A student investigates an electrical circuit. A power supply with negligible internal resistance is connected to six resistors, each of resistance $Z$, and a resistor of resistance $R$, as shown in Fig. 2.1. The current measured by the ammeter is $I$. The experiment is repeated for different values of $R$. It is suggested that $I$ and $R$ are related by the equation $$ E = 3IR + 4IZ $$ where $E$ is the electromotive force (e.m.f.) of the power supply. (a) A graph is plotted…

(a) gradient = 3/E, y-intercept = 4Z/E · (b) 1/I = (4440±100), (5410±150), (6250±200), (7140±260), (8000±320), (8700±380) A^-1 · (c)(i) See graph · (c)(ii) See graph with best-fit and worst acceptable lines · (c)(iii) (6.5×10^2 ± 7×10^1) A^-1 kΩ^-1 · (c)(iv) (3.7×10^3 ± 3×10^2) A^-1 · (d)(i) E = 4.6 V, Z = 4.3×10^3 Ω · (d)(ii) 19% · (e) 9.7×10^3 Ω

Physics 9702/52 — February/March 2024

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

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q115MMedium-HardPlanningPlanning

Fig. 1.1

Fig. 1.1 shows a thin cylindrical metal rod of length $L$ .

One end of the rod is hit with a hammer. A stationary sound wave is set up within the rod. The rod vibrates at its resonant frequency $f$ .

A microphone placed at the other end of the rod detects the sound wave emitted from the rod. The frequency of the detected sound is also $f$ .

A number of rods of different length are available.

It is suggested that $f$ is related to $L$ by the relationship

2fL^n = \sqrt{\frac{E}{\rho}}

where $\rho$ is the density of the metal, and $E$ and $n$ are constants.

Plan a laboratory experiment to test the relationship between $f$ and $L$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $E$ and $n$ .

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

Procedure / measurements

Choose rods of the same metal and (as far as possible) the same diameter so that $\rho$ and $E$ are constant; vary only $L$ .
Measure $L$ of each rod using a metre rule/vernier calipers (same method each time).
Support the rod at its centre (light clamp/soft supports) so the ends are free.
Strike one end with a small hammer in the same way each time.
Place a microphone close to the other end and connect to an oscilloscope / data-logger / frequency counter.
Record the resonant frequency $f$ from the microphone signal (e.g. from oscilloscope time-base or FFT peak).
Repeat strikes (e.g. 3 times) for each rod and take the mean $f$ .

Control of variables

Same material (constant $\rho$ and $E$ ) and similar diameter (to avoid changing modes/damping).
Same support position (centre), same microphone position, and similar striking method.
Keep temperature as constant as possible.

Determination of $\rho$

Measure $\rho$ for the metal (or use a data-book value):

Measure mass $m$ on a balance.
Measure diameter $d$ with a micrometer and use $A = \pi d^2/4$ .
Compute volume $V = AL$ and then $\rho = m/V$ .

Analysis to determine $n$ and $E$

Given

2fL^n = \sqrt{\frac{E}{\rho}}

Take base-10 logs:

\lg f = -n\,\lg L + \frac{1}{2}\lg\left(\frac{E}{\rho}\right) - \lg 2

Calculate $\lg f$ and $\lg L$ for each rod.
Plot $\lg f$ (y-axis) against $\lg L$ (x-axis).
Gradient $m = -n \Rightarrow n = -m$ .
Intercept $c = \frac{1}{2}\lg\left(\frac{E}{\rho}\right) - \lg 2$ .
Hence

\frac{1}{2}\lg\left(\frac{E}{\rho}\right) = c + \lg 2

\lg\left(\frac{E}{\rho}\right) = 2(c+\lg 2) \quad\Rightarrow\quad E = \rho\,10^{\,2(c+\lg 2)}

Safety

Wear eye protection; ensure the rod is securely supported so it cannot fly out.
Keep fingers away from impact point; use a small hammer and controlled ضرب.
Consider ear protection / keep sound level moderate.

Final answer

See working

Detailed explanation

Background Concept

A metal rod can support longitudinal stationary (standing) waves. At resonance, the rod vibrates strongly at a particular frequency $f$ determined by its length and the wave speed in the rod. The wave speed $v$ in a solid (for longitudinal waves) depends on material properties, and is commonly related to Young modulus $E$ and density $\rho$ by a relationship of the form

v \propto \sqrt{\frac{E}{\rho}}

The question gives the suggested relationship

2 f L^n = \sqrt{\frac{E}{\rho}}

where $E$ and $n$ are constants for the situation. To test such a model experimentally, you vary $L$ and measure $f$ , while keeping all other relevant factors constant.

A key idea in Paper 5 is: when a relationship involves a power of a variable (here $L^n$ ), it is usually best tested by taking logarithms to turn it into a straight-line form.

Understanding the Question

You have several rods with different lengths $L$ . You excite a resonance by striking one end, and a microphone at the other end measures the frequency $f$ emitted.

You must:

give a practical method to measure $f$ reliably for each $L$ ,
identify what variables must be controlled so that changes in $f$ are due only to changes in $L$ ,
explain a graph/data method that allows you to find both unknown constants $n$ and $E$ .

Because $E$ appears together with $\rho$ in the expression, you also need a value of $\rho$ (either measured or taken from reliable reference data).

Approach

Choose variables: $L$ is the independent variable; $f$ is the dependent variable.
Collect data: for each rod length, excite resonance and measure $f$ using an electronic method (oscilloscope/FFT/frequency counter). Repeat to obtain a mean.
Control variables: ensure the rods are the same material and similar geometry; keep the measurement arrangement and conditions the same.
Linearise: take logs of the given equation to get a straight-line graph of $\lg f$ vs $\lg L$ .
Extract constants: gradient gives $n$ ; intercept gives $E$ (after using $\rho$ ).

Step-by-Step Reasoning

diagram-1

1) Data collection (getting $f$ )

When the rod is struck, the microphone produces a voltage signal that oscillates at the sound frequency.
An oscilloscope can measure the period $T$ from the trace, then $f = 1/T$ .
Alternatively, many data loggers/oscilloscopes can display an FFT spectrum, where the resonance shows as a peak; the peak frequency is $f$ .
Repeating the strike reduces random variation (different strike strengths and background noise), so you average several values of $f$ .

2) Controlling variables

To test the relationship between $f$ and $L$ , you should keep constant anything else that could affect the resonant frequency:

Material: $E$ and $\rho$ depend on the material, so all rods should be the same metal.
Cross-sectional area/diameter: large changes can affect damping and which modes are most easily excited/detected; keeping diameter similar makes the comparison fair.
Support condition: boundary conditions affect resonant frequencies. Supporting each rod in the same way (e.g. at its centre) keeps the mode pattern comparable.
Microphone position: keeping the microphone distance/orientation fixed keeps signal strength comparable.
Temperature: material properties (especially $E$ ) can change slightly with temperature; keep room conditions steady.

3) Determining $\rho$

Since the final goal is $E$ , you need $\rho$ .

If the metal is known, you may use a data-book value.
A good experimental approach is to measure:
- mass $m$ with a balance,
- diameter $d$ with a micrometer (best for small diameters),
- compute cross-sectional area $A = \pi d^2/4$ ,
- volume $V = AL$ ,
- density $\rho = m/V$ .
  Measuring $\rho$ on one representative rod is often sufficient if all rods are from the same stock.

4) Linearising to find $n$ and $E$

Start with the given model:

2fL^n = \sqrt{\frac{E}{\rho}}

Take logs (base 10, but natural logs would work similarly):

\lg(2fL^n) = \lg\left(\sqrt{\frac{E}{\rho}}\right)

Use log laws: $\lg(ab)=\lg a+\lg b$ , $\lg(L^n)=n\lg L$ , and $\lg(\sqrt{x})=\frac{1}{2}\lg x$ :

\lg 2 + \lg f + n\lg L = \frac{1}{2}\lg\left(\frac{E}{\rho}\right)

Rearrange into $y = mx + c$ form with $y=\lg f$ and $x=\lg L$ :

\lg f = -n\lg L + \frac{1}{2}\lg\left(\frac{E}{\rho}\right) - \lg 2

So, plotting $\lg f$ (y) against $\lg L$ (x):

the gradient $m = -n$ so $n=-m$ ,
the intercept $c = \frac{1}{2}\lg\left(\frac{E}{\rho}\right) - \lg 2$ .

Then solve for $E$ :

\frac{1}{2}\lg\left(\frac{E}{\rho}\right) = c + \lg 2

\lg\left(\frac{E}{\rho}\right) = 2(c+\lg 2)

\frac{E}{\rho} = 10^{2(c+\lg 2)} \quad\Rightarrow\quad E = \rho\,10^{2(c+\lg 2)}

This uses the straight-line graph to determine both constants from experimental data.

Key Takeaways

In a planning question, clearly state independent, dependent, and controlled variables.
Use appropriate sensors (microphone + oscilloscope/FFT) for reliable frequency measurement.
For relationships involving powers, use a log-log plot: gradient gives the power, intercept gives the constant.
If a constant includes material properties (here $E$ and $\rho$ ), explain how $\rho$ will be obtained.

Common Mistakes

Not controlling the material: using different metals would change $\rho$ and $E$ , invalidating the test.
Measuring frequency vaguely (“listen for pitch”) rather than using an instrument (oscilloscope/FFT/counter).
Plotting $f$ against $L$ directly; this will not give a straight line if $n \neq 1$ .
Forgetting to explain how to obtain $E$ from the intercept (many stop at finding $n$ ).
Giving generic safety (“be careful”) instead of specific hazards: hammer impact, rod slipping, loud sound.

Things to Be Careful About

Ensure the log graph is correctly defined: axes must be $\lg f$ vs $\lg L$ (not $f$ vs $\lg L$ , etc.).
Keep significant figures consistent in calculated log values and in the final constants.
Check units conceptually: $E$ should come out in pascals ( $\text{Pa}$ ).
Resonance detection: the largest peak may include overtones; use a consistent criterion (e.g. dominant peak) and repeat readings.
Rod support: clamping too tightly can damp vibrations and shift resonance; use light support at the same position each time.

Techniques used

identify independent, dependent and controlled variablesmeasure frequency using a microphone with an oscilloscope or FFT/data loggermeasure density from mass and dimensionslinearise a power law using logarithmsdetermine constants from the gradient and intercept of a log-log graph

Working

Given

E = 3IR + 4IZ

Divide by $I$ :

\frac{E}{I} = 3R + 4Z

\frac{1}{I} = \frac{3}{E}R + \frac{4Z}{E}

Answer

gradient $= \dfrac{3}{E}$

$y$ -intercept $= \dfrac{4Z}{E}$

Final answer

gradient = 3/E, y-intercept = 4Z/E

Detailed explanation

Background Concept

To analyse experimental data with a straight-line graph, we try to rewrite the given relationship in the form

y = mx + c

where:

$m$ is the gradient (slope),
$c$ is the $y$ -intercept.

Here, the student plots $\frac{1}{I}$ on the $y$ -axis and $R$ on the $x$ -axis, so we want an expression of the form

\frac{1}{I} = (\text{something})\,R + (\text{something}).

Understanding the Question

You are told that

E = 3IR + 4IZ

and the graph is $\frac{1}{I}$ against $R$ . So you must rearrange the equation to make $\frac{1}{I}$ the subject, and then read off:

the coefficient of $R$ as the gradient,
the constant term as the $y$ -intercept.

Approach

Divide the whole equation by $I$ to create $\frac{E}{I}$ .
Divide by $E$ to isolate $\frac{1}{I}$ .
Compare with $y = mx + c$ .

Step-by-Step Reasoning

Start with

E = 3IR + 4IZ.

Divide every term by $I$ :

\frac{E}{I} = 3R + 4Z.

Now divide by $E$ :

\frac{1}{I} = \frac{3}{E}R + \frac{4Z}{E}.

Comparing with $y = mx + c$ where $y = \frac{1}{I}$ and $x = R$ :

gradient $m = \frac{3}{E}$ ,
intercept $c = \frac{4Z}{E}$ .

Key Takeaways

Rearrange into $y = mx + c$ using the variables on the axes.
Gradient is the coefficient of the $x$ -variable; intercept is the constant term.

Common Mistakes

Making $I$ the subject instead of $\frac{1}{I}$ .
Stating gradient as $\frac{E}{3}$ (inverting incorrectly).
Missing that the intercept contains $Z$ .

Things to Be Careful About

If later the graph uses $R$ in $\text{k}\Omega$ , the numerical gradient changes by a factor of $1000$ (because $R = 1000\,(R/\text{k}\Omega)$ ).

Techniques used

rearrange an equation into straight-line formidentify gradient and intercept by comparison with $y = mx + c$

(b)

Values of $R$ and $I$ are given in Table 2.1.

Table 2.1

$R / \text{k}\Omega$	$I / \mu\text{A}$	$\frac{1}{I} / \text{A}^{-1}$
1.25	$225 \pm 5$
2.55	$185 \pm 5$
3.90	$160 \pm 5$
5.25	$140 \pm 5$
6.55	$125 \pm 5$
7.80	$115 \pm 5$

Calculate and record values of $\frac{1}{I} / \text{A}^{-1}$ in Table 2.1.

Include the absolute uncertainties in $\frac{1}{I}$ .

2M Similar

DifficultyMedium

Worked solution

Working

For $y = \dfrac{1}{I}$ ,

\frac{\Delta y}{y} = \frac{\Delta I}{I} \quad\Rightarrow\quad \Delta\left(\frac{1}{I}\right)=\frac{1}{I}\left(\frac{\Delta I}{I}\right).

(1) $I = (225 \pm 5)\,\mu\text{A} = (225 \pm 5)\times 10^{-6}\,\text{A}$

\frac{1}{I}=4.44\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=1.0\times 10^{2}\,\text{A}^{-1}

(2) $I = (185 \pm 5)\,\mu\text{A}$

\frac{1}{I}=5.41\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=1.5\times 10^{2}\,\text{A}^{-1}

(3) $I = (160 \pm 5)\,\mu\text{A}$

\frac{1}{I}=6.25\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=2.0\times 10^{2}\,\text{A}^{-1}

(4) $I = (140 \pm 5)\,\mu\text{A}$

\frac{1}{I}=7.14\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=2.6\times 10^{2}\,\text{A}^{-1}

(5) $I = (125 \pm 5)\,\mu\text{A}$

\frac{1}{I}=8.00\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=3.2\times 10^{2}\,\text{A}^{-1}

(6) $I = (115 \pm 5)\,\mu\text{A}$

\frac{1}{I}=8.70\times 10^{3}\,\text{A}^{-1},\quad \Delta\left(\frac{1}{I}\right)=3.8\times 10^{2}\,\text{A}^{-1}

Answer

Record in Table 2.1:

$4440 \pm 100\ \text{A}^{-1}$
$5410 \pm 150\ \text{A}^{-1}$
$6250 \pm 200\ \text{A}^{-1}$
$7140 \pm 260\ \text{A}^{-1}$
$8000 \pm 320\ \text{A}^{-1}$
$8700 \pm 380\ \text{A}^{-1}$

Final answer

1/I = (4440±100), (5410±150), (6250±200), (7140±260), (8000±320), (8700±380) A^-1

Detailed explanation

Background Concept

When a measured quantity is transformed, its uncertainty must also be transformed.

For a reciprocal:

y = \frac{1}{I} = I^{-1}

the fractional (percentage) uncertainty stays the same in magnitude:

\left|\frac{\Delta y}{y}\right| = \left|\frac{\Delta I}{I}\right|.

So the absolute uncertainty in $\frac{1}{I}$ is

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

Understanding the Question

The table gives $I$ in $\mu\text{A}$ with an absolute uncertainty of $\pm 5\,\mu\text{A}$ each time. You must:

Convert each $I$ into amperes.
Calculate $\frac{1}{I}$ in $\text{A}^{-1}$ .
Calculate and record the absolute uncertainty in $\frac{1}{I}$ .

Approach

For each row:

Compute $y = 1/I$ .
Compute fractional uncertainty $\Delta I/I$ .
Multiply: $\Delta y = y(\Delta I/I)$ .
Round $y$ to match the precision implied by $\Delta y$ .

Step-by-Step Reasoning

Example (first row):

I = 225\,\mu\text{A} = 225\times 10^{-6}\,\text{A}

\frac{1}{I} = \frac{1}{225\times 10^{-6}} = 4444\,\text{A}^{-1}

Fractional uncertainty:

\frac{\Delta I}{I} = \frac{5}{225} = 0.0222

\Delta\left(\frac{1}{I}\right)=4444\times 0.0222 \approx 99\,\text{A}^{-1}.

You repeat the same method for every row.

Rounding: since the uncertainty is around $10^2\,\text{A}^{-1}$ , quoting $1/I$ to the nearest $10\,\text{A}^{-1}$ is sensible (e.g. $4440 \pm 100\,\text{A}^{-1}$ ).

Key Takeaways

For $y = 1/x$ , fractional uncertainty in $y$ equals fractional uncertainty in $x$ .
Always convert to SI units (here $\mu\text{A} \to \text{A}$ ) before calculating.
Round the value to be consistent with the uncertainty.

Common Mistakes

Forgetting to convert $\mu\text{A}$ to $\text{A}$ , giving answers off by $10^6$ .
Using $\Delta(1/I)=1/\Delta I$ (incorrect).
Rounding $\Delta(1/I)$ to too many significant figures.

Things to Be Careful About

Keep uncertainties as absolute uncertainties in $\text{A}^{-1}$ (needed for error bars later).
Ensure the unit in the table is $\text{A}^{-1}$ , not $\mu\text{A}^{-1}$ .

Techniques used

convert microampere readings to amperescalculate reciprocal quantitiespropagate absolute uncertainty through a reciprocal using fractional uncertaintiesround values to match uncertainty precision

(c)

(i)

Plot a graph of $\frac{1}{I} / \text{A}^{-1}$ against $R / \text{k}\Omega$ . Include error bars for $\frac{1}{I}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the six points of $\left(R/\text{k}\Omega,\ 1/I\right)$ from Table 2.1 on Graph 2.1.

Label axes: $R/\text{k}\Omega$ (x-axis) and $1/I\ /\ \text{A}^{-1}$ (y-axis).

Draw vertical error bars for each point using the absolute uncertainties in $1/I$ .

Final answer

See graph

Detailed explanation

Background Concept

A good physics graph must:

have axes labelled with quantity and unit,
use a sensible linear scale (usually),
plot points accurately,
show uncertainty using error bars.

Error bars represent the range within which the true value could lie given the measurement uncertainty. Here, only $1/I$ has a stated uncertainty, so only vertical error bars are required.

Understanding the Question

You have calculated $1/I$ and its absolute uncertainty for each $R$ . You must transfer this data onto the provided grid:

$x$ -coordinate: $R$ in $\text{k}\Omega$ ,
$y$ -coordinate: $1/I$ in $\text{A}^{-1}$ ,
vertical error bar half-length: $\pm\,\Delta(1/I)$ .

Approach

Mark each point from the table.
For each point, draw an error bar extending up and down by the absolute uncertainty.
Check that plotted points match the table values and lie within the axis limits.

Step-by-Step Reasoning

Choose the given axes (already provided) and ensure your plotting is consistent.
For example, at $R=1.25\,\text{k}\Omega$ , if $1/I = 4440\,\text{A}^{-1}$ with uncertainty $\pm 100\,\text{A}^{-1}$ , the error bar should extend from $4340$ to $4540\,\text{A}^{-1}$ .
Repeat for all six points.

Key Takeaways

Error bars must be drawn using absolute (not percentage) uncertainty.
Only draw error bars in the direction(s) where uncertainty is provided.

Common Mistakes

Drawing horizontal error bars even though no uncertainty in $R$ is given.
Plotting $I$ instead of $1/I$ .
Omitting units on axes.

Things to Be Careful About

Use a sharp pencil and small crosses/dots so the position is unambiguous.
Ensure the full error bar is centred on the plotted point.

Techniques used

choose sensible axis scales that use most of the gridplot data points accurately from a tableadd vertical error bars from absolute uncertaintieslabel axes with quantity and unit

(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

Draw a straight line of best fit through the plotted points.

Draw one worst acceptable straight line (steepest or shallowest) that is consistent with all the error bars.

Label the lines clearly as “best fit” and “worst acceptable”.

Final answer

See graph with best-fit and worst acceptable lines

Detailed explanation

Background Concept

A best-fit line represents the overall trend of the data. When uncertainties are shown with error bars, we can estimate uncertainty in derived quantities (like gradient) by drawing a “worst acceptable” line.

A worst acceptable line is a straight line that:

is still consistent with the data within uncertainty, meaning it passes through all (or at least is consistent with) the error bars,
has the most extreme gradient (either steepest or shallowest) compared with the best-fit line.

Understanding the Question

You have already plotted $1/I$ vs $R$ with vertical error bars. Now you must:

Draw and label the line of best fit.
Draw and label a worst acceptable line (used later for gradient and intercept uncertainties).

Approach

Best-fit line: balance the points so that roughly equal scatter lies above and below the line.
Worst acceptable line: pivot the line to make the gradient as different as possible while still being consistent with the error bars.

Step-by-Step Reasoning

c-ii-best-worst

Draw the best-fit line as a single straight line (do not join dot-to-dot).
For the worst acceptable line, choose either:
- steepest: make the line as steep as possible while still passing within each vertical error bar, or
- shallowest: make the line as shallow as possible while still passing within each vertical error bar.
Label both lines on the graph.

Key Takeaways

The worst acceptable line must be justified by the error bars.
Labelling matters: examiners need to see which line is which.

Common Mistakes

Drawing a curve instead of a straight line.
Drawing a worst line that ignores one or more error bars.
Forgetting to label the lines.

Things to Be Careful About

Use long ruler lines that extend across most of the plotted range.
The worst line should be noticeably different from the best-fit line but still acceptable given the uncertainties.

Techniques used

draw a straight line of best fit through scattered pointsconstruct 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 the best-fit line (large triangle):

\text{gradient } m \approx 6.5\times 10^{2}\ \text{A}^{-1}\text{k}\Omega^{-1}

Using the worst acceptable line:

m_{\text{worst}} \approx 7.2\times 10^{2}\ \text{A}^{-1}\text{k}\Omega^{-1}

Absolute uncertainty:

\Delta m = |m_{\text{worst}}-m| \approx 7\times 10^{1}\ \text{A}^{-1}\text{k}\Omega^{-1}

Answer

m = (6.5\times 10^{2} \pm 7\times 10^{1})\ \text{A}^{-1}\text{k}\Omega^{-1}

Final answer

(6.5×10^2 ± 7×10^1) A^-1 kΩ^-1

Detailed explanation

Background Concept

For a straight line graph of $y$ against $x$ , the gradient is

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

To reduce random reading error, you should use two points far apart on the line (a large triangle).

When error bars are present, the uncertainty in the gradient is estimated by comparing the best-fit gradient with the gradient of a worst acceptable line:

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

Understanding the Question

You have drawn two straight lines on your graph:

best fit,
worst acceptable.

You must calculate the gradient of the best-fit line and quote an absolute uncertainty using the difference between the two gradients.

Approach

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

Step-by-Step Reasoning

c-iii-gradient-triangle

Suppose your large triangle on the best-fit line gives approximately $m_{\text{best}} \approx 6.5\times 10^{2}\,\text{A}^{-1}\text{k}\Omega^{-1}$ .
Using two widely spaced points on the worst acceptable line gives about $m_{\text{worst}} \approx 7.2\times 10^{2}\,\text{A}^{-1}\text{k}\Omega^{-1}$ (a steeper line).
Then

\Delta m \approx |7.2\times 10^{2} - 6.5\times 10^{2}| = 0.7\times 10^{2} = 7\times 10^{1}.

So you report

m = (6.5\times 10^{2} \pm 7\times 10^{1})\ \text{A}^{-1}\text{k}\Omega^{-1}.

(Values may differ slightly depending on your drawn lines; ECF is normally applied.)

Key Takeaways

Use a large triangle for gradient.
Gradient uncertainty comes from best-fit vs worst acceptable line.

Common Mistakes

Calculating gradient using two neighbouring points, giving a large percentage error.
Using points that are not on the line (mixing in raw data points).
Taking uncertainty as half the difference (unless the mark scheme specifically instructs it; Paper 5 typically uses the full difference).

Things to Be Careful About

Include correct units: here $\text{A}^{-1}\text{k}\Omega^{-1}$ .
Read $\Delta x$ in $\text{k}\Omega$ because the axis is $R/\text{k}\Omega$ .

Techniques used

determine gradient from two well-separated points on a best-fit lineuse a large triangle to reduce percentage reading errorestimate gradient uncertainty from best-fit versus worst acceptable line

(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

From the best-fit line, $y$ -intercept

c \approx 3.7\times 10^{3}\ \text{A}^{-1}

From the worst acceptable line,

c_{\text{worst}} \approx 3.4\times 10^{3}\ \text{A}^{-1}

Absolute uncertainty:

\Delta c = |c_{\text{worst}}-c| \approx 3\times 10^{2}\ \text{A}^{-1}

Answer

c = (3.7\times 10^{3} \pm 3\times 10^{2})\ \text{A}^{-1}

Final answer

(3.7×10^3 ± 3×10^2) A^-1

Detailed explanation

Background Concept

The $y$ -intercept is the value of $y$ when $x=0$ . For a straight line

y = mx + c,

$c$ is found by extending the line to the $y$ -axis (at $x=0$ ) and reading off the value.

With error bars, you can estimate uncertainty in $c$ by comparing best-fit and worst acceptable lines:

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

Understanding the Question

You must read the intercept from your best-fit line on the $1/I$ vs $R$ graph, and then find an absolute uncertainty using the worst acceptable line.

Approach

Extend the best-fit line back to $R=0$ and read $c_{\text{best}}$ .
Extend the worst acceptable line back to $R=0$ and read $c_{\text{worst}}$ .
Take the absolute difference as the uncertainty.

Step-by-Step Reasoning

From the best-fit line, the intercept is about $3.7\times 10^{3}\,\text{A}^{-1}$ .
From a steepest worst acceptable line, the intercept is about $3.4\times 10^{3}\,\text{A}^{-1}$ .
So

\Delta c \approx |3.4\times 10^{3} - 3.7\times 10^{3}| = 0.3\times 10^{3} = 3\times 10^{2}\,\text{A}^{-1}.

Therefore

c = (3.7\times 10^{3} \pm 3\times 10^{2})\ \text{A}^{-1}.

(Exact values depend on your drawn lines; ECF is normally applied.)

Key Takeaways

The intercept is read at $x=0$ (extend the line if necessary).
Intercept uncertainty comes from the spread between acceptable lines.

Common Mistakes

Reading the intercept where the line crosses the first grid line rather than the axis.
Using a data point as the intercept.
Giving a percentage uncertainty when an absolute uncertainty is requested.

Things to Be Careful About

Quote $c$ with unit $\text{A}^{-1}$ .
Use the same worst acceptable line as you used for the gradient uncertainty (consistent method).

Techniques used

read the y-intercept from a straight-line graphestimate intercept uncertainty by comparing best-fit and worst acceptable lines

(d)

(i)

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

$E$ = ______
$Z$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a):

\frac{1}{I} = \frac{3}{E}R + \frac{4Z}{E}

But the graph uses $x = R/\text{k}\Omega$ , so $R = 1000x$ and

\frac{1}{I} = \frac{3000}{E}x + \frac{4Z}{E}.

Hence

\text{gradient } m = \frac{3000}{E} \Rightarrow E = \frac{3000}{m}

Using $m = 6.5\times 10^{2}\ \text{A}^{-1}\text{k}\Omega^{-1}$ :

E = \frac{3000}{6.5\times 10^{2}} = 4.6\ \text{V}

And intercept $c = \dfrac{4Z}{E}$ :

Z = \frac{cE}{4}

Using $c = 3.7\times 10^{3}\ \text{A}^{-1}$ and $E = 4.6\ \text{V}$ :

Z = \frac{(3.7\times 10^{3})(4.6)}{4} \approx 4.3\times 10^{3}\ \Omega

Answer

E = 4.6\ \text{V}

Z = 4.3\times 10^{3}\ \Omega

Final answer

E = 4.6 V, Z = 4.3×10^3 Ω

Detailed explanation

Background Concept

When you plot a straight-line graph, the gradient and intercept can be used to determine constants in the original equation.

From part (a) the linear form in SI units ( $R$ in $\Omega$ ) is:

\frac{1}{I} = \frac{3}{E}R + \frac{4Z}{E}.

However, if the horizontal axis uses $R$ in $\text{k}\Omega$ , then $R = 1000x$ where $x = R/\text{k}\Omega$ . This multiplies the coefficient of $x$ by $1000$ .

Understanding the Question

You must use your measured gradient and intercept from the graph (parts (c)(iii) and (c)(iv)) to find:

the e.m.f. $E$ (in volts),
the resistance $Z$ (in ohms).

The key subtlety is the unit on the $x$ -axis: $R/\text{k}\Omega$ .

Approach

Rewrite the straight-line equation in terms of $x = R/\text{k}\Omega$ .
Match graph gradient and intercept to that equation.
Solve for $E$ and $Z$ , then add correct units.

Step-by-Step Reasoning

Start from

\frac{1}{I} = \frac{3}{E}R + \frac{4Z}{E}.

Let $x = R/\text{k}\Omega$ . Then $R = 1000x$ (in $\Omega$ ), so

\frac{1}{I} = \frac{3}{E}(1000x) + \frac{4Z}{E} = \frac{3000}{E}x + \frac{4Z}{E}.

So:

gradient $m = \frac{3000}{E}$ ,
intercept $c = \frac{4Z}{E}$ .

Calculate $E$ :

E = \frac{3000}{m}.

With $m \approx 6.5\times 10^{2}\,\text{A}^{-1}\text{k}\Omega^{-1}$ ,

E \approx 4.6\,\text{V}.

Then calculate $Z$ from $c = \frac{4Z}{E}$ :

Z = \frac{cE}{4}.

With $c \approx 3.7\times 10^{3}\,\text{A}^{-1}$ and $E \approx 4.6\,\text{V}$ ,

Z \approx 4.3\times 10^{3}\,\Omega.

Key Takeaways

Always account for axis units when linking gradient to constants.
Use $m$ to find $E$ , then use $c$ and $E$ to find $Z$ .

Common Mistakes

Using $m = 3/E$ even though the graph uses $R/\text{k}\Omega$ .
Giving $Z$ in $\text{k}\Omega$ without converting to $\Omega$ when asked.
Mixing best-fit and worst-line values for constants (unless instructed).

Things to Be Careful About

Units: $E$ in $\text{V}$ , $Z$ in $\Omega$ .
Consistency: use the best-fit gradient and intercept for the best estimates of $E$ and $Z$ .

Techniques used

relate gradient and intercept to physical constants by comparing equationsconvert between kΩ and Ω when interpreting a graph gradientsubstitute gradient and intercept to calculate constants with units

(ii)

Determine the percentage uncertainty in $Z$ .

percentage uncertainty in $Z$ = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Z = \frac{cE}{4}, \quad E = \frac{3000}{m}

\frac{\Delta Z}{Z} = \frac{\Delta c}{c} + \frac{\Delta E}{E}, \quad \frac{\Delta E}{E} = \frac{\Delta m}{m}

Using $c = 3.7\times 10^{3}$ , $\Delta c = 3\times 10^{2}$ and $m = 6.5\times 10^{2}$ , $\Delta m = 7\times 10^{1}$ :

\frac{\Delta Z}{Z} = \frac{3\times 10^{2}}{3.7\times 10^{3}} + \frac{7\times 10^{1}}{6.5\times 10^{2}} = 0.081 + 0.108 \approx 0.19

Answer

percentage uncertainty in $Z$ $\approx 19\%$

Final answer

19%

Detailed explanation

Background Concept

For quantities combined by multiplication/division, fractional uncertainties add:

If $Q = ab$ , then

\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b}.

If $Q = \frac{a}{b}$ , then

\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + \frac{\Delta b}{b}.

Constants such as $4$ or $3000$ are exact, so they contribute no uncertainty.

Understanding the Question

You found:

gradient $m$ with absolute uncertainty $\Delta m$ ,
intercept $c$ with absolute uncertainty $\Delta c$ .

From these you calculated $Z$ . Now you must find the percentage uncertainty in $Z$ .

Approach

Write $Z$ in terms of quantities that have uncertainties (here $c$ and $m$ via $E$ ).
Convert absolute uncertainties to fractional uncertainties.
Add fractional uncertainties and multiply by $100$ to get percentage.

Step-by-Step Reasoning

You have

Z = \frac{cE}{4} \quad\text{and}\quad E = \frac{3000}{m}.

So $Z \propto cE$ and $E \propto 1/m$ .

Thus

\frac{\Delta Z}{Z} = \frac{\Delta c}{c} + \frac{\Delta E}{E} \quad\text{with}\quad \frac{\Delta E}{E} = \frac{\Delta m}{m}.

Using the representative values:

\frac{\Delta c}{c} \approx \frac{3\times 10^{2}}{3.7\times 10^{3}} \approx 0.081

\frac{\Delta m}{m} \approx \frac{7\times 10^{1}}{6.5\times 10^{2}} \approx 0.108

Add them:

\frac{\Delta Z}{Z} \approx 0.081 + 0.108 = 0.189.

Percentage uncertainty:

0.189 \times 100\% \approx 19\%.

Key Takeaways

For products/quotients, add fractional uncertainties.
For $E = \text{constant}/m$ , fractional uncertainty in $E$ equals fractional uncertainty in $m$ .

Common Mistakes

Subtracting uncertainties for a quotient (they still add).
Using absolute uncertainties directly without converting to fractional form.
Forgetting to multiply by $100$ to convert to percent.

Things to Be Careful About

Use the absolute uncertainties you obtained from best-fit vs worst acceptable lines.
Keep consistent significant figures; percentage uncertainty is typically quoted to 1–2 significant figures.

Techniques used

propagate fractional uncertainties through multiplication and divisioncombine percentage uncertainties additively for products

(e)

The experiment is repeated. Determine the resistance $R$ that gives a value of $I$ of $0.10\ \text{mA}$ .

$R$ = ______ $\Omega$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Use

E = 3IR + 4IZ

R = \frac{E - 4IZ}{3I}.

With $I = 0.10\ \text{mA} = 1.0\times 10^{-4}\ \text{A}$ , $E = 4.6\ \text{V}$ and $Z = 4.3\times 10^{3}\ \Omega$ :

R = \frac{4.6 - 4(1.0\times 10^{-4})(4.3\times 10^{3})}{3(1.0\times 10^{-4})} \approx 9.7\times 10^{3}\ \Omega

Answer

R = 9.7\times 10^{3}\ \Omega

Final answer

9.7×10^3 Ω

Detailed explanation

Background Concept

Once you have determined constants ( $E$ and $Z$ ), you can use the model equation to predict values for new conditions.

Here the relationship is linear in $R$ :

E = 3IR + 4IZ.

If $E$ and $Z$ are constant for the power supply and the resistors, changing $R$ changes the current $I$ .

Understanding the Question

You are asked: if the current is set (or measured) to be $0.10\,\text{mA}$ , what value of $R$ would produce that current (using the same circuit and constants)?

Given/previously found:

$I = 0.10\,\text{mA}$ ,
$E$ from the graph,
$Z$ from the graph.

Unknown:

$R$ in ohms.

Approach

Convert $I$ to amperes.
Rearrange the equation to make $R$ the subject.
Substitute values and calculate.

Step-by-Step Reasoning

Convert current:

0.10\,\text{mA} = 1.0\times 10^{-4}\,\text{A}.

Rearrange:

E = 3IR + 4IZ \Rightarrow 3IR = E - 4IZ \Rightarrow R = \frac{E - 4IZ}{3I}.

Substitute (using best estimates):

$E \approx 4.6\,\text{V}$ ,
$Z \approx 4.3\times 10^{3}\,\Omega$ ,
$I = 1.0\times 10^{-4}\,\text{A}$ .

Compute $4IZ$ first:

4IZ = 4(1.0\times 10^{-4})(4.3\times 10^{3}) = 1.72\,\text{V}.

Then

R = \frac{4.6 - 1.72}{3\times 10^{-4}} \approx \frac{2.88}{3\times 10^{-4}} \approx 9.6\times 10^{3}\,\Omega.

So $R$ is about $9.7\times 10^{3}\,\Omega$ (depending on the rounded $E$ and $Z$ you use).

Key Takeaways

Always convert mA to A before substitution.
Rearranging first avoids algebra mistakes during substitution.

Common Mistakes

Using $I = 0.10\,\text{A}$ instead of $0.10\,\text{mA}$ .
Forgetting the factor $4$ in $4IZ$ .
Giving the answer in $\text{k}\Omega$ when the question asks for $\Omega$ .

Things to Be Careful About

Unit consistency: $E$ in volts, $I$ in amperes, $R$ and $Z$ in ohms.
Rounding: use a sensible number of significant figures consistent with $E$ and $Z$ .

Techniques used

substitute target values into an empirical relationshiprearrange an equation to solve for an unknown resistancemaintain consistent SI units throughout

Physics 9702/52 — February/March 2024

Procedure / measurements

Control of variables

Determination of ρ\rhoρ

Analysis to determine nnn and EEE

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Data collection (getting fff)

2) Controlling variables

3) Determining ρ\rhoρ

4) Linearising to find nnn and EEE

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

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

Determination of $\rho$

Analysis to determine $n$ and $E$

1) Data collection (getting $f$ )

3) Determining $\rho$

4) Linearising to find $n$ and $E$