m
a
min
max 2
min V =
b
aE
max4
minmin 2
×
=
m
a
max
min 2
Absolute uncertainty = V
b
b
E
E
a
a
V
m
m
a
a





 ∆
+
∆
+
∆
=




 ∆
+
∆ 22

9702 PHYSICS
maximum raw mark 30
12

P h is the independent variable or vary h. [1]
P Q is the dependent variable or measure Q (allow t). [1]
P Keep l constant. [1]
M Labelled diagram of apparatus: including labelled measuring cylinder/calibrated beaker to
receive water. (Measurement may be credited in the text.) [1]
M Vary position of vertical/larger tube. [1]
M Measure h and l with a rule/caliper. [1]
M Measure d with a travelling microscope or vernier calipers. [1]
M Measure t with stopwatch. [1]
A Plot a graph of Q against h. [Allow lg Q against lg h] [1]
A
gradient
2 4
×
=
l
gdπρ
η
Must include gradient and η must be subject of formula. [1]
S Reasoned method to prevent spills, e.g. use tray/sink/cloths on floor. Reasoned method to
prevent injury when adjusting metal/glass tubes by wearing protective gloves. Mark may not
be scored from the diagram. [1]
1 Repeat experiment for same h and average
2 Method to determine the density of water including method to measure mass and volume
and equation
3 Take many readings of d and average
4 Relationship is valid if straight line passing through origin [if lg-lg graph allow straight line with
gradient = 1]
5 Method to check that tube is horizontal
6 Detail on measuring h to the centre of the horizontal tube e.g. add radius of tube
7 Keep temperature of water constant
Do not allow vague computer methods.
[Total: 15]

(a) A1 Gradient = 1 / EW
y-intercept = 1 / E
(b) T1
V
1
/ V–1 Column heading. Allow equivalent unit.
e.g. V–1
/ V–1
or 1/V / 1/V or 1/V (V–1
)
T2 0.20 or 0.196
0.22 or 0.222
0.25 or 0.250
0.30 or 0.303
0.33 or 0.333
0.37 or 0.370
(c) (i) G1 Six points plotted correctly Must be less than half a small square. Ecf allowed
from table. Penalise ‘blobs’.
U1 All error bars in C plotted
correctly
Must be within half a small square. Ecf allowed
from table. Horizontal.
(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line
should pass between (0.75, 0.200) and (0.75,
0.205) and upper end of line should pass between
(3.0, 0.352) and (3.0, 0.358). Allow ecf from points
plotted incorrectly – examiner judgement.
G3 Worst acceptable straight
line.
Steepest or shallowest
possible line that passes
through all the error bars.
Line should be clearly labelled or dashed. Should
pass from top of top error bar to bottom of bottom
error bar or bottom of top error bar to top of bottom
error bar. Mark scored only if error bars are plotted.
(c) (iii) C1 Gradient of best fit line The triangle used should be at least half the length
of the drawn line. Check the read offs. Work to half
a small square. Do not penalise POT.
(c) (iv) C2 y-intercept Expect to see point substituted into y = mx + c
FOX does not score. Do not penalise POT.
Should be about 0.15.
U3 Uncertainty in y-intercept Difference in worst y-intercept and y-intercept.
FOX does not score. Allow ecf from (c)(iv).

(d) (i) C3 E = 1/ y-intercept and V Method required. Do not check calculation.
Allow ecf from (c)(iv).
U4 Absolute uncertainty in E
(d) (ii) C4 Between 2.00 × 10–3
F and
2.40 × 10–3
F and given to 2
or 3 s.f.
Must be in range. Allow use of mF.
(d) (iii) U5 Percentage uncertainty in W %uncertainty in E + %uncertainty in gradient
[Total: 15]
(c) (iv) y-intercept [U3]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (i) [U4]
Absolute uncertainty = max E – E = E – min E =
2
minmax EE −
Absolute uncertainty = E
c
c
×
∆
(d) (iii) [U5]
W
W∆
∆W = max W – W = W – min W =
2
minmax WW −
max W =
mE minmin
1
×
min W=
mE maxmax
1
×
Percentage uncertainty = 100×





+
E
E
m
m ∆∆
= 100×





+
c
c
m
m ∆∆

9702 s13 ms_all

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