A student investigates the refraction of white light entering a transparent rectangular block. A narrow beam of light enters the block at the midpoint of one of the shorter sides. The angle of incidence $\theta$ is measured, as shown in Fig. 2.1. The distance $d$ between the corner of the block and the point where the beam of light touches the boundary of the block is measured. The experiment is repeated for different values of $\theta$. It is suggested that $d$ and $\theta$…

(a) gradient = Bn^2, y-intercept = -B · (b) 615±10, 458±8.6, 292±6.8, 216±5.9, 166±5.2, 139±4.7 (all in cm^2) · (c)(i) See graph · (c)(ii) See graph · (c)(iii) (152 ± 5) cm^2 · (c)(iv) (-51 ± 10) cm^2 · (d)(i) B = 51 cm^2, n = 1.73 · (d)(ii) 12% · (e) 23.6°

Physics 9702/52 — October/November 2024

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 coil made from resistance wire.

Fig. 1.1

The coil is placed in cooking oil of mass $m$ . The total length of the resistance wire in the oil is $L$ .

A potential difference $V$ is applied to the coil. The temperature of the oil increases by $\Delta\theta$ in time $t$ .

It is suggested that $\Delta\theta$ is related to $L$ by the relationship

\frac{AtV^2}{L} = mK\Delta\theta + Z

where $A$ is the cross-sectional area of the wire, and $K$ and $Z$ are constants.

Plan a laboratory experiment to test the relationship between $\Delta\theta$ and $L$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine values for $K$ and $Z$ .

In your plan you should include:

the procedure to be followed
the measurements to be taken
the control of variables
the analysis of the data
any safety precautions to be taken.

DifficultyMedium-Hard

Worked solution

diagram-1

Method (procedure and measurements)

Put cooking oil in an insulated container (e.g. polystyrene cup / calorimeter) and measure its mass $m$ using a balance.
Immerse a resistance-wire coil fully in the oil. Arrange the wire so the length in the oil $L$ can be changed (e.g. marked wire with crocodile clips / adjustable immersion depth). Measure $L$ with a metre rule (or measure total immersed length directly).
Measure wire diameter $d$ with a micrometer and calculate

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

Connect the coil to a d.c. power supply with a switch. Use a voltmeter across the coil to set and keep $V$ constant.
Place a thermometer/temperature probe in the oil and stir continuously (or use a magnetic stirrer).
For a chosen $L$ , record the initial temperature $\theta_1$ . Switch on and heat for a fixed time $t$ (stopwatch). Switch off and record the final temperature $\theta_2$ .

\Delta\theta = \theta_2 - \theta_1.

Repeat for at least 5–6 different values of $L$ over a wide range. Repeat readings and average $\Delta\theta$ .

Control of variables

Keep $m$ constant (same mass of oil each run).
Keep $V$ constant (monitor with voltmeter) and keep $t$ constant.
Use the same wire throughout so $A$ is constant.
Keep initial temperature the same for each run (allow oil to cool back to the same starting temperature).
Reduce heat loss consistently: lid/insulation, same container, continuous stirring, coil not touching sides.

Analysis to test relationship and find $K$ and $Z$

For each run calculate

y = \frac{AtV^2}{L}.

Plot a graph of $y$ (vertical) against $\Delta\theta$ (horizontal).
If the suggested relationship is correct, the graph is a straight line with

\frac{AtV^2}{L} = mK\Delta\theta + Z.

Gradient $= mK$ so

K = \frac{\text{gradient}}{m}.

$y$ -intercept $= Z$ .

Safety

Hot oil can cause burns: wear eye protection, use a heatproof mat, do not overheat, handle container carefully.
Electrical safety: use low voltage, keep hands dry, switch off power before adjusting the coil/clips.

Final answer

See working (plan + graph gives K from gradient and Z from intercept).

Detailed explanation

Background Concept

The coil is a resistive heater. When a potential difference $V$ is applied, electrical energy is transferred to thermal energy in the oil. The temperature rise depends on how much energy is delivered and on the heat capacity of the oil.

The suggested model is

\frac{AtV^2}{L} = mK\Delta\theta + Z,

where:

$A$ is the cross-sectional area of the wire (constant if the same wire is used),
$t$ is the heating time,
$V$ is the p.d. across the wire,
$L$ is the length of wire actually in the oil (this affects resistance and hence heating),
$m$ is the oil mass,
$K$ and $Z$ are constants to be found.

This is already in a linear form: it matches $y = mx + c$ if we define

y = \frac{AtV^2}{L}, \qquad x = \Delta\theta.

Then the gradient is $mK$ and the intercept is $Z$ .

Understanding the Question

You must propose a practical experiment that changes $L$ and measures the resulting temperature rise $\Delta\theta$ after a known time $t$ at a known p.d. $V$ , while keeping other factors (especially $m$ , $t$ , $V$ , and $A$ ) controlled.

You also have to explain how to process the measurements so that you can obtain numerical values for $K$ and $Z$ from your results.

Approach

Choose variables: make $L$ the independent variable (you deliberately change it) and measure $\Delta\theta$ as the dependent variable.
Keep key controls fixed: $m$ , $V$ , $t$ , and wire properties ( $A$ ) must be constant so that any change in $\Delta\theta$ is due to changing $L$ .
Collect a range of readings: take several different values of $L$ to allow a meaningful straight-line test.
Linear analysis: for each reading compute $y = AtV^2/L$ and plot $y$ against $\Delta\theta$ . Extract $K$ and $Z$ from gradient and intercept.

Step-by-Step Reasoning

diagram-1

Apparatus set-up

Put oil in an insulated container to reduce heat losses to the surroundings. Heat loss is a major reason data do not fit a straight line.
Put the wire coil fully in the oil and ensure it does not touch the container walls (otherwise heat can conduct directly into the container).
Insert a thermometer/temperature probe and stir continuously so the measured temperature represents the whole oil volume (oil develops temperature gradients if not stirred).

How to vary and measure $L$

The key experimental challenge is changing the length of wire in the oil in a controlled, measurable way.
A practical method is to use a long wire with length marks and crocodile clips so only a chosen section is in circuit and immersed, or to adjust the immersion depth of a marked wire.
Each time you change $L$ , measure it with a ruler/metre rule. You need several values (typically 5–6+) spread over a good range so the graph gradient and intercept are reliable.

Measurements of $m$ , $V$ , $t$ , and $A$

Measure $m$ on a balance (and keep it constant for all trials).
Measure $V$ with a voltmeter across the coil and adjust the supply so $V$ is the same for every run.
Use a stopwatch to keep $t$ constant. Starting and stopping at the right times matters because heating power is continuous.
Determine $A$ by measuring diameter $d$ with a micrometer:

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

Using the same wire ensures $A$ is not a changing variable.

Determining $\Delta\theta$

Record the initial temperature $\theta_1$ just before switching on.
Heat for exactly time $t$ , then record the final temperature $\theta_2$ .

\Delta\theta = \theta_2 - \theta_1.

A good control is to begin each trial from the same $\theta_1$ (let the oil cool back, or use fresh oil at the same starting temperature).

Data analysis to test the model

For each trial compute

y = \frac{AtV^2}{L}.

Plot $y$ (vertical axis) against $\Delta\theta$ (horizontal axis). If the model is correct, the points should lie close to a straight line.
Draw a best-fit straight line.

Finding $K$ and $Z$
Comparing

\frac{AtV^2}{L} = mK\Delta\theta + Z

with $y = (\text{gradient})x + (\text{intercept})$ gives:

gradient $= mK \Rightarrow K = \dfrac{\text{gradient}}{m}$ ,
intercept $= Z$ .

So reading the gradient and intercept from your best-fit line directly provides $K$ and $Z$ .

(If you also draw a worst acceptable line, you can estimate uncertainties in gradient/intercept and hence uncertainties in $K$ and $Z$ .)

Key Takeaways

In planning questions, marks come from a workable method, clear measurements, and controls.
Turn the given relationship into a straight-line graph so constants can be read from gradient/intercept.
Ensure the experiment really varies only the intended independent variable ( $L$ ) while keeping $m$ , $V$ , $t$ , and $A$ fixed.

Common Mistakes

Plotting the wrong variables (e.g. $\Delta\theta$ vs $L$ without linearising/without using the given equation), which prevents $K$ and $Z$ being found.
Not stating how $L$ is changed and measured (you must explain a practical way to vary immersed length).
Failing to control $V$ (assuming the supply setting is constant without measurement).
Not stirring the oil, leading to unreliable temperature readings.
Ignoring heat loss and not using insulation/lid, causing poor linearity.

Things to Be Careful About

Units and consistency: use SI units in calculations ( $A$ in $\text{m}^2$ , $L$ in $\text{m}$ , $t$ in $\text{s}$ , $V$ in $\text{V}$ , $m$ in $\text{kg}$ ).
Starting temperature: if $\theta_1$ differs between runs, heat losses to surroundings can differ too, reducing the straight-line quality.
Electrical heating changes resistance: the wire’s resistance changes with temperature; keeping temperature rises moderate and using a constant $V$ helps consistency.
Avoid superheating oil: keep temperature safely below smoking point and handle hot apparatus carefully.
Graph quality: use enough points and a wide range of $\Delta\theta$ (or $L$ ) so gradient and intercept are determined with smaller percentage uncertainty.

Techniques used

identify independent, dependent and controlled variables for a heating experimentkeep potential difference and heating time constant while varying immersed wire lengthmeasure temperature rise and compute derived quantities from the given modellinearise the suggested relationship and plot a straight-line graphdetermine constants from the gradient and y-intercept of the best-fit line

Working

Given

\frac{B}{B+d^2}=\frac{\sin^2\theta}{n^2}

Bn^2=(B+d^2)\sin^2\theta

d^2\sin^2\theta=B(n^2-\sin^2\theta)

Divide by $\sin^2\theta$ :

d^2=B\left(\frac{n^2}{\sin^2\theta}-1\right)=Bn^2\left(\frac{1}{\sin^2\theta}\right)-B

So with $y=d^2$ and $x=\dfrac{1}{\sin^2\theta}$ ,

y=(Bn^2)x+(-B)

Answer

gradient $=Bn^2$

$y$ -intercept $=-B$

Final answer

gradient = Bn^2, y-intercept = -B

Detailed explanation

Background Concept

To use a graph to find constants, we try to rewrite the given relationship into a straight-line form

y=mx+c

If we decide what to plot on the axes (here $y=d^2$ and $x=\dfrac{1}{\sin^2\theta}$ ), then we rearrange the physics equation until it looks like a constant multiplied by $x$ , plus a constant term. The constant multiplying $x$ is the gradient $m$ , and the constant term is the intercept $c$ .

Understanding the Question

You are told that

\frac{B}{B+d^2}=\frac{\sin^2\theta}{n^2}

and that a graph of $d^2$ (vertical axis) against $\dfrac{1}{\sin^2\theta}$ (horizontal axis) is plotted. The task is to express the gradient and the intercept in terms of the constants $B$ and $n$ .

Approach

Start with the given equation.
Rearrange to make $d^2$ the subject.
Ensure your final expression is in the form

d^2 = (\text{constant})\left(\frac{1}{\sin^2\theta}\right) + (\text{constant})

Read off gradient and intercept by comparison with $y=mx+c$ .

Step-by-Step Reasoning

Start with

\frac{B}{B+d^2}=\frac{\sin^2\theta}{n^2}

Cross-multiply:

Bn^2=(B+d^2)\sin^2\theta

Expand the right-hand side:

Bn^2=B\sin^2\theta + d^2\sin^2\theta

Make $d^2\sin^2\theta$ the subject:

d^2\sin^2\theta=Bn^2-B\sin^2\theta=B(n^2-\sin^2\theta)

Divide by $\sin^2\theta$ :

d^2=B\left(\frac{n^2}{\sin^2\theta}-1\right)=Bn^2\left(\frac{1}{\sin^2\theta}\right)-B

Now compare with $y=mx+c$ where $y=d^2$ and $x=\dfrac{1}{\sin^2\theta}$ :

gradient $m=Bn^2$
intercept $c=-B$

Key Takeaways

Linearisation means rearranging to match $y=mx+c$ using the chosen axes.
Gradient is the coefficient of the plotted $x$ -quantity.
Intercept is the constant term (value of $y$ when $x=0$ ).

Common Mistakes

Giving gradient as $\dfrac{B}{n^2}$ (from not rearranging fully).
Missing the negative sign in the intercept.
Treating the intercept as $+B$ .

Things to Be Careful About

The graph uses $\dfrac{1}{\sin^2\theta}$ , not $\sin^2\theta$ .
Keep bracket/expansion steps clear so the coefficient of $\dfrac{1}{\sin^2\theta}$ is unambiguous.

Techniques used

rearrange the given relationship into the form y = mx + cmatch plotted variables to identify gradient and interceptcompare coefficients to extract constants

(b)

Values of $\theta$ , $\frac{1}{\sin^2 \theta}$ and $d$ are given in Table 2.1.

Table 2.1

$\theta / ^\circ$	$\frac{1}{\sin^2 \theta}$	$d / \text{cm}$
28.5	4.39	$24.8 \pm 0.2$
33.5	3.28	$21.4 \pm 0.2$
42.5	2.19	$17.1 \pm 0.2$
50.0	1.70	$14.7 \pm 0.2$
57.5	1.41	$12.9 \pm 0.2$
63.5	1.25	$11.8 \pm 0.2$

Calculate and record values of $d^2 / \text{cm}^2$ in Table 2.1.
Include the absolute uncertainties in $d^2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $d^2$ , absolute uncertainty:

\Delta(d^2)=2d\,\Delta d

(using $\Delta d = 0.2\ \text{cm}$ ).

$d=24.8\ \text{cm}$ :

d^2=(24.8)^2=615\ \text{cm}^2,\qquad \Delta(d^2)=2(24.8)(0.2)=9.9\approx 10\ \text{cm}^2

$d=21.4\ \text{cm}$ :

d^2=(21.4)^2=458\ \text{cm}^2,\qquad \Delta(d^2)=2(21.4)(0.2)=8.6\ \text{cm}^2

$d=17.1\ \text{cm}$ :

d^2=(17.1)^2=292\ \text{cm}^2,\qquad \Delta(d^2)=2(17.1)(0.2)=6.8\ \text{cm}^2

$d=14.7\ \text{cm}$ :

d^2=(14.7)^2=216\ \text{cm}^2,\qquad \Delta(d^2)=2(14.7)(0.2)=5.9\ \text{cm}^2

$d=12.9\ \text{cm}$ :

d^2=(12.9)^2=166\ \text{cm}^2,\qquad \Delta(d^2)=2(12.9)(0.2)=5.2\ \text{cm}^2

$d=11.8\ \text{cm}$ :

d^2=(11.8)^2=139\ \text{cm}^2,\qquad \Delta(d^2)=2(11.8)(0.2)=4.7\ \text{cm}^2

Answer

$d^2 / \text{cm}^2$ (with absolute uncertainties):

$615 \pm 10$
$458 \pm 8.6$
$292 \pm 6.8$
$216 \pm 5.9$
$166 \pm 5.2$
$139 \pm 4.7$

Final answer

615±10, 458±8.6, 292±6.8, 216±5.9, 166±5.2, 139±4.7 (all in cm^2)

Detailed explanation

Background Concept

When you calculate a new (derived) quantity from a measured quantity, you should propagate the uncertainty.

Here the derived quantity is

y=d^2

If $d$ has absolute uncertainty $\Delta d$ , then for a power law $y=d^n$ the fractional uncertainty multiplies by $n$ :

\frac{\Delta y}{y} = n\frac{\Delta d}{d}

For $y=d^2$ this gives

\frac{\Delta(d^2)}{d^2} = 2\frac{\Delta d}{d} \quad\Rightarrow\quad \Delta(d^2)=2d\Delta d

Understanding the Question

The table gives $d$ values, each with $\pm 0.2\ \text{cm}$ . You must:

calculate $d^2$ for each row;
calculate the absolute uncertainty in $d^2$ for each row.

These uncertainties are then used as vertical error bars later.

Approach

For each row:

Compute $d^2$ by squaring.
Use

\Delta(d^2)=2d\Delta d

Quote $d^2$ and $\Delta(d^2)$ in $\text{cm}^2$ .

Step-by-Step Reasoning

Take the first row as an example.

Measured: $d=24.8\pm 0.2\ \text{cm}$ .
Square the central value:

d^2=(24.8)^2=615\ \text{cm}^2

Propagate uncertainty:

\Delta(d^2)=2d\Delta d=2(24.8)(0.2)=9.92\ \text{cm}^2\approx 10\ \text{cm}^2

Repeat exactly the same two-step process for each $d$ in the table.

Key Takeaways

For $y=d^2$ , the absolute uncertainty is $\Delta(d^2)=2d\Delta d$ .
Units must be squared too: $\text{cm} \rightarrow \text{cm}^2$ .
These uncertainties become the lengths of the graph’s $y$ -error bars.

Common Mistakes

Writing $\Delta(d^2)= (\Delta d)^2$ (incorrect for propagation here).
Forgetting to square the unit.
Rounding $d^2$ too aggressively so it loses consistency with the uncertainty.

Things to Be Careful About

Keep the uncertainty to 1–2 significant figures and round $d^2$ to a consistent precision that matches it.
Use the given $\Delta d=0.2\ \text{cm}$ for every row (since it is stated constant).

Techniques used

square a measured quantity to obtain a derived quantitypropagate uncertainty through a power using fractional uncertaintiescalculate absolute uncertainty from a percentage uncertainty

(c)

(i)

Plot a graph of $d^2 / \text{cm}^2$ against $\frac{1}{\sin^2 \theta}$ . Include error bars for $d^2$ .

Graph 2.1

2M Similar

DifficultyMedium

Worked solution

Answer

Plot the six points $(x,y)$ where

x=\frac{1}{\sin^2\theta},\qquad y=d^2

with vertical error bars of $\pm\Delta(d^2)$ :

$(4.39,\ 615)$ with $\pm 10$
$(3.28,\ 458)$ with $\pm 8.6$
$(2.19,\ 292)$ with $\pm 6.8$
$(1.70,\ 216)$ with $\pm 5.9$
$(1.41,\ 166)$ with $\pm 5.2$
$(1.25,\ 139)$ with $\pm 4.7$

Axes labelled $d^2 / \text{cm}^2$ and $\left(1/\sin^2\theta\right)$ .

graph-c-i

Final answer

See graph

Detailed explanation

Background Concept

A good physics graph must:

have correctly labelled axes (quantity and unit);
use a sensible scale (use at least half the grid in each direction);
plot points accurately;
include error bars when uncertainties are given.

Error bars represent the range within which the true value may lie. Here only $d$ (hence $d^2$ ) has an uncertainty provided, so we draw vertical error bars for $d^2$ .

Understanding the Question

You are provided with $x=\dfrac{1}{\sin^2\theta}$ and you calculated $y=d^2$ and $\Delta y=\Delta(d^2)$ in part (b). You must place these on the provided grid and show each $\pm\Delta(d^2)$ as a vertical error bar through the point.

Approach

Put $x=\dfrac{1}{\sin^2\theta}$ on the horizontal axis and $y=d^2$ on the vertical axis.
Choose scales that spread the data out.
Plot each point.
For each point, draw a vertical error bar from $y-\Delta y$ to $y+\Delta y$ .

Step-by-Step Reasoning

Label axes clearly: $d^2 / \text{cm}^2$ on $y$ and $\left(1/\sin^2\theta\right)$ on $x$ (dimensionless).
Plot the coordinates from the table (six points).
Add error bars using your computed uncertainties, e.g. for $y=615\ \text{cm}^2$ with $\Delta y=10\ \text{cm}^2$ , the bar runs from $605$ to $625$ .

graph-c-i

Key Takeaways

Error bars should be centred on each plotted point.
Only plot error bars for quantities with uncertainties (here, $d^2$ ).
Clear axis labels and sensible scales are essential for marks.

Common Mistakes

Missing units on the $y$ -axis.
Drawing error bars horizontally (there is no stated uncertainty in $x$ ).
Using error bar total length $\Delta y$ instead of $\pm\Delta y$ .

Things to Be Careful About

Plotting accuracy: points should be within about half a small square.
Use the same scale all along each axis; avoid awkward scales that are hard to read accurately.

Techniques used

choose suitable axis scales and labels with unitsplot data points with vertical error barsuse derived uncertainties to set error bar lengths

(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 single straight line of best fit through the trend of the plotted points.

Draw a worst acceptable straight line (steepest or shallowest) that still passes through all the error bars.

Label the lines "best fit" and "worst".

graph-c-ii

Final answer

See graph

Detailed explanation

Background Concept

A best-fit line is drawn to represent the underlying linear relationship, balancing the scatter so that roughly equal numbers of points lie above and below (and the distances are overall minimised).

A worst acceptable line is used to estimate uncertainty in gradient/intercept: it is the steepest or shallowest straight line that is still consistent with the data within their error bars.

Understanding the Question

After plotting points and error bars in (c)(i), you must:

add the best-fit straight line;
add one worst acceptable straight line;
label both, because later parts require gradients and intercepts from them.

Approach

Draw the best-fit line as a straight line through the middle of the error bars.
Then try drawing an extreme line (either maximum slope or minimum slope) that still intersects every vertical error bar.
Choose the extreme that is furthest from the best-fit line (gives the largest difference in gradient/intercept).

Step-by-Step Reasoning

For the best-fit line: do not join dot-to-dot. Use a ruler and make it a single straight line.
For the worst acceptable line: rotate the ruler to make the line as steep (or as shallow) as possible, but check it still goes through each error bar range. If it misses even one error bar, it is not acceptable.

graph-c-ii

Key Takeaways

Best-fit: "most representative" straight line.
Worst acceptable: "most extreme" line still consistent with error bars.
These two lines allow you to estimate uncertainties in gradient and intercept.

Common Mistakes

Drawing the worst line through the points rather than through the error bars.
Drawing two worst lines (usually only one is required).
Forgetting to label the lines.

Things to Be Careful About

The worst line must remain straight.
It must intersect all error bars, not necessarily the central points.
Use long spans on the grid so later gradient measurements are accurate.

Techniques used

draw a straight line of best fit through scattered datadraw a worst acceptable line consistent with all error barslabel lines clearly for later measurements

(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.

( x_1, y_1 )=(4.39,615),\qquad ( x_2, y_2 )=(1.25,139)

\text{gradient } m=\frac{\Delta y}{\Delta x}=\frac{615-139}{4.39-1.25}=\frac{476}{3.14}=1.52\times 10^2

Worst acceptable line (example):

m_{\text{worst}}\approx 1.57\times 10^2

Absolute uncertainty:

\Delta m = \left|m_{\text{worst}}-m\right|\approx \left|157-152\right|=5

Answer

\text{gradient } m = (152 \pm 5)\ \text{cm}^2

Final answer

(152 ± 5) cm^2

Detailed explanation

Background Concept

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

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

To reduce percentage reading error, choose two points that are far apart on the line (large triangle). Do not use two adjacent points.

To estimate uncertainty in gradient in Paper 5, you compare the gradient of your best-fit line with the gradient of a worst acceptable line:

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

Understanding the Question

You must read the gradient of the best-fit line from your graph and include an absolute uncertainty obtained using the worst acceptable line you drew in (c)(ii).

Approach

Pick two widely separated points on the best-fit line (they can be convenient grid intersections; they do not have to be original data points).
Compute $m_{\text{best}}=\dfrac{\Delta y}{\Delta x}$ .
Repeat using two points on the worst acceptable line to get $m_{\text{worst}}$ .
Use $\Delta m=|m_{\text{worst}}-m_{\text{best}}|$ .

Step-by-Step Reasoning

Suppose the best-fit line passes close to points like $(4.39,615)$ and $(1.25,139)$ .

Compute:

\Delta y = 615-139=476\ \text{cm}^2

\Delta x = 4.39-1.25=3.14

m_{\text{best}}=\frac{476}{3.14}=1.52\times 10^2\ \text{cm}^2

From the worst acceptable line, you might obtain a slightly different slope, e.g. $m_{\text{worst}}\approx 1.57\times 10^2\ \text{cm}^2$ .

Uncertainty:

\Delta m = |157-152|=5\ \text{cm}^2

So quote $m=(152\pm 5)\ \text{cm}^2$ .

Key Takeaways

Use a large triangle for gradients.
Uncertainty comes from how much the slope could plausibly change while still fitting within error bars.

Common Mistakes

Using $\dfrac{\Delta x}{\Delta y}$ by accident.
Using two data points not on the best-fit line (instead of points on the line itself).
Halving the difference between best and worst gradients (Paper 5 typically uses the full difference).

Things to Be Careful About

Units: here $x$ is dimensionless, so gradient has the same unit as $y$ , i.e. $\text{cm}^2$ .
Choose points read cleanly from the line to avoid large rounding error.

Techniques used

determine gradient from two widely separated points on a best-fit linedetermine a worst-case gradient from a worst acceptable linecalculate absolute uncertainty as the difference between 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

Using $y=mx+c$ with $m=152\ \text{cm}^2$ and a point on the best-fit line, e.g. $(x,y)=(1.25,139)$ :

c = y-mx = 139 - 152(1.25)=139-190=-51\ \text{cm}^2

Worst acceptable line gives (example):

c_{\text{worst}}\approx -61\ \text{cm}^2

Absolute uncertainty:

\Delta c = |c_{\text{worst}}-c|=|-61-(-51)|=10\ \text{cm}^2

Answer

\text{$y$-intercept } = (-51 \pm 10)\ \text{cm}^2

Final answer

(-51 ± 10) cm^2

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 either:

read it directly by extending the line to the $y$ -axis, or
calculate it using one point on the line and the gradient:

c=y-mx

Uncertainty in $c$ is estimated similarly to gradient: compare $c$ from the best-fit line to $c$ from the worst acceptable line.

Understanding the Question

You must find the intercept of the best-fit line on the $d^2$ vs $\dfrac{1}{\sin^2\theta}$ graph and include an absolute uncertainty based on your worst acceptable line.

Approach

Determine $c$ from the best-fit line (direct read-off at $x=0$ , or using $c=y-mx$ ).
Determine $c_{\text{worst}}$ from the worst acceptable line.
Use

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

Step-by-Step Reasoning

Using the algebra method with a clearly read point on the best-fit line:

Suppose $m=152\ \text{cm}^2$ .
Take a point on the best-fit line, e.g. $(1.25,139)$ .

Then

c=139-152(1.25)=139-190=-51\ \text{cm}^2

Repeat for the worst acceptable line to get a different intercept, e.g. $c_{\text{worst}}\approx -61\ \text{cm}^2$ .

\Delta c = |-61-(-51)|=10\ \text{cm}^2

Key Takeaways

Intercept can be found by reading at $x=0$ or by $c=y-mx$ .
Use best vs worst line to estimate intercept uncertainty.

Common Mistakes

Forgetting that the intercept may be negative.
Using a point that is not on the drawn best-fit line.
Quoting an intercept with no unit.

Things to Be Careful About

Extrapolation to $x=0$ can amplify reading errors; using $c=y-mx$ with a well-chosen point often improves accuracy.
Units of intercept are the same as $y$ (here $\text{cm}^2$ ).

Techniques used

determine y-intercept by extrapolating the best-fit line to x = 0use y = mx + c to calculate the intercept from one point and a gradientestimate intercept uncertainty using the worst acceptable line

(d)

(i)

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

$B$ = ______
$n$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a): gradient $=Bn^2$ and $y$ -intercept $=-B$ .

Using $y$ -intercept $c=-51\ \text{cm}^2$ :

B=-c=51\ \text{cm}^2

Using gradient $m=152\ \text{cm}^2$ :

Bn^2=m \Rightarrow n^2=\frac{m}{B}=\frac{152}{51}=2.98

n=\sqrt{2.98}=1.73

Answer

B = 51\ \text{cm}^2

n = 1.73

Final answer

B = 51 cm^2, n = 1.73

Detailed explanation

Background Concept

Once you have a straight-line form

y=mx+c

you can identify constants in the original physics equation by matching your rearranged expression.

From part (a) we found:

d^2 = (Bn^2)\left(\frac{1}{\sin^2\theta}\right) - B

So, if the graph is $y=d^2$ against $x=\dfrac{1}{\sin^2\theta}$ :

m = Bn^2,\qquad c = -B

Also, refractive index $n$ is dimensionless.

Understanding the Question

You have numerical values for gradient and intercept from the graph. You must convert these into the constants $B$ and $n$ using the expressions you derived in (a). You must include appropriate units for $B$ .

Approach

Use $c=-B$ to get $B$ .
Use $m=Bn^2$ to get $n^2=\dfrac{m}{B}$ .
Take the square root to find $n$ .
Check units: because $d^2$ has units of $\text{cm}^2$ , $B$ must also be $\text{cm}^2$ .

Step-by-Step Reasoning

If the intercept is $c=-51\ \text{cm}^2$ , then

B=-c=51\ \text{cm}^2

If the gradient is $m=152\ \text{cm}^2$ , then

Bn^2=m \Rightarrow n^2=\frac{m}{B}=\frac{152}{51}=2.98

Hence

n=\sqrt{2.98}=1.73

Key Takeaways

Intercept gives $B$ directly (with a sign change).
Gradient and $B$ together give $n$ .
Units are checked by comparing with the linear equation.

Common Mistakes

Taking $B$ equal to the intercept instead of the negative of the intercept.
Forgetting to take the square root and giving $n^2$ instead of $n$ .
Giving units for $n$ (it is dimensionless).

Things to Be Careful About

Use consistent significant figures (usually 2–3 s.f. based on graph readings).
If your intercept is negative, $B$ should come out positive (since it represents a squared-length scale here).

Techniques used

use gradient and intercept to determine constants from a linear modelhandle sign correctly when relating intercept to a physical constantcompute a refractive index from a ratio and square root

(ii)

Determine the percentage uncertainty in $n$ .

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

1M Similar

DifficultyMedium

Worked solution

Working

n=\left(\frac{m}{B}\right)^{1/2}

So fractional uncertainty:

\frac{\Delta n}{n}=\frac{1}{2}\left(\frac{\Delta m}{m}+\frac{\Delta B}{B}\right)

Using $m=152\pm 5\ \text{cm}^2$ and $B=51\pm 10\ \text{cm}^2$ :

\frac{\Delta n}{n}=\frac{1}{2}\left(\frac{5}{152}+\frac{10}{51}\right)=\frac{1}{2}(0.033+0.196)=0.115

Percentage uncertainty:

0.115\times 100\% = 11.5\% \approx 12\%

Answer

\text{percentage uncertainty in } n \approx 12\%

Final answer

12%

Detailed explanation

Background Concept

When quantities are multiplied or divided, fractional (or percentage) uncertainties add:

\frac{\Delta Q}{Q}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\quad\text{for }Q=\frac{A}{B}

If you then take a power, $Q=A^p$ , the fractional uncertainty multiplies by $|p|$ :

\frac{\Delta Q}{Q}=|p|\frac{\Delta A}{A}

Here

n=\left(\frac{m}{B}\right)^{1/2}

so the factor of $\tfrac{1}{2}$ appears.

Understanding the Question

You already obtained a value of $n$ from the gradient $m$ and intercept (giving $B$ ). You must now find the percentage uncertainty in $n$ , using the uncertainties in $m$ and $B$ that come from the best and worst acceptable lines.

Approach

Write $n$ in terms of $m$ and $B$ .
Write the fractional uncertainty rule for $n=(m/B)^{1/2}$ .
Substitute $\Delta m, m, \Delta B, B$ .
Convert to percentage.

Step-by-Step Reasoning

Start from

n=\left(\frac{m}{B}\right)^{1/2}

For the inside ratio $m/B$ :

\frac{\Delta(m/B)}{m/B}=\frac{\Delta m}{m}+\frac{\Delta B}{B}

Then applying the square root (power $1/2$ ):

\frac{\Delta n}{n}=\frac{1}{2}\left(\frac{\Delta m}{m}+\frac{\Delta B}{B}\right)

Substitute the values from the graph and evaluate, then multiply by $100\%$ .

Key Takeaways

Add fractional uncertainties for division.
Multiply by $\tfrac{1}{2}$ because $n$ is a square root.
Convert to percentage at the end.

Common Mistakes

Forgetting the factor $\tfrac{1}{2}$ .
Subtracting uncertainties when dividing (they add).
Using absolute uncertainties directly without first forming fractional uncertainties.

Things to Be Careful About

Use the uncertainties you obtained from the best vs worst lines (not the scatter of points).
Keep enough significant figures during intermediate steps so the final percentage is not distorted by rounding.

Techniques used

propagate uncertainties through a quotientpropagate uncertainties through a square rootcombine fractional uncertainties additively for multiplication or division

(e)

The experiment is repeated. Determine the angle $\theta$ that gives a value of $d$ of $30.0\ \text{cm}$ .

$\theta$ = ______ $^\circ$

1M Similar

DifficultyMedium

Worked solution

Working

From (a):

d^2 = Bn^2\left(\frac{1}{\sin^2\theta}\right)-B

\frac{1}{\sin^2\theta}=\frac{d^2+B}{Bn^2}

Using $d=30.0\ \text{cm}$ , $d^2=900\ \text{cm}^2$ , $B=51\ \text{cm}^2$ and $Bn^2=\text{gradient}=152\ \text{cm}^2$ :

\frac{1}{\sin^2\theta}=\frac{900+51}{152}=6.26

\sin^2\theta=\frac{1}{6.26}=0.160 qquad\Rightarrow\qquad \sin\theta=0.400

\theta=\sin^{-1}(0.400)=23.6^\circ

Answer

\theta \approx 23.6^\circ

Final answer

23.6°

Detailed explanation

Background Concept

Once constants are found from a linear graph, you can use the model equation to predict values. Here the relationship rearranges to

d^2 = Bn^2\left(\frac{1}{\sin^2\theta}\right)-B

So if you know $d$ , you can solve for $\sin^2\theta$ and then find $\theta$ using an inverse sine.

Understanding the Question

You repeat the experiment and want the incident angle $\theta$ that would produce a distance $d=30.0\ \text{cm}$ . Since $d$ is larger than any in the table, you should expect $\theta$ to be smaller than the smallest table value (because smaller $\theta$ gives smaller $\sin\theta$ , larger $1/\sin^2\theta$ , hence larger $d^2$ ).

Approach

Square $d$ to get $d^2$ .
Rearrange the straight-line form to make $\dfrac{1}{\sin^2\theta}$ the subject.
Invert to get $\sin^2\theta$ .
Take the square root to get $\sin\theta$ , then use $\sin^{-1}$ .

Step-by-Step Reasoning

Start from

d^2 = Bn^2\left(\frac{1}{\sin^2\theta}\right)-B

Rearrange:

d^2 + B = Bn^2\left(\frac{1}{\sin^2\theta}\right)

\frac{1}{\sin^2\theta}=\frac{d^2+B}{Bn^2}

Substitute $d=30.0\ \text{cm}$ so $d^2=900\ \text{cm}^2$ , and use your graph-derived constants (often easiest is $Bn^2 =$ gradient):

\frac{1}{\sin^2\theta}=\frac{900+B}{\text{gradient}}

Compute $\sin^2\theta$ , then $\sin\theta$ , then $\theta$ .

Key Takeaways

Use the straight-line rearranged equation for predictions.
Using $Bn^2$ as the gradient avoids carrying $n$ separately.
Check your answer is physically sensible (here $\theta$ should be less than the smallest measured angle).

Common Mistakes

Using $d$ instead of $d^2$ in the equation.
Forgetting to add $B$ to $d^2$ .
Taking $\theta=\sin^{-1}(\sin^2\theta)$ instead of first taking the square root.

Things to Be Careful About

Calculator mode: ensure degrees.
Rounding too early can shift $\theta$ by a noticeable amount; keep at least 3 s.f. until the final step.

Techniques used

substitute a target value into a model equationrearrange to solve for a trigonometric functiontake an inverse trigonometric function to determine an angle

Physics 9702/52 — October/November 2024

Method (procedure and measurements)

Control of variables

Analysis to test relationship and find KKK and ZZZ

Safety

Background Concept

Understanding the Question

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Step-by-Step Reasoning

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Key Takeaways

Analysis to test relationship and find $K$ and $Z$