9702 w16 ms_all

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice October/November 2016
MARK SCHEME
Maximum Mark: 40
Published
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 will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the October/November 2016 series for most
Cambridge IGCSE®
, Cambridge International A and AS Level components and some Cambridge O Level
components.

Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 11
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B

PHYSICS 9702/12
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
Question
Number
Key
Question
Number
Key
1 B 21 D
2 D 22 B
3 D 23 A
4 C 24 B
5 A 25 A
6 D 26 B
7 B 27 D
8 B 28 C
9 D 29 C
10 C 30 B
11 A 31 A
12 D 32 C
13 B 33 D
14 D 34 B
15 C 35 B
16 C 36 B
17 C 37 C
18 B 38 A
19 C 39 C
20 C 40 A

PHYSICS 9702/13
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B

PHYSICS 9702/21
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) (density =) mass/volume B1 [1]
(b) (i) d = [(6 × 7.5)/(π × 8100)]1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)/3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)/3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10–2
/3.7 × 107
= 1.6 × 10–9
s (1.59 × 10–9
s) A1 [1]
(ii) E = V/d C1
= 2500 / 4.0 × 10–2
= 6.3 × 104
NC–1
(6.25 × 104
or 62500NC–1
) A1 [2]
(iii) a = Eq/m or F = ma and F = Eq C1
= (6.3 × 104
× 1.60 × 10–19
)/9.11 × 10–31
= 1.1 × 1016
ms–2
A1 [2]
(iv) s = ut + ½at2
= ½ × 1.1 × 1016
× (1.6 × 10–9
)2
C1
= 1.4 × 10–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) vX–t graph: horizontal line at a non-zero value of vX B1
vY–t graph: straight line through the origin with positive gradient B1 [2]

© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F/x or k = gradient C1
k = 600Nm–1
A1 [2]
(ii) (W =) ½kx2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)2
= 0.48J or (W =) 0.5 × 24 × 0.040 = 0.48J A1 [2]
(iii) 1. (EK =) ½mv2
C1
= ½ × 0.025 × 6.02
= 0.45J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)J] C1
average resistive force = 0.030/0.040 C1
= 0.75N A1 [3]
(iv) efficiency = [useful energy out/total energy in] (×100) C1
= [0.45/0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1/(6.25 × 10–4
) or 1/(2.5 × 250 × 10–6
) = 1600Hz A1 [2]
(c) (i) for maximum frequency: fo = fsv/(v – vs)
1640 = (1600 × 330) / (330 – vs) C1
vs = 8(.0)ms–1
(8.049ms–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)

© UCLES 2016
5 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) nλ = d sinθ C1
substitution of θ = 90° or sinθ = 1 C1
4 × 500 × 10–9
= d × sin90°
line spacing = 2.0 × 10–6
m A1 [3]
(c) wavelength of red light is longer (than 500nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms)othertoelectrical(fromed)(transformenergyordonework
B1 [1]
(b) (i) 1. V = IR or E = IR C1
I = 14/6.0
= 2.3 (2.33)A A1 [2]
2. total resistance of parallel resistors = 8.0Ω C1
current = 14/(6.0 + 8.0)
= 1.0A A1 [2]
(ii) P = EI (allow P = VI) or P = V2
/R or P = I2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (142
/ 6.0) – (142
/14)
or (2.332
× 6.0) – (1.02
× 14)
= 19W (18W if 2.3A used) A1 [2]
(c) I = Anvq
ratio = (0.50n/n) × (1.8A/A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]

© UCLES 2016
7 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i) proton: up, up, down/uud B1
neutron: up, down, down/udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down/6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]

PHYSICS 9702/22
Paper 2 AS Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) (i) force/area (normal to the force) B1 [1]
(ii) (p = F/A so) units: kgms–2
/m2
= kg m–1
s–2
A1 [1]
allow use of other correct equations:
e.g. (∆p = ρg∆h so) kg m–3
ms–2
m = kg m–1
s–2
e.g. (p = W/∆V so) kg ms–2
m/m3
= kg m–1
s–2
(b) units for m: kg, t: s and ρ: kgm–3
C1
units of C: kg/s (kg m–3
kg m–1
s–2
)1/2
or
units of C2
: kg2
/s2
kg m–3
kg m–1
s–2
C1
units of C: m2
A1 [3]
2 (a) ∆E = mg∆h C1
= 0.030 × 9.81 × (–)0.31
= (–)0.091 J A1 [2]
(b) E = ½mv 2
C1
(initial) E = ½ × 0.030 × 1.32
(= 0.0254) C1
0.5 × 0.030 × v 2
= (0.5 × 0.030 × 1.32
) + (0.030 × 9.81 × 0.31) so v = 2.8ms–1
or
0.5 × 0.030 × v 2
= (0.0254) + (0.091) so v = 2.8ms–1
A1 [3]
(c) (i) 0.096 = 0.030(v + 2.8) C1
v = 0.40ms–1
A1 [2]
(ii) F = ∆p/(∆)t or F = ma
= 0.096/20×10–3
or 0.030 (0.40 + 2.8)/20×10–3
C1
= 4.8 N A1 [2]
(d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic
or
(relative) speed of approach (of ball and wall) not equal to/greater than (relative)
speed of separation, so inelastic. B1 [1]
(e) force = work done/distance moved
= (0.091 – 0.076)/0.60 C1
= 0.025N A1 [2]

© UCLES 2016
3 (a) resultant force (in any direction) is zero B1
resultant moment/torque (about any point) is zero B1 [2]
(b) (i) force = 33 sin 52° or 33 cos 38°
= 26N A1 [1]
(ii) 26 × 0.30 or W × 0.20 or 12 × 0.40 C1
26 × 0.30 = (W × 0.20) + (12 × 0.40) C1
W = 15N A1 [3]
(c) (i) E = ∆σ /∆ε or E = σ /ε C1
∆σ = 2.0 × 1011
× 7.5 × 10–4
= 1.5 × 108
Pa A1 [2]
(ii) ∆σ = ∆F/A or σ = F/A C1
A = 78/1.5×108
(= 5.2 × 10–7
m2
) C1
5.2 × 10–7
= πd 2
/4
d = 8.1 × 10–4
m A1 [3]
(b) (i) waves (from slits) overlap (at point X) B1
path difference (from slits to X) is zero/
phase difference (between the two waves) is zero
(so constructive interference gives bright fringe) B1 [2]
(ii) difference in distances = λ/2 = 580/2
= 290nm A1 [1]
(iii) λ = ax/D C1
D = [0.41 × 10–3
× (2 × 2.0 × 10–3
)]/580 × 10–9
C1
= 2.8m A1 [3]
(iv) same separation/fringe width/number of fringes
bright fringe(s)/central bright fringe/(fringe at) X less bright
dark fringe(s)/(fringe at) Y/(fringe at) Z brighter
contrast between fringes decreases
Any two of the above four points, 1 mark each B2 [2]

© UCLES 2016
5 (a) total/sum of electromotive forcesore.m.f.s
= total/sum of potential differences or p.d.s M1
around a loop/(closed) circuit A1 [2]
(b) (i) (current in battery =) current in A + current in B or IA + IB C1
(I =) 0.14 + 0.26 = 0.40A A1 [2]
(ii) E = V + Ir
6.8 = 6.0 + 0.40r or 6.8 = 0.40(15 + r) C1
r = 2.0 Ω A1 [2]
(iii) R = V/I C1
ratio (= RA /RB) = (6.0/0.14)/(6.0/0.26)
= 42.9/23.1 or 0.26/0.14
= 1.9 (1.86) A1 [2]
(iv) 1. P = EI or VI or P = I 2
R or P = V 2
/R C1
= 6.8 × 0.40 = 0.402
× 17 = 6.82
/17
= 2.7W (2.72W) A1 [2]
2. output power = VI
= 6.0 × 0.40 (= 2.40W) C1
efficiency = (6.0 × 0.40)/(6.8 × 0.40) = 2.40/2.72
= 0.88 or 88% (allow 0.89 or 89%) A1 [2]

© UCLES 2016
or
or
(b) (i) e0
1
)(+
or )+(
β0
1 B1
)(e
0
0ν B1 [2]
(ii) weak (nuclear force/interaction) B1 [1]
(iii) • mass-energy
• momentum
• proton number
• nucleon number
• charge
Any three of the above quantities, 1 mark each B3 [3]
(c) (quark structure of proton is) up, up, down or uud B1
up/u (quark charge) is (+)⅔(e), down/d (quark charge) is –⅓(e) C1
⅔e + ⅔e – ⅓e = (+)e A1 [3]

PHYSICS 9702/23
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) (density =) mass/volume B1 [1]
(b) (i) d = [(6 × 7.5)/(π × 8100)]1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)/3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)/3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10–2
/3.7 × 107
= 1.6 × 10–9
s (1.59 × 10–9
s) A1 [1]
(ii) E = V/d C1
= 2500 / 4.0 × 10–2
= 6.3 × 104
NC–1
(6.25 × 104
or 62500NC–1
) A1 [2]
(iii) a = Eq/m or F = ma and F = Eq C1
= (6.3 × 104
× 1.60 × 10–19
)/9.11 × 10–31
= 1.1 × 1016
ms–2
A1 [2]
(iv) s = ut + ½at2
= ½ × 1.1 × 1016
× (1.6 × 10–9
)2
C1
= 1.4 × 10–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) vX–t graph: horizontal line at a non-zero value of vX B1
vY–t graph: straight line through the origin with positive gradient B1 [2]

© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F/x or k = gradient C1
k = 600Nm–1
A1 [2]
(ii) (W =) ½kx2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)2
= 0.48J or (W =) 0.5 × 24 × 0.040 = 0.48J A1 [2]
(iii) 1. (EK =) ½mv2
C1
= ½ × 0.025 × 6.02
= 0.45J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)J] C1
average resistive force = 0.030/0.040 C1
= 0.75N A1 [3]
(iv) efficiency = [useful energy out/total energy in] (×100) C1
= [0.45/0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1/(6.25 × 10–4
) or 1/(2.5 × 250 × 10–6
) = 1600Hz A1 [2]
(c) (i) for maximum frequency: fo = fsv/(v – vs)
1640 = (1600 × 330) / (330 – vs) C1
vs = 8(.0)ms–1
(8.049ms–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)

© UCLES 2016
(b) nλ = d sinθ C1
substitution of θ = 90° or sinθ = 1 C1
4 × 500 × 10–9
= d × sin90°
line spacing = 2.0 × 10–6
m A1 [3]
(c) wavelength of red light is longer (than 500nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms)othertoelectrical(fromed)(transformenergyordonework
B1 [1]
(b) (i) 1. V = IR or E = IR C1
I = 14/6.0
= 2.3 (2.33)A A1 [2]
2. total resistance of parallel resistors = 8.0Ω C1
current = 14/(6.0 + 8.0)
= 1.0A A1 [2]
(ii) P = EI (allow P = VI) or P = V2
/R or P = I2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (142
/ 6.0) – (142
/14)
or (2.332
× 6.0) – (1.02
× 14)
= 19W (18W if 2.3A used) A1 [2]
(c) I = Anvq
ratio = (0.50n/n) × (1.8A/A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]

© UCLES 2016
or
or
(b) (i) proton: up, up, down/uud B1
neutron: up, down, down/udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down/6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 October/November 2016
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (b) Value for T in the range 0.60s to 0.80s with unit. [1]
Evidence of repeat timings (at least two recordings of nT where n ⩾ 5). [1]
(c) Correct calculation of k. [1]
Value of k must be given to the same number of s.f. as (or one more than) the s.f.
in the values of the raw times. [1]
(e) Six sets of readings of (h – h1) and T (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
∆(h – h1) ⩾ 30.0cm.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. T/s or T (s).
Consistency: [1]
All values of (h – h1) must be given to the nearest mm.
(f) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
are not allowed.
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 must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
Quality: [1]
All points in the table must be plotted on the grid for this mark to be awarded.
All points must be within 0.01s on the y-axis of 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. There must be at least five points left after the anomalous
point is disregarded.
Line must not be kinked or thicker than half a small square.

© UCLES 2016
(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the
drawn line.
The method of calculation must be correct. Do not allow ∆x/∆y.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check 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:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(g) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Do not allow fractions.
Units for P (e.g. sm–1
or scm–1
or smm–1
) and Q (s) correct. [1]
2 (a) (i) Value for D with unit in the range 0.14mm to 0.16mm. [1]
(ii) Percentage uncertainty in D based on an absolute uncertainty of 0.01mm.
Correct method of calculation to obtain percentage uncertainty. [1]
(c) (iii) Value of I in range 10mA ⩽ I ⩽ 200mA with unit (collected without help from
Supervisor). [1]
(iv) Value of V in range 0.4V ⩽ V ⩽ 1.0V with unit (collected without help from
Supervisor). [1]
(d) (i) Value of d > D and d < 1mm. [1]
(ii) Correct calculation of G. [1]
(iii) Justification for s.f. in G linked to s.f. in D and d. [1]
(f) (i) Second value of V. [1]
Quality: second value of V less than first value of V. [1]
(ii) Second value of d. [1]

© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
numerical criterion. [1]
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw a
conclusion
Take many readings (for
different diameters) and plot a
graph/
take more readings and
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Difficult to measure diameter(s)
with reason e.g. awkward placing
micrometer round wire/
only one direction to measure
diameter
Provide separate lengths of wire
C Meter readings changed in a
particular direction over time/
repeat readings of I or V were
often different/
contact resistance varies
Use power supply/
allow reading to reach steady
value/
method of cleaning crocodile
clips or wires
D Wire is very thin introducing a large
percentage error in the diameter
Use thicker wire/
use digital micrometer
E Rheostat movement not precise
enough – overshot I reading
Method to ensure exact current
easier to produce e.g. use of
screw thread adjustment

PHYSICS 9702/33
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) (v) Value of I in range 20 mA ⩽ I ⩽ 200mA with unit. [1]
(c) Six sets of readings of R and y (with correct trend and without help from Supervisor)
scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Values of R must include 33Ω.
Each column heading must contain a quantity and a unit where appropriate.
convention, e.g. 1/y/m–1
or 1/y (m–1
).
Consistency: [1]
All values of raw y must be given to the nearest mm.
Significant figures: [1]
Every value of 1/R must be given to 2 or 3 significant figures.
Calculation: [1]
1/y calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
are not allowed.
All observations must be plotted on the grid.
Plotted points must be accurate to half a small square.
Quality: [1]
All points in the table must be plotted on the grid for this mark to be awarded.
All points must be within 0.001cm–1
(0.1m–1
) (to scale) on the y-axis of a
straight line.
the candidate. There must be at least five points left after the anomalous
point is disregarded.

© UCLES 2016
(iii) Gradient: [1]
drawn line.
directions.
y-intercept: [1]
Either:
Or:
small square.
(e) Value of P = –candidate’s gradient and value of Q =candidate’s intercept.
Do not allow fractions. [1]
Units for P (e.g. Ωm–1
) and Q (e.g. m–1
) correct. [1]
(f) (ii) Value of X in the range 5.0Ω to 20.0Ω from correct calculation. [1]
2 (b) (iii) All raw readings stated to at least 0.1s. Value for T with unit in the range
0.50s to 0.90s. [1]
Evidence of repeats (at least two recordings of nT where n ⩾ 5). [1]
(c) (i) Correct calculation of l. [1]
(ii) Justification of s.f. in l linked to s.f. in time readings (e.g. time, raw time, nT). [1]
(d) (ii) Value of d (l ± 0.050m). [1]
(iii) Absolute uncertainty in d in range 2mm–8mm.
If repeated readings have been taken, then the uncertainty can be half the
range (but not zero) if the working is clearly shown.
(e) (ii) Value for t in the range 2.0s–8.0s. [1]
(f) Second value of T. [1]
Second value of t. [1]
Quality: Second value of t > first value of t. [1]

© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
A Two readings not enough to draw a
conclusion
Take many readings (for
different masses) and plot a
graph/
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
calculate average k
B Difficult to set d with reason e.g.
knot takes some string/knot or
weight slips
Improved method to set d e.g.
trial and error/glue knots to stop
slipping
C Difficult to measure d with reason
e.g. parallax error/cannot get ruler
close/hands move holding ruler in
place/difficult to judge centre of
mass(es)/rule not vertical
Clamped metre rule/
measure to top and bottom of
masses and find the average/
detailed method to check rule
vertical/
detailed method to reduce
parallax error
Eyes level
Eyes perpendicular
Reference to set square
without detail
D Difficult to measure (n)T with
reason e.g. difficult to judge
complete oscillation
Use fiducial marker at centre of
oscillation/
use a motion sensor placed
under masses
Reaction time error
Force on release
E Difficult to see when coil separation
is constant
or
Separation of coils constant for a
short time
Video with timer/
video and view frame by frame/
screen behind spring
Video timer
F Movement of spring along rod Method to fix spring to rod e.g.
Blu-Tack or groove in rod
Air conditioning/draughts
Unwanted modes of
oscillation

PHYSICS 9702/34
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (b) (ii) Value for x in range 24.0cm to 26.0cm, with unit. [1]
(iv) Value for T in range 0.30s to 1.00s. [1]
Evidence of repeat readings (at least two recordings of nT where n ⩾ 5). [1]
(c) Six sets of values for x and T (with correct trend and without help from Supervisor)
Range: [1]
xmin ⩽ 20.0cm and xmax ⩾ 30.0cm.
Each column heading must contain a quantity and an appropriate unit. The
presentation of the quantity and unit must conform to accepted scientific convention
e.g. 1/T2
(s–2
) or 1/T2
/1/s2
.
Consistency: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
Every value of 1/T2
must be given to the same number of s.f. as (or one greater than)
the number of s.f. in the corresponding times.
Calculation: [1]
Values of 1/T2
calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) are not allowed.
grid in both x and y directions
Scale markings must be no more than three large squares apart.
All observations in the table must be plotted on the grid.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
All points must be within ±1.0cm (to scale) of a straight line in the x direction.
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
Allow one anomalous plot if clearly indicated (i.e. circled or labelled) by the
candidate. There must be at least five points left after the anomalous point is
disregarded.

© UCLES 2016
(iii) Gradient: [1]
drawn line.
directions.
y-intercept: [1]
Either:
Or:
small square.
(e) Value of p = candidate’s gradient and value of q = candidate’s intercept.
Units for p (e.g. cm–1
s–2
) and q (e.g. s–2
) correct. [1]
2 (b) L in range 19.0cm to 21.0cm, with unit. [1]
(c) (iv) Values for x1 and x2 to nearest mm and x2 > x1. [1]
Evidence of repeat readings of x1 and x2. [1]
(v) Correct calculation of X. [1]
(d) Absolute uncertainty in X in range 2mm to 10mm.
If repeated readings have been taken, then absolute uncertainty can be half the
range (but not zero) if working is clearly shown.
(e) Second value for L. [1]
Second values for x1 and x2. [1]
Quality: X smaller for larger L. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Justification of s.f. in k based on the s.f. in L, x1 and x2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a stated
(g) Value for X = 50cm. [1]

© UCLES 2016
A Two readings not enough to draw
a conclusion
Take more readings and plot
graph/
obtain more k values and
compare
Two readings not
enough for accurate
results
Repeat readings
Few readings
calculate average k
B Metre rule is not parallel to
bench/horizontal
Use a second rule and measure
at both ends/
use a (spirit) level
C Difficult to move stands with
reason e.g. friction/
bench is rough/
stands tend to stick
Guide for stands (fixed to
bench)/
mount stands on rollers/
put wheels on stands/
method to reduce friction e.g.
sand bench with sandpaper
Use a smooth(er) bench
Use lubricant
D Difficulty with rule
e.g. rule skewed/
moves sideways
Use V-shaped rods/
groove in rods/
guide for ruler with some details
Falls off
E Difficult to measure x with reason
e.g parallax error/
difficult to tell point where rod
touches ruler
Scale on vertical edge of rule/
draw a line on the rod/
use a thinner rod/
replace rods with sharp edges
e.g. prisms
Large contact area
Do not allow ‘use a computer to improve the experiment’.

PHYSICS 9702/35
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) (i) Value of VF in the range 260.0cm3
to 400.0cm3
. [1]
(c) (v) All values of raw y to the nearest mm and less than 20cm. [1]
(d) (ii) Six sets of readings of V and y (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Maximum value of V ⩾ 200cm3
.
Each column heading must contain a quantity and an appropriate unit.
convention, e.g. y/m or y (cm), V/cm3
, 2VF /3/cm3
.
Consistency: [1]
All values of raw V must be given to the nearest cm3
.
(e) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) are not allowed.
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 on the grid (at least 5) for this mark to
be awarded.
All points must be within ± 0.25cm (to scale) in the y direction from a straight line.
the candidate.
There must be at least four points left after disregarding the anomalous point.

© UCLES 2016
(iii) Gradient: [1]
drawn line.
The method of calculation must be correct. Do not allow ∆x/∆y. Sign of
gradient on answer line must match graph drawn.
directions.
y-intercept: [1]
Either:
Or:
small square.
(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
Units for P (e.g. mm–2
, cm–2
, m–2
but not e.g. cm/cm3
) and Q (e.g. m) correct. [1]
(g) (i) V calculated correctly to the number of significant figures given by the candidate.
Sign of answer must be consistent with P and Q values in (f). [1]
(ii) Valid comment with a comparison of volumes e.g. V is greater than VF. [1]
2 (a) (ii) Value of L with unit in range 0.790m to 0.810m. [1]
(iii) Absolute uncertainty in L either 1mm or 2mm.
If repeated readings have been taken, then the uncertainty can be half the
range (but not zero) if the working is clearly shown.
(b) (ii) All values of raw d1 with unit to the nearest mm. [1]
(c) (ii) d2 > d1 showing scale on rule is used correctly. [1]
(iii) Correct calculation of |d2 – d1|. [1]
(iv) Correct calculation of M with consistent unit. Do not allow answers to 1 s.f. [1]
(d) Justification for s.f. in M linked to s.f. in m and L. [1]

© UCLES 2016
(e) Second value of d1. [1]
Second value of d2. [1]
Quality: Second value of |d2 – d1| < first value of |d2 – d1|. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [1]
(g) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw
a conclusion
Take more readings and plot a
graph/
obtain more k values and
compare
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings
and calculate average
k
B Difference between d2 and d1 is
small/
(d2 – d1) is small/
large % uncertainty in (d2 – d1)
Improved method to measure
(d2 – d1) values e.g. travelling
microscope/larger masses/clamp
vernier calipers above strip
Laser
Ultrasonic position
sensor
Longer strip
Change strip
Deflection of strip
C Difficult to distribute mass evenly/
gaps between masses vary/
masses do not lie in a straight
line/
masses have slots
Improved method of distribution
e.g. use marker or scale on
wooden strip/
use smaller masses to produce
the same m/
continuous strip of named material
e.g. modelling clay
Masses inaccurate
Stick masses to strip
Masses falling
Replace strip with
metre rule
Masses with no slots
D Difficult to measure d values with
reason e.g. rule not vertical/
difficult to hold second rule still
as a pointer/
parallax error/
rule not held stationary
Improved method to measure d
e.g. clamp metre rule/
place a set square on the floor
next to ruler (to ensure rule is
vertical)/
use of pointer (to metre rule)
Strip oscillates when
masses are put on it
Use a set square
Effects of wind
Reference to L
E Strip becomes permanently bent
or deformed
Method to overcome deformation
e.g. check d without masses
before and between readings/
lay weights on strip to flatten
Elastic properties of
wood vary
Use a new strip

PHYSICS 9702/36
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (b) (i) Value for x in range 45.0cm to 55.0cm. [1]
(iii) Value for I in range 500µA to 1500µA (or 0.50mA to 1.50mA), with unit. [1]
(c) Six sets of values for x and I (with correct trend and without help from Supervisor)
Range: [1]
x values must include 20cm or less and 80cm or more.
Each column heading must contain a quantity and an appropriate unit.
convention e.g. I /µA.
Consistency: [1]
All values of raw x must be given to the nearest mm.
(d) (i) Axes: [1]
are not allowed.
Scale markings must be no more than three large squares apart.
Plotting of points must be accurate to half a small square.
Quality: [1]
All points in the table (at least 5) must be plotted on the grid.
All points must be within ±20 µA (±0.02 mA) of a straight line in the y (I) direction.
Judge by balance of all points on the grid about the candidate's line (at least five
points). There must be an even distribution of points either side of the line along
the full length.
One anomalous plot is allowed if clearly indicated (i.e. circled or labelled). There
must be at least five points left after disregarding the anomalous point.
Lines must not be kinked or thicker than half a small square.

© UCLES 2016
(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the drawn
line.
directions.
y-intercept: [1]
Either:
Or:
small square.
(e) Value of S = candidate’s gradient and value of T = candidate’s intercept.
Consistent units for S (e.g. µAcm–1
) and T (e.g. µA). [1]
(f) Calculation: r calculated correctly to the s.f. given by the candidate. [1]
Significant figures: r given to 2 or 3 s.f. [1]
2 (b) (ii) x1 in range 10.0cm to 40.0cm. [1]
(c) Value of x2 < x1. [1]
(d) (i) Second value of x1. [1]
(ii) Value of x2 given to nearest mm and all other raw values of x in (b), (c) and (d)
are to the nearest mm. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Justification of the s.f. in k based on the s.f. in x1 and the s.f in x2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [1]
(f) (i) Raw values of D to nearest 0.001cm and in range 1.400cm to 2.200cm. [1]
Evidence of repeated readings for D. [1]
(ii) Absolute uncertainty in D of 0.001cm or 0.002cm.
If repeated readings have been taken, then absolute uncertainty could be
half the range (but not zero) if working is clearly shown.

© UCLES 2016
(iii) V calculated correctly. [1]
(iv) Quality: M in range 3g to 13g. [1]
(g) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings are not enough
to draw a valid conclusion
Take more readings and plot
graph/
compare k values
Two readings not enough
for accurate results
Repeat readings
Few readings
calculate average k
B Empty beaker moves on
bench
Fix beaker with Blu-Tack/tape/
glue
C Difficult to balance rule:
rule slips on pivot/
wind disturbs balance
Make groove in rule (under
50cm mark)/
other practical method e.g.
hinge/nail through rule
Blu-tack
Tape
Switch off fans
String slips on rule
D Spheres/string/tape still wet
after immersion so mass
changes
or
string/tape adds to mass of
sphere
Use waterproof string/
use wire
Dry the spheres
Waterproof tape
E Difficult to measure x with
reason, e.g. string too thick
(so it covers graduations on
rule)
Use thin(ner) string Parallax problems
F Marble not round Improved method of finding V
(e.g. liquid displacement)
Repeat readings and
average

PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm/r2
= mv2
/r or GMm/r2
= mrω2
and v = 2πr/T or ω = 2π/T M1
with algebra to T2
= 4π2
r3
/GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM/r2
= v2
/r or GM/r2
= rω2
and v = 2πr/T or ω = 2π/T (M1)
with algebra to T2
= 4π2
r3
/GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24 × 3600)2
= 4π2
r3
/(6.67 × 10–11
× 6.0 × 1024
) C1
r3
= 7.57 × 1022
r = 4.2 × 107
m A1 [2]
(c) (T/24)2
= {(2.64 × 107
)/(4.23 × 107
)}3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T2
/r3
) = 242
/(4.23 × 107
)3
(B1)
k = 7.61 × 10–21
T2
(= kr3
) = 7.61 × 10–21
× (2.64 × 107
)3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV/T = constant or pV = nRT C1
T (= 5 × 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0 × 105
× 4.0 × 10–4
= n × 8.31 × 300
or
5.0 × 105
× 4.0 × 10–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]

© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240J A1
2. same value as given in 1. (= 240J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m = ω2
M1
ω = 2πf M1
(2 × 64/0.810) = (2π × f)2
leading to f = 2.0 Hz A1 [3]
(b) v0 = ωx0 or v0 = 2πfx0
or
v = ω(x0
2
– x2
)1/2
and x = 0 C1
v0 = 2π × 2.0 × 1.6 × 10–2
= 0.20ms–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]

© UCLES 2016
(b) (i) gain/dB = 10lg(P2 /P1)
32 = 10lg[PMIN /(0.38 × 10–6
)]
or
–32 = 10lg(0.38 × 10–6
/PMIN) C1
PMIN = 6.0 × 10–4
W A1 [2]
(ii) attenuation = 10lg[(9.5 × 10–3
)/(6.02 × 10–4
)] C1
= 12dB
attenuation per unit length (= 12/58) = 0.21dBkm–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, EA = (3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] (= 10.79NC–1
) M1
at P, EB = (12 × 10–12
)/[4πε0(10 × 10–2
)2
] (= 10.79NC–1
) M1
or
(3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] – (12 × 10–12
)/[4πε0(10 × 10–2
)2
] = 0
or
(3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] = (12 × 10–12
)/[4πε0(10 × 10–2
)2
] (M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99 × 109
{(3.0 × 10–12
)/(5.0 × 10–2
) + (12 × 10–12
)/(10 × 10–2
)} C1
= 1.62V A1 [2]
(d) ½mv2
= qV
EK = ½ × 107 × 1.66 × 10–27
× v2
C1
qV = 47 × 1.60 × 10–19
× 1.62 C1
v2
= 1.37 × 108
v = 1.2 × 104
ms–1
A1 [3]

© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00/3.00 = 1.50/R C1
or
V+ = (3.00 × 4.5)/(2.00 + 3.00) = 2.7
2.7 = 4.5 × R/(R + 1.50) (C1)
resistance = 2.25kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between VOUT and earth A1 [2]
2. below 24°C, RT > 1.5kΩ or resistance of thermistor increases/high B1
V– < V+ or V– decreases/low (must not contradict initial statement) M1
VOUT is positive/+5 (V) and LED labelled as ‘pointing’ from VOUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv2
/r = Bqv B1
momentum or p or mv = Bqr B1 [2]

© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) flux linkage = BAN
= π × 10–3
× 2.8 × π × (1.6 × 10–2
)2
× 85 = 6.0 × 10–4
Wb B1 [1]
(c) e.m.f. = ∆NΦ/∆t
e.m.f. = (6.0 × 10–4
× 2)/0.30 C1
e.m.f. = 4.0mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3s, 0.6s → 1.0s, 1.6s → 2.0s B1
E = 4 mV for t = 0.3s → 0.6s (either polarity) B1
E = 2 mV for t = 1.0s → 1.6s B1
with opposite polarity B1 [4]

© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /λ
= (6.63 × 10–34
× 3.00 × 108
)/(436 × 10–9
) C1
= 4.56 × 10–19
J (4.6 × 10–19
J) A1 [2]
(c) (i) Φ = hc/λ0
λ0 = (6.63 × 10–34
× 3.00 × 108
)/(1.4 × 1.60 × 10–19
) C1
= 890nm A1 [2]
(ii) λ0 = (6.63 × 10–34
× 3.00 × 108
)/(4.5 × 1.60 × 10–19
)
= 280nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890nm > 436nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280nm < 436nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]

© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A0 exp(–λt) and either t = t½, A = ½A0 or ½A0 = A0 exp(–λt½) M1
so ln2 = λt½ (and ln2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A = λN
N = 200/(2.1 × 10–6
) C1
= 9.52 × 107
C1
mass = (9.52 × 107
× 222 × 10–3
)/(6.02 × 1023
)
or
mass = 9.52 × 107
× 222 × 1.66 × 10–27
C1
= 3.5 × 10–17
kg A1 [4]

PHYSICS 9702/42
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) force per unit mass B1 [1]
(b) (i) radius/diameter/size (of Proxima Centauri) ≪ /is much less than
4.0 × 1013
km/separation (of Sun and star)
or
(because) it is a uniform sphere B1 [1]
(ii) 1. field strength = GM/x2
= (6.67 × 10–11
× 2.5 × 1029
)/(4.0 × 1013
× 103
)2
C1
= 1.0 × 10–14
Nkg–1
A1 [2]
2. force = field strength × mass
= 1.0 × 10–14
× 2.0 × 1030
C1
or
force = GMm/x2
= (6.67 × 10–11
× 2.5 × 1029
× 2.0 × 1030
)/(4.0 × 1013
× 103
)2
(C1)
= 2.0 × 1016
N A1 [2]
(c) force (of 2 × 1016
N) would have little effect on (large) mass of Sun B1
would cause an acceleration of Sun of 1.0 × 10–14
ms–2
/very small/negligible
acceleration B1 [2]
or
many stars all around the Sun (B1)
net effect of forces/fields is zero (B1)
2 (a) (i) number of moles/amount of substance B1 [1]
(ii) kelvin temperature/absolute temperature/thermodynamic temperature B1 [1]
(b) pV = nRT
4.9 × 105
× 2.4 × 103
× 10–6
= n × 8.31 × 373 B1
n = 0.38 (mol) C1
number of molecules or N = 0.38 × 6.02 × 1023
= 2.3 × 1023
A1 [3]

© UCLES 2016
or
pV = NkT (C1)
4.9 × 105
× 2.4 × 103
× 10–6
= N × 1.38 × 10–23
× 373 (M1)
number of molecules or N = 2.3 × 1023
(A1)
(c) volume occupied by one molecule = (2.4 × 103
) / (2.3 × 1023
) C1
= 1.04 × 10–20
cm3
mean spacing = (1.04 × 10–20
)1/3
C1
= 2.2 × 10–7
cm (allow 1 s.f.) A1 [3]
(allow other dimensionally correct methods e.g. V = (4/3)πr3
)
3 (a) (sum of/total) potential energy and kinetic energy of (all) molecules/particles M1
reference to random (distribution) A1 [2]
(b) (i) no heat enters (gas)/leaves (gas)/no heating (of gas) B1
work done by gas (against atmosphere as it expands) M1
internal energy decreases A1 [3]
(ii) volume decreases so work done on ice/water B1
(allow work done negligible because ∆V small)
heating of ice (to break rigid forces/bonds) M1
internal energy increases A1 [3]
4 (a) (i) 0.225s and 0.525s A1 [1]
(ii) period or T = 0.30s and ω = 2π/T C1
ω = 2π/0.30
ω = 21rads–1
A1 [2]
(iii) speed = ωx0 or ω(x0
2
– x2
)1/2
and x = 0 C1
= 20.9 × 2.0 × 10–2
= 0.42ms–1
A1 [2]

© UCLES 2016
or
use of tangent method:
correct tangent shown on Fig. 4.2 (C1)
working e.g. ∆y/∆x leading to maximum speed in range (0.38–0.46)ms–1
(A1)
(b) sketch: reasonably shaped continuous oval/circle surrounding (0,0) B1
curve passes through (0, 0.42) and (0, –0.42) B1
curve passes through (2.0, 0) and (–2.0, 0) B1 [3]
5 (a) transducer/transmitter can be also be used as the receiver
or
transducer both transmits and receives
receives reflected pulses between the emitted pulses
(needs to be pulsed) in order to measure/determine depth(s)
(needs to be pulsed) to determine nature of boundaries
Any three of the above marking points, 1 mark each B2 [2]
(b) (i) product of speed of (ultra)sound and density (of medium) M1
reference to speed of sound in medium A1 [2]
(ii) if Z1 and Z2 are (nearly) equal, IT/I0 (nearly) equal to 1/unity/(very) little
reflection/mostly transmission B1
if Z1 ≫ Z2 or Z1 ≪ Z2 or the difference between Z1 and Z2 is (very) large,
then IT/I0 is small/zero/mostly reflection/little transmission B1 [2]
6 (a) E = 0 or EA = (–)EB (at x = 11cm) B1
QA/x2
= QB /(20 – x)2
= 112
/92
C1
QA /QB or ratio = 1.5 A1 [3]
or
E ∝ Q because r same or E = Q/4πε0r2
and r same (B1)
QA/QB = 48/32 (C1)
QA /QB or ratio = 1.5 (A1)

© UCLES 2016
(b) (i) for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V C1
q∆V = ½mv2
2 × (1.60 × 10–19
) × 0.58 = ½ × 4 × 1.66 × 10–27
× v2
C1
v2
= 5.59 × 107
v = 7.5 × 103
ms–1
A1 [3]
(ii) ∆V = 0.22 V C1
2 × (1.60 × 10–19
) × 0.22 = ½ × 4 × 1.66 × 10–27
× v2
v2
= 2.12 × 107
v = 4.6 × 103
ms–1
A1 [2]
7 (a) (i) charge/potential (difference) or charge per (unit) potential (difference) B1 [1]
(ii) (V = Q/4πε0r and C = Q/V)
for sphere, C (= Q/V) = 4πε0r C1
C = 4π × 8.85 × 10–12
× 12.5 × 10–2
= 1.4 × 10–11
F A1 [2]
(b) (i) 1/CT = 1/3.0 + 1/6.0
CT = 2.0µF A1 [1]
(ii) total charge = charge on 3.0µF capacitor = 2.0 (µ) × 9.0 = 18 (µC) C1
potential difference = Q/C = 18 (µ)C/3.0 (µ)F = 6.0V A1 [2]
or
argument based on equal charges:
3.0 × V = 6.0 × (9.0 – V) (C1)
V = 6.0 V (A1)
(iii) potential difference (= 9.0 – 6.0) = 3.0V C1
charge (= 3.0 × 2.0 (µ)) = 6.0µC A1 [2]

© UCLES 2016
8 (a) P shown between earth symbol and voltmeter B1 [1]
(b) (i) gain = (50 × 103
)/100 = 500 C1
VIN (= 5.0/500) = 0.010V A1 [2]
(ii) VIN (= 5.0/5.0) = 1.0V A1 [1]
(c) e.g. multi-range (volt)meter
c.r.o. sensitivity control
amplifier channel selector B1 [1]
9 (a) (by Newton’s third law) force on wire is up(wards) M1
by (Fleming’s) left-hand rule/right-hand slap rule to give current A1
in direction left to right shown on diagram A1 [3]
(b) force ∝ current or F = BIL or B (= 0.080/6.0L) = 1/75L C1
maximum current = 2.5 × √2 C1
= 3.54A
maximum force in one direction = (3.54/6.0) × 0.080 C1
= 0.047N
difference (= 2 × 0.047) = 0.094N
or
force varies from 0.047N upwards to 0.047N downwards A1 [4]
10 nuclei emitting r.f. (pulse) B1
Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic
field B1
nuclei can be located (within a slice) B1
changing field enables position of detection (slice) to be changed B1 [4]
11 (a) (induced) e.m.f. proportional/equal to rate M1
(b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1
e.m.f induced in solenoid is greater (for same current) M1
induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3]

© UCLES 2016
or
same supply so same induced e.m.f. balancing it (B1)
(rate of change of) flux linkage is same (M1)
smaller current for same flux when core present (A1)
(c) e.g. (heating due to) eddy currents in core
(heating due to current in) resistance of coils
hysteresis losses/losses due to changing magnetic field in core
Any two of the above marking points, 1 mark each B2 [2]
12 (a) (i) electron diffraction/electron microscope (allow other sensible suggestions) B1 [1]
(ii) photoelectric effect/Compton scattering (allow other sensible suggestions) B1 [1]
(b) (i) arrow clear from –0.54 eV to –3.40 eV B1 [1]
(ii) E = hc/λ or E = hf and c = fλ C1
λ = (6.63 × 10–34
× 3.00 × 108
)/[(3.40 – 0.54) × 1.60 × 10–19
] = 4.35 × 10–7
m A1 [2]
(c) (i) wavelength associated with a particle M1
that is moving/has momentum/has speed/has velocity A1 [2]
(ii) λ = h/mv
v = (6.63 × 10–34
) / (9.11 × 10–31
× 4.35 × 10–7
) C1
= 1.67 × 103
ms–1
A1 [2]
13 X-ray image of a (single) slice/cross-section (through the patient) M1
taken from different angles/rotating X-ray (beam) A1
computer is used to form/process/build up/store image B1
2D image (of the slice) B1
repeated for many/different (neighbouring) slices M1
to build up 3D image A1 [6]

© UCLES 2016
14 (a) (i) He4
2 or α4
2 B1 [1]
(ii) n1
0 B1 [1]
(b) (i) ∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) C1
= 30.98697 – 30.98413
= 2.84 × 10–3
u C1 [2]
(ii) E = c2
∆m or mc2
C1
= (3.0 × 108
)2
× 2.84 × 10–3
× 1.66 × 10–27
= 4.2 × 10–13
J A1 [2]
(c) mass of products is greater than mass of Al plus α
or
reaction causes (net) increase in (rest) mass (of the system) B1
α-particle must have at least this amount of kinetic energy B1 [2]

PHYSICS 9702/43
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm/r2
= mv2
/r or GMm/r2
= mrω2
and v = 2πr/T or ω = 2π/T M1
with algebra to T2
= 4π2
r3
/GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM/r2
= v2
/r or GM/r2
= rω2
and v = 2πr/T or ω = 2π/T (M1)
with algebra to T2
= 4π2
r3
/GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24 × 3600)2
= 4π2
r3
/(6.67 × 10–11
× 6.0 × 1024
) C1
r3
= 7.57 × 1022
r = 4.2 × 107
m A1 [2]
(c) (T/24)2
= {(2.64 × 107
)/(4.23 × 107
)}3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T2
/r3
) = 242
/(4.23 × 107
)3
(B1)
k = 7.61 × 10–21
T2
(= kr3
) = 7.61 × 10–21
× (2.64 × 107
)3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV/T = constant or pV = nRT C1
T (= 5 × 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0 × 105
× 4.0 × 10–4
= n × 8.31 × 300
or
5.0 × 105
× 4.0 × 10–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]

© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240J A1
2. same value as given in 1. (= 240J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m = ω2
M1
ω = 2πf M1
(2 × 64/0.810) = (2π × f)2
leading to f = 2.0 Hz A1 [3]
(b) v0 = ωx0 or v0 = 2πfx0
or
v = ω(x0
2
– x2
)1/2
and x = 0 C1
v0 = 2π × 2.0 × 1.6 × 10–2
= 0.20ms–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]

© UCLES 2016
(b) (i) gain/dB = 10lg(P2 /P1)
32 = 10lg[PMIN /(0.38 × 10–6
)]
or
–32 = 10lg(0.38 × 10–6
/PMIN) C1
PMIN = 6.0 × 10–4
W A1 [2]
(ii) attenuation = 10lg[(9.5 × 10–3
)/(6.02 × 10–4
)] C1
= 12dB
attenuation per unit length (= 12/58) = 0.21dBkm–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, EA = (3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] (= 10.79NC–1
) M1
at P, EB = (12 × 10–12
)/[4πε0(10 × 10–2
)2
] (= 10.79NC–1
) M1
or
(3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] – (12 × 10–12
)/[4πε0(10 × 10–2
)2
] = 0
or
(3.0 × 10–12
)/[4πε0(5.0 × 10–2
)2
] = (12 × 10–12
)/[4πε0(10 × 10–2
)2
] (M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99 × 109
{(3.0 × 10–12
)/(5.0 × 10–2
) + (12 × 10–12
)/(10 × 10–2
)} C1
= 1.62V A1 [2]
(d) ½mv2
= qV
EK = ½ × 107 × 1.66 × 10–27
× v2
C1
qV = 47 × 1.60 × 10–19
× 1.62 C1
v2
= 1.37 × 108
v = 1.2 × 104
ms–1
A1 [3]

© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00/3.00 = 1.50/R C1
or
V+ = (3.00 × 4.5)/(2.00 + 3.00) = 2.7
2.7 = 4.5 × R/(R + 1.50) (C1)
resistance = 2.25kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between VOUT and earth A1 [2]
2. below 24°C, RT > 1.5kΩ or resistance of thermistor increases/high B1
V– < V+ or V– decreases/low (must not contradict initial statement) M1
VOUT is positive/+5 (V) and LED labelled as ‘pointing’ from VOUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv2
/r = Bqv B1
momentum or p or mv = Bqr B1 [2]

© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
(b) flux linkage = BAN
= π × 10–3
× 2.8 × π × (1.6 × 10–2
)2
× 85 = 6.0 × 10–4
Wb B1 [1]
(c) e.m.f. = ∆NΦ/∆t
e.m.f. = (6.0 × 10–4
× 2)/0.30 C1
e.m.f. = 4.0mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3s, 0.6s → 1.0s, 1.6s → 2.0s B1
E = 4 mV for t = 0.3s → 0.6s (either polarity) B1
E = 2 mV for t = 1.0s → 1.6s B1
with opposite polarity B1 [4]

© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /λ
= (6.63 × 10–34
× 3.00 × 108
)/(436 × 10–9
) C1
= 4.56 × 10–19
J (4.6 × 10–19
J) A1 [2]
(c) (i) Φ = hc/λ0
λ0 = (6.63 × 10–34
× 3.00 × 108
)/(1.4 × 1.60 × 10–19
) C1
= 890nm A1 [2]
(ii) λ0 = (6.63 × 10–34
× 3.00 × 108
)/(4.5 × 1.60 × 10–19
)
= 280nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890nm > 436nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280nm < 436nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]

© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A0 exp(–λt) and either t = t½, A = ½A0 or ½A0 = A0 exp(–λt½) M1
so ln2 = λt½ (and ln2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A = λN
N = 200/(2.1 × 10–6
) C1
= 9.52 × 107
C1
mass = (9.52 × 107
× 222 × 10–3
)/(6.02 × 1023
)
or
mass = 9.52 × 107
× 222 × 1.66 × 10–27
C1
= 3.5 × 10–17
kg A1 [4]

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
Question Answer Marks
1 Defining the problem
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Methods of data collection
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v0 = ey-intercept
1

© UCLES 2016
Additional detail including safety considerations 6
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –λB + ln v0
9. Relationship is valid if the graph is a straight line.
10. Method to vary B, e.g. re-magnetise in a coil.

© UCLES 2016
2 (a)
gradient =
1
YE
y-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”.
1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical.
1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
Read-offs must be accurate to half a small square.
1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y-intercept determined correctly by substitution into y = mx + c.
1

© UCLES 2016
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆ ∆ 
= + × 
 
100
m c
m c
or
∆ ∆ 
= + × 
 
100
m E
m E
or
∆
= ×100
Y
Y
Maximum/minimum methods:
gradientmin
interceptmax
gradientminmin
1
max
−
=
×
=
y
E
Y
gradientmax
interceptmin
gradientmaxmax
1
min
−
=
×
=
y
E
Y
1

PHYSICS 9702/52
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
p is the independent variable and B is the dependent variable, or vary p and
measure B.
1
Keep the current/I (in the electromagnet) constant. 1
Labelled diagram showing Hall probe correctly positioned (along p) and
ruler correctly positioned and either Hall probe or rule supported.
1
Correct circuit diagram to include d.c. power supply in series with coil and
ammeter. Must be a workable circuit diagram to measure current through
the coil.
1
Measure p with ruler. 1
Method to determine an accurate value of p.
Examples include:
Height of P above bench – height of electromagnet
Height of P measured from ruler across the top of the electromagnet
1
Method of analysis
Plot a graph of ln B against p. 1
α = – gradient 1
IN
k
y intercept
e −
=
1

© UCLES 2016
1. Keep the number of turns/N constant.
2. Use large number of turns/current (to increase B).
3. Avoid overheating the coil/do not touch hot coil.
4. Use of variable resistor to keep ammeter reading constant.
5. Method to ensure that Hall probe is equidistant from the poles, e.g.
determine centre of electromagnet and use of plumb line/ruler and spirit
level/set square.
6. Adjust Hall probe until maximum reading obtained/perpendicular to field.
7. Repeat each experiment for the same value of p and reverse the
current/Hall probe and average
8. ln B = –αp + ln kNI
10. Calibrate Hall probe using a known field.

© UCLES 2016
2 (a) gradient = q
y-intercept = lg p
1
(b)
2.80 or 2.799 or 2.7993 0.28 or 0.279
2.79 or 2.792 or 2.7924 0.30 or 0.301
2.77 or 2.771 or 2.7709 0.36 or 0.362
2.72 or 2.716 or 2.7160 0.49 or 0.491
2.69 or 2.690 or 2.6902 0.57 or 0.568
2.67 or 2.672 or 2.6721 0.61 or 0.613
All first column correct – either 2 and 3 decimal places or 3 and 4 decimal
places.
All second column correct. Allow a mixture of decimal places.
Uncertainties in lg (V/V) from ± 0.02 to ± 0.01. Allow more than one
significant figure.
1
1
1
1
All error bars in lg (V/V) plotted correctly.
All error bars to be plotted. Total length of bar must be accurate to less than
half a small square and symmetrical.
1
Line must not be drawn from top point to bottom point unless points are
balanced.
Upper end of line should pass between (2.694, 0.55) and (2.700, 0.55) and
lower end of line should pass between (2.770, 0.35) and (2.776, 0.35).
1
1
(iii) Gradient determined with a triangle that is at least half the length of the drawn
line.
Gradient must be negative.
1
or
uncertainty = ½(steepest worst line gradient – shallowest worst line gradient)
1

© UCLES 2016
(iv) y-intercept determined by substitution into y = mx + c.
1
line
or
y-intercept)
1
(d) Use of p = 10answer to 2(c)(iv)
or
lg p = answer to 2(c)(iv)
1
q = gradient and in the range –2.50 to –2.70 and given to 2 or 3 s.f. 1
(e) Use of V = p × 950q
or
lg V = q lg 950 + lg p
or
lg V = q lg 950 + y-intercept
Correct substitution of numbers must be seen to give V.
1

PHYSICS 9702/53
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

© UCLES 2016
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v0 = ey-intercept
1

© UCLES 2016
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –λB + ln v0
10. Method to vary B, e.g. re-magnetise in a coil.

© UCLES 2016
2 (a)
gradient =
1
YE
y-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical.
1
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
1
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y-intercept determined correctly by substitution into y = mx + c.
1

© UCLES 2016
line
or
y-intercept)
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆ ∆ 
= + × 
 
100
m c
m c
or
∆ ∆ 
= + × 
 
100
m E
m E
or
∆
= ×100
Y
Y
Maximum/minimum methods:
gradientmin
interceptmax
gradientminmin
1
max
−
=
×
=
y
E
Y
gradientmax
interceptmin
gradientmaxmax
1
min
−
=
×
=
y
E
Y
1

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