Physics 9702/51 — May/June 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

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.

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

\frac{ES}{V} = \frac{KAN^2f}{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.

DifficultyMedium-Hard

Worked solution

diagram-1

Answer

Variables

Independent variable: frequency $f$ of the applied a.c. signal.
Dependent variable: peak p.d. $V$ across resistor $S$ .
Control variables: peak applied voltage $E$ (keep constant), resistor $S$ , coil geometry ( $A$ , $l$ ) and number of turns $N$ (same coil), temperature (to keep resistances constant).

Apparatus

Signal generator, thin coil (air-cored), known resistor $S$ (and DMM to check $S$ ), dual-channel oscilloscope, connecting leads, metre rule, vernier calipers/micrometer.

Procedure and measurements

Connect coil in series with resistor $S$ to the signal generator.
Connect oscilloscope channel 1 across the resistor to measure peak p.d. $V$ .
Connect oscilloscope channel 2 across the signal generator output (terminals) to measure peak applied voltage $E$ .
Set the signal generator to a sinusoid. Choose a value of $E$ (e.g. a few volts peak) and keep it constant for all readings (monitor channel 2 and adjust amplitude if necessary).
Vary $f$ over a suitable range with at least 6 values.
For each $f$ $f$ , measure:
- $f$ from the signal generator display or from the oscilloscope period $T$ ( $f=1/T$ ),
- $V$ from the oscilloscope trace (measure $V_{pp}$ and use $V=\frac{1}{2}V_{pp}$ ),
- $E$ similarly (or confirm it remains constant).
  Repeat readings and average.
Measure coil dimensions once: measure coil diameter $d$ with calipers and calculate $A=\pi(d/2)^2$ ; measure coil length $l$ with a ruler. Determine/record $N$ (from coil specification or by counting turns).

Analysis of data (test of relationship and finding $K$ )

Rearrange

\frac{ES}{V}=\frac{KAN^2}{l}f

Calculate $y=\dfrac{ES}{V}$ for each $f$ and plot a graph of $y$ (vertical) against $f$ (horizontal).

Expect a straight line through the origin.
Gradient $m$ is

m=\frac{KAN^2}{l}

K=\frac{ml}{AN^2}.

Safety

Use low-voltage a.c. output from the signal generator.
Use a resistor with adequate power rating; switch off if resistor/coil heats up.
Avoid short circuits and exposed conductors; keep leads tidy.
Take care with oscilloscope earth reference (ensure the generator output is not inadvertently shorted to ground).

Final answer

See working

Detailed explanation

Background Concept

A coil has inductance, so with an alternating current it produces a back e.m.f. that depends on how quickly the current changes. In an a.c. circuit with a coil in series with a resistor, the current (and therefore the resistor p.d.) depends on frequency. The question provides a suggested relationship:

\frac{ES}{V} = \frac{KAN^2 f}{l}

Here:

$E$ is the peak applied voltage (across the supply terminals),
$V$ is the peak p.d. across the resistor $S$ ,
$A$ is the coil cross-sectional area,
$l$ is the coil length,
$N$ is number of turns,
$K$ is a constant to be determined experimentally.

The important experimental skill is to turn this into a linear form so you can test it with a straight-line graph and use the gradient to find $K$ .

Understanding the Question

You are given a series circuit (coil + resistor) driven by an a.c. source. You can vary the frequency $f$ and you can measure the peak resistor voltage $V$ using an oscilloscope. You must:

design a method to take readings of $V$ for different $f$ while keeping other quantities controlled,
show how to process the readings to test whether the proposed proportionality is correct,
explain clearly how $K$ is obtained from your graph.

Because the relationship involves $E$ and $S$ , you either need to measure $E$ as well or keep $E$ constant and known. Similarly, $S$ , $A$ , $l$ , and $N$ must be known/constant.

Approach

Choose variables: vary $f$ (independent variable) and measure $V$ (dependent variable).
Control key factors: keep $E$ constant because it appears in the equation; keep the same resistor $S$ and the same coil so $A$ , $l$ , $N$ remain constant.
Collect data carefully: for each $f$ , read peak values from the oscilloscope (use peak-to-peak then halve).
Linearise: compute $ES/V$ and plot it against $f$ so that the graph should be a straight line.
Extract $K$ from gradient: use the gradient relationship to calculate $K$ .

Step-by-Step Reasoning

diagram-1

Circuit and measurements
- Put the coil and resistor $S$ in series with a signal generator.
- Use oscilloscope channel 1 across the resistor to measure $V$ .
- Use oscilloscope channel 2 across the generator output to measure $E$ (this is also a good check that $E$ is being kept constant).
Choosing the frequency range and number of readings
- Pick at least 6 different frequencies spanning a wide range (enough to show a clear trend and allow a best-fit line).
- Ensure the waveforms remain sinusoidal and the oscilloscope trace is stable.
Keeping $E$ constant (control of variables)
- When you change $f$ , the generator amplitude may drift. The relationship is only being tested properly if $E$ is constant.
- Monitor $E$ on channel 2; after each frequency change, adjust the generator amplitude so that the peak (or peak-to-peak) value is the same as your chosen value.
Reading peak values from the oscilloscope
- Measure peak-to-peak voltage from the number of vertical divisions multiplied by the volts-per-division.
- Convert to peak using $V = \frac{1}{2}V_{pp}, \qquad E = \frac{1}{2}E_{pp}.$
- Repeat the measurement at each frequency and average to reduce random reading error.
Measuring coil parameters
- Measure the coil diameter $d$ with vernier calipers (take several readings around the coil and average).
- Calculate area $A = \pi\left(\frac{d}{2}\right)^2.$
- Measure coil length $l$ with a ruler (or calipers if short).
- Record $N$ (count turns or use manufacturer’s stated value).
Data processing and graph
- For each frequency, calculate $y = \frac{ES}{V}.$
- Plot $y$ (vertical axis) against $f$ (horizontal axis).
- If the suggested relationship is correct, the graph should be a straight line with (approximately) zero intercept because it predicts direct proportionality.
Determining $K$ from the gradient
- From $\frac{ES}{V} = \left(\frac{KAN^2}{l}\right)f$ the gradient $m$ of the $\left(\frac{ES}{V}\right)$ vs $f$ graph is $m = \frac{KAN^2}{l}.$
- Hence $K = \frac{ml}{AN^2}.$
Uncertainties (good practice in Paper 5 plans)
- State likely dominant uncertainties: reading oscilloscope amplitudes, keeping $E$ constant, measurement of $d$ (affects $A$ ), and $l$ .
- To estimate uncertainty in $K$ , you can use the uncertainty in the gradient (from a worst acceptable line) and combine with percentage uncertainties in $A$ , $l$ , and $N$ (if $N$ is not exact).

Key Takeaways

A planning question is scored for: clear variables, workable method, correct measurements, control of variables, and a graph/analysis that directly tests the proposed equation.
Linearisation is crucial: plotting $\dfrac{ES}{V}$ against $f$ makes the test “straight line through origin” and gives a clean route to $K$ via the gradient.
Measuring peak quantities on an oscilloscope is typically done via peak-to-peak then halving.

Common Mistakes

Varying frequency but not keeping $E$ constant (then changes in $V$ are not solely due to $f$ ).
Plotting the wrong graph (e.g. $V$ vs $f$ ) without using the provided linear form.
Using RMS values for one voltage and peak values for the other (mixing definitions).
Forgetting to state how $A$ and $l$ are measured, or not stating how $N$ is known.
Not stating how to obtain $K$ from the graph (must link gradient to $K$ explicitly).

Things to Be Careful About

Oscilloscope readings: always convert divisions correctly and be consistent about peak vs peak-to-peak.
The oscilloscope ground clip can create a short if connected incorrectly; ensure the generator and oscilloscope grounds are compatible.
Heating: if current is too large, $S$ (and coil resistance) may change with temperature, affecting $V$ . Use modest voltages and allow cooling time if needed.
Graph quality: use a wide frequency range, enough points, and a best-fit line; if discussing uncertainty, mention worst acceptable line for the gradient uncertainty.

Techniques used

identify independent, dependent and control variablesmeasure peak p.d. using an oscilloscope from peak-to-peak amplitudekeep the signal generator output amplitude constant while varying frequencylinearise the relationship and plot a straight-line graphuse the gradient of the graph to determine the constant

Answer

Gradient $= \dfrac{R}{E}$

$y$ -intercept $= \dfrac{Z}{E}$

Final answer

Gradient = R/E, y-intercept = Z/E

Detailed explanation

Background Concept

A straight-line graph has the form

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept. In many experiments, you are given a physical relationship that is not obviously in this form, so you rearrange it until the plotted variables match $y$ and $x$ .

For resistors in parallel, the equivalent resistance is reduced. For $n$ identical resistors each of resistance $R$ in parallel,

R_{\text{parallel}} = \frac{R}{n}

Understanding the Question

The circuit has a fixed series resistor $Z$ and a variable parallel combination $R/n$ between $P$ and $Q$ . The suggested relationship is

E = I\left(\frac{R}{n} + Z\right)

You are told to plot $\frac{1}{I}$ (vertical axis) against $\frac{1}{n}$ (horizontal axis). So we want an equation with $\frac{1}{I}$ on the left and $\frac{1}{n}$ appearing linearly on the right.

Approach

Divide the given equation by $I$ to make $1/I$ .
Expand the brackets so it looks like (constant), $\times (1/n)$ + (constant).
Compare with $y = mx + c$ to read off gradient and intercept.

Step-by-Step Reasoning

Start with

E = I\left(\frac{R}{n} + Z\right)

Divide both sides by $EI$ :

\frac{1}{I} = \frac{1}{E}\left(\frac{R}{n} + Z\right)

Split the bracketed term:

\frac{1}{I} = \frac{R}{E}\cdot \frac{1}{n} + \frac{Z}{E}

Now match to $y = mx + c$ with $y = 1/I$ and $x = 1/n$ .

So:

gradient $m = R/E$
intercept $c = Z/E$ .

Key Takeaways

Linearise by rearranging into $y = mx + c$ .
The coefficient of the plotted $x$ variable is the gradient.
The constant term is the $y$ -intercept.

Common Mistakes

Leaving the equation as $E = I(\cdots)$ and trying to read gradient directly.
Using $n$ instead of $1/n$ when matching to the graph axes.
Mixing up which constant is gradient and which is intercept.

Things to Be Careful About

Ensure the plotted variables exactly match the rearranged equation: here it must be $1/I$ vs $1/n$ .
Keep $E$ outside the bracket correctly when dividing through: both terms must be divided by $E$ .

Techniques used

rearrange the given equation into the form y = mx + cidentify the gradient and intercept by matching coefficientsinterpret gradient and intercept in terms of circuit parameters

(b)

Values of $n$ , $\frac{1}{n}$ and $I$ are given in Table 2.1.

Table 2.1

$n$	$\frac{1}{n}$	$I / \mu\text{A}$
5	0.200	$455 \pm 5$
6	0.167	$525 \pm 5$
7	0.143	$580 \pm 5$
8	0.125	$635 \pm 5$
9	0.111	$685 \pm 5$
11	0.0909	$765 \pm 5$

Calculate and record values of $\frac{1}{I} / 10^3\ \text{A}^{-1}$ in Table 2.1. Include the absolute uncertainties in $\frac{1}{I}$ .

2M Similar

DifficultyMedium

Worked solution

For $I$ in $\mu\text{A}$ ,

\frac{1}{I} = \frac{1}{I\times 10^{-6}}\ \text{A}^{-1}

and

\Delta\left(\frac{1}{I}\right)=\frac{\Delta I}{I^2}

Example ( $I=455\pm 5\ \mu\text{A}$ ):

\frac{1}{I}=\frac{1}{455\times 10^{-6}}=2.20\times 10^{3}\ \text{A}^{-1}

\Delta\left(\frac{1}{I}\right)=\frac{5\times 10^{-6}}{(455\times 10^{-6})^2}=2.42\times 10^{1}\ \text{A}^{-1}=0.024\times 10^{3}\ \text{A}^{-1}

Answer (values for Table 2.1)

$\frac{1}{I} / 10^{3}\ \text{A}^{-1}$ :

$n=5$ : $2.20 \pm 0.024$
$n=6$ : $1.90 \pm 0.018$
$n=7$ : $1.72 \pm 0.015$
$n=8$ : $1.57 \pm 0.012$
$n=9$ : $1.46 \pm 0.011$
$n=11$ : $1.31 \pm 0.0085$

Final answer

1/I / 10^3 A^-1: 2.20±0.024, 1.90±0.018, 1.72±0.015, 1.57±0.012, 1.46±0.011, 1.31±0.0085

Detailed explanation

Background Concept

If a measured quantity $I$ has an absolute uncertainty $\Delta I$ , and you calculate a new quantity

y = \frac{1}{I}

then for small uncertainties the absolute uncertainty in $y$ is found using differentiation:

\Delta y \approx \left|\frac{\mathrm{d}}{\mathrm{d}I}\left(\frac{1}{I}\right)\right|\Delta I = \frac{\Delta I}{I^2}

This is the standard method used in Paper 5 for a reciprocal.

Understanding the Question

You are given $I$ (in $\mu\text{A}$ ) with $\pm 5\ \mu\text{A}$ uncertainty for each reading. The table requires you to calculate $1/I$ and record it in the scaled unit $1/I\,/\,10^3\ \text{A}^{-1}$ , and also include the absolute uncertainty in $1/I$ .

Approach

For each row:

Convert $I$ from $\mu\text{A}$ to $\text{A}$ .
Compute $1/I$ .
Compute $\Delta(1/I)=\Delta I/I^2$ using $\Delta I = 5\ \mu\text{A}$ converted to amperes.
Divide both the value and uncertainty by $10^3$ to match the table heading.

Step-by-Step Reasoning

Take $I = 455\pm 5\ \mu\text{A}$ .

Convert to amperes: $I = 455\times 10^{-6}\ \text{A}$ and $\Delta I = 5\times 10^{-6}\ \text{A}$ .
Reciprocal:

\frac{1}{I} = \frac{1}{455\times 10^{-6}} = 2.1978\times 10^{3}\ \text{A}^{-1}

Uncertainty:

\Delta\left(\frac{1}{I}\right) = \frac{5\times 10^{-6}}{(455\times 10^{-6})^2} = 24.2\ \text{A}^{-1}

Put into the table scale by dividing by $10^3$ :

\frac{1}{I}/10^3 = 2.1978,\quad \Delta\left(\frac{1}{I}/10^3\right)=0.0242

Repeat the same steps for each current value.

Key Takeaways

Always convert prefixes ( $\mu$ ) before doing uncertainty propagation.
For $y=1/I$ , the absolute uncertainty grows like $1/I^2$ .
If the table uses a scale factor (here $10^3$ ), apply it to both the value and its uncertainty.

Common Mistakes

Forgetting to convert $\mu\text{A}$ to $\text{A}$ before taking the reciprocal.
Using percentage uncertainty rules without first finding the correct absolute uncertainty for $1/I$ .
Scaling the value by $10^3$ but not scaling the uncertainty by the same factor.

Things to Be Careful About

Keep units consistent: $1/I$ has units $\text{A}^{-1}$ .
Quote uncertainties to 1–2 significant figures and match the decimal place of the value sensibly.
Use $I^2$ in the denominator for $\Delta(1/I)$ (not just $I$ ).

Techniques used

convert microampere readings into amperescalculate reciprocal valuespropagate absolute uncertainty through a reciprocal functionpresent values with consistent significant figures and units

(c)

(i)

Plot a graph of $\frac{1}{I} / 10^3\ \text{A}^{-1}$ against $\frac{1}{n}$ . Include error bars for $\frac{1}{I}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Points plotted ( $x=1/n$ , $y=1/I\,/10^3\ \text{A}^{-1}$ ) with vertical error bars $\pm \Delta y$ :

$(0.200,\ 2.20)\pm 0.024$

$(0.167,\ 1.90)\pm 0.018$

$(0.143,\ 1.72)\pm 0.015$

$(0.125,\ 1.57)\pm 0.012$

$(0.111,\ 1.46)\pm 0.011$

$(0.0909,\ 1.31)\pm 0.0085$

Final answer

Graph plotted with the six points and vertical error bars using the calculated ± uncertainties in 1/I.

Detailed explanation

Background Concept

When plotting experimental data, you should:

label axes with the quantity and unit,
use a sensible scale that uses a large fraction of the grid,
plot points as small crosses/dots,
include error bars when uncertainties are given.

An error bar shows the range of possible true values based on the uncertainty. Here the uncertainty is in $1/I$ , so the error bars are vertical.

Understanding the Question

You must plot $y = 1/I\,/10^3\ \text{A}^{-1}$ against $x = 1/n$ using the values from your completed table, and include the error bars for $1/I$ from part (b).

Approach

Put $1/n$ on the horizontal axis and $1/I\,/10^3\ \text{A}^{-1}$ on the vertical axis (as stated).
Plot each coordinate pair from the table.
For each point, draw a vertical error bar of length $\pm \Delta(1/I\,/10^3)$ .

Step-by-Step Reasoning

Using the calculated values:

At $x=0.200$ , plot $y\approx 2.20$ and draw an error bar from $2.20-0.024$ to $2.20+0.024$ .
Repeat similarly for the remaining five points.

Because $1/n$ is given to 3–4 s.f. and has no uncertainty stated, no horizontal error bars are required.

Key Takeaways

Error bars go on the axis corresponding to the uncertain quantity.
Always match the plotted quantities to the axis labels given in the question.

Common Mistakes

Plotting $I$ instead of $1/I$ .
Drawing horizontal error bars when only $1/I$ has an uncertainty.
Forgetting the $10^3$ scale on the $y$ -axis and plotting values 1000 times too large.

Things to Be Careful About

Make sure the points are plotted with consistent rounding (don’t over-round before plotting).
Error bars should be centred on the plotted point and drawn neatly with small caps.

Techniques used

choose appropriate axis scales and labels for a graphplot data points accurately from a tableadd vertical error bars using absolute uncertainties

(ii)

Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines.

2M Similar

DifficultyMedium

Worked solution

Answer

A straight line of best fit is drawn through the data.

A worst acceptable straight line is drawn such that it passes through all error bars.

Both lines are labelled.

Final answer

Best-fit line and worst acceptable line drawn and labelled.

Detailed explanation

Background Concept

A best-fit line is the straight line that best represents the trend of the data (roughly equal scatter of points above and below). When uncertainties are shown with error bars, the “true” line could lie anywhere within those bars.

A worst acceptable line is an extreme line (steepest or shallowest) that is still consistent with the data within their error bars. This is used to estimate the uncertainty in the gradient and intercept.

Understanding the Question

You have already plotted points with vertical error bars. Now you must:

draw the best-fit straight line,
draw one worst acceptable straight line,
label both on the graph.

Approach

Draw the best-fit line: do not join point-to-point; draw one straight line.
Draw a worst acceptable line: choose an extreme slope but ensure the line still passes through every vertical error bar (or at least is consistent with them as required by the paper).
Label each line clearly (e.g. “best fit” and “worst”).

Step-by-Step Reasoning

Best fit: aim for a balanced distribution of points above and below the line, and do not force it through the origin unless the data demand it.
Worst acceptable: try drawing the steepest possible line (high on the left, low on the right) that still intersects each error bar; if that is not the greatest deviation from the best-fit gradient, try the shallowest possible line instead.

Key Takeaways

Best-fit line uses the central trend.
Worst acceptable line uses the uncertainty limits to estimate how much the gradient/intercept could vary.

Common Mistakes

Drawing a “worst line” that ignores some error bars.
Drawing two worst lines when only one is asked for.
Not labelling the two lines (costs marks).

Things to Be Careful About

Use a ruler; the lines must be straight.
The worst acceptable line should be noticeably different in slope from the best-fit line, but still acceptable given the error bars.

Techniques used

draw a straight line of best fit by balancing scatterdraw a worst acceptable line consistent with all error barslabel lines clearly on a graph

(iii)

Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.

gradient = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using two well-separated points on the best-fit line, e.g.
$(0.200,\ 2.20)$ and $(0.0909,\ 1.31)$ :

\text{gradient}_{\text{best}}=\frac{2.20-1.31}{0.200-0.0909}=8.15

Worst acceptable line (extreme):

\text{gradient}_{\text{worst}}\approx 8.46

\Delta(\text{gradient})=|8.46-8.15|=0.31

Answer

\text{gradient} = 8.15 \pm 0.31

(in units of $10^3\ \text{A}^{-1}$ )

Final answer

8.15 ± 0.31 (in units of 10^3 A^-1)

Detailed explanation

Background Concept

The gradient of a straight-line graph is

m = \frac{\Delta y}{\Delta x}

To reduce the percentage reading error, you choose two points far apart on the drawn line (large triangle). When using a worst acceptable line, you do the same and then compare gradients.

The absolute uncertainty in the gradient (Paper 5 method) is typically taken as

\Delta m = |m_{\text{best}} - m_{\text{worst}}|

where $m_{\text{worst}}$ is the gradient of the worst acceptable line.

Understanding the Question

You must find:

the gradient of the best-fit straight line on the graph of $y=1/I\,/10^3\ \text{A}^{-1}$ against $x=1/n$ ,
and its absolute uncertainty using the worst acceptable line.

Approach

Pick two widely separated points that lie on the best-fit line (not necessarily data points).
Compute $m_{\text{best}} = \Delta y / \Delta x$ .
Do the same for the worst acceptable line to get $m_{\text{worst}}$ .
Take $\Delta m = |m_{\text{best}} - m_{\text{worst}}|$ .

Step-by-Step Reasoning

From the best-fit line, taking approximately $(0.200, 2.20)$ and $(0.0909, 1.31)$ :

\Delta y = 2.20-1.31 = 0.89

\Delta x = 0.200-0.0909 = 0.1091

m_{\text{best}} = \frac{0.89}{0.1091} \approx 8.15

For a steepest acceptable line, you would use the extreme ends of the error bars to maximise slope (high $y$ at high $x$ , low $y$ at low $x$ ), giving a gradient around $8.46$ . Then

\Delta m = |8.46-8.15|=0.31

So the gradient is $8.15 \pm 0.31$ (in the plotted units).

Key Takeaways

Use a large triangle on the line for gradient.
Uncertainty comes from comparing best and worst acceptable lines.

Common Mistakes

Using two adjacent points, giving a large reading uncertainty.
Using $\Delta x/\Delta y$ by accident.
Finding the gradient from the raw data points rather than from the drawn line.

Things to Be Careful About

Make sure you are reading coordinates from the line itself.
Keep consistent units: here the plotted $y$ values are already in $10^3\ \text{A}^{-1}$ .

Techniques used

determine gradient from a large triangle on a best-fit linedetermine gradient of a worst acceptable linecalculate absolute uncertainty from best and worst gradients

(iv)

Determine the $y$ -intercept of the line of best fit. Include the absolute uncertainty in your answer.

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

For best-fit line, using $y = mx + c$ with $m=8.15$ and a point on the line, e.g. $(0.200,\ 2.20)$ :

c_{\text{best}} = 2.20 - (8.15)(0.200) = 0.56

Worst acceptable line gives intercept $c_{\text{worst}} \approx 0.600$ .

\Delta c = |0.600-0.558| \approx 0.042

Answer

\text{$y$-intercept} = 0.56 \pm 0.04

(in units of $10^3\ \text{A}^{-1}$ )

Final answer

0.56 ± 0.04 (in units of 10^3 A^-1)

Detailed explanation

Background Concept

For a straight line

y = mx + c

the $y$ -intercept $c$ is the value of $y$ when $x=0$ . On a graph you can often read it directly by extending the line to the $y$ -axis, or calculate it using a point on the line:

c = y - mx

Uncertainty in intercept is found similarly to gradient, by comparing the best-fit and worst acceptable lines:

\Delta c = |c_{\text{best}} - c_{\text{worst}}|

Understanding the Question

You need the intercept of the best-fit line for the plot of $1/I\,/10^3\ \text{A}^{-1}$ against $1/n$ , and the absolute uncertainty using the worst acceptable line.

Approach

Find $c_{\text{best}}$ either by reading off at $x=0$ or by using $c=y-mx$ .
Find $c_{\text{worst}}$ from the worst acceptable line in the same way.
Take $\Delta c = |c_{\text{best}}-c_{\text{worst}}|$ .

Step-by-Step Reasoning

Using $m_{\text{best}}\approx 8.15$ and a point on the best-fit line such as $(x,y)=(0.200,2.20)$ :

c_{\text{best}} = 2.20 - (8.15)(0.200) \approx 0.56

For the worst acceptable line, the intercept might be about $0.600$ (from the line you drew). Then

\Delta c \approx |0.600-0.558| \approx 0.042

So quote $c = 0.56 \pm 0.04$ in the plotted units.

Key Takeaways

Intercept can be found by extending to $x=0$ or by $c=y-mx$ .
Uncertainty is from best vs worst acceptable line.

Common Mistakes

Reading the intercept from the wrong axis or at the wrong $x$ value.
Using a data point rather than a point on the drawn best-fit line.
Forgetting that the axis is scaled ( $10^3\ \text{A}^{-1}$ ).

Things to Be Careful About

Choose a point on the line with clearly readable coordinates.
Keep enough significant figures so that subtracting $mx$ from $y$ is not dominated by rounding.

Techniques used

read the y-intercept from the best-fit linedetermine the intercept using y = mx + cestimate intercept uncertainty using a worst acceptable line

(d)

The e.m.f. $E$ of the battery is determined twice during the experiment. The values obtained are $5.6\ \text{V}$ and $6.0\ \text{V}$ .

(i)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $R$ and $Z$ . Include appropriate units.

$R$ = ______
$Z$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

Mean e.m.f.:

E = \frac{5.6+6.0}{2}=5.8\ \text{V}

From (a):

\text{gradient (for }1/I\text{ vs }1/n) = \frac{R}{E},\quad \text{intercept} = \frac{Z}{E}

But graph uses $\left(\frac{1}{I}\right)/10^3$ , so

\frac{1}{I} = (8.15\times 10^3)\left(\frac{1}{n}\right) + (0.558\times 10^3)

Hence

R = E(8.15\times 10^3)=5.8\times 8.15\times 10^3=4.73\times 10^4\ \Omega

Z = E(0.558\times 10^3)=5.8\times 0.558\times 10^3=3.24\times 10^3\ \Omega

Answer

R = 4.7\times 10^4\ \Omega

Z = 3.2\times 10^3\ \Omega

Final answer

R = 4.7×10^4 Ω, Z = 3.2×10^3 Ω

Detailed explanation

Background Concept

From the linearised form

\frac{1}{I} = \frac{R}{E}\left(\frac{1}{n}\right) + \frac{Z}{E}

we can identify:

gradient (of a graph of $1/I$ against $1/n$ ) is $R/E$ ,
intercept is $Z/E$ .

If the graph’s $y$ -axis is scaled (here it plots $(1/I)/10^3$ ), then the numerical gradient and intercept you read must be converted back to the unscaled $1/I$ values by multiplying by $10^3$ .

Understanding the Question

You have measured $E$ twice (5.6 V and 6.0 V), and from the graph you have the gradient and intercept (with the $10^3$ scaling). You must use these to calculate the actual resistor values $R$ and $Z$ .

Approach

Use the mean of the two $E$ readings as the best estimate.
Convert the gradient and intercept from plotted units back to actual $1/I$ units by multiplying by $10^3$ .
Use $R = E\times (\text{gradient})$ and $Z = E\times (\text{intercept})$ .

Step-by-Step Reasoning

Mean e.m.f.:

E = \frac{5.6+6.0}{2} = 5.8\ \text{V}

Suppose the best-fit line has gradient $8.15$ and intercept $0.558$ on the axis labelled $1/I\,/10^3\ \text{A}^{-1}$ . This means

actual gradient for $1/I$ is $8.15\times 10^3\ \text{A}^{-1}$ ,
actual intercept is $0.558\times 10^3\ \text{A}^{-1}$ .

Now apply the identifications:

\frac{R}{E} = 8.15\times 10^3 \quad\Rightarrow\quad R = E(8.15\times 10^3)

R = 5.8\times 8.15\times 10^3 \approx 4.73\times 10^4\ \Omega

Similarly,

\frac{Z}{E} = 0.558\times 10^3 \quad\Rightarrow\quad Z = E(0.558\times 10^3)

Z \approx 5.8\times 0.558\times 10^3 \approx 3.24\times 10^3\ \Omega

Key Takeaways

Convert graph readings back to true units if the axis uses a scale factor.
Gradient gives $R/E$ and intercept gives $Z/E$ .
Multiplying by $E$ returns resistances in ohms.

Common Mistakes

Forgetting the $10^3$ factor from the $y$ -axis and obtaining resistances 1000 times too small.
Using only one of the $E$ readings without any justification.
Swapping $R$ and $Z$ (mixing up gradient and intercept).

Things to Be Careful About

Use consistent significant figures (typically 2–3 s.f.).
Ensure you treat $E$ in volts and $1/I$ in $\text{A}^{-1}$ so that $R$ and $Z$ come out in $\Omega$ .

Techniques used

use mean value from repeated measurementsrelate gradient and intercept to physical constantsapply scaling factors from graph axis units

(ii)

Determine the percentage uncertainty in your value of $R$ .

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium

Worked solution

Working

For $E=5.6\ \text{V}$ and $6.0\ \text{V}$ :

E = 5.8\ \text{V},\quad \Delta E = \frac{6.0-5.6}{2}=0.2\ \text{V}

\%\Delta E = \frac{0.2}{5.8}\times 100 = 3.4\%

From (c)(iii): $m = 8.15 \pm 0.31$

\%\Delta m = \frac{0.31}{8.15}\times 100 = 3.8\%

Since $R \propto Em$ ,

\%\Delta R = \%\Delta E + \%\Delta m = 3.4+3.8 = 7.2\%

Answer

\text{percentage uncertainty in }R \approx 7.2\%

Final answer

7.2%

Detailed explanation

Background Concept

If a quantity is found by multiplying two measured quantities, e.g.

R = Em

then the percentage (fractional) uncertainties add:

\frac{\Delta R}{R} = \frac{\Delta E}{E} + \frac{\Delta m}{m}

To estimate uncertainty in a mean from two repeated readings, a common Paper 5 method is to take half the range:

\Delta E = \frac{E_{\max}-E_{\min}}{2}

Understanding the Question

You found $R$ from the gradient and $E$ . The gradient has an uncertainty from the best/worst line method, and $E$ was measured twice (5.6 V and 6.0 V). You must find the percentage uncertainty in $R$ .

Approach

Use the mean $E$ as the best estimate and half the range as $\Delta E$ .
Convert both $\Delta E$ and $\Delta m$ into percentage uncertainties.
Add them because $R$ is proportional to the product $Em$ .

Step-by-Step Reasoning

Mean e.m.f.:

E = \frac{5.6+6.0}{2} = 5.8\ \text{V}

Uncertainty in $E$ (half-range):

\Delta E = \frac{6.0-5.6}{2} = 0.2\ \text{V}

Percentage uncertainty in $E$ :

\%\Delta E = \frac{0.2}{5.8}\times 100 \approx 3.4\%

From the graph, suppose

m = 8.15 \pm 0.31

Then

\%\Delta m = \frac{0.31}{8.15}\times 100 \approx 3.8\%

Since $R = Em$ (after handling the axis scaling consistently),

\%\Delta R = 3.4\% + 3.8\% = 7.2\%

Key Takeaways

For products, percentage uncertainties add.
Two repeated readings can give an uncertainty estimate using half the range.
Use the uncertainty in the gradient from best vs worst lines.

Common Mistakes

Using the full range $0.4\ \text{V}$ instead of half-range $0.2\ \text{V}$ .
Combining absolute uncertainties directly (must use percentage for multiplication).
Ignoring the uncertainty in $E$ even though repeat values are given.

Things to Be Careful About

Quote the final percentage uncertainty to a sensible number of significant figures (usually 2 s.f.).
Make sure you use the same best estimate of $E$ in the uncertainty calculation as you used when calculating $R$ .

Techniques used

determine uncertainty in a mean value from repeated readingscombine percentage uncertainties for quantities multiplied togethercalculate percentage uncertainty from absolute uncertainty

(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.

$I$ = ______ $\text{A}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $E=5.8\ \text{V}$ , $R=4.73\times 10^4\ \Omega$ , $Z=3.24\times 10^3\ \Omega$ and $n=20$ :

I = \frac{E}{\frac{R}{n}+Z}

\frac{R}{n} = \frac{4.73\times 10^4}{20} = 2.37\times 10^3\ \Omega

I = \frac{5.8}{(2.37\times 10^3 + 3.24\times 10^3)} = 1.04\times 10^{-3}\ \text{A}

Answer

I = 1.04\times 10^{-3}\ \text{A}

Final answer

1.04×10^-3 A

Detailed explanation

Background Concept

The total resistance in series adds. Here, the circuit has:

a parallel network of $n$ identical resistors of resistance $R$ , giving equivalent resistance $R/n$ ,
in series with a fixed resistor $Z$ .

So the total resistance is

R_{\text{total}} = \frac{R}{n} + Z

and with negligible internal resistance the current is

I = \frac{E}{R_{\text{total}}}

Understanding the Question

You increase the number of parallel resistors to $n=20$ (keeping the same $R$ for each resistor). Using your previously determined values of $R$ and $Z$ and the battery e.m.f. $E$ , you must predict the new total current $I$ .

Approach

Calculate the equivalent parallel resistance $R/n$ for $n=20$ .
Add $Z$ to get the total series resistance.
Use $I = E/(R/n + Z)$ .

Step-by-Step Reasoning

With $n=20$ :

\frac{R}{n} = \frac{4.73\times 10^4}{20} = 2.37\times 10^3\ \Omega

Total resistance:

R_{\text{total}} = 2.37\times 10^3 + 3.24\times 10^3 = 5.61\times 10^3\ \Omega

Current:

I = \frac{5.8}{5.61\times 10^3} = 1.04\times 10^{-3}\ \text{A}

Key Takeaways

More parallel resistors reduces $R/n$ , increasing current.
Use series-plus-parallel simplification before substituting into $I=E/R$ .

Common Mistakes

Using $nR$ instead of $R/n$ for parallel resistors.
Forgetting that $Z$ is in series and must be added to $R/n$ .
Mixing $\text{k}\Omega$ and $\Omega$ units.

Things to Be Careful About

Keep values in $\Omega$ when adding resistances.
Quote $I$ in amperes; check order of magnitude (here about $10^{-3}\ \text{A}$ matches the earlier $\mu\text{A}$ – $\text{mA}$ scale).

Techniques used

apply the circuit model equation to a new value of ncalculate equivalent resistance of identical resistors in parallelsubstitute numerical values to predict a current

Physics 9702/51 — May/June 2025

Answer

Variables

Apparatus

Procedure and measurements

Analysis of data (test of relationship and finding KKK)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer (values for Table 2.1)

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Analysis of data (test of relationship and finding $K$ )