Physics 9702/52 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Two coils, P and Q, are placed close to each other, as shown in Fig. 1.1.
R
coil P coil Q
Fig. 1.1
A resistor of resistance R is connected in series with coil P.
A changing magnetic flux of frequency f in coil P causes an electromotive force (e.m.f.) E to be induced across the terminals of coil Q.
It is suggested that E is related to R by the relationship
E = 2πf M ( R V+ k ) where V is the potential difference across the resistor and coil P, and k and M are constants.
Plan a laboratory experiment to test the relationship between E and R.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine values for k and M.
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
R is the independent variable and E is the dependent variable or vary R and measure E 1
keep V constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
coil P placed close to coil Q
separate workable circuit for coil Q
(a.c.) voltmeter or oscilloscope connected across coil Q
(Do not accept a power supply connected to coil Q.)
a.c. power supply/signal generator connected to resistor and coil P in series 1
workable circuit with power supply and (a.c.) voltmeter/oscilloscope in parallel with resistor and coil P or across terminals of 1
power supply/signal generator
method to determine R, e.g. measure current in R and p.d. across R and use R = VR/I or measure R using an ohmmeter 1
1 Method of Analysis
1 1 1
plot a graph of against R or equivalent (e.g. R against )
E E
Do not accept logarithms.
1 1
M=
2πfV × gradient
1 gradient
(for R against : M= )
E 2 πfV
k = 2πfVM × y -intercept 1
or
y -intercept
k=
gradient
1
(for R against : k = − y-intercept)
E
1 Additional detail including safety considerations 6
D1 precaution linked to hot coil (P) / hot resistor, e.g. use of (heat-proof) gloves, wait until circuit cools down
or
precaution linked to shocks from high voltages e.g. use of (insulating) gloves or switch off supply before touching the
circuit (to change R)
D2 keep the number of turns on (both) coils constant
D3 keep f constant
D4 keep distance between the coils constant
D5 method to keep distance between the coils constant, e.g. fix/clamp coils to bench
D6 method to measure f, e.g. read from signal generator or use of oscilloscope
D7 method to determine f from oscilloscope, e.g. period from oscilloscope T = time-base × horizontal distance and
f = 1/T
D8 method to determine V or E from oscilloscope, e.g. V = y-gain × vertical distance
D9 method to increase E e.g. use iron core/more turns on coil Q/high frequency/high p.d. (across R and coil P)
D10 relationship valid if a straight line is produced (not passing through the origin)
Do not accept straight line passing through the origin.
A student investigates how the volume of a gas varies with its temperature. Air is trapped in a transparent cylinder of diameter d with a movable piston as shown in Fig. 2.1.
d
cylinder
movable piston
trapped air h
Fig. 2.1
The distance between the base of the cylinder and the bottom of the piston is h.
The trapped air is heated by placing the cylinder in water of temperature θ. The increase in temperature of the trapped air causes the piston to move. When the piston stops moving, the value of h is measured.
For each value of h, the volume V of the trapped air is calculated.
The experiment is repeated for different values of θ.
It is suggested that V and θ are related by the equation
pV = Yk (θ + Z )
where k is the Boltzmann constant, p is the atmospheric pressure, and Y and Z are constants.
(a) A graph is plotted of V on the y-axis against θ on the x-axis.
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
(b) Values of θ and h are given in Table 2.1.
Table 2.1
θ / °C h / mm V / 10–5 m3
23 62.4 ± 0.1
35 65.2 ± 0.1
48 68.1 ± 0.1
62 70.9 ± 0.1
73 73.3 ± 0.1
88 76.1 ± 0.1
The value of d is (27.9 ± 0.1) mm.
The volume V is calculated using the relationship πd 2h
V= .
4
Calculate and record values of V / 10–5 m3 in Table 2.1.
Include the absolute uncertainties in V. [2]
(c) (i) Plot a graph of V / 10–5 m3 against θ / °C. Include error bars for V. [2]
(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]
4.8
4.7
V / 10–5 m3
4.6
4.5
4.4
4.3
4.2
4.1
4.0
3.9
3.8
3.7
20 30 40 50 60 70 80 90 θ / °C
(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 Y and Z. Include appropriate units.
Data: p = (1.01 ± 0.01) × 105 Pa k = 1.38 × 10 –23 J K–1
Y = ……
Z = ……
[2]
(ii) Determine the percentage uncertainty in Y.
percentage uncertainty in Y = …… % [1]
(e) The experiment is repeated. Determine the temperature θ that gives a value of h of 60.0 mm.
θ = …… °C [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) Yk 1
gradient =
p
YkZ
y-intercept =
p
2(b) 1
V / 10–5 m3 absolute uncertainty
3.81 or 3.815 ± 0.03
3.99 or 3.986 ± 0.03
4.16 or 4.163 ± 0.04
4.33 or 4.335 ± 0.04
4.48 or 4.481 ± 0.04
4.65 or 4.652 ± 0.04
Values of V correct as shown above.
Absolute uncertainties in V correct as shown above. 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 V 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 (27.5, 3.90) and (29.5, 3.90) and between (82.0, 4.60) and (84.0, 4.60).
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 of worst acceptable line determined 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) Y determined using gradient and Y and Z given to 2 or 3 significant figures. 1
p × gradient
Y = = 7.3188 × 10 27 × gradient
k
Z determined using y-intercept and Y and Z given with SI units. 1
p × y -intercept y -intercept
Z= or Z =
Yk gradient
Units:
Y: no unit
Z: °C
2(d)(ii) Percentage uncertainty in Y with method shown. 1
Δp Δgradient
percentage uncertainty = + × 100
p gradient
or
Correct substitution for max/min methods.
2(e) θ determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1
powers of ten.
π × ( 0.0279 ) × 0.0600
2
V= = 3.67 × 10 −5
4
and
pV V V − y -intercept
θ= − Z or θ = − Z or θ =
Yk gradient gradient
or using h directly:
πd 2 h
pπd h 2 2 − y -intercept
πd h
θ= − Z or θ = − Z or θ = 4
4Yk 4 × gradient gradient
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On this page, free: the mark scheme for AS & A Level Physics 9702/52, May/June 2023 (9702_s23_ms_52), and the question paper (9702_s23_qp_52), to read or download, with each question's marking points written out underneath.
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Physics 9702/52 for May/June 2023 is out of 30 marks, with 1 hour 15 minutes to answer it.
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