A student investigates the cooling of a liquid in a beaker. The temperature $\theta_R$ of the laboratory is measured using a thermometer. Hot water is added to an insulated beaker, as shown in Fig. 2.1. The thermometer measures the temperature of the water. At time $t$ the temperature of the water is $\theta$. A series of readings of $t$ and $\theta$ are taken. It is suggested that $\theta$ and $t$ are related by the equation $$ \theta = \theta_R + (\theta_0 -…

(a) gradient = -1/K; y-intercept = ln(θ0 − θR) · (b) See working · (c)(i) Graph plotted with y-error bars · (c)(ii) Best-fit line and worst acceptable line drawn and labelled · (c)(iii) (-3.10 ± 0.24) × 10^-2 min^-1 · (c)(iv) 4.21 ± 0.04 · (d)(i) K = 32.3 min; θ0 = 85.6 °C · (d)(ii) 3.2 °C · (e) 75.5 min

Physics 9702/52 — February/March 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

Fig. 1.1 shows two identical cylindrical metal conductors P and Q, each of length $L$ and cross-sectional area $A$ .

The conductors are placed parallel to each other. The perpendicular distance from the midpoint of P to point X is $p$ . The perpendicular distance from the midpoint of Q to point X is $q$ .

The two conductors are electrically connected in parallel. This parallel combination is connected in series to a power supply and a resistor. The potential difference $V$ between the ends of P is the same as the potential difference between the ends of Q.

The magnetic flux density at X due to the currents in the conductors is $B$ .

It is suggested that $B$ is related to $p$ by the relationship

B = \frac{YAV}{Lp} + \frac{YZAV}{Lq}

where $Y$ and $Z$ are constants.

Plan a laboratory experiment to test the relationship between $B$ and $p$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $Y$ 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.

DifficultyMedium-Hard

Worked solution

Variables

Independent variable: $p$ (perpendicular distance from midpoint of conductor P to point X).
Dependent variable: magnetic flux density $B$ at X.
Control variables: $q$ (keep fixed), $V$ across P and Q (keep constant), same conductors so $L$ and $A$ constant, point X on the perpendicular line through the midpoints, orientation of Hall probe, ambient magnetic field.

Apparatus

Two identical straight cylindrical conductors P and Q on insulating clamps, low-voltage d.c. power supply, series resistor, switch, ammeter, voltmeter across the parallel combination, Hall probe / gaussmeter, metre rule / vernier scale to measure $p$ and $q$ , micrometer screw gauge for diameter (to find $A$ ).

diagram-1

Procedure and measurements

Mount Q fixed. Mark the midpoint of Q and place the Hall probe at point X a fixed perpendicular distance $q$ from this midpoint (measure $q$ with a rule and keep the probe position fixed).
Mount P parallel to Q with its midpoint on the same perpendicular line through X. Adjust the separation so that the perpendicular distance from P’s midpoint to X is $p$ .
Connect P and Q in parallel. Connect this parallel pair in series with the resistor, ammeter and power supply.
Set the supply so that the potential difference across P (and Q) is a chosen constant value $V$ (monitor with the voltmeter). Keep $V$ the same for all readings.
For each value of $p$ $p$ :
- measure and record $p$ (and check $q$ unchanged),
- measure $B_{\text{on}}$ with current flowing,
- measure background field $B_{\text{off}}$ with switch open,
- take $B = B_{\text{on}} - B_{\text{off}}$ .
Take at least 6 values of $p$ over a wide range, repeating readings and averaging.
Measure $L$ of a conductor and diameter $d$ to obtain $A = \pi d^2/4$ .

Analysis of data (to find $Y$ and $Z$ )

With $q$ constant, the suggested relationship

B = \frac{YAV}{L}\left(\frac{1}{p}\right) + \frac{YZAV}{Lq}

So plot a graph of $B$ (y-axis) against $1/p$ (x-axis).

Gradient $m = YAV/L$
Intercept $c = YZAV/(Lq)$

Hence

Y = \frac{mL}{AV}

and

Z = \frac{cq}{m}

Use a best-fit line and worst acceptable line to estimate uncertainties in $m$ and $c$ , and propagate to uncertainties in $Y$ and $Z$ .

Safety

Use low voltage and a series resistor to limit current.
Conductors may heat up: switch off between readings / keep current small and avoid touching metal.
Keep leads insulated and secure to prevent short circuits.

Final answer

See working

Detailed explanation

Background Concept

A current in a long straight conductor produces a magnetic field (and hence magnetic flux density $B$ ) around it. In many experiments, we do not calculate $B$ from first principles; instead we measure $B$ with a Hall probe and test whether a suggested mathematical relationship matches the data.

Here the suggested model is

B = \frac{YAV}{Lp} + \frac{YZAV}{Lq}

where:

$B$ is the magnetic flux density at point X,
$p$ and $q$ are perpendicular distances from the midpoints of P and Q to X,
$V$ is the p.d. across each conductor (same because P and Q are in parallel),
$A$ and $L$ are the cross-sectional area and length of each conductor,
$Y$ and $Z$ are constants to be found experimentally.

A key skill in Paper 5 planning is to (i) decide what to vary and what to keep constant, (ii) decide what to measure and how, and (iii) choose a graph that gives a straight line so constants can be extracted from a gradient and intercept.

Understanding the Question

You must plan an experiment to test how $B$ depends on $p$ for two identical parallel conductors connected in parallel, and then explain how your results can be used to determine $Y$ and $Z$ .

Important details from the stem:

P and Q are identical (same $L$ and $A$ ), and each has p.d. $V$ across it.
$B$ at X is due to the currents in both conductors.
The proposed equation contains both $p$ and $q$ , so if you vary $p$ you should ideally keep $q$ constant; otherwise the second term changes too and the test becomes ambiguous.

Approach

Make $p$ the independent variable and measure $B$ at X for many values of $p$ .
Keep $q$ constant by fixing the position of X relative to conductor Q (e.g. keep Q and the Hall probe fixed) and moving conductor P to change $p$ .
Keep $V$ constant and monitor it with a voltmeter across the parallel pair.
Remove systematic effects:
- subtract the background magnetic field (Earth field and nearby equipment) by measuring with current off.
- keep the Hall probe orientation fixed.
Linearise: with $q$ constant, $B$ should be linear in $1/p$ . A straight-line graph allows $Y$ and $Z$ to be extracted from gradient and intercept.

Step-by-Step Reasoning

1) Set up a workable geometry

You need P and Q straight, parallel, and with their midpoints aligned so that $p$ and $q$ are genuinely the perpendicular distances from the midpoints to point X.

A practical way:

Clamp Q fixed on insulating supports.
Place the Hall probe at X a fixed distance below Q (or to the side—any perpendicular direction is fine as long as you measure it consistently). Measure this distance once as $q$ .
Clamp P parallel to Q, and adjust its position so its midpoint lies on the same perpendicular line to X. Then changing the separation between P and Q changes $p$ while leaving $q$ unchanged.

diagram-1

2) Circuit arrangement and keeping $V$ constant

Because P and Q are in parallel, they share the same p.d. $V$ . Put the parallel combination in series with:

a resistor (current limiting),
an ammeter (optional but useful for checking current is not drifting),
a switch,
the d.c. supply.

Place a voltmeter across the parallel combination so you can set and maintain a constant $V$ for all readings.

3) Measurements to take

For each $p$ :

Measure $p$ with a ruler/vernier (record uncertainty).
Measure $B_{\text{on}}$ with the Hall probe when current flows.
Measure $B_{\text{off}}$ with the switch open to get the background field.
Calculate $B = B_{\text{on}} - B_{\text{off}}$ .

Also measure the constants needed for later:

$L$ (length of each conductor) with a rule.
Diameter $d$ with a micrometer; then compute

A = \frac{\pi d^2}{4}.

Take multiple readings of $B$ for each $p$ and average, because $B$ can fluctuate due to probe noise and small movements.

4) Control of variables

To make it a fair test of “ $B$ vs $p$ ”:

Keep $q$ fixed (do not move Q or X).
Keep $V$ constant (adjust the supply if needed and record $V$ each time).
Keep the Hall probe at the same orientation (Hall probes measure the component of $B$ along a specific axis).
Keep conductors straight and parallel; keep midpoints aligned with X.
Reduce heating (temperature rise changes resistance and could change currents even if supply settings drift). Use low current and switch off between readings.

5) Linearising and extracting $Y$ and $Z$

Start from the given model:

B = \frac{YAV}{Lp} + \frac{YZAV}{Lq}.

If $q$ , $A$ , $L$ , $V$ are constants during the experiment, rewrite as

B = \left(\frac{YAV}{L}\right)\left(\frac{1}{p}\right) + \left(\frac{YZAV}{Lq}\right).

This is in the straight-line form $y = mx + c$ with:

$y \to B$
$x \to 1/p$
gradient

m = \frac{YAV}{L}

intercept

c = \frac{YZAV}{Lq}.

So:

Y = \frac{mL}{AV}

and, using $m = YAV/L$ to eliminate $Y$ ,

Z = \frac{cq}{m}.

A straight line within uncertainty supports the suggested relationship.

6) Uncertainties (what you would write in Paper 5)

Add error bars in $B$ (from probe resolution and repeat scatter) and in $1/p$ (from uncertainty in $p$ ).
Find uncertainty in gradient and intercept using a worst acceptable line.
Then propagate:
- fractional uncertainty in $Y$ is the sum of fractional uncertainties in $m$ , $L$ , $A$ , $V$ as appropriate (often $A$ is from $d$ so $\Delta A/A \approx 2\Delta d/d$ ).
- fractional uncertainty in $Z$ comes mainly from uncertainties in $c$ , $q$ , and $m$ .

Key Takeaways

In planning, make one variable change (here $p$ ) while keeping all others in the equation constant (especially $q$ ).
Use background subtraction ( $B_{\text{on}}-B_{\text{off}}$ ) to reduce systematic error.
Turn a proposed relationship into a straight-line graph; use gradient and intercept to determine constants ( $Y$ and $Z$ ).
For Paper 5, you must describe measurements, controls, analysis, and safety clearly and specifically.

Common Mistakes

Varying $p$ by moving point X, causing $q$ to change as well (then you are not testing the stated dependence cleanly).
Forgetting that a Hall probe measures one component of $B$ and rotating it between readings.
Not measuring/subtracting the background magnetic field.
Plotting $B$ against $p$ instead of against $1/p$ (loses the linear method for finding constants).
Giving a graph method but not stating explicitly how $Y$ and $Z$ are calculated from gradient/intercept.

Things to Be Careful About

Keep the conductors and the probe away from magnetic materials and other current-carrying wires (stray fields).
Ensure the “midpoint” distances $p$ and $q$ are measured from the correct locations (mark midpoints).
Avoid very small $p$ where the Hall probe might be too close to the conductor (risk of contact, poor geometry, very large gradients in $B$ ).
Use consistent SI units (convert cm to m before calculating $1/p$ if needed).
Heating: if temperature rises, resistance changes and currents may drift; keep current low and take readings quickly.

Techniques used

identify independent, dependent and controlled variablesmeasure magnetic flux density with a Hall probe and subtract background fieldvary one geometric parameter while keeping the other fixedlinearise the relationship and plot a straight-line graphdetermine constants from gradient and intercept (including uncertainties)

Working

Given

\theta = \theta_R + (\theta_0-\theta_R)e^{-t/K}

\theta-\theta_R = (\theta_0-\theta_R)e^{-t/K}

Take ln:

\ln(\theta-\theta_R)=\ln(\theta_0-\theta_R)-\frac{t}{K}

Comparing with $y = mx + c$ for a graph of $y=\ln(\theta-\theta_R)$ against $x=t$ :

Answer

gradient $= -\dfrac{1}{K}$

$y$ -intercept $= \ln(\theta_0-\theta_R)$

Final answer

gradient = -1/K; y-intercept = ln(θ0 − θR)

Detailed explanation

Background Concept

Cooling that follows Newton’s law of cooling can produce an exponential approach to room temperature. An exponential form like

\theta = \theta_R + (\theta_0-\theta_R)e^{-t/K}

is hard to analyse directly on a straight-line graph. A standard technique is to linearise by taking natural logs so that we can use

y = mx + c

where the gradient $m$ and intercept $c$ can be read from a straight line.

Understanding the Question

You are told to plot a graph of $\ln(\theta-\theta_R)$ (vertical axis) against $t$ (horizontal axis). The question asks: once you have rewritten the given cooling equation into the form $\ln(\theta-\theta_R)=mt+c$ , what are the expressions for:

the gradient $m$
the intercept $c$ .

Approach

Subtract $\theta_R$ from both sides so the exponential is by itself.
Take $\ln$ of both sides.
Rearrange so it looks like $\ln(\theta-\theta_R)=(-1/K)t + \ln(\theta_0-\theta_R)$ .
Read off gradient and intercept.

Step-by-Step Reasoning

Start with:

\theta = \theta_R + (\theta_0-\theta_R)e^{-t/K}

Subtract $\theta_R$ :

\theta-\theta_R = (\theta_0-\theta_R)e^{-t/K}

Now take natural logs:

\ln(\theta-\theta_R) = \ln(\theta_0-\theta_R) + \ln\left(e^{-t/K}\right)

and since $\ln(e^a)=a$ :

\ln(\theta-\theta_R)=\ln(\theta_0-\theta_R)-\frac{t}{K}

Compare with $y=mx+c$ where $y=\ln(\theta-\theta_R)$ and $x=t$ :

$m=-1/K$
$c=\ln(\theta_0-\theta_R)$ .

Key Takeaways

Exponentials are linearised by taking logs.
After linearisation, gradient and intercept come from direct comparison with $y=mx+c$ .

Common Mistakes

Forgetting to subtract $\theta_R$ before taking $\ln$ .
Writing gradient as $-K$ or $-K^{-1}$ without showing it is specifically $-1/K$ .
Giving intercept as $\theta_0-\theta_R$ instead of $\ln(\theta_0-\theta_R)$ .

Things to Be Careful About

The log is of $(\theta-\theta_R)$ , not of $\theta$ .
When plotting $\ln((\theta-\theta_R)/\text{°C})$ , the division by $\text{°C}$ is just to make the argument dimensionless; the algebraic intercept still corresponds to $\ln(\theta_0-\theta_R)$ (numerically the same as $\ln((\theta_0-\theta_R)/\text{°C})$ if you treat the temperature difference in °C as a pure number).

Techniques used

take natural logarithms to linearise an exponential relationshiprearrange into the form $y = mx + c$identify gradient and intercept by comparison

(b)

Values of $t$ and $\theta$ are given in Table 2.1.

Table 2.1

$t$ / min	$\theta$ / °C	$(\theta - \theta_R)$ / °C	$\ln((\theta - \theta_R) / \text{°C})$
6.0	75.0 $\pm$ 0.5
12.0	64.5 $\pm$ 0.5
18.0	57.0 $\pm$ 0.5
24.0	50.0 $\pm$ 0.5
30.0	44.5 $\pm$ 0.5
36.0	41.0 $\pm$ 0.5

The value of $\theta_R$ is $(18.5 \pm 0.5)\ \text{°C}$ .

Calculate and record values of $(\theta - \theta_R) / \text{°C}$ and $\ln((\theta - \theta_R) / \text{°C})$ in Table 2.1. Include the absolute uncertainties in $(\theta - \theta_R)$ and $\ln((\theta - \theta_R) / \text{°C})$ .

2M Similar

DifficultyMedium

Worked solution

Working

For $(\theta-\theta_R)$ :

(\theta-\theta_R)=\theta-18.5

Absolute uncertainty:

\Delta(\theta-\theta_R)=0.5+0.5=1.0\ \text{°C}

For $y=\ln((\theta-\theta_R)/\text{°C})$ with $x=(\theta-\theta_R)$ :

\Delta y \approx \frac{\Delta x}{x}

Completed values (to suitable d.p.):

$t$ / min	$(\theta-\theta_R)$ / °C	$\ln((\theta-\theta_R)/\text{°C})$	abs unc in $\ln$
$6.0$	$56.5 \pm 1.0$	$4.034$	$0.018$
$12.0$	$46.0 \pm 1.0$	$3.829$	$0.022$
$18.0$	$38.5 \pm 1.0$	$3.651$	$0.026$
$24.0$	$31.5 \pm 1.0$	$3.450$	$0.032$
$30.0$	$26.0 \pm 1.0$	$3.258$	$0.039$
$36.0$	$22.5 \pm 1.0$	$3.113$	$0.044$

Answer

Values recorded in Table 2.1 as above.

Final answer

See working

Detailed explanation

Background Concept

When you calculate a new quantity from measurements, you must also estimate its uncertainty.

Two key rules used here:

Addition/subtraction: absolute uncertainties add.

If $z = a-b$ , then typically

\Delta z = \Delta a + \Delta b

Natural log: for $y=\ln x$ , a small change satisfies

\Delta y \approx \frac{\Delta x}{x}

This comes from differentiation: $\mathrm{d}(\ln x)=\mathrm{d}x/x$ .

Understanding the Question

You are given $t$ and $\theta$ (each $\theta$ has $\pm 0.5\ \text{°C}$ ) and $\theta_R = 18.5 \pm 0.5\ \text{°C}$ . You must fill two extra columns:

$(\theta-\theta_R)$ and its absolute uncertainty
$\ln((\theta-\theta_R)/\text{°C})$ and its absolute uncertainty (for plotting error bars later).

Approach

For each row:

Compute $x = \theta-\theta_R$ .
Uncertainty in $x$ is $\Delta x = 0.5+0.5$ .
Compute $y = \ln(x/\text{°C})$ (numerically this is just $\ln x$ if $x$ is in °C).
Compute $\Delta y \approx \Delta x/x$ .

Step-by-Step Reasoning

Example for $t=6.0\ \text{min}$ :

Difference:

x = 75.0-18.5 = 56.5\ \text{°C}

Absolute uncertainty in difference:

\Delta x = 0.5+0.5=1.0\ \text{°C}

Log value:

y=\ln\left(\frac{56.5}{\text{°C}}\right)=\ln(56.5)=4.034

Uncertainty in log:

\Delta y \approx \frac{\Delta x}{x}=\frac{1.0}{56.5}=0.0177\approx 0.018

Repeat the same process for each row; the uncertainty in $\ln$ grows as the water cools because $x$ gets smaller, so $\Delta x/x$ increases.

Key Takeaways

Subtraction: add absolute uncertainties.
For $\ln x$ , the absolute uncertainty is approximately the fractional uncertainty in $x$ .
Error bars should reflect these calculated $\Delta(\ln x)$ values.

Common Mistakes

Using $\Delta(\theta-\theta_R)=\pm 0.5$ instead of $\pm 1.0$ .
Calculating $\ln(\theta)$ instead of $\ln(\theta-\theta_R)$ .
Giving uncertainty in $\ln$ as $\pm 1.0$ (log uncertainties are unitless and usually much smaller).

Things to Be Careful About

Keep consistent decimal places for $(\theta-\theta_R)$ (here to $0.1\ \text{°C}$ is appropriate because inputs are to $0.1\ \text{°C}$ ).
$\ln$ values should be to at least 3 decimal places so plotting is accurate.
The argument of a logarithm should be dimensionless; writing $\ln((\theta-\theta_R)/\text{°C})$ makes this explicit.

Techniques used

propagate absolute uncertainties through subtractioncompute a derived quantity column from raw datapropagate uncertainties through a logarithm using $\Delta(\ln x) \approx \Delta x/x$

(c)

(i)

Plot a graph of $\ln((\theta - \theta_R) / \text{°C})$ against $t / \text{min}$ . Include error bars for $\ln((\theta - \theta_R) / \text{°C})$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Points plotted on Graph 2.1 at:

(6.0,4.034),\ (12.0,3.829),\ (18.0,3.651),\ (24.0,3.450),\ (30.0,3.258),\ (36.0,3.113)

Vertical error bars drawn for $\ln((\theta-\theta_R)/\text{°C})$ of:

\pm 0.018,\ \pm 0.022,\ \pm 0.026,\ \pm 0.032,\ \pm 0.039,\ \pm 0.044

respectively.

Final answer

Graph plotted with y-error bars

Detailed explanation

Background Concept

A straight-line graph is used to test whether a linear relationship holds and to extract constants. When there are measurement uncertainties, points should be plotted with error bars so you can judge how well a straight line fits within uncertainty.

Understanding the Question

You must use your calculated $\ln((\theta-\theta_R)/\text{°C})$ values and their absolute uncertainties from (b), then:

plot $y$ against $t$ on the provided grid
include vertical error bars of size $\pm \Delta y$ for each point.

Approach

Put each point at its $(t,y)$ coordinate.
For each, draw a vertical line from $y-\Delta y$ to $y+\Delta y$ (horizontal “caps” are optional unless specifically required).
Ensure the plotted points are small, neat crosses or dots so the best-fit line can be judged.

Step-by-Step Reasoning

Using the completed table:

At $t=6.0$ , plot $y=4.034$ .
Draw an error bar from $4.034-0.018$ to $4.034+0.018$ .

Repeat for all six points. Because the $\Delta(\ln)$ values increase as temperature difference decreases, the later points (larger $t$ ) should have visibly larger error bars.

Key Takeaways

Plot the transformed variable actually requested ( $\ln((\theta-\theta_R)/\text{°C})$ ).
Error bars are based on the uncertainty in the transformed variable, not the original temperature directly.

Common Mistakes

Plotting $\theta$ against $t$ instead of $\ln((\theta-\theta_R)/\text{°C})$ against $t$ .
Using the same-sized error bar for every point (here they are different).
Plotting using too few decimal places so points are shifted noticeably.

Things to Be Careful About

Axes: $t/\text{min}$ on $x$ and $\ln((\theta-\theta_R)/\text{°C})$ on $y$ .
Error bars must be centred on the plotted point.
Do not join the points dot-to-dot; a best-fit line is drawn in (c)(ii).

Techniques used

choose appropriate axis scales and labelsplot data points with correct coordinatesadd vertical error bars using calculated $\Delta(\ln)$ values

(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 drawn through the plotted points.

A worst acceptable straight line drawn (steepest/shallowest that still passes through all error bars) and both lines labelled clearly.

Final answer

Best-fit line and worst acceptable line drawn and labelled

Detailed explanation

Background Concept

When data should be linear, experimental scatter means points will not lie exactly on a straight line. The line of best fit represents the most reasonable overall trend.

To estimate uncertainty in gradient/intercept from a graph, Paper 5 often uses a worst acceptable line: the steepest (or shallowest) straight line that is still consistent with the plotted points within their error bars.

Understanding the Question

After plotting points and vertical error bars, you must draw:

one best-fit straight line
one worst acceptable straight line
and label them.

Approach

Best-fit line: balance the points above and below; do not force it through every point.
Worst acceptable line: pivot to make the line as different as possible (steepest or shallowest) while still intersecting every error bar.
Label each line so it is clear which is which.

Step-by-Step Reasoning

Draw the best-fit line first: it should run through the middle of the scatter and follow the overall downward trend.
Then draw the worst acceptable line:
- Choose whether you will draw the shallowest or steepest line relative to the best fit (either is acceptable; you normally pick whichever gives the larger change in gradient).
- Make sure your line still passes through (or at least touches) every vertical error bar.

This second line is not “another best fit”; its purpose is to create a plausible extreme so that the difference from the best-fit gradient/intercept can be used as an uncertainty.

Key Takeaways

Worst acceptable line must be consistent with all error bars.
Labelling is essential so the examiner can see which line you used for uncertainties.

Common Mistakes

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

Things to Be Careful About

Use a ruler.
Extend lines sufficiently so you can read intercept at $t=0$ and choose widely separated points for gradient in (c)(iii)/(c)(iv).

Techniques used

draw a single straight line of best fit through scattered pointsdraw a worst acceptable line consistent with all error barsuse best and worst lines to estimate uncertainty

(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 widely spaced points on the best-fit line (e.g. at $t=6.0$ and $t=36.0$ ):

m = \frac{\Delta y}{\Delta x} = \frac{3.090-4.021}{36.0-6.0} = \frac{-0.931}{30.0} = -3.10 \times 10^{-2}\ \text{min}^{-1}

Worst acceptable gradient (from worst acceptable line) e.g.

m_{\text{worst}} \approx -2.86 \times 10^{-2}\ \text{min}^{-1}

Absolute uncertainty:

\Delta m = \left|m-m_{\text{worst}}\right| \approx \left| -3.10\times 10^{-2} - (-2.86\times 10^{-2}) \right| = 2.4\times 10^{-3}\ \text{min}^{-1}

Answer

\text{gradient} = (-3.10 \pm 0.24)\times 10^{-2}\ \text{min}^{-1}

Final answer

(-3.10 ± 0.24) × 10^-2 min^-1

Detailed explanation

Background Concept

For a straight-line graph, the gradient is

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

Using points far apart reduces percentage reading error.

To estimate uncertainty graphically:

find $m_{\text{best}}$ from the best-fit line
find $m_{\text{worst}}$ from the worst acceptable line
take an absolute difference as the uncertainty (some schemes use half-range if you draw both steepest and shallowest; if only one worst line is drawn, use the difference from best).

Understanding the Question

You must read the gradient of your best-fit line from the graph of $\ln((\theta-\theta_R)/\text{°C})$ against $t$ , and include an absolute uncertainty using your worst acceptable line.

Approach

Pick two points on the best-fit line (not necessarily plotted data points) that are widely separated.
Calculate $m_{\text{best}}$ .
Repeat using the worst acceptable line to get $m_{\text{worst}}$ .
Uncertainty is the difference between them.

Step-by-Step Reasoning

Suppose from the best-fit line you can read two convenient points around $t=6$ and $t=36$ . Compute

m_{\text{best}}=\frac{y_2-y_1}{t_2-t_1}

The gradient should be negative because $\ln(\theta-\theta_R)$ decreases with time.

Then read two points from the worst acceptable line in the same way to obtain $m_{\text{worst}}$ . The uncertainty is

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

and quote the gradient as $m_{\text{best}} \pm \Delta m$ with unit $\text{min}^{-1}$ .

Key Takeaways

Use large triangles / widely spaced points for gradients.
Uncertainty comes from the range of acceptable lines given the error bars.

Common Mistakes

Using $\Delta x/\Delta y$ (inverted gradient).
Choosing two points very close together.
Forgetting the gradient is negative.
Quoting no unit; here it must be $\text{min}^{-1}$ .

Things to Be Careful About

Use points on the drawn line, not the raw data points unless they lie on the line.
Keep consistent significant figures: uncertainty typically to 1–2 s.f., gradient to match.

Techniques used

determine gradient from two well-separated points on a straight linecalculate uncertainty using best-fit and worst acceptable gradientsuse correct sign and units for a graph gradient

(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 best-fit line, extrapolate to $t=0$ :

c \approx 4.21

From worst acceptable line, intercept e.g.

c_{\text{worst}} \approx 4.25

Absolute uncertainty:

\Delta c = |c-c_{\text{worst}}| = |4.21-4.25| = 0.04

Answer

\text{$y$-intercept} = 4.21 \pm 0.04

Final answer

4.21 ± 0.04

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 find it by extending the line to cross the $y$ -axis (or by calculating $c=y-mx$ from a point on the line).

Uncertainty in the intercept can be estimated by comparing the intercepts from the best-fit and worst acceptable lines.

Understanding the Question

You need the intercept of your best-fit line on the graph of $\ln((\theta-\theta_R)/\text{°C})$ vs $t$ , and you must include an absolute uncertainty.

Approach

Extend the best-fit line back to $t=0$ and read off $c$ .
Extend the worst acceptable line back to $t=0$ and read off $c_{\text{worst}}$ .
Take $\Delta c = |c-c_{\text{worst}}|$ .

Step-by-Step Reasoning

The intercept is often outside the plotted range of $t$ values, so you may need to extrapolate.
Use a ruler and extend carefully; small angle changes can cause noticeable changes in $c$ .
Once you have both intercepts, the difference is your absolute uncertainty.

Key Takeaways

Intercept is a read-off at $t=0$ .
Worst acceptable line provides the uncertainty estimate.

Common Mistakes

Reading the intercept at the left edge of the graph instead of at $t=0$ .
Forgetting to include uncertainty.
Mixing up intercept with the value at $t=6\ \text{min}$ .

Things to Be Careful About

Quote $c$ to match the precision justified by the graph scale (typically 2 d.p. at most here).
Ensure you use the same $t$ origin as the axis label (minutes).

Techniques used

read the intercept from a straight-line graph by extrapolation to $t=0$calculate intercept uncertainty using best-fit and worst acceptable linesstate intercept to appropriate significant figures

(d)

(i)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $K$ and $\theta_0$ . Include appropriate units.

$K$ = ______ $\text{min}$
$\theta_0$ = ______ $\text{°C}$

2M Similar

DifficultyMedium

Worked solution

Working

From (a): gradient $m=-1/K$ and intercept $c=\ln(\theta_0-\theta_R)$ .

With $m = -3.10\times 10^{-2}\ \text{min}^{-1}$ :

K = -\frac{1}{m} = -\frac{1}{-3.10\times 10^{-2}} = 32.3\ \text{min}

With $c=4.21$ :

\theta_0-\theta_R = e^{c} = e^{4.21} = 67.1\ \text{°C}

\theta_0 = \theta_R + 67.1 = 18.5 + 67.1 = 85.6\ \text{°C}

Answer

K = 32.3\ \text{min}

\theta_0 = 85.6\ \text{°C}

Final answer

K = 32.3 min; θ0 = 85.6 °C

Detailed explanation

Background Concept

After linearising an exponential relationship, the straight-line graph constants map back to the physical constants.

Here:

\ln(\theta-\theta_R)=\ln(\theta_0-\theta_R)-\frac{t}{K}

So on a plot of $y=\ln(\theta-\theta_R)$ vs $x=t$ :

gradient $m=-1/K$
intercept $c=\ln(\theta_0-\theta_R)$ .

Understanding the Question

You have already found (from your graph):

gradient $m$ with unit $\text{min}^{-1}$
intercept $c$ (dimensionless)

Using these and the expressions from (a), you must calculate numerical values for:

$K$ (in minutes)
$\theta_0$ (in °C).

Approach

Rearrange $m=-1/K$ to get $K=-1/m$ .
Rearrange $c=\ln(\theta_0-\theta_R)$ by exponentiating: $\theta_0-\theta_R=e^c$ .
Add $\theta_R$ to obtain $\theta_0$ .

Step-by-Step Reasoning

Since $m$ is negative, $K$ will come out positive because of the minus sign:

K=-\frac{1}{m}

For the intercept:

\theta_0-\theta_R=e^c

This gives the initial temperature difference above room temperature. Finally:

\theta_0 = \theta_R + (\theta_0-\theta_R)

Key Takeaways

Gradient gives the time constant: $K= -1/m$ .
Intercept gives the initial excess temperature: $\theta_0-\theta_R = e^c$ .

Common Mistakes

Using $K= -m$ instead of $K=-1/m$ .
Forgetting to exponentiate the intercept.
Forgetting to add $\theta_R$ to get $\theta_0$ .

Things to Be Careful About

Units: $K$ must be in minutes because $t$ was in minutes.
Make sure the calculator is in natural log/exponential mode (use $e^x$ , not $10^x$ ).

Techniques used

use gradient-intercept relationships to identify model parameterssolve for the time constant from the gradientuse the intercept to find an exponential parameter and then add the ambient temperature

(ii)

Determine the absolute uncertainty in your value of $\theta_0$ .

absolute uncertainty = ______ $\text{°C}$

1M Similar

DifficultyMedium-Hard

Worked solution

Working

From $\theta_0 = \theta_R + e^{c}$ .

With $c = 4.21 \pm 0.04$ :

A=e^{c}=67.1\ \text{°C}

\Delta A \approx A\,\Delta c = 67.1\times 0.04 = 2.7\ \text{°C}

Combine with $\Delta\theta_R = 0.5\ \text{°C}$ :

\Delta\theta_0 = \Delta\theta_R + \Delta A = 0.5 + 2.7 = 3.2\ \text{°C}

Answer

absolute uncertainty $= 3.2\ \text{°C}$

Final answer

3.2 °C

Detailed explanation

Background Concept

Uncertainty propagation depends on the mathematical operation.

For a sum $z=a+b$ , a common Paper 5 method is to add absolute uncertainties:

\Delta z = \Delta a + \Delta b

For an exponential $A=e^c$ , use the small-change approximation from differentiation:

\frac{\Delta A}{A} \approx \Delta c

\Delta A \approx A\Delta c

Understanding the Question

You found $\theta_0$ in (d)(i), and now you must find the absolute uncertainty in $\theta_0$ .

Since

\theta_0 = \theta_R + e^{c}

the uncertainty comes from both $\theta_R$ and the intercept $c$ .

Approach

Convert intercept uncertainty $\Delta c$ into uncertainty in $A=e^c$ using $\Delta A \approx A\Delta c$ .
Add the room-temperature uncertainty $\Delta\theta_R$ to get $\Delta\theta_0$ .

Step-by-Step Reasoning

First compute $A=e^c$ (this is $\theta_0-\theta_R$ ).
Then compute

\Delta A \approx A\Delta c

because a small change in $c$ changes $A$ proportionally.

Finally, because $\theta_0$ is a sum,

\Delta\theta_0 = \Delta\theta_R + \Delta A

This gives an absolute uncertainty in °C.

Key Takeaways

Exponential uncertainty: $\Delta(e^c) \approx e^c\Delta c$ .
Sum uncertainty: add absolute uncertainties.

Common Mistakes

Treating $\Delta A$ as equal to $\Delta c$ (wrong units and wrong magnitude).
Forgetting that $\theta_R$ has its own uncertainty.
Quoting the uncertainty to too many significant figures.

Things to Be Careful About

$\Delta c$ is dimensionless, but $\Delta A$ must be in °C.
If your $c$ uncertainty is found using a different worst acceptable line, your numerical uncertainty will differ; method marks come from using the correct propagation.

Techniques used

propagate uncertainty through an exponential using $\Delta(e^c) \approx e^c\Delta c$combine absolute uncertainties for a sumquote an absolute uncertainty to appropriate precision

(e)

Determine the time $t$ for the temperature to reach $25.0\ \text{°C}$ .

$t$ = ______ $\text{min}$

1M Similar

DifficultyMedium

Worked solution

Working

For $\theta=25.0\ \text{°C}$ ,

\theta-\theta_R = 25.0-18.5=6.5\ \text{°C}

y=\ln\left(\frac{\theta-\theta_R}{\text{°C}}\right)=\ln(6.5)=1.87

Using $y = mt + c$ with $m=-3.10\times 10^{-2}\ \text{min}^{-1}$ and $c=4.21$ :

t = \frac{y-c}{m} = \frac{1.87-4.21}{-3.10\times 10^{-2}} = 75.5\ \text{min}

Answer

t = 75.5\ \text{min}

Final answer

75.5 min

Detailed explanation

Background Concept

Once you have a model (from your straight-line fit), you can predict values by substitution.

From the linearised equation:

\ln(\theta-\theta_R)=c+mt

you can find the time for any temperature by:

computing $y=\ln(\theta-\theta_R)$ for that temperature
rearranging for $t$ :

t=\frac{y-c}{m}

Alternatively, you can rearrange the original exponential directly with logs; both approaches are equivalent.

Understanding the Question

You are asked for the time when the water reaches $25.0\ \text{°C}$ , using the relationship you have established (i.e. using your determined constants from the graph).

Approach

Compute the temperature difference above room: $\theta-\theta_R$ .
Take the natural log to find the corresponding $y$ value on the graph.
Use the straight-line equation with your best-fit gradient and intercept to solve for $t$ .

Step-by-Step Reasoning

At $25.0\ \text{°C}$ , the excess above room is $6.5\ \text{°C}$ .
The plotted variable is

y=\ln\left(\frac{\theta-\theta_R}{\text{°C}}\right)=\ln(6.5)

Substitute into $y=mt+c$ and rearrange:

t=\frac{y-c}{m}

Because $m$ is negative and $y<c$ , the time comes out positive, as expected.

Key Takeaways

You can use the linear graph form to predict times/temperatures quickly.
Always use $\theta-\theta_R$ , not $\theta$ .

Common Mistakes

Using $\ln(25.0)$ instead of $\ln(25.0-18.5)$ .
Sign error when dividing by a negative gradient.
Using base-10 logarithms instead of natural logs.

Things to Be Careful About

Ensure consistent units: $t$ in minutes because the gradient is in $\text{min}^{-1}$ .
If you use rounded values of $m$ and $c$ , your $t$ may differ slightly; the key is correct method and sensible rounding.

Techniques used

substitute a target value into a fitted modelsolve an exponential equation using logarithmsuse the straight-line form to determine the corresponding time

Physics 9702/52 — February/March 2025

Variables

Apparatus

Procedure and measurements

Analysis of data (to find YYY and ZZZ)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

1) Set up a workable geometry

2) Circuit arrangement and keeping VVV constant

3) Measurements to take

4) Control of variables

5) Linearising and extracting YYY and ZZZ

6) Uncertainties (what you would write in Paper 5)

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

Analysis of data (to find $Y$ and $Z$ )

2) Circuit arrangement and keeping $V$ constant

5) Linearising and extracting $Y$ and $Z$