Physics 9702/52 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Fig. 1.1 shows a coil made from resistance wire.
Fig. 1.1
The coil is placed in cooking oil of mass m. The total length of the resistance wire in the oil is L.
A potential difference V is applied to the coil. The temperature of the oil increases by Δθ in time t.
It is suggested that Δθ is related to L by the relationship
AtV 2
= mKΔθ + Z
L where A is the cross-sectional area of the wire, and K and Z are constants.
Plan a laboratory experiment to test the relationship between Δθ and L.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine values for K and Z.
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
L is the independent variable and Δθ or temperature change/increase is the dependent variable 1
or
vary L and measure Δθ or temperature change/increase
keep t constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
• oil in a beaker/container (on a bench)
• coil fully submerged in oil
• (bulb of) thermometer in the oil
• at least three labels from thermometer, coil or resistance wire, oil, beaker/container, clamp/stand, bench
Do not accept other heating sources.
method to determine V – diagram of workable circuit including: 1
• power supply connected to wire
• voltmeter positioned to measure V across the coil
measure the initial and final temperature and find the difference Δθ 1
method to determine t, e.g. use stopwatch/timer 1
and
method to determine L e.g. use a rule(r) to measure L / length of wire or e.g. using number of turns and measure the
diameter of the coil with rule(r) / calipers
1 Method of Analysis
1 1 1
plot a graph of Δθ against or equivalent, e.g. against Δθ
L L
Do not accept logarithms.
1
1 1
for Δθ against for against Δθ
L L
AtV 2 AtV 2 × gradient
K= K=
m × gradient m
1
1 1
for Δθ against for against Δθ
L L
Z = −mK × y -intercept Z = AtV 2 × y -intercept
or
AtV × y -intercept
2
Z=−
gradient
1 Additional detail including safety considerations 6
D1 precaution linked to hot oil / beaker / wire, e.g. use of gloves to prevent burns from oil
or
precaution linked to spillage of oil, e.g. perform experiment in a tray
D2 keep A and m and V constant
πd 2
D3 use a micrometer to measure the diameter (d) of the wire and A =
4
D4 repeat measurements of d along the wire and average
D5 method to reduce heat loss e.g. add insulation around the container / add a lid to the container
D6 method to keep V constant, e.g. adjust / change a variable resistor / power supply to keep V or voltmeter reading
constant
D7 use a balance to determine the mass of the oil
and
mass of oil = mass of (beaker + oil) − mass of beaker
or
place beaker on balance and zero balance, then add oil and read balance
D8 stir the oil for uniform temperature
or
keep the initial temperature (of oil) constant
D9 repeat the experiment for the same value of L and average Δθ / average temperature change
Z
D10 relationship valid if a straight line is produced (passing through − )
mK
Do not accept line passing through the origin.
D11 method to determine L accurately, e.g. measure length of unwound coil
A student investigates the refraction of white light entering a transparent rectangular block. A narrow beam of light enters the block at the midpoint of one of the shorter sides. The angle of incidence θ is measured, as shown in Fig. 2.1.
block
beam of light
θ
d
Fig. 2.1 (not to scale)
The distance d between the corner of the block and the point where the beam of light touches the boundary of the block is measured.
The experiment is repeated for different values of θ.
It is suggested that d and θ are related by the equation
B sin2 θ
=
B + d2 n2 where B and n are constants.
1
(a) A graph is plotted of d 2 on the y-axis against on the x-axis. sin2 θ
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
1
(b) Values of θ, and d are given in Table 2.1. sin2 θ
Table 2.1
1 θ /° d / cm d 2 / cm2 sin2 θ
28.5 4.39 24.8 ± 0.2
33.5 3.28 21.4 ± 0.2
42.5 2.19 17.1 ± 0.2
50.0 1.70 14.7 ± 0.2
57.5 1.41 12.9 ± 0.2
63.5 1.25 11.8 ± 0.2
Calculate and record values of d 2 / cm2 in Table 2.1.
Include the absolute uncertainties in d 2. [2]
1
(c) (i) Plot a graph of d 2 / cm2 against . Include error bars for d 2. [2]
sin2 θ
(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]
650
600
550
d 2 / cm2
500
450
400
350
300
250
200
150
100
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5
1 sin2 θ
(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer.
y-intercept = …… [2]
(d) (i) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of B and n. Include appropriate units.
B = ……
n = ……
[2]
(ii) Determine the percentage uncertainty in n.
percentage uncertainty in n = …… % [1]
θ = …… ° [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) gradient = Bn2 1
y-intercept = − B
2(b) 1
d2 / cm2
615 or 615.0
458 or 458.0
292 or 292.4
216 or 216.1
166 or 166.4
139 or 139.2
Values correct as shown above.
Uncertainties in d2 decreasing from ±10 to ±4 or ±5. 1
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.
Error bars in d2 plotted correctly. 1
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
Do not accept line from top point to bottom point.
Points must be balanced.
Line must pass between (1.90, 250) and (2.00, 250) and between (3.55, 500) and (3.65, 500).
Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1
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 x 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) B determined using y-intercept (B = – y-intercept) and B and n given to 2 or 3 or 4 significant figures. 1
n determined using gradient 1
and
B and n given with SI units with correct powers of ten.
gradient gradient
n= or n =
B −y -intercept
Unit for B: cm2
No unit for n.
2(d)(ii) Percentage uncertainty in n determined with method shown. 1
1 Δy -intercept Δgradient
percentage uncertainty = + × 100
2 y -intercept gradient
or
correct substitution for max/min methods.
2(e) θ determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1
power of ten.
gradient
θ = sin−1
−y -intercept + 900
or
n 2B
θ = sin−1
B + 900
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Physics 9702/52 for October/November 2024 is out of 30 marks, with 1 hour 15 minutes to answer it.
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