A student investigates how the temperature of the surroundings affects the rate of cooling of water. (i) The student: - pours $100\ \text{cm}^3$ of hot water into a $250\ \text{cm}^3$ beaker - uses a thermometer to take the temperature of the water at time $t = 0\ \text{s}$. The thermometer reading at time $t = 0\ \text{s}$ is shown in Fig. 2.1. Record the temperature of the water at time $t = 0\ \text{s}$ in Table 2.1. Table 2.1 $t / \text{s}$ $\theta / ^\circ\text{C}$ ---…

(a)(i) 72 · (a)(ii) to ensure an even temperature throughout the water · (a)(iii) 0.067 °C/s · (a)(iv) 0.033 °C/s · (a)(v) C1 is greater than C2 because the temperature difference between the water and the surroundings is larger at the start. · (b)(i) 0.20 °C/s · (b)(ii) C3 is greater than C1 because the surroundings are colder, resulting in a larger temperature difference and faster heat loss. · (b)(iii) The initial temperatures are different (72 °C and 75 °C), so the…

Physics 5054/41 — May/June 2024

Cambridge O-Level · Alternative to Practical · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Experimental Contexts · Analysis, Conclusions and Evaluation · Observations and Measurements · Planning Experiments and Investigations

Q110MMediumExperimental ContextsAnalysis, Conclusions and Evaluation

A student finds the volume of a small glass ball (marble) by two different methods.

(a)

method 1

The student:

places six small glass balls by the side of a metre rule as shown in Fig. 1.1
makes sure that there are no gaps between the balls.

Fig. 1.1

(i)

Take readings on the metre rule of the positions of points A and B shown in Fig. 1.1.

Give your readings to the nearest $0.1\ \text{cm}$ .

position of point A = ______ $\text{cm}$
position of point B = ______ $\text{cm}$

1M Similar

(ii)

The length $l$ on Fig. 1.1 is the distance between points A and B.
The average diameter $d$ of one ball can be found using the equation:

l = 6d

Use your answers to (a)(i) to calculate the length $l$ and diameter $d$ . Give your answers to the nearest $0.1\ \text{cm}$ .

$l$ = ______ $\text{cm}$
$d$ = ______ $\text{cm}$

2M Similar

(iii)

The average volume $V$ of one glass ball using this method, is given by the equation:

V = \frac{3.14d^3}{6}

Calculate $V$ .

$V$ = ______ $\text{cm}^3$

1M Similar

(b)

method 2

The student pours water into a measuring cylinder.

(i)

The volume of water in the measuring cylinder $V_1$ is shown in Fig. 1.2.

Fig. 1.2

Write down the reading $V_1$ .

$V_1$ = ______ $\text{cm}^3$

1M Similar

(ii)

The six glass balls are carefully added to the water in the measuring cylinder.

The new reading on the measuring cylinder $V_2$ is shown in Fig. 1.3.

Fig. 1.3

The volume $V_T$ of the six balls is given by

V_T = V_2 - V_1

Calculate $V_T$ . Show your working.

$V_T$ = ______ $\text{cm}^3$

1M Similar

(iii)

Calculate the average volume $V$ of one ball using this method. Give your answer to the nearest $0.1\ \text{cm}^3$ .

$V$ = ______ $\text{cm}^3$

1M Similar

(c)

Suggest whether method 1 or method 2 gives the more accurate value for the volume of the ball.

Explain your answer.

method giving more accurate value ______
explanation ______

1M Similar

DifficultyMedium

Worked solution

Answer

method giving more accurate value: method 1
explanation: The measuring cylinder can only measure to the nearest $1.0\text{ cm}^3$ (or $0.5\text{ cm}^3$ ), whereas measuring the diameter with a metre rule to the nearest $0.1\text{ cm}$ is more accurate.

Final answer

Method 1 is more accurate because the measuring cylinder has a lower precision (larger division size) than the metre rule used to measure the diameter.

Detailed explanation

Walkthrough

Method 1 uses a metre rule to measure the total length of six balls. The rule is graduated to $1\text{ mm}$ ( $0.1\text{ cm}$ ), allowing a precise measurement of the diameter. Method 2 uses a measuring cylinder. The scale in Fig. 1.2 and 1.3 shows divisions of $1\text{ cm}^3$ . The volume of water is read to the nearest $1\text{ cm}^3$ (or perhaps $0.5\text{ cm}^3$ with estimation). A difference of $12\text{ cm}^3$ from a reading of $27\text{ cm}^3$ to $39\text{ cm}^3$ means the uncertainty is relatively large (around $\pm 1\text{ cm}^3$ ). Measuring the diameter to $0.1\text{ cm}$ gives a much smaller relative uncertainty. Therefore, method 1 is more accurate.

Key Takeaways

Accuracy depends on the precision of the apparatus. A smaller division size on a measuring instrument generally leads to more accurate results.

Common Mistakes

Saying method 2 is more accurate because it measures volume directly. While it measures volume directly, the precision of the measuring cylinder is poor, making the result less accurate. Saying "method 1 is better because it is faster" is not a valid physics reason.

Things to Be Careful About

The mark scheme specifically mentions the limitation of the measuring cylinder ( $0.5\text{ cm}^3$ or $1.0\text{ cm}^3$ precision) or the accuracy of measuring the diameter to the nearest mm. Ensure your explanation links the apparatus limitation to the accuracy of the final volume.

Techniques used

compare the precision of the two measurement methods

(d)

The student now uses the six glass balls to find the average mass of one glass ball using a small beaker and a top pan (electronic) balance.

Describe the method the student uses.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Place the empty beaker on the top pan balance and tare it (set to zero).
Add the six glass balls to the beaker.
Read the total mass of the six balls from the balance.
Divide the total mass by 6 to find the average mass of one ball.

(Alternatively: Record the mass of the empty beaker, add the six balls, record the new total mass, subtract the empty beaker mass, and divide by 6.)

Final answer

Place the beaker on the balance and tare it (or record its mass). Add the six balls, read the total mass, and divide by 6.

Detailed explanation

Walkthrough

To find the average mass of one ball, you need the total mass of all six balls. Since the balls are small, their individual masses might be too small to measure accurately on a top pan balance, or the balance might not have enough precision. By measuring six together, you get a larger, more readable mass. You must account for the mass of the beaker. The best method is to place the beaker on the balance and use the 'tare' function to reset the display to zero. Then add the six balls and read the mass directly. Finally, divide by 6. If the balance does not have a tare function, you record the mass of the empty beaker, add the balls, record the new mass, subtract the beaker's mass, and divide by 6.

Key Takeaways

When measuring the mass of small objects, measure multiple at once to improve precision and readability. Always account for the container's mass by taring or subtracting.

Common Mistakes

Forgetting to mention taring or subtracting the beaker's mass. Measuring the mass of just one ball (which might be below the balance's precision). Not dividing by 6 at the end.

Things to Be Careful About

The mark scheme awards one mark for the beaker on balance and tare/record, and one mark for adding 6 balls, finding total mass, and dividing by 6. Ensure both steps are clearly described in your method.

Techniques used

describe a method to find the average mass of multiple identical objects using a balance

Q210MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationPlanning Experiments and Investigations

A student investigates how the temperature of the surroundings affects the rate of cooling of water.

(a)

(i)

The student:

pours $100\ \text{cm}^3$ of hot water into a $250\ \text{cm}^3$ beaker
uses a thermometer to take the temperature of the water at time $t = 0\ \text{s}$ .

The thermometer reading at time $t = 0\ \text{s}$ is shown in Fig. 2.1.

Fig. 2.1

Record the temperature of the water at time $t = 0\ \text{s}$ in Table 2.1.

Table 2.1

$t / \text{s}$	$\theta / ^\circ\text{C}$
0	______
30	69
60	67
90	66
120	65
150	64
180	63
210	62
240	61

1M Similar

(ii)

The student then records the temperature $\theta$ of the water every $30\ \text{s}$ for $240\ \text{s}$ . The results are recorded in Table 2.1.

Before taking each temperature reading, the student carefully stirs the water in the beaker. Explain why.

1M Similar

(iii)

Calculate the average cooling rate $C_1$ of the water for the first $90\ \text{s}$ of the experiment. Use the readings in Table 2.1 and the equation:

C_1 = \frac{\theta_0 - \theta_{90}}{t}

where $\theta_0$ is the temperature of the water at $0\ \text{s}$ , $\theta_{90}$ is the temperature at $90\ \text{s}$ and $t$ is the time of $90\ \text{s}$ .

Give the unit for $C_1$ .

$C_1$ = ______ unit ______

2M Similar

(iv)

Calculate the average cooling rate $C_2$ of the water for the final $90\ \text{s}$ of the experiment. Use the equation:

C_2 = \frac{\theta_{150} - \theta_{240}}{t}

where $\theta_{150}$ is the temperature of the water at $150\ \text{s}$ , $\theta_{240}$ is the temperature of the water at $240\ \text{s}$ and $t$ is the time of $90\ \text{s}$ .

$C_2$ = ______ unit ______

1M Similar

(v)

Compare your values of $C_1$ and $C_2$ . Explain any difference in these values.

1M Similar

(b)

The student repeats the procedure described in (a)(i) but this time he places the $250\ \text{cm}^3$ beaker inside a larger beaker containing iced water. The arrangement is shown in Fig. 2.2.

Fig. 2.2

The student reads the temperature $\theta$ of the hot water, records the reading and immediately starts the stop-watch.

Table 2.2. shows the temperature $\theta$ at times $t = 0\ \text{s}$ , $30\ \text{s}$ , $60\ \text{s}$ , and $90\ \text{s}$ .

Table 2.2

$t / \text{s}$	$\theta / ^\circ\text{C}$
0	75
30	68
60	62
90	57

(i)

Calculate the average cooling rate of the hot water for the $90\ \text{s}$ . Use the readings in Table 2.2 and the equation:

C_3 = \frac{\theta_0 - \theta_{90}}{t}

$C_3$ = ______ unit ______

1M Similar

(ii)

Describe how $C_3$ differs from $C_1$ . Explain your answer.

1M Similar

(iii)

The recorded readings show that this experiment is not a valid comparison of $C_1$ and $C_3$ .

By referring to the results recorded in Table 2.1 and Table 2.2, explain why this is not a valid comparison.

1M Similar

(iv)

State one other variable that should be kept constant to make a valid comparison.

1M Similar

Q314MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and Evaluation

A student measures the focal length of a lens.

Fig. 3.1 shows the apparatus she uses and the position of the lens when a clearly focused image is formed on the screen.

Fig. 3.1

The student:

places the screen a distance $D = 60.0\ \text{cm}$ from the illuminated object
places the lens between the object and the screen so that the lens is very close to the illuminated object
moves the lens slowly away from the illuminated object until a clearly focused image is formed on the screen.

(a)

(i)

Measure and record distance $x$ on Fig. 3.1.

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

1M Similar

(ii)

Fig. 3.1 is drawn to a scale of one fifth full size.

Use your answer to (a)(i) to calculate the actual object distance $u$ from the lens.

$u$ = ______ $\text{cm}$

1M Similar

(iii)

Deduce the image distance $v$ , the distance from the lens to the screen when a clear image is seen.

$v$ = ______ $\text{cm}$

1M Similar

(b)

Fig. 3.2 shows the shape of the illuminated object and Fig. 3.3 shows the image seen on the screen.

Fig. 3.2

Fig. 3.3

Describe two differences between the illuminated object and its image on the screen.

______
______

2M Similar

(c)

(i)

The student moves the screen away from the illuminated object and repeats the procedure for values of $D = 70.0\ \text{cm}$ , $80.0\ \text{cm}$ , $90.0\ \text{cm}$ and $100.0\ \text{cm}$ .

Table 3.1 shows the values recorded.

Add your values for $u$ and $v$ from (a)(ii) and (a)(iii) to Table 3.1 on page 12.

Complete Table 3.1 by calculating the value of $(u \times v)$ for each value of $D$ .

Give your answers to 3 significant figures.

Table 3.1

$D / \text{cm}$	$u / \text{cm}$	$v / \text{cm}$
60.0
70.0	21.0	49.0
80.0	19.5	60.5
90.0	18.6	71.4
100.0	17.8	82.2

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $D = 60.0\text{ cm}$ :

u = 24.0\text{ cm}, \quad v = 36.0\text{ cm}, \quad u \times v = 24.0 \times 36.0 = 864 \rightarrow 864\text{ cm}^2 \text{ (or } 860\text{ cm}^2 \text{ if using } u=24, v=36\text{)}

Calculating $(u \times v)$ for all rows to 3 significant figures:

For $D = 60.0\text{ cm}$ : $24.0 \times 36.0 = 864\text{ cm}^2$ (or $860\text{ cm}^2$ )
For $D = 70.0\text{ cm}$ : $21.0 \times 49.0 = 1029 \rightarrow 1030\text{ cm}^2$
For $D = 80.0\text{ cm}$ : $19.5 \times 60.5 = 1179.75 \rightarrow 1180\text{ cm}^2$
For $D = 90.0\text{ cm}$ : $18.6 \times 71.4 = 1328.04 \rightarrow 1330\text{ cm}^2$
For $D = 100.0\text{ cm}$ : $17.8 \times 82.2 = 1463.16 \rightarrow 1460\text{ cm}^2$

Answer

$D / \text{cm}$	$u / \text{cm}$	$v / \text{cm}$	$(u \times v) / \text{cm}^2$
60.0	24.0	36.0	864
70.0	21.0	49.0	1030
80.0	19.5	60.5	1180
90.0	18.6	71.4	1330
100.0	17.8	82.2	1460

Final answer

Values of (u × v): 864 (or 860), 1030, 1180, 1330, 1460

Detailed explanation

Walkthrough

Insert the calculated values of $u = 24.0\text{ cm}$ and $v = 36.0\text{ cm}$ into the first row corresponding to $D = 60.0\text{ cm}$ .
For each row in Table 3.1, calculate the product $u \times v$ $u \times v$ :
- $D = 60.0\text{ cm}$ : $24.0 \times 36.0 = 864\text{ cm}^2$
- $D = 70.0\text{ cm}$ : $21.0 \times 49.0 = 1029\text{ cm}^2 \rightarrow 1030\text{ cm}^2$
- $D = 80.0\text{ cm}$ : $19.5 \times 60.5 = 1179.75\text{ cm}^2 \rightarrow 1180\text{ cm}^2$
- $D = 90.0\text{ cm}$ : $18.6 \times 71.4 = 1328.04\text{ cm}^2 \rightarrow 1330\text{ cm}^2$
- $D = 100.0\text{ cm}$ : $17.8 \times 82.2 = 1463.16\text{ cm}^2 \rightarrow 1460\text{ cm}^2$
Round every calculated value to exactly 3 significant figures as explicitly instructed in the question.

Key Takeaways

Always follow rounding instructions in table completion questions precisely (here, 3 significant figures).

Common Mistakes

Writing raw unrounded numbers (e.g. $1029$ , $1179.75$ ).

Things to Be Careful About

Trailing zeros are required to show 3 significant figures (e.g. $1030$ , $1180$ , $1330$ , $1460$ ).

Techniques used

calculate products of measured quantitiesexpress values to three significant figures

(ii)

Use the grid provided in Fig. 3.4 on page 13 to plot a graph of $(u \times v) / \text{cm}^2$ on the y-axis against $D / \text{cm}$ on the x-axis.

You do not need to start your axes at the origin (0,0).

Draw the straight line of best fit.

Fig. 3.4

4M Similar

DifficultyMedium

Worked solution

Answer

Plot the graph using the following criteria:

Axes: Label the horizontal x-axis as $D / \text{cm}$ and the vertical y-axis as $(u \times v) / \text{cm}^2$ .
Scales: Linear, sensible scales using more than half the grid in both directions. (e.g. x-axis starting at $50$ or $60\text{ cm}$ , y-axis starting at $800\text{ cm}^2$ ).
Plotting: Plot the five points precisely to within half a small square:
- $(60.0, 864)$
- $(70.0, 1030)$
- $(80.0, 1180)$
- $(90.0, 1330)$
- $(100.0, 1460)$
Line of best fit: Draw a single, clean, thin straight line that passes evenly through the plotted points.

Final answer

Graph plotted with correctly labelled axes, sensible non-origin scales, accurately plotted points, and a thin straight line of best fit.

Detailed explanation

Walkthrough

To construct the graph:

Axes and Labels: Clearly write $(u \times v) / \text{cm}^2$ on the y-axis and $D / \text{cm}$ on the x-axis, including the full quantity and unit.
Choosing Scales:
- $D$ ranges from $60.0$ to $100.0\text{ cm}$ . A good scale starts at $50$ or $60\text{ cm}$ with $10\text{ cm}$ represented by $2\text{ cm}$ or $4\text{ cm}$ on the grid.
- $(u \times v)$ ranges from $864$ to $1460\text{ cm}^2$ . A good scale starts at $800\text{ cm}^2$ with $100\text{ cm}^2$ represented by $2\text{ cm}$ or $4\text{ cm}$ of grid.
- Do not use awkward scale multiples (such as units of 3 or 7).
Plotting Points: Mark each data pair with a small neat cross ( $+$ or $\times$ ) to within half a small square of accuracy.
Line of Best Fit: Using a long ruler, draw a single straight line that balances the points on either side without forcing it through any single outlier or origin.

Key Takeaways

The plotted area should occupy more than half the grid in both directions.
Always use small, precise crosses for data points.

Common Mistakes

Forcing the line to start from $(0,0)$ when non-zero origins were specified.
Drawing thick, 'feathered', or double lines.

Things to Be Careful About

Ensure units are included in the axis labels.

Techniques used

choose suitable linear scales occupying more than half the gridplot experimental data points accuratelydraw a thin straight line of best fit

(iii)

The focal length $f$ of the lens is numerically equal to the gradient of the line.

Calculate the gradient of the line. Show all working and indicate on your graph in Fig. 3.4 the values you use.

$f$ = ______ $\text{cm}$

2M Similar

(d)

The lens manufacturer states that the focal length of the lens is $15.0\ \text{cm} \pm 10\%$ .

Decide, with a calculation, whether your value of $f$ agrees with this statement and tick the box that shows your answer.

calculation:

[ ] My value for $f$ agrees with the manufacturer's statement.
[ ] My value for $f$ does not agree with the manufacturer's statement.

1M Similar

Q46MMedium-HardPlanning Experiments and InvestigationsExperimental ContextsAnalysis, Conclusions and Evaluation

Plan an experiment to investigate how the thickness of a metal wire affects its resistance.

The resistance of a wire can be found using the equation:

\text{resistance of wire} = \frac{\text{potential difference (p.d.) across wire}}{\text{current in the wire}}

The following apparatus is available:

six lengths of metal wire, each of different thickness
an ammeter
a voltmeter
a power supply
several connecting leads
a micrometer.

Other apparatus normally available in a school laboratory can also be used.

In your plan, you should:

draw a circuit diagram to show how you will use the apparatus
explain briefly how to carry out the investigation
state the key variables to keep constant
draw a table, with column headings, to show how to display readings (you are not required to enter any readings in the table)
explain how to use these readings to reach a conclusion.

DifficultyMedium-Hard

Worked solution

Circuit diagram

Method

Use the micrometer to measure the diameter (thickness) of each wire.
Connect the circuit as shown, placing one test wire in the series circuit at a time.
Record the ammeter reading (current $I$ ) and the voltmeter reading (p.d. $V$ ) for that wire.
Calculate the resistance of the wire using

R = \frac{V}{I}

Repeat for each wire of different thickness.

Variables to keep constant

the length of each wire
the material of each wire

Results table

Wire	Thickness (diameter) / mm	p.d. / V	current / A	Resistance / Ω

Conclusion

Compare the calculated resistances for the different wire thicknesses. If the resistance values are different, then the thickness of the wire affects its resistance. If they are all the same, then thickness does not affect the resistance in this experiment.

Final answer

Circuit diagram; method using micrometer and measuring V and I; length and material kept constant; table with thickness/diameter and resistance with units; conclusion comparing resistance values for different thicknesses.

Detailed explanation

Walkthrough

This is a planning question, so you need to give a complete experimental plan. Each bullet in the question is a different marking point.

First, draw the circuit. The ammeter must be in series with the wire, because it measures the same current that passes through the wire. The voltmeter must be in parallel with the wire only, because it measures the p.d. across just the wire.

Second, describe the method. The thickness of the wire is too small to measure with a ruler, so use a micrometer. For each wire, connect it in the circuit, record the current and p.d., then calculate the resistance using the given equation $R = V/I$ .

Third, state the variables to keep constant. To make a fair test, only the thickness should change. Therefore the length and the material of every wire must be the same.

Fourth, draw a results table. The table must have a column for the independent variable (thickness/diameter) and a column for the dependent result (resistance). Every column heading must include its unit.

Finally, write the conclusion. The conclusion must be based on the readings: compare the resistance values. If they are different for different thicknesses, then thickness affects resistance; if they are the same, it does not.

Key Takeaways

A plan question is marked by separate components: apparatus, circuit, method, controlled variables, table and conclusion.
The ammeter is placed in series; the voltmeter is placed in parallel with the component being measured.
A micrometer is used for small lengths such as wire thickness.
A fair test requires only one variable to change; all other relevant variables must be kept constant.
A results table should have column headings that include both the quantity and its unit.
The conclusion must be justified by the readings, not just stated as a known fact.

Common Mistakes

Putting the voltmeter in series or the ammeter in parallel.
Measuring wire thickness with a ruler, which is not accurate enough.
Failing to control the length or material of the wire.
Writing table headings without units, such as just "resistance" instead of "resistance / Ω".
Giving a conclusion such as "thinner wire has more resistance" without explaining that it comes from comparing the readings.

Things to Be Careful About

Use the micrometer reading to the precision it allows, often to 0.01 mm.
The voltmeter must be connected across the test wire only, not across the whole circuit.
State clearly that length and material are kept constant so that only the thickness is changing.
In the table, units belong in the column headings, not repeated in every cell.
If you choose to plot a graph, plot resistance against thickness/diameter so the relationship can be seen clearly.

Techniques used

draw a circuit with the ammeter in series and the voltmeter in parallel with the wiremeasure the wire diameter with a micrometerkeep the length and material of the wire constantcalculate resistance using R = V/Icompare the resistance values for different wire thicknesses