Physics 9702/33 — February/March 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate an electrical circuit.

(a)

• Connect the circuit shown in Fig. 1.1.
E, F, G and H are crocodile clips.

Fig. 1.1

• Using the long connecting lead, clip crocodile clips F and G on the resistance wire so that they are approximately $0.45\ \text{m}$ apart.

• The length of resistance wire between F and G is $x$ .
Measure and record $x$ .

$x$ = ______ $\text{m}$

• The current in the circuit is $I$ .
Close S and record $I$ .

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

• Open S.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example readings (student-dependent):

x = 0.450\ \text{m}

I = 0.290\ \text{A}

Final answer

See working (student-dependent)

Detailed explanation

Background Concept

In a simple d.c. circuit, the ammeter measures the current $I$ flowing in series. The length of resistance wire included in the circuit affects the total resistance, so changing a length on the wire can change the current.

A metre rule is used to measure the separation $x$ between two clip positions along the wire. The key practical skills here are:

making a clear, parallax-free length measurement,
ensuring the ammeter is in series and reads steadily,
recording values with appropriate precision and units.

Understanding the Question

You are asked to:

set up the circuit of Fig. 1.1,
set clips $F$ and $G$ about $0.45\ \text{m}$ apart,
measure and record the wire length between them, $x$ (in $\text{m}$ ),
close the switch and record the current $I$ (in $\text{A}$ ), then open the switch.

The actual numerical values depend on your apparatus, so the mark is for correct measurement technique, sensible values, and correct recording (unit, precision).

Approach

Build the circuit exactly as in the figure.
Place $F$ and $G$ on the wire and measure their separation using the metre rule scale.
Close switch $S$ briefly, wait for the ammeter reading to settle, and record $I$ .
Open $S$ to prevent heating of the wire (which would change its resistance).

Step-by-Step Reasoning

circuit-bypass

Set up: Ensure the cell, switch $S$ , and ammeter $A$ are all in series with the resistance wire.
Set $x$ : Move crocodile clips $F$ and $G$ so they are roughly $0.45\ \text{m}$ apart.
Measure $x$ : Read positions of $F$ and $G$ on the metre rule and calculate $x$ as the difference. Record to a sensible precision (e.g. $0.001\ \text{m}$ or $0.01\ \text{m}$ depending on how well you can align the clip).
Measure $I$ : Close $S$ , let the current settle, record $I$ (typically to $0.001\ \text{A}$ or $0.01\ \text{A}$ depending on meter resolution), then open $S$ .

Key Takeaways

Measure lengths using end readings on a scale and take the difference.
Ammeter must be in series; record $I$ with unit $\text{A}$ .
Keep switch closed only briefly to reduce heating.

Common Mistakes

Measuring $x$ from the end of the wire instead of between $F$ and $G$ .
Recording $x$ without units or in $\text{cm}$ when the answer space expects $\text{m}$ .
Leaving the switch closed for long periods so the wire heats up and $I$ drifts.

Things to Be Careful About

Avoid parallax when reading the metre rule.
Ensure crocodile clips make good electrical contact (otherwise $I$ may be unstable).
Record $x$ and $I$ to consistent, sensible precision that matches your instruments.

Techniques used

assemble the circuit as shown in the diagrammeasure a length using a metre rule with appropriate precisiontake a current reading from an ammeter after closing the switch

(b)

Change $x$ and record $x$ and $I$ .
Repeat until you have six sets of values of $x$ and $I$ .

Record your results in a table. Include values of $\frac{1}{I}$ in your table.

9M Similar

DifficultyMedium

Worked solution

Answer

Record six pairs of $x$ and $I$ , and calculate $1/I$ .

Example of a suitable table (values are illustrative):

$x\ /\ \text{m}$	$I\ /\ \text{A}$	$\left(\frac{1}{I}\right)\ /\ \text{A}^{-1}$
0.10	0.200	5.00
0.20	0.238	4.20
0.30	0.286	3.50
0.40	0.357	2.80
0.50	0.435	2.30
0.60	0.556	1.80

Final answer

See working (student-dependent)

Detailed explanation

Background Concept

When investigating a relationship experimentally, you choose an independent variable (here $x$ ) and measure a dependent variable (here $I$ ). To test a proposed linear model of the form

\frac{1}{I} = ax + b,

you must calculate $1/I$ for each measured current and present the results clearly.

Good data-handling practice in Paper 3 includes:

a suitable range of $x$ values,
six sets of readings (as requested),
clear table headings with quantity and unit,
consistent precision in each column,
correct calculation of derived quantities.

Understanding the Question

You must change $x$ several times and, each time, measure $I$ . You then make a results table containing:

$x$ (in $\text{m}$ ),
$I$ (in $\text{A}$ ),
$1/I$ (in $\text{A}^{-1}$ ),
for six different values of $x$ .

Approach

Choose six different values of $x$ spread across a wide practical range (not all clustered near $0.45\ \text{m}$ ).
For each $x$ , close the switch briefly, record $I$ , then open the switch.
Compute $1/I$ for each row.
Build one clear table with correct headings and units.

Step-by-Step Reasoning

Choosing $x$ values: pick values that are well separated (e.g. steps of $0.05$ to $0.15\ \text{m}$ ), staying within the wire length.
Taking $I$ readings: for each $x$ , close switch $S$ briefly and wait for a steady ammeter reading. Open $S$ after recording to limit heating.
Repeat/quality: if one reading looks anomalous or unstable, repeat it immediately; keep contact points clean and firm.
Calculating $1/I$ : for each measured current, compute

\frac{1}{I}

and record it to consistent significant figures (typically matching the precision of $I$ ).
5) Table presentation: include units in the headings (not in the body). Use consistent decimal places within each column where reasonable.

Key Takeaways

Six well-spaced readings improve the reliability of the graph.
Derived quantities must be calculated correctly and presented with units.
Consistent precision and clear headings are essential for marks.

Common Mistakes

Missing the $1/I$ column or giving it without unit $\text{A}^{-1}$ .
Writing units in every cell instead of in the heading.
Using too narrow a range of $x$ (gives a poor graph and uncertain gradient).
Inconsistent precision (e.g. mixing $0.2\ \text{A}$ with $0.2374\ \text{A}$ without justification).

Things to Be Careful About

Do not let the wire heat up (it changes resistance and therefore changes $I$ ).
Ensure $I$ is not so large that it exceeds the ammeter range.
If $I$ is very small, $1/I$ becomes very large and uncertainties become significant; choose $x$ so readings remain sensible.

Techniques used

vary the independent variable over a suitable rangerecord paired readings in a single results table with headings and unitscalculate a derived quantity using recorded datause consistent significant figures within each column

(c)

(i)

Plot a graph of $\frac{1}{I}$ on the $y$ -axis against $x$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $\left(\frac{1}{I}\right)$ on the $y$ -axis (unit $\text{A}^{-1}$ ) against $x$ on the $x$ -axis (unit $\text{m}$ ) using a suitable scale and all six data points.

Final answer

Graph plotted

Detailed explanation

Background Concept

A graph is used to reveal and test relationships between quantities. If the suggested model is

\frac{1}{I} = ax + b,

then a plot of $y = 1/I$ against $x$ should produce a straight line.

Marks are typically awarded for:

correct axes (right variables on the right axes),
correct labels including units,
sensible scales (not cramped; not awkward like 3 squares = 0.07),
accurate plotting of all points.

Understanding the Question

You must use your table from (b) and create a graph with:

horizontal axis: $x$ (in $\text{m}$ ),
vertical axis: $1/I$ (in $\text{A}^{-1}$ ).

Approach

Decide the minimum and maximum of your $x$ and $1/I$ data.
Choose scales that spread the data over at least half of the graph paper in both directions.
Label axes as $x/\text{m}$ and $(1/I)/\text{A}^{-1}$ (or equivalent clear labelling).
Plot each point with a small, neat cross.

Step-by-Step Reasoning

From the table, identify the range of $x$ and $1/I$ .
Mark a scale along each axis that covers the full range.
Add axis labels with units.
Plot each pair $(x, 1/I)$ carefully; do not join dots.

Key Takeaways

The variables on the axes must match the instruction.
Labels must include units.
Good scaling and accurate points reduce uncertainty in the gradient.

Common Mistakes

Plotting $I$ instead of $1/I$ .
Missing units on axes.
Using a tiny section of the grid (poor scale choice).
Plotting blobs too large to judge accuracy.

Things to Be Careful About

Check that you have not inverted a value incorrectly (e.g. using $1/0.25 = 4$ ).
Ensure each plotted point corresponds to the correct row of the table.

Techniques used

choose suitable axis scales to use at least half the gridlabel axes with quantities and unitsplot points accurately from a table of values

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Choose two well-separated points on the best-fit line, e.g.

(x_1, y_1) = (0.10\ \text{m},\ 5.00\ \text{A}^{-1}),\quad (x_2, y_2) = (0.60\ \text{m},\ 1.82\ \text{A}^{-1})

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{1.82 - 5.00}{0.60 - 0.10} = -6.36\ \text{A}^{-1}\ \text{m}^{-1}

Read the $y$ -intercept from the graph:

y\text{-intercept} = 5.64\ \text{A}^{-1}

Answer

Gradient $= -6.36\ \text{A}^{-1}\ \text{m}^{-1}$

$y$ -intercept $= 5.64\ \text{A}^{-1}$

Final answer

Gradient = -6.36 A^-1 m^-1, y-intercept = 5.64 A^-1 (illustrative)

Detailed explanation

Background Concept

For a straight-line graph, the gradient (slope) is

\text{gradient} = \frac{\Delta y}{\Delta x},

and the $y$ -intercept is the value of $y$ when $x=0$ (where the line crosses the $y$ -axis).

Here:

$y = 1/I$ with unit $\text{A}^{-1}$ ,
$x$ has unit $\text{m}$ ,
so the gradient unit must be $\text{A}^{-1}\ \text{m}^{-1}$ .

Understanding the Question

You must extract two numerical features from your best-fit line:

its gradient,
its $y$ -intercept.

These are then used later to find constants $a$ and $b$ .

Approach

Use two points on the best-fit line (not necessarily measured points).
Make the points as far apart as possible to reduce percentage uncertainty.
Compute $\Delta y/\Delta x$ .
Read the intercept by extending the line to the $y$ -axis and reading the value.

Step-by-Step Reasoning

Select two points: pick grid intersections where you can read $x$ and $y$ accurately.
Calculate gradient:

\text{gradient} = \frac{y_2-y_1}{x_2-x_1}

Check that $\Delta x$ is in $\text{m}$ and $\Delta y$ in $\text{A}^{-1}$ .
3) Find $y$ -intercept: find where the line crosses the $y$ -axis ( $x=0$ ). If the intercept is off the graph, extend the line carefully.
4) Units: gradient $\text{A}^{-1}\ \text{m}^{-1}$ ; intercept $\text{A}^{-1}$ .

Key Takeaways

Use a large triangle on the line for an accurate gradient.
Gradient is always $\Delta y/\Delta x$ , never $\Delta x/\Delta y$ .
Units come directly from the axes.

Common Mistakes

Using two experimental points rather than points on the best-fit line.
Choosing points too close together (large uncertainty).
Getting the sign wrong: if the line slopes downwards, gradient is negative.
Forgetting units or using the wrong ones.

Things to Be Careful About

Read values to the precision allowed by the graph scale.
Make sure you identify $y$ as $1/I$ , not $I$ .
If you extend the line, keep the ruler aligned with the original best-fit line.

Techniques used

calculate the gradient from two widely spaced points on the best-fit lineread the y-intercept where the line crosses the y-axisinclude correct units for gradient and intercept

(d)

It is suggested that the quantities $I$ and $x$ are related by

\frac{1}{I} = ax + b

where $a$ and $b$ are constants.

Using your answers in (c)(iii), determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For the graph of $y = \frac{1}{I}$ against $x$ :

\frac{1}{I} = ax + b

a = \text{gradient},\qquad b = y\text{-intercept}

Using (c)(iii):

a = -6.36\ \text{A}^{-1}\ \text{m}^{-1}

b = 5.64\ \text{A}^{-1}

Answer

a = -6.36\ \text{A}^{-1}\ \text{m}^{-1}

b = 5.64\ \text{A}^{-1}

Final answer

a = gradient, b = y-intercept (with units)

Detailed explanation

Background Concept

A straight-line graph has form

y = mx + c,

where $m$ is the gradient and $c$ is the intercept on the $y$ -axis.

In this experiment, the suggested relationship is

\frac{1}{I} = ax + b.

Comparing directly:

$y \equiv 1/I$ ,
$m \equiv a$ ,
$c \equiv b$ .

Units:

$1/I$ has unit $\text{A}^{-1}$ ,
$x$ has unit $\text{m}$ ,
therefore

[a] = \frac{\text{A}^{-1}}{\text{m}} = \text{A}^{-1}\ \text{m}^{-1},\qquad [b] = \text{A}^{-1}.

Understanding the Question

You are not asked to re-plot anything. You simply use your gradient and intercept from (c)(iii) to state $a$ and $b$ , including appropriate units.

Approach

Set $a$ equal to the gradient.
Set $b$ equal to the $y$ -intercept.
Attach units based on the graph axes.

Step-by-Step Reasoning

Write the linear model and compare with $y=mx+c$ .
Replace $m$ with your measured gradient and $c$ with your intercept.
Confirm units are consistent with axis units.

Key Takeaways

Straight-line identification: gradient $\rightarrow$ coefficient of $x$ ; intercept $\rightarrow$ constant term.
Units come from the variables plotted.

Common Mistakes

Swapping $a$ and $b$ .
Quoting $a$ without the $\text{m}^{-1}$ .
Using the gradient of the wrong graph (e.g. if axes were swapped).

Things to Be Careful About

Keep the sign of $a$ (negative if the line slopes downward).
Ensure $b$ is read at $x=0$ (not at your smallest measured $x$ ).

Techniques used

match the plotted graph variables to y = mx + cidentify constants from gradient and interceptdeduce units from the relationship between axes variables

(e)

Theory suggests that

a = -\frac{P}{Q}

where $P$ is the resistance per unit length of the resistance wire and $Q$ is $1.5\ \text{V}$ .

Use your value of $a$ to calculate the value of $P$ .

$P$ = ______ $\Omega\ \text{m}^{-1}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

a = -\frac{P}{Q}

P = -aQ

With $Q = 1.5\ \text{V}$ and $a = -6.36\ \text{A}^{-1}\ \text{m}^{-1}$ :

P = -(-6.36)(1.5) = 9.54\ \Omega\ \text{m}^{-1}

Answer

P = 9.54\ \Omega\ \text{m}^{-1}

Final answer

P = 9.54 Ω m^-1 (illustrative)

Detailed explanation

Background Concept

The gradient $a$ of a suitable graph can be linked to a physical constant. Here, theory gives

a = -\frac{P}{Q},

where:

$P$ is resistance per unit length of the wire ( $\Omega\ \text{m}^{-1}$ ),
$Q$ is a voltage ( $\text{V}$ ).

Because $1\ \Omega = 1\ \text{V A}^{-1}$ , multiplying $a$ (unit $\text{A}^{-1}\ \text{m}^{-1}$ ) by $Q$ (unit $\text{V}$ ) gives $\Omega\ \text{m}^{-1}$ , as required.

Understanding the Question

You must use your experimental value of $a$ (from the graph) and the known value $Q=1.5\ \text{V}$ to calculate $P$ .

Approach

Rearrange the theoretical equation to make $P$ the subject:

P = -aQ.

Then substitute your value of $a$ and $Q=1.5\ \text{V}$ .

Step-by-Step Reasoning

Start with

a = -\frac{P}{Q}.

Multiply both sides by $Q$ :

aQ = -P.

Multiply by $-1$ :

P = -aQ.

Substitute values, keeping the sign correct (if $a$ is negative, $P$ comes out positive).
Quote $P$ with unit $\Omega\ \text{m}^{-1}$ .

Key Takeaways

Always rearrange carefully and track negative signs.
Check units: $\text{V}\times \text{A}^{-1}=\Omega$ .

Common Mistakes

Forgetting the minus sign and getting a negative $P$ .
Using $Q=1.5$ without units and then writing an incorrect unit for $P$ .
Using $b$ instead of $a$ .

Things to Be Careful About

Use your own measured $a$ from (d), not the illustrative value shown here.
Use a sensible number of significant figures (often matching the precision of the gradient from the graph).

Techniques used

rearrange an equation to make the required quantity the subjectsubstitute the gradient-derived constant into the theoretical expressioncheck unit consistency in a derived constant

Q2Medium-EasyManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationPresentation of Data and Observations

In this experiment, you will investigate the thermal properties of plastic pipe.

(a)

• You are provided with two lengths of plastic pipe.
• Select one of them and, using the waterproof pen, label one end A.
• Make a mark approximately $12\ \text{cm}$ from end A, as shown in Fig. 2.1.

Fig. 2.1

• The distance from end A to the mark is $L$ , as shown in Fig. 2.1.
Measure and record $L$ .

$L$ = ______

• Use the thermometer to measure the room temperature $T_0$ .
Record $T_0$ .

$T_0$ = ______ $^{\circ}\text{C}$

2M Similar

DifficultyEasy

Worked solution

Answer

$L$ measured with a ruler to the nearest $0.1\ \text{cm}$ (example):

$L = 12.0\ \text{cm}$

Room temperature (example):

$T_0 = 25.0\ ^{\circ}\text{C}$

Final answer

Example readings: L = 12.0 cm, T0 = 25.0 °C

Detailed explanation

Background Concept

In Paper 3, marks for measurements are awarded mainly for good technique and appropriate recording. A length should be measured with a suitable instrument (e.g. a metre rule) and recorded to a precision consistent with the smallest scale division. A temperature should be measured with the thermometer provided and recorded with a sensible resolution (often $0.5\ ^{\circ}\text{C}$ or $0.1\ ^{\circ}\text{C}$ depending on the scale).

Understanding the Question

You must (1) mark a point about $12\ \text{cm}$ from end A, then measure the actual distance from end A to the mark (this is $L$ ), and (2) measure the room temperature $T_0$ . These are raw measurements used later to test the proposed relationship.

Approach

Measure $L$ along the pipe from the end face at A to the drawn mark.
Measure $T_0$ with the thermometer at room conditions.
Record both with units and appropriate precision.

Step-by-Step Reasoning

Place the ruler alongside the pipe so the zero is aligned with end A (or measure between two clear reference points if the end cannot be aligned exactly).
Read the position of the mark at eye level to reduce parallax; record $L$ to the nearest small division (commonly $0.1\ \text{cm}$ ).
For $T_0$ , ensure the thermometer bulb is in air (not touching your hand or a warm surface), wait for the reading to stabilise, and read at eye level.
Record $T_0$ with the unit $^{\circ}\text{C}$ .

Key Takeaways

Raw readings must be recorded clearly, with units and appropriate precision.
Parallax and poor alignment are common causes of systematic error in length measurements.

Common Mistakes

Measuring $L$ from the wrong end (not from end A).
Writing $T_0$ with no unit or an unrealistic precision (e.g. $25.000\ ^{\circ}\text{C}$ ).
Reading the ruler/thermometer at an angle (parallax).

Things to Be Careful About

The mark is “approximately” $12\ \text{cm}$ : you must measure the actual $L$ you made, not assume $12\ \text{cm}$ .
Ensure the reference is the end face at A (not the start of a label line drawn on the pipe).

Techniques used

measure a length using a metre rule with appropriate precisionread a thermometer at eye level to avoid parallaxrecord raw measurements with units and sensible significant figures

(b)

(i)

• Set up the apparatus as shown in Fig. 2.2.
The point Z is the lower edge of the end of the wooden strip.

Fig. 2.2

• Ensure that end A is at the bottom of the measuring cylinder.
• The distance between the centre of the bolt and the centre of the nail is $s$ .
The distance between the centre of the nail and the end of the wooden strip is $d$ , as shown in Fig. 2.2.
• Measure and record $s$ and $d$ .

$s$ = ______
$d$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Measure between the stated centres (example readings):

$s = 7.5\ \text{cm}$

$d = 22.5\ \text{cm}$

Final answer

Example readings: s = 7.5 cm, d = 22.5 cm

Detailed explanation

Background Concept

This set-up uses a wooden strip as a lever about the nail (pivot). The distances $s$ and $d$ are lever arms measured from the pivot to (i) the bolt connection and (ii) the end point Z. Because the later calculation uses a ratio $\frac{s}{d}$ , identifying the correct reference points (centres) is essential.

Understanding the Question

You are told:

$s$ is the distance between the centre of the bolt and the centre of the nail.
$d$ is the distance between the centre of the nail and the end of the wooden strip (point Z).
You must measure and record $s$ and $d$ .

Approach

Use the diagram to identify the nail (pivot), the bolt (where the pipe is attached), and point Z (the lower edge of the strip end).
Measure centre-to-centre distances with a ruler, keeping the ruler parallel to the strip.

Step-by-Step Reasoning

diagram-1

Locate the nail: it passes through the wooden strip and is held in a boss (this is the pivot).
Locate the bolt: it is fixed to the strip and connects to the pipe/masses.
Measure $s$ along the strip from the centre of the nail to the centre of the bolt.
Identify Z as the lower edge at the end of the strip and measure $d$ from the nail centre to Z along the strip.
Record both distances in $\text{cm}$ to a consistent precision (typically $0.1\ \text{cm}$ ).

Key Takeaways

Always measure the quantity exactly as defined (centre-to-centre here).
Lever problems are sensitive to the correct lever arm lengths.

Common Mistakes

Measuring to the edge of the nail/bolt rather than the centre.
Measuring $d$ to the wrong end of the strip (not to point Z).
Measuring off the strip (diagonal measurement) rather than along the line of the strip.

Things to Be Careful About

If the nail or bolt head obscures the centre, estimate the centre by measuring diameter and halving, or use the apparent midpoint.
Keep the strip still while measuring to reduce random error.

Techniques used

set up apparatus according to a diagram while maintaining correct alignmentmeasure distances between specified centres using a rulerrecord measurements with consistent precision

(ii)

• The distance from the bench to Z is $H_1$ , as shown in Fig. 2.2.
Measure and record $H_1$ .

$H_1$ = ______

• Slowly and carefully pour very hot water into the measuring cylinder until the water level reaches the mark on the pipe.
The distance between the bench and Z will change. When Z reaches its lowest position the new distance from the bench to Z is $H_2$ .
Measure and record $H_2$ .

$H_2$ = ______

• The temperature of the water in the measuring cylinder is $T$ .
Measure and record $T$ .

$T$ = ______ $^{\circ}\text{C}$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example readings:

$H_1 = 18.4\ \text{cm}$

$H_2 = 17.6\ \text{cm}$

$T = 85.0\ ^{\circ}\text{C}$

Final answer

Example readings: H1 = 18.4 cm, H2 = 17.6 cm, T = 85.0 °C

Detailed explanation

Background Concept

When the pipe is heated by hot water, it expands and changes the geometry of the lever system, causing point Z to move. The experiment uses changes in height ( $H_1$ to $H_2$ ) as the measurable effect of thermal expansion. Temperature $T$ is needed to relate expansion to temperature rise.

Understanding the Question

You must measure:

$H_1$ : the distance from the bench to point Z before adding hot water.
$H_2$ : the new distance from the bench to Z when Z reaches its lowest position after filling the cylinder to the mark.
$T$ : the temperature of the hot water in the cylinder.

Approach

Use a ruler or metre rule to measure vertical distances from the bench to Z.
Add hot water carefully until the level reaches the mark, then watch Z move and record the minimum height.
Measure water temperature with the thermometer once the water has settled.

Step-by-Step Reasoning

Before adding hot water, hold the ruler vertically on the bench near Z and read the height of Z at eye level: this is $H_1$ .
Pour hot water slowly until the water surface reaches the mark on the pipe. Pouring slowly reduces overshoot and splashing.
Observe Z as the pipe heats; it will move and then stop changing. Record the lowest position reached as $H_2$ .
Place the thermometer bulb in the hot water (not touching the pipe or cylinder wall), wait for a steady reading, and record $T$ .

Key Takeaways

Identify the correct moment to take $H_2$ : at the lowest (final) position.
Temperature readings require time to stabilise.

Common Mistakes

Recording $H_2$ too early (before Z stops moving).
Measuring from the wrong reference level (not from the bench).
Measuring $T$ while the thermometer bulb is touching the container wall (can give an unrepresentative reading).

Things to Be Careful About

Avoid parallax when reading the ruler and thermometer.
Ensure the water level is exactly at the mark; if above/below, the heated length is not the intended $L$ and affects the result.

Techniques used

measure vertical heights from a reference level with a ruleridentify the lowest position of a moving point and record itmeasure temperature after allowing the thermometer to reach equilibrium

(iii)

• Calculate $H_1 - H_2$ .

$H_1 - H_2$ = ______

• Estimate the percentage uncertainty in your value of $H_1 - H_2$ .
Show your working.

percentage uncertainty = ______ $\%$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

H_1 - H_2 = 18.4 - 17.6 = 0.8\ \text{cm}

Assume each height reading has uncertainty $\pm 0.1\ \text{cm}$ .

Absolute uncertainty in $H_1 - H_2$ :

\Delta(H_1 - H_2) = 0.1 + 0.1 = 0.2\ \text{cm}

Percentage uncertainty:

\frac{0.2}{0.8} \times 100 = 25\%

Answer

$H_1 - H_2 = 0.8\ \text{cm}$

percentage uncertainty $= 25\%$

Final answer

H1 − H2 = 0.8 cm; percentage uncertainty = 25%

Detailed explanation

Background Concept

When you calculate a difference of two measured values, the absolute uncertainty in the result is found by adding the absolute uncertainties of the two readings (worst-case). Then percentage uncertainty is

\%\ \text{uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100.

This is especially important when the difference is small, because the percentage uncertainty can become large.

Understanding the Question

You must:

Compute the change in height $H_1 - H_2$ .
Estimate the percentage uncertainty in this change and show working.
The uncertainty must be based on how precisely you can read $H_1$ and $H_2$ .

Approach

Subtract your measured heights to find $H_1 - H_2$ .
Decide the reading uncertainty for each height (often half a smallest division; many candidates use $\pm 0.1\ \text{cm}$ for a mm-scale ruler).
For a subtraction, add the absolute uncertainties.
Convert to percentage.

Step-by-Step Reasoning

Using the example values, subtract: $18.4\ \text{cm} - 17.6\ \text{cm} = 0.8\ \text{cm}$ .
Estimate reading uncertainty. If the ruler is marked in $1\ \text{mm}$ divisions, a common estimate is $\pm 1\ \text{mm} = \pm 0.1\ \text{cm}$ per reading.
Combine for a difference (worst-case): $0.1 + 0.1 = 0.2\ \text{cm}$ .
Percentage uncertainty: $\frac{0.2}{0.8} \times 100 = 25\%$ .

Key Takeaways

Differences of similar numbers can have large percentage uncertainties.
For addition/subtraction, add absolute uncertainties.

Common Mistakes

Using percentage uncertainties of $H_1$ and $H_2$ instead of absolute uncertainties.
Subtracting uncertainties ( $0.1 - 0.1$ ) instead of adding.
Forgetting to multiply by $100$ to convert to percent.

Things to Be Careful About

Your uncertainty must match your instrument (don’t claim $\pm 0.01\ \text{cm}$ if you used a normal ruler).
Keep units consistent: if you use $\text{mm}$ for uncertainty, convert to $\text{cm}$ (or vice versa) before forming the percentage.

Techniques used

subtract two measured quantities to obtain a changeestimate absolute uncertainty from instrument resolutioncombine absolute uncertainties for a differencecalculate percentage uncertainty

(iv)

• Calculate $\Delta L$ where $\Delta L = \frac{s(H_1 - H_2)}{d}$ .

$\Delta L$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta L = \frac{s(H_1 - H_2)}{d} = \frac{7.5\times 0.8}{22.5} = 0.266\overline{6}\ \text{cm}

Answer

$\Delta L = 0.267\ \text{cm}$

Final answer

ΔL = 0.267 cm

Detailed explanation

Background Concept

The expression

\Delta L = \frac{s(H_1 - H_2)}{d}

comes from lever geometry: the movement at one point on a rigid bar is proportional to its distance from the pivot. Here, a measured vertical displacement $H_1-H_2$ at Z is converted into the extension of the pipe $\Delta L$ using the ratio of lever arms $\frac{s}{d}$ .

Understanding the Question

You are given the formula for $\Delta L$ and must calculate it using your measured $s$ , $d$ , and $H_1-H_2$ .

Approach

Ensure $s$ , $d$ , and $H_1-H_2$ are in the same length unit.
Substitute into the formula.
Quote $\Delta L$ to a sensible number of significant figures based on the raw data.

Step-by-Step Reasoning

Using example values in cm keeps the calculation simple and consistent.
Compute numerator: $s(H_1-H_2)=7.5\times 0.8=6.0\ \text{cm}^2$ .
Divide by $d$ : $6.0/22.5=0.266\overline{6}\ \text{cm}$ .
Round appropriately (typically 2–3 s.f. depending on measurement precision): $0.267\ \text{cm}$ .

Key Takeaways

Keep units consistent; the lever ratio is dimensionless.
Calculated quantities should not be quoted to unrealistic precision.

Common Mistakes

Mixing mm and cm (e.g. $H_1-H_2$ in mm but $s,d$ in cm).
Writing too many decimal places (false precision).
Using $d/s$ instead of $s/d$ .

Things to Be Careful About

Because $H_1-H_2$ may be small, rounding it too aggressively can change $\Delta L$ noticeably; keep an extra digit during working and round only at the end.

Techniques used

substitute measured values into a given formulamaintain consistent units through a calculationround a calculated result to appropriate significant figures

(c)

(i)

• Remove the pipe from the measuring cylinder and empty the water into the beaker.
• Select the other pipe and label one end of this pipe B. Make a mark approximately $19\ \text{cm}$ from end B.
• Measure and record the value of $L$ for this pipe.

$L$ = ______

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(ii)

• Place this pipe in the measuring cylinder as shown in Fig. 2.2.
• Ensure that end B is at the bottom of the measuring cylinder.
• Repeat (b)(ii) and (b)(iv).

$H_1$ = ______
$H_2$ = ______
$T$ = ______ $^{\circ}\text{C}$
$\Delta L$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Example readings:

$H_1 = 18.0\ \text{cm}$

$H_2 = 16.9\ \text{cm}$

$T = 80.0\ ^{\circ}\text{C}$

\Delta L = \frac{s(H_1-H_2)}{d}=\frac{7.5(18.0-16.9)}{22.5}=\frac{7.5\times 1.1}{22.5}=0.366\overline{6}\ \text{cm}

Answer

$\Delta L = 0.367\ \text{cm}$

Final answer

Example: H1 = 18.0 cm, H2 = 16.9 cm, T = 80.0 °C, ΔL = 0.367 cm

Detailed explanation

Background Concept

Repeating the same measurement process with a different pipe length tests the proposed proportionality to $L$ . Consistency of method is crucial: you want any difference in $\Delta L$ to be due to physics (different $L$ and possibly different $T$ ) rather than a changed measuring technique.

Understanding the Question

For the second pipe, you must repeat part (b)(ii) (measure $H_1$ , $H_2$ , and $T$ ) and part (b)(iv) (calculate $\Delta L$ ). The same apparatus geometry ( $s$ and $d$ ) is used.

Approach

Set the second pipe in the cylinder with end B at the bottom.
Measure $H_1$ before heating and $H_2$ at the lowest point after heating.
Measure $T$ of the hot water.
Compute $\Delta L$ from the given formula.

Step-by-Step Reasoning

With the pipe at room temperature, record $H_1$ using the same ruler position/reference as before.
Add hot water to the mark and wait until Z reaches its lowest stable position; record $H_2$ .
Measure the water temperature $T$ .
Compute $H_1-H_2$ and substitute into

\Delta L = \frac{s(H_1-H_2)}{d}.

Round the final $\Delta L$ appropriately.

Key Takeaways

Repeatability depends on keeping the method and reference points consistent.
$\Delta L$ is a derived quantity; its quality depends on careful readings of $H_1$ and $H_2$ .

Common Mistakes

Forgetting to ensure end B is at the bottom (heated length becomes wrong).
Using a different reference point for height measurements between trials.
Not waiting for Z to reach its minimum before recording $H_2$ .

Things to Be Careful About

If $s$ and $d$ are unchanged, you should reuse the same measured values; do not re-measure sloppily and introduce extra variation unless instructed.
Hot water temperature may drift while you wait; measure $T$ promptly and, if possible, stir for uniform temperature.

Techniques used

repeat measurements under similar conditions for a second trialmeasure vertical heights consistently from the same referenceapply a given formula to compute a derived quantity

(d)

It is suggested that $\Delta L$ , $L$ , $T$ and $T_0$ are related by

\Delta L = kL(T - T_0)

where $k$ is a constant.

Using your data, calculate two values for $k$ .

first value of $k$ = ______
second value of $k$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From

\Delta L = kL(T-T_0)

k = \frac{\Delta L}{L(T-T_0)}

First pipe:

k_1 = \frac{0.267}{12.0(85.0-25.0)}=\frac{0.267}{720}=3.71\times 10^{-4}\ ^{\circ}\text{C}^{-1}

Second pipe:

k_2 = \frac{0.367}{19.0(80.0-25.0)}=\frac{0.367}{1045}=3.51\times 10^{-4}\ ^{\circ}\text{C}^{-1}

Answer

first value of $k = 3.71\times 10^{-4}\ ^{\circ}\text{C}^{-1}$

second value of $k = 3.51\times 10^{-4}\ ^{\circ}\text{C}^{-1}$

Final answer

k1 = 3.71×10^-4 °C^-1, k2 = 3.51×10^-4 °C^-1

Detailed explanation

Background Concept

The suggested model

\Delta L = kL(T-T_0)

states that thermal extension $\Delta L$ is proportional to the original heated length $L$ and the temperature rise $\Delta T = T-T_0$ . The constant $k$ plays the role of a coefficient of linear expansion (units of inverse temperature).

Understanding the Question

You have two sets of measurements (one for each pipe length). For each set, you know (or have calculated) $\Delta L$ , you measured $L$ , you measured the hot-water temperature $T$ , and you measured room temperature $T_0$ . You must calculate $k$ twice and report both values.

Approach

Rearrange the given equation to make $k$ the subject.
Use each data set separately.
Keep temperature differences in $^{\circ}\text{C}$ (same size as in K for differences) and keep $L$ and $\Delta L$ in the same length unit (cm cancels).

Step-by-Step Reasoning

Rearrange:

\Delta L = kL(T-T_0) \Rightarrow k=\frac{\Delta L}{L(T-T_0)}.

For trial 1, compute $\Delta T_1=85.0-25.0=60.0\ ^{\circ}\text{C}$ and substitute with $\Delta L_1$ and $L_1$ .
For trial 2, compute $\Delta T_2=80.0-25.0=55.0\ ^{\circ}\text{C}$ and substitute.
Quote $k$ in $^{\circ}\text{C}^{-1}$ (or $\text{K}^{-1}$ ; both acceptable for a temperature difference).

Key Takeaways

Temperature differences are what matter in thermal expansion.
Use consistent units so cancellations happen cleanly.

Common Mistakes

Using $T$ instead of $(T-T_0)$ .
Mixing mm and cm for $L$ and $\Delta L$ .
Reporting $k$ with no unit.

Things to Be Careful About

Even if you convert to kelvin, $T-T_0$ is unchanged numerically; do not add 273 twice or create an incorrect difference.
If your $\Delta L$ is small, rounding too early can change $k$ noticeably; keep extra digits until the final step.

Techniques used

rearrange a proportionality equation to solve for a constantsubstitute experimental values into the rearranged equationcompute and compare two values of a constant from separate trials

(e)

It is suggested that the percentage uncertainty in the values of $k$ is $30\%$ .
Using this uncertainty, explain whether your results support the relationship in (d).

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Mean value:

\bar{k}=\frac{3.71+3.51}{2}\times 10^{-4}=3.61\times 10^{-4}\ ^{\circ}\text{C}^{-1}

Percentage difference:

\frac{|3.71-3.51|\times 10^{-4}}{3.61\times 10^{-4}}\times 100 = 5.5\%

Answer

Since $5.5\% < 30\%$ , the two values of $k$ agree within the uncertainty, so the results support the relationship in (d).

Final answer

Yes — values of k agree within 30%, so results support the relationship.

Detailed explanation

Background Concept

Two experimental values are considered consistent if their difference is no larger than what you would expect from the stated uncertainty. A common method is to calculate the percentage difference between the two values and compare it with the percentage uncertainty.

Understanding the Question

You are told that the percentage uncertainty in $k$ is $30\%$ . Using that, you must decide whether your two calculated values of $k$ are consistent enough to support the model in (d).

Approach

Compute how far apart the two $k$ values are (difference).
Express this difference as a percentage (often relative to the mean).
If the percentage difference is less than (or comparable to) $30\%$ , conclude that the values agree within uncertainty.

Step-by-Step Reasoning

Find the mean $\bar{k}$ to use as a reference scale.
Find the absolute difference $|k_1-k_2|$ .
Convert to percentage difference:

\%\ \text{difference} = \frac{|k_1-k_2|}{\bar{k}}\times 100.

Compare with $30\%$ :

If smaller, the discrepancy is within the uncertainty, supporting the relationship.
If larger, the discrepancy is too big, and the data would not support the relationship.

Key Takeaways

“Support” here means “consistent within uncertainty,” not “exactly equal.”
Always link your final statement to the numerical comparison.

Common Mistakes

Saying “supports” with no quantitative comparison.
Comparing the difference to $30\%$ of the larger/smaller value without stating what you did.
Using $30\%$ uncertainty in $T$ or $\Delta L$ instead of in $k$ (the question specifies uncertainty in $k$ ).

Things to Be Careful About

The uncertainty is large ( $30\%$ ), so quite a big spread in $k$ could still be acceptable; don’t over-interpret small differences.
Make sure your percentage is computed correctly (difference divided by a reference value, then $\times 100$ ).

Techniques used

compare two experimental values using percentage differenceapply an uncertainty criterion to decide consistencystate a conclusion linked to the comparison

(f)

(i)

Describe four sources of uncertainty or limitations of the procedure for this experiment.
For any uncertainties in measurement that you describe, you should state the quantity being measured and the reason for the uncertainty.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Uncertainty in $H_1$ and $H_2$ : Z is not a sharp pointer and readings can suffer from parallax when using a ruler/metre rule.
Uncertainty in $H_2$ : difficult to judge the exact lowest position of Z because it moves while the pipe is heating and may oscillate slightly.
Uncertainty in $T$ : thermometer has limited resolution/response time, and the water temperature may not be uniform (temperature gradients if not stirred).
Uncertainty in the heated length: the water level may not be exactly at the mark (meniscus/overshoot), so the effective length at temperature $T$ is not exactly $L$ .

Final answer

Four limitations/uncertainties listed (see solution).

Detailed explanation

Background Concept

An uncertainty/limitation statement scores well when it (i) names the quantity affected and (ii) explains the physical reason the reading is hard to take or the method is not ideal. Good answers are specific (e.g. “parallax in reading $H$ ”) rather than vague (e.g. “human error”).

Understanding the Question

You must give four sources of uncertainty or limitations in this experiment. If you mention a measurement uncertainty, you must state the quantity being measured and why it is uncertain.

Approach

Think through the procedure and identify where:

readings are difficult (small changes, moving parts, unclear reference points),
conditions are not controlled (temperature not uniform, water level not exactly at mark),
the system may not behave ideally (friction at pivot, slipping of connections).
Then write four distinct, clearly explained points.

Step-by-Step Reasoning

Possible high-quality points include:

Heights $H_1$ and $H_2$ : Z may be a broad wooden edge with no fine pointer; judging its exact level against a ruler introduces parallax and alignment error.
Lowest height $H_2$ : while heating, the lever can move continuously; deciding the exact minimum is subjective, especially if Z overshoots or vibrates.
Temperature $T$ : hot water can cool during the measurement; if not stirred, the water near the pipe may be cooler/warmer than where the thermometer bulb sits; thermometer lag adds further uncertainty.
Water level at the mark (heated length): reading a meniscus in a cylinder and stopping exactly at the mark is difficult; if the level is not exactly at the mark, the length of pipe at temperature $T$ is not the intended $L$ .
Other creditable limitations (if needed) could be friction at the pivot, slipping of the pipe/attachments, or heating of the cylinder/pipe not reaching thermal equilibrium.

Key Takeaways

State the quantity and the reason.
Choose distinct limitations that affect the key calculated value $k$ .

Common Mistakes

Writing vague points (“heat loss”, “human error”) without linking to what is measured and how it affects results.
Repeating the same idea four times (e.g. four variants of “parallax”).
Listing improvements instead of limitations (that belongs in (f)(ii)).

Things to Be Careful About

Avoid claiming impossible precision (e.g. “exactly measure the lowest point”).
Make sure each of the four points is clearly different and refers to a different weakness in the method or measurement.

Techniques used

identify random and systematic uncertainties in a procedurelink each uncertainty to a specific measurement and a physical causestate limitations that affect validity or reliability

(ii)

Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Add a thin pointer (or attach a fiducial marker) at Z and use a set square to read $H_1$ and $H_2$ vertically to reduce parallax.
Use a video recording (or a travelling microscope/pointer scale) to identify the minimum position of Z more precisely.
Stir the hot water before measuring $T$ and use a digital temperature probe/data logger for faster response and better resolution.
Repeat measurements for each pipe several times and take mean values of $H_1-H_2$ (and hence $\Delta L$ and $k$ ) to reduce random uncertainty.

Final answer

Four improvements listed (see solution).

Detailed explanation

Background Concept

Improvements should be actionable changes that reduce uncertainty, control variables better, or increase reliability. The strongest improvements explicitly target the limitations identified in (f)(i).

Understanding the Question

You must propose four improvements to the experiment. You may use other apparatus or change procedures. Each improvement should plausibly make the measured $k$ more reliable/accurate.

Approach

For each limitation, ask: “What change would make that measurement easier or more precise?” Then propose a practical solution (pointer, better instruments, repetition, better temperature control, etc.).

Step-by-Step Reasoning

Improve height readings: a sharp pointer at Z gives a clear reference; a set square ensures the measurement is vertical and reduces parallax.
Improve detection of minimum $H_2$ : video allows frame-by-frame identification of the lowest point; alternatively, a scale and pointer arrangement can make the minimum clearer.
Improve temperature measurement: stirring reduces gradients; a digital probe/data logger reduces reaction time and can track cooling, letting you record a representative temperature.
Improve reliability: repeating each run and averaging reduces random scatter and makes the comparison of two $k$ values more meaningful.
Other valid improvements could include insulating the cylinder to reduce cooling, using a lid to reduce heat loss, or increasing $L$ (larger expansion gives a larger $H_1-H_2$ and smaller percentage uncertainty).

Key Takeaways

Improvements must be specific and practical.
The best improvements clearly reduce a named uncertainty.

Common Mistakes

Giving vague suggestions (“use better equipment”) without stating what and how it helps.
Repeating the same improvement in different words.
Suggesting changes that alter the physics being tested without control (e.g. changing materials) rather than improving measurement quality.

Things to Be Careful About

Ensure each improvement is distinct.
Make sure the improvement is feasible with typical lab equipment and genuinely addresses the measurement challenges in this set-up.

Techniques used

propose procedural changes that reduce uncertaintymatch an improvement to a specific limitationincrease reliability by repetition and averaging