Physics 9702/53 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Fig. 1.1 shows a thin coil of cross-sectional area A and length l connected to a resistor of resistance
S and two terminals.
l
S
Fig. 1.1
An alternating voltage is applied to the terminals. The peak value of the alternating voltage is E and the frequency is f. The peak value of the potential difference V across the resistor is determined using an oscilloscope.
It is suggested that V is related to f by the relationship
ES KAN 2f
=
V l
where N is the number of turns on the coil and K is a constant.
Plan a laboratory experiment to test the relationship between V and f.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine a value for K.
In your plan you should include:
• the procedure to be followed
• the measurements to be taken
• the control of variables
• the analysis of the data
• any safety precautions to be taken.
Diagram
…… [15]
Mark scheme
1 Defining the problem
vary f and measure V or f is the independent variable and V is the dependent variable 1
keep E constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
• circuit with a.c. supply
• oscilloscope connected in parallel with the resistor
• workable circuit
• oscilloscope and a.c. supply labelled
labelled signal generator or variable frequency power supply connected across the terminals 1
method to determine V or E from oscilloscope, e.g. multiply amplitude / height of wave by y-gain on oscilloscope 1
method to determine f from oscilloscope, e.g. determine period T by multiplying number of divisions in 1 cycle or horizontal 1
distance in 1 cycle by the time base and f = 1 / T
Method of Analysis
1 1 1
plot a graph of against f or equivalent, e.g. f against
V V
Allow logarithms e.g. lg V against lg f.
relationship valid if a straight line is produced passing through the origin 1
(for lg V against lg f: relationship valid if a straight line is produced with gradient = −1)
1 1
1 1
against f f against
V V
l ES l ES 1
K= × gradient K= 2
×
AN 2 AN gradient
El S
(for lg V against lg f: K = 2
× 10− y -intercept ).
AN
Additional detail including safety considerations 6
D1 precaution linked to hot coil or hot resistor or prevention of burns from coil or resistor, e.g. use gloves / switch off
power supply when not measuring V to prevent burns from coil / resistor
D2 keep N and A and l and S constant
D3 method to keep S constant, e.g. switch off power supply between readings to prevent heating of resistor or to allow
resistor to cool
πd 2
D4 method to determine A, e.g. use calipers / micrometer to measure diameter (of coil) / d and A=
4
D5 repeat measurements of diameter d along the length of the coil / in different directions and determine the average
value of d
D6 method to determine the value of S, e.g. separate circuit diagram showing resistor connected to ohmmeter, or
circuit diagram showing resistor connected to a power supply with an ammeter and voltmeter and S = V / I
D7 measure l with a ruler / calipers
D8 oscilloscope drawn connected across terminals / across signal generator and description to determine E
D9 adjust y-gain for maximum amplitude
or
adjust time base for length of one wave or measure n waves and divide measured time by n
1 D10 method to keep E constant, e.g. check p.d. and alter supply
or
method to keep l constant, e.g. tape coil
or
method to keep A constant, e.g. wind wire on a cylinder
A student investigates an electrical circuit.
The circuit is set up as shown in Fig. 2.1.
Z
A
P Q
Fig. 2.1
A battery of negligible internal resistance is connected to a resistor of resistance Z. Five resistors, each of resistance R, are connected in parallel between P and Q.
The switch is closed. The total current I in the circuit is measured using the ammeter.
The experiment is then repeated by changing the number n of resistors, each of resistance R, connected in parallel between P and Q.
It is suggested that I and n are related by the equation
E=I (Rn + Z) where E is the electromotive force (e.m.f.) of the battery.
1 1
(a) A graph is plotted of on the y-axis against on the x-axis.
I n
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
1
(b) Values of n, and I are given in Table 2.1. n
Table 2.1
1 1 n I / μA / 103 A–1 n I
5 0.200 455 ± 5
6 0.167 525 ± 5
7 0.143 580 ± 5
8 0.125 635 ± 5
9 0.111 685 ± 5
11 0.0909 765 ± 5
1 1
Calculate and record values of / 103 A–1 in Table 2.1. Include the absolute uncertainties in .
I I
[2]
1 1 1
(c) (i) Plot a graph of / 103 A–1 against . Include error bars for . [2]
I n I
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines. [2]
(iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.
gradient = …… [2]
2.3
2.2
2.1
2.0
1
/ 103A–1
I
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22
1 n
(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer.
y-intercept = …… [2]
(d) The e.m.f. E of the battery is determined twice during the experiment. The values obtained are 5.6 V and 6.0 V.
(i) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of R and Z. Include appropriate units.
R = ……
Z = ……
[2]
(ii) Determine the percentage uncertainty in your value of R.
percentage uncertainty = …… % [1]
(e) The experiment is repeated with 20 resistors, each of resistance R, connected in parallel between P and Q. Determine the total current I in the circuit.
I = …… A [1]
[Total: 15]
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series.
Mark scheme
2(a) R 1
gradient =
E
Z
y-intercept =
E
2(b) 1
1
/ 103 A–1
I
2.20 or 2.198
1.90 or 1.905
1.72 or 1.724
1.57 or 1.575
1.46 or 1.460
1.31 or 1.307
1
Values of / 103 A–1 correct as shown above.
I
2(b) 1 1
Uncertainties in / 103 A–1 from ± 0.02 or ± 0.03 decreasing to ± 0.01.
I
2(c)(i) Six points from (b) plotted correctly. 1
Must be within half a small square. Diameter of points must be less than half a small square.
1 1
Error bars in plotted correctly.
I
All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
2(c)(ii) Straight line of best fit drawn. 1
Thickness of the line must be less than half a small square.
Do not accept line from top point to bottom point.
Line must pass between (0.101, 1.40) and (0.104, 1.40) and between (0.189, 2.10) and (0.194, 2.10)
Worst acceptable straight line drawn (steepest or shallowest possible line that passes through all the error bars). 1
Thickness of the line must be less than half a small square.
All error bars must be plotted.
2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1
Distance between data points must be greater than half the length of the drawn line.
Gradient determined of worst acceptable line with clear substitution of data points into Δy / Δx. 1
uncertainty = (gradient of line of best fit – gradient of worst acceptable line)
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1
y-intercept of worst acceptable line determined by substitution into y = mx + c. 1
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)
Do not accept ECF from false origin method.
2(d)(i) R determined using gradient and R and Z given to 2 or 3 significant figures. 1
R = gradient × 5.8
Z determined using y-intercept and R and Z given with units with appropriate powers of ten. 1
Z = y-intercept × 5.8
unit of R: Ω or V A–1
unit of Z: Ω or V A–1
2(d)(ii) Percentage uncertainty determined using ΔE = 0.2 (V) with method shown. 1
ΔE Δgradient
ΔR% = + × 100
E gradient
or
0.2 Δgradient
ΔR% = + × 100
5.8 gradient
2(e) I determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1
1
I=
gradient
+y -intercept
20
or
E
I=
R
20 +Z
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9702/53 May/June 2025 mark scheme?+
On this page, free: the mark scheme for AS & A Level Physics 9702/53, May/June 2025 (9702_s25_ms_53), and the question paper (9702_s25_qp_53), to read or download, with each question's marking points written out underneath.
How many marks is 9702/53 out of?+
Physics 9702/53 for May/June 2025 is out of 30 marks, with 1 hour 15 minutes to answer it.
Can I get my 9702/53 answers marked?+
Yes. Upload your answers to 9702/53/M/J/25 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.