Physics 5054/32 — October/November 2025
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Experimental Contexts · Observations and Measurements · Analysis, Conclusions and Evaluation · Use of Techniques, Apparatus and Materials · Planning Experiments and Investigations
In this experiment you will investigate a light dependent resistor (LDR).
You are provided with:
- a power supply
- a switch
- a voltmeter with two leads that may be connected between different points in the circuit
- a light dependent resistor (LDR)
- a 560 resistor
- a piece of card.
The supervisor has constructed a series circuit consisting of the power supply, the LDR, the resistor and the switch. The circuit has three points labelled P, Q and S.
Draw a diagram of the circuit arrangement using the correct symbols for the components in the circuit.
Choose from the symbols shown in Fig. 1.1.
You do not need to label points P, Q and S on your diagram.
Answer
A single-loop series circuit containing:
- the d.c. power supply ( terminals)
- an open (or closed) switch
- an LDR (resistor rectangle with two incoming arrows pointing towards it)
- a fixed resistor (plain rectangle)
Series circuit with d.c. power supply, switch, LDR, and fixed resistor
Walkthrough
To draw the correct circuit diagram:
- Identify the components required from the stem:
- Power supply: two terminals labelled and
- Switch: standard switch symbol
- LDR (Light-Dependent Resistor): rectangular box with two arrows pointing inwards/towards it
- Fixed resistor: plain rectangular box ()
- Connect all four components in a single closed loop (series circuit) with straight connecting lines representing wires.
- Ensure no parallel branches or extra components (such as a permanently connected voltmeter) are included in this basic circuit diagram.
Key Takeaways
- In a series circuit, all components are connected one after another in a single loop.
- An LDR symbol is a fixed resistor with arrows pointing inward (light shining on it).
Common Mistakes
- Confusing the LDR symbol with an LED (arrows pointing outwards) or a thermistor (diagonal line with a foot).
- Drawing the voltmeter in series into the main loop.
Things to Be Careful About
- Ensure neat component symbols and clear connecting lines without unintended gaps or short circuits.
Connect the voltmeter across the LDR between points P and Q.
Close the switch.
Record , the voltmeter reading across P and Q.
This is under normal lighting conditions.
= ______
Open the switch.
Answer
(typical value recorded to at least , )
1.8 V
Walkthrough
- The voltmeter is connected in parallel across terminals P and Q (across the LDR).
- Close the switch to complete the circuit and read the potential difference under ambient room light.
- The reading should be recorded to at least precision and must be within the supply voltage limit (typically for a standard lab cell pack).
- Open the switch immediately after taking the reading to prevent unnecessary draining of the power supply.
Key Takeaways
- Voltmeters are always connected in parallel across the component being measured.
- Readings on digital or analogue meters must be recorded with their appropriate decimal precision.
Common Mistakes
- Omitting the decimal place (e.g. writing instead of ).
- Leaving the switch closed between measurements.
Things to Be Careful About
- Ensure the voltmeter polarity matches the circuit polarity if using an analogue or polarized digital meter.
Disconnect the voltmeter from points P and Q.
Reconnect the voltmeter across the 560 resistor between points Q and S.
Close the switch.
Record , the voltmeter reading across Q and S.
= ______
Open the switch.
Answer
(typical value , recorded to at least )
1.2 V
Walkthrough
- Disconnect the voltmeter leads from across the LDR (P and Q) and connect them across the fixed resistor (between points Q and S).
- Close the switch and record the potential difference .
- Under normal room lighting, the resistance of the LDR is typically greater than or comparable to , so .
- Open the switch after recording the value.
Key Takeaways
- In a series circuit, the total voltage of the supply is shared between the components in proportion to their resistances.
Common Mistakes
- Swapping the readings or taking the reading with an open switch (which would give ).
Things to Be Careful About
- Ensure the voltmeter leads make good electrical contact at points Q and S.
The current in the circuit is calculated using the equation:
where .
Use your reading in (b)(ii) to calculate the current .
= ______
Working
Answer
(or )
0.0021 A
Walkthrough
- Use the provided formula:
- Substitute the measured value of from part (b)(ii) (e.g., ) and the known fixed resistance :
- Express the final value to a sensible number of significant figures (2 or 3 s.f.).
Key Takeaways
- Current in a single series loop is the same through all components.
- Ohm's law relates current, potential difference, and resistance: .
Common Mistakes
- Using instead of in the formula.
- Arithmetic rounding errors when handling small decimal numbers.
Things to Be Careful About
- Keep the current in amperes (A) as specified by the response line, or ensure standard scientific notation is clear.
Calculate the resistance of the LDR under normal lighting conditions using the equation shown.
= ______
Working
Answer
840 Ω
Walkthrough
- Use the formula for the resistance of the LDR:
- Substitute the value of obtained in (b)(i) and the current calculated in (c):
- Check that the calculated resistance has a sensible physical magnitude (typically hundreds to thousands of ohms for an LDR under normal room light).
Key Takeaways
- The resistance of any component can be found from the p.d. across it divided by the current flowing through it: .
Common Mistakes
- Inverting the fraction (calculating instead of ).
- Using instead of .
Things to Be Careful About
- Carry forward unrounded intermediate values of current to prevent rounding errors in .
Disconnect the voltmeter from points Q and S.
Reconnect the voltmeter across the LDR between points P and Q.
Place the piece of card on top of the LDR.
Close the switch.
Record a new value of for the LDR in the dark.
= ______
Open the switch.
Answer
(typical value recorded such that and )
2.7 V
Walkthrough
- Move the voltmeter leads back to points P and Q (across the LDR).
- Cover the LDR with the opaque card to simulate dark conditions.
- Close the switch and record the new reading for .
- In the dark, the LDR's resistance rises significantly, so it takes a larger fraction of the total supply voltage. The reading must be greater than or equal to in (b)(i), up to a maximum of the supply voltage ().
- Open the switch.
Key Takeaways
- Reducing light intensity on an LDR increases its resistance, causing a larger share of the series circuit voltage to drop across it.
Common Mistakes
- Forgetting to place the card directly over the LDR.
- Leaving gaps where light can still enter, resulting in minimal change in reading.
Things to Be Careful About
- Ensure the card completely covers the light-sensitive surface of the LDR.
Compare your reading for with the LDR under normal lighting conditions in (b)(i) with with the LDR covered by card in (e).
Suggest what causes the change in the voltmeter readings as the intensity of the light reaching the LDR decreases.
Answer
The resistance of the LDR increases as the light intensity reaching it decreases.
Resistance of the LDR increases as light intensity decreases
Walkthrough
- Compare the two readings: in the dark (part (e)) is larger than in the light (part (b)(i)).
- In a series potential divider circuit, the potential difference across a component is directly proportional to its resistance ().
- Since increases when the card covers the LDR, the resistance of the LDR must have increased due to the lower light intensity.
Key Takeaways
- LDR property: higher light intensity lower resistance; lower light intensity (darkness) higher resistance (LURD: Light Up, Resistance Down).
Common Mistakes
- Stating that current increases (current actually decreases because total circuit resistance goes up).
- Stating that the power supply voltage changed.
Things to Be Careful About
- Clearly state that the resistance of the LDR increases, not just that 'the voltage increased'.
Close the switch.
Hold the card horizontally about 50 cm above the LDR.
Slowly move the card towards the LDR until it rests on top of the LDR.
Observe the reading on the voltmeter as you move the card.
Open the switch.
Describe the changes you see to the voltmeter reading as the card is moved downwards.
Answer
- The voltmeter reading is initially (more or less) constant / does not change much as the distance decreases from .
- The reading then increases as the card gets close to the LDR.
Initially remains constant, then increases as the card gets close to the LDR
Walkthrough
- At a large distance (around ), ambient light from all sides still reaches the LDR easily, so the card casts very little shadow and the reading remains essentially constant.
- As the card comes very close to the LDR, it blocks ambient light and casts a dense shadow, significantly decreasing the light intensity reaching the LDR surface.
- As a result, the resistance of the LDR rises sharply, causing the voltmeter reading across the LDR to increase noticeably until the card rests on top.
- Describing both phases (constant at large distance, then increasing close up) earns full marks.
Key Takeaways
- Ambient light is multi-directional; an obstacle far away has minimal shading effect, but close up it blocks significant light.
- The voltmeter responds directly to the local illuminance at the sensor surface.
Common Mistakes
- Claiming the voltage increases uniformly throughout the entire distance.
- Stating that the voltage decreases instead of increases.
Things to Be Careful About
- Mention both parts of the observation: the initial plateau / small change and the subsequent rise when close.
In this experiment, you will investigate the absorption of thermal radiation by different coloured surfaces.
You are provided with:
- a lamp connected to a power supply
- a thermometer with a piece of white card attached to its bulb
- a thermometer with a piece of black card attached to its bulb
- a clamp, boss and stand
- a stopwatch
- a 30 cm ruler.
The supervisor has arranged the thermometer which has a piece of white card attached to its bulb so that the bulb is level with the filament of a lamp.
The lamp is switched off.
Fig. 2.1 shows the apparatus.
Adjust the distance of the white card attached to the thermometer bulb from the lamp until it is approximately 1 cm.
Answer
Record the initial room temperature, e.g. 20.0 °C.
20.0 °C (example reading)
Walkthrough
The candidate must record the starting temperature of the white card before the lamp is switched on. This is the room temperature. The thermometer should be read to the precision of its smallest scale division, which is typically 0.1 °C or 0.2 °C for a standard laboratory thermometer. A sensible room temperature is between 18 °C and 22 °C.
Key Takeaways
Always record an initial reading before starting an experiment. Read instruments to their specified precision, including a trailing zero if the scale allows it (e.g., 20.0, not 20).
Common Mistakes
- Recording the reading to the nearest whole degree (e.g., 20 °C) when the scale allows 0.1 °C.
- Recording a temperature that is unreasonably high or low (e.g., 35 °C for room temperature).
Things to Be Careful About
Ensure the thermometer has had time to stabilise at room temperature before recording the initial reading. Include the unit °C.
Switch on the lamp, and at the same time, start the stopwatch.
Record, in Table 2.1, the reading on the thermometer every 60 s for 300 s.
Switch off the lamp.
Table 2.1
| time | white card temperature | black card temperature |
|---|---|---|
| 0 | ||
| 60 | ||
| 120 | ||
| 180 | ||
| 240 | ||
| 300 |
Answer
Record a full set of 5 temperature readings at 60, 120, 180, 240, and 300 s. The temperature must show a steady increase. Example readings: 20.5, 21.0, 21.5, 22.0, 22.5 °C.
Full set of 5 readings showing a steady increase, e.g. 20.5, 21.0, 21.5, 22.0, 22.5 °C at 60, 120, 180, 240, 300 s
Walkthrough
The candidate switches on the lamp and starts the stopwatch simultaneously. They must read the thermometer every 60 s for a total of 300 s, giving 5 additional readings. Because the lamp emits thermal radiation, the white card absorbs it and its temperature rises. The readings must therefore be strictly increasing. The table must be filled with values to the appropriate precision (e.g., 1 decimal place).
Key Takeaways
When recording data over time, ensure the values reflect the expected physical process. Here, thermal radiation causes heating, so temperature must increase with time.
Common Mistakes
- Recording temperatures that decrease or stay the same (this would be physically incorrect for a heating experiment).
- Using inconsistent decimal places in the table.
- Forgetting to record the 300 s reading.
Things to Be Careful About
Read the thermometer to the nearest 0.1 °C. Ensure the stopwatch is started at the exact same time the lamp is switched on. The readings should be consistent with a gradual heating curve.
Carefully remove the thermometer from the clamp and place it on the bench.
Place the thermometer which has a piece of black card attached to its bulb in the clamp.
Make sure that the black card is facing the lamp and that the bulb of the thermometer is level with the filament of the lamp.
Adjust the distance between the lamp and the black card so that it is approximately 1 cm.
Repeat the procedure in (a)(i) and (a)(ii) and record, in Table 2.1, the temperatures for the black card.
Answer
Record a full set of 5 temperature readings for the black card at 60, 120, 180, 240, and 300 s. The final temperature must be greater than or equal to the white card's final temperature, and the total temperature increase must be greater. Example readings: 20.8, 21.5, 22.2, 22.8, 23.0 °C.
Full set of 5 readings showing a greater temperature increase than the white card, e.g. 20.8, 21.5, 22.2, 22.8, 23.0 °C
Walkthrough
The experiment is repeated with the black card. Black surfaces are better absorbers of thermal radiation than white surfaces, so the black card will heat up faster. The candidate must record 5 readings at the same time intervals (60, 120, 180, 240, 300 s). The temperature increase for the black card (Δθ_B) must be greater than the temperature increase for the white card (Δθ_W). The final temperature θ_B at 300 s should be greater than or equal to θ_W at 300 s.
Key Takeaways
When comparing two materials or surfaces in an experiment, ensure the data reflects the known physical properties. Black surfaces absorb more thermal radiation than white surfaces.
Common Mistakes
- Recording a smaller temperature increase for the black card than for the white card (physically incorrect).
- Not using the same time intervals for both experiments.
Things to Be Careful About
Maintain the same distance d between the lamp and the card. Ensure the initial temperature is the same as in part (a) or allow the thermometer to cool back to room temperature. Record to 1 decimal place.
Determine the temperature increase between and for each card.
for white card = ______
for black card = ______
Working
Using example data:
Answer
for white card = 2.5 °C
for black card = 3.0 °C
Δθ_W = 2.5 °C, Δθ_B = 3.0 °C (example values)
Walkthrough
The temperature increase Δθ is calculated by subtracting the initial temperature (at t = 0 s) from the final temperature (at t = 300 s) for each card. The candidate must do this for both the white and black cards using their recorded data. The unit is °C.
Key Takeaways
Δθ is a difference in temperature, not an absolute temperature. Always subtract the initial value from the final value.
Common Mistakes
- Calculating the difference between two intermediate readings instead of the first and last.
- Forgetting the unit °C.
Things to Be Careful About
Use the exact values recorded in the table. Do not round intermediate values until the final answer. Include the unit °C.
Calculate the average rate of increase of temperature of each card. Use the equation:
where .
Include the unit in your answers.
average rate of temperature increase of white card = ______ unit ______
average rate of temperature increase of black card = ______ unit ______
Working
where .
For white card:
For black card:
Answer
average rate of temperature increase of white card = 0.0083 unit °C/s
average rate of temperature increase of black card = 0.010 unit °C/s
White: 0.0083 °C/s, Black: 0.010 °C/s (example values)
Walkthrough
The candidate uses the given equation to calculate the average rate of temperature increase. The total time is 300 s. The rate is the temperature increase divided by the time. The unit is °C per second (°C/s). The candidate must calculate this for both cards and explicitly state the unit.
Key Takeaways
Rate of change is calculated as the change in quantity divided by the time taken. Always include the correct derived unit.
Common Mistakes
- Forgetting to divide by the total time (300 s) instead of using a single 60 s interval.
- Forgetting to include the unit °C/s.
- Using incorrect significant figures (e.g., 0.008333... instead of 0.0083).
Things to Be Careful About
The unit is required for a mark. Write it as °C/s, not °C s⁻¹ (5054 prefers the slash form). Ensure the division is done correctly: Δθ / 300, not 300 / Δθ.
Use your answers to (c)(ii) to deduce a conclusion which compares the absorption of thermal radiation by the two different coloured cards.
State your conclusion.
Answer
The black card has a greater rate of temperature increase than the white card. Therefore, the black card is a better absorber of thermal radiation (or heat) than the white card.
The black card is a better absorber of thermal radiation than the white card.
Walkthrough
The candidate compares the average rates of temperature increase calculated in part (c)(ii). Since the black card heats up faster (greater rate of temperature increase) under the same conditions, it must be absorbing thermal radiation from the lamp more effectively. The conclusion must state that the black card is a better absorber than the white card.
Key Takeaways
A faster temperature rise under identical irradiation conditions indicates a higher rate of absorption of thermal radiation. Dark, matte surfaces are generally better absorbers (and emitters) than light, shiny surfaces.
Common Mistakes
- Stating that the black card 'absorbs more heat' without mentioning thermal radiation.
- Concluding that the white card is a better absorber (contradicting the data).
- Not comparing the two cards explicitly.
Things to Be Careful About
Use the word 'absorber' or 'absorbs'. Do not say 'the black card has more heat'. The conclusion must be consistent with the data (black card rate > white card rate).
State two variables that should be controlled in this experiment to ensure a valid conclusion.
controlled variable 1 ______
controlled variable 2 ______
Answer
Any two of the following:
- Same distance of the card from the lamp.
- Same brightness / intensity / power of the lamp (same voltage / current from the power supply).
- Same area / thickness / material / size of the card.
- Same initial / room temperature.
- Same total time (300 s).
- Same height of the thermometer above the bench.
Same distance d from the lamp; same lamp brightness/power (or same card area/material).
Walkthrough
To ensure a fair test comparing the absorption of thermal radiation by the two cards, all other factors that could affect the temperature rise must be kept constant (controlled variables). The intensity of radiation falls off with distance, so the distance must be the same. The power output of the lamp must be the same. The physical properties of the cards (area, thickness, material) must be identical except for colour. The initial temperature and total time must also be the same.
Key Takeaways
In a comparative experiment, only the independent variable (colour of the card) should change. All other variables that influence the dependent variable (temperature increase) must be controlled.
Common Mistakes
- Listing the independent variable (colour) or dependent variable (temperature) as a controlled variable.
- Giving vague answers like 'be more careful' or 'use the same equipment' without specifying what is being controlled.
- Not realising that the lamp's power output must be constant.
Things to Be Careful About
The answer must be specific. 'Same lamp' is not enough; specify 'same brightness' or 'same voltage'. 'Same card' is not enough; specify 'same area and thickness'.
In this experiment you will use a balancing method to determine the mass of a metre rule.
You have been provided with:
- a metre rule with a millimetre scale
- a triangular block to act as a pivot
- slotted masses making a total of 100 g.
Place the metre rule on the pivot.
Place a mass on the metre rule with its centre at the 5.0 cm mark.
Adjust the position of the metre rule on the pivot until the metre rule is as close to balance as possible.
The mass must stay at the 5.0 cm mark.
Fig. 3.1 shows the balanced metre rule.
Read the position of the pivot on the metre rule when the metre rule is balanced. The position of the pivot is the distance between the 0 cm mark on the rule and the tip of the pivot.
Record the position of the pivot, in centimetres to the nearest millimetre, in Table 3.1 on page 10.
Answer
Record the position of the pivot on the metre rule to the nearest millimetre (one decimal place in , e.g. ).
Pivot position recorded to the nearest 0.1 cm (e.g. 42.5 cm)
Walkthrough
The student places a mass at the mark of a metre rule and adjusts the balance point on a triangular pivot.
The reading on the metre rule directly above the apex of the pivot gives the pivot position.
Because the metre rule has a millimetre scale, measurements in centimetres must be recorded to the nearest (e.g., ).
Key Takeaways
- Always record metre rule measurements to the precision of the instrument scale, which is or .
Common Mistakes
- Forgetting the decimal place (e.g. writing instead of ).
Things to Be Careful About
- Ensure the rule is balanced horizontally before taking the scale reading.
Calculate and record, in Table 3.1 on page 10:
- the distance between the 5.0 cm mark and the pivot
- the distance between the pivot and the 50.0 cm mark.
Working
For a representative pivot position of :
Answer
Record values of and in Table 3.1.
a = 37.5 cm, b = 7.5 cm (consistent with pivot position)
Walkthrough
From Fig. 3.1:
- is the distance from the mark to the pivot: .
- is the distance from the pivot to the mark (the centre of gravity of the rule): .
Both values are calculated to one decimal place and recorded in the table.
Key Takeaways
- Distance and distance sum to .
Common Mistakes
- Inverting the subtraction or adding instead of subtracting.
Things to Be Careful About
- Ensure both and are given with consistent decimal places.
Answer
Complete the table for .
Representative completed entries:
- As increases, the pivot must be moved closer to the mark to balance the increased anticlockwise moment.
- Trend: pivot position decreases, decreases, increases.
Complete set of readings with pivot and a decreasing, b increasing
Walkthrough
By the principle of moments for a balanced rule:
As the mass placed at increases, the moment of the mass increases. To maintain balance, distance must decrease and distance must increase. Therefore, the pivot moves closer to the mark, which means:
- The recorded pivot position decreases.
- decreases.
- increases.
All entries in Table 3.1 must be recorded to the nearest .
Key Takeaways
- Increasing the applied load on one side requires a smaller moment arm () to balance the ruler's weight.
Common Mistakes
- Reversing the trend or failing to balance the ruler properly for larger masses.
Things to Be Careful About
- Keep the mass centered strictly at the mark for all trials.
Describe how you ensure that the centre of each mass placed on the metre rule is directly above the 5.0 cm mark.
Answer
Take the scale readings at both outer edges of the mass and find their mean (or look through the central slot / hole in the mass to align it with the line).
Take the scale reading at both edges of the mass and find the mean
Walkthrough
Slotted masses have a finite diameter, so their centre cannot be read directly unless aligned carefully. Standard acceptable techniques include:
- Reading the position of the left and right edges on the metre rule and averaging them to ensure the midpoint is at .
- Measuring the diameter/width of the mass, dividing by 2, and placing the edges that distance to the left and right of .
- Viewing through the central slot/hole of the mass to align it directly over the mark.
Key Takeaways
- To align the centre of a cylindrical or slotted mass on a rule, use symmetry by placing the outer edges equidistant from the mark.
Common Mistakes
- Vague statements like "just look at it carefully" or "place it in the middle".
Things to Be Careful About
- Ensure you clearly describe a method based on measurement or visual alignment of the mass edges/slot.
Calculate the ratio for each value of . Record your answers in Table 3.1.
Table 3.1
| position of pivot / | ||||
|---|---|---|---|---|
| 20 | ||||
| 40 | ||||
| 60 | ||||
| 80 | ||||
| 100 |
Working
Calculate for each row in Table 3.1 to 2 or 3 significant figures.
Representative table:
Answer
| position of pivot / | ||||
|---|---|---|---|---|
| 20 | 42.5 | 37.5 | 7.5 | 0.20 |
| 40 | 36.8 | 31.8 | 13.2 | 0.42 |
| 60 | 32.5 | 27.5 | 17.5 | 0.64 |
| 80 | 29.1 | 24.1 | 20.9 | 0.87 |
| 100 | 26.5 | 21.5 | 23.5 | 1.09 |
All values of r = b/a correctly calculated
Walkthrough
For each row in Table 3.1, divide the value in the column by the value in the column:
Ensure values of are recorded to a consistent and appropriate number of significant figures (usually 2 or 3 s.f.).
Key Takeaways
- Ratios are dimensionless quantities (no units).
- Keep significant figures consistent across the calculated column.
Common Mistakes
- Inverting the ratio (calculating instead of ).
Things to Be Careful About
- Do not round excessively to 1 significant figure.
On the grid provided in Fig. 3.2 on page 11, plot a graph of on the -axis against on the -axis.
Start your axes from the origin (0, 0).
Draw the straight line of best fit.
Answer
- Axes: Horizontal axis labelled from 0 to at least ; vertical axis labelled from 0 to at least 1.2. Both axes start at .
- Scales: Sensible, linear scales using more than half the grid in both directions (e.g. on -axis, on -axis).
- Plotting: All five points plotted accurately to within half a small square with neat crosses or encircled dots.
- Line of best fit: A single thin, straight line drawn with a ruler passing evenly through or balancing the plotted points.
Graph of r against m plotted with linear scales from (0,0), accurate points, and best-fit straight line
Walkthrough
To score all 4 marks on graph plotting:
- Axes & Labels: Label the -axis with '' (no unit as it is a ratio) and the -axis with ''. Ensure the origin is explicitly as instructed.
- Scale Choice: Choose convenient, non-awkward multiples (e.g. 1, 2, 5 or multiples of 10) so that the points occupy more than half the grid area along both axes.
- Plotting: Mark each point with a sharp pencil as a small 'x' or a small dot with a circle around it. Every plotted point must be within small square of its true coordinate.
- Best-Fit Line: Use a transparent ruler to draw a straight line that balances the points on either side with an even distribution, avoiding kinks or thick double lines.
Key Takeaways
- Always label axes with quantity and unit (if applicable).
- Never use awkward scales such as units of 3 or 7.
- The best-fit line must be a single, sharp line.
Common Mistakes
- Not starting the axes from when explicitly instructed.
- Using scales that occupy less than half the available grid.
- Drawing point-to-point lines instead of a best-fit straight line.
Things to Be Careful About
- Ensure the line is drawn cleanly with a single stroke of a sharp pencil.
Calculate the gradient of your graph.
Show clearly on the graph how you obtained the numbers you use for your calculation, and show your working.
= ______
Working
Choose two points on the line of best fit separated by , for example and :
Answer
0.0110 g^-1 (value in range 0.0075 - 0.012)
Walkthrough
- Mark a large gradient triangle on the graph line so that (at least half the line length).
- Read off the coordinates of the two points on the best-fit line (do not use raw data points unless they lie exactly on the line).
- Compute the gradient:
For a standard wooden metre rule with mass around , typically falls within the range to .
Key Takeaways
- Always use a large triangle spanning more than half the drawn line.
- Read coordinates from the best-fit line itself, not the data table.
Common Mistakes
- Using a triangle that is too small (e.g. ).
- Inverting the gradient (calculating ).
Things to Be Careful About
- Check read-off precision against the grid scale divisions.
The mass of the metre rule is given by the equation shown.
Determine the mass of the metre rule to the nearest gram.
= ______
Working
Rounding to the nearest whole gram:
Answer
91 g
Walkthrough
From the principle of moments:
Comparing this to , the gradient is , so:
Substitute the calculated value of into this formula and round the result to the nearest gram (whole number) as requested.
Key Takeaways
- Pay attention to rounding instructions ("to the nearest gram" means no decimal places).
Common Mistakes
- Leaving decimals when the question specifies "to the nearest gram".
- Arithmetic errors when inverting a decimal value.
Things to Be Careful About
- Ensure the final value is physically reasonable for a metre rule (typically between and ).
A student says that the centre of gravity of the metre rule is at the 50.0 cm mark.
Describe how you use the apparatus provided to check that this statement is correct.
Answer
Place the metre rule on the pivot without any masses attached, and check if it balances horizontally when the pivot is positioned at the mark.
Check if the metre rule balances on the pivot at the 50.0 cm mark without any added mass
Walkthrough
The centre of gravity of an object is the point at which its entire weight may be considered to act. If a ruler is supported at its centre of gravity with no external loads applied, the clockwise and anticlockwise moments of its weight about the pivot are equal and opposite (zero net moment), so the ruler balances horizontally.
To check if the centre of gravity is at the mark:
- Remove all added masses.
- Place the ruler on the triangular pivot at the mark.
- Observe whether the ruler balances horizontally on its own.
Key Takeaways
- An unloaded object balances in equilibrium when pivoted directly beneath its centre of gravity.
Common Mistakes
- Keeping the masses on the ruler.
- Suggesting using a balance/scales to weigh it, which does not find the centre of gravity.
Things to Be Careful About
- State clearly that no masses are attached and that the pivot is placed at the mark.
A student investigates the rate of cooling of hot water in a beaker.
Plan an experiment to investigate the relationship between the thickness of the cardboard insulation wrapped around the beaker and the rate of cooling of the hot water in the beaker.
The apparatus available includes:
- a supply of hot water
- a beaker
- a thermometer
- a supply of 1 mm thick cardboard sheets.
You are not required to do this experiment.
In your plan include:
- any other apparatus needed
- a brief description of the method, including what you will measure and how you make sure that your measurements are accurate
- the variables you will control
- a results table to record your measurements (you are not required to enter any readings in the table)
- how you will process your results to reach a conclusion.
Additional apparatus
- a stopwatch (stop clock / timer)
- a measuring cylinder (to measure the same volume of water each time)
Method
- Wrap the beaker with a known number of cardboard sheets (e.g. 1 sheet).
- Measure a fixed volume of hot water (e.g. 200 cm3) and pour it into the beaker.
- Record the initial temperature of the water (e.g. 80 °C) and start the stopwatch.
- Time how long the temperature takes to fall to a fixed value (e.g. 60 °C). This time is a measure of the rate of cooling – a longer time means a slower rate.
- Repeat steps 1–4 using 2, 3 and 4 sheets of cardboard, keeping the same volume and initial temperature each time.
- Repeat each run two or three times and average the times to make the readings more reliable. Read the thermometer at eye level to avoid parallax.
Variables to control
- same volume / mass of hot water
- same initial temperature of the hot water
- same temperature range (e.g. always 80 °C to 60 °C)
- same thermometer, beaker and surroundings
Results table
| Number of cardboard sheets | Thickness of cardboard / mm | Time for temperature to fall from 80 °C to 60 °C / s |
|---|---|---|
| 1 | 1 | |
| 2 | 2 | |
| 3 | 3 | |
| 4 | 4 |
(No readings are needed in this blank.)
Processing and conclusion
Plot a graph of number of cardboard sheets (or thickness) on the x-axis against the time to cool (or the derived rate of cooling) on the y-axis, or draw a bar chart. If the time to cool increases as the number of sheets increases, the rate of cooling becomes slower. The conclusion is that the thicker the cardboard, the slower the rate of cooling – the cardboard insulates better.
See working
Walkthrough
This is a Plan an investigation question from Paper 3. The mark scheme rewards six separate components: an additional apparatus, a method that ‘times the hot water cooling’, at least two different thicknesses, a control variable, a results table with units, and a way to process the results to reach a conclusion.
First, identify the variables. The independent variable is the thickness of the cardboard (or the number of sheets). The dependent variable is the rate of cooling of the water. The rate of cooling is not measured directly; instead we make it a fixed temperature drop (e.g. 80 °C to 60 °C) and time how long that takes.
The only additional apparatus needed is a stopwatch (or stopclock / timer), which earns MP1. A measuring cylinder can help to measure the same volume of water, but it is not required.
Second, the method must at least ‘time hot water cooling’ (MP2). Wrap the beaker in a known number of sheets, pour a fixed volume of hot water, record the initial temperature, start the stopwatch and time the fall from 80 °C to 60 °C. Repeating for 2, 3 and 4 sheets gives at least two different thicknesses (MP3). Repeating each run and averaging the times increases accuracy; reading the thermometer at eye level avoids parallax.
Third, the controlled variables (MP4). The volume / mass of the water, its initial temperature, the temperature range, and the same thermometer/beaker/surroundings must stay the same. Without these, a shorter or longer time could be caused by something other than the cardboard thickness.
Fourth, the results table (MP5) must have a column for the thickness / number of sheets and a column for temperature and/or time, all with units. The example table has ‘Thickness / mm’ and ‘Time for temperature to fall error / s’.
Fifth, the analysis (MP6): plotting a graph of number of sheets against time (or making a bar chart) makes the relationship clear. If the time to cool increases when the thickness increases, the rate of cooling decreases. So a conclusion such as ‘thicker cardboard means a slower rate of cooling – the cardboard insulates better’ is fully justified.
More formally, if is the temperature fall and the time taken, the average rate of cooling is
Keeping the same means a longer corresponds to a smaller rate of cooling.
Key Takeaways
- A good plan includes the apparatus, a basic procedure, at least two different thicknesses, the variables to control, a table with units, and a way to process results to a conclusion.
- Rate of cooling can be measured as a time for a fixed temperature drop.
- All table headings do include the physical quantity and its unit (e.g. time / s, thickness / mm).
- Repeating and averaging readings, and reading the thermometer without parallax, makes the experiment more accurate.
- A graph/bar chart is the characteristic of conclusion supported by the data.
Common Mistakes
- Forgetting to include a stopwatch / stopclock / timer – this is the only additional piece of apparatus the mark scheme rewards (MP1).
- Using only one thickness of cardboard, so there is no comparison (MP3).
- Missing the control variables such as the same volume or initial temperature of water (MP4).
- Drawing a table without units, or without a time/temperature column (MP5).
- Reaching a conclusion without a graph or without comparing the measured values (MP6).
- Giving vague accuracy improvements such as ‘be careful’ – 5054 wants specific techniques like repeating, averaging and avoiding parallax.
Things to Be Careful About
- Use the same temperature range for every run (e.g. always 80 °C to 60 °C) so each measurement is the same temperature drop.
- Every column in the results table must carry a unit (mm, s, °C).
- If the time to cool increases with a thicker, that is a slower rate of cooling, not a faster one.
- Write the conclusion in the form ‘thicker cardboard reduces the rate of cooling’ – not just ‘thickness affects the cooling’.
Part 3 Practical notes
Because the mark scheme is based on six explicit marks, a candidate’s plan should display all six clearly as separate points. There is no single correct set of numbers; the whole plan is evaluated. It is vital that the method includes a stopwatch (or timer) and ‘timing the hot water’. The choice of thicknesses (1, 2, 3, 4 sheets) is a sensible range. Controlled variables are the volume, initial temperature and same temperature range. The table must have all the units. The graph/conclusion step must use the data or a comparison.