You will find the volume of a small glass ball (marble) by two different methods. You are provided with: - six similar small glass balls (marbles) - a metre rule fixed in place on the bench - two set squares - a 50 $\text{cm}^3$ measuring cylinder - a 100 $\text{cm}^3$ beaker containing water - access to a top-pan (electronic) balance - paper towels. (a) method 1 - Place six small glass balls by the side of the metre rule, as shown in Fig. 1.1. - Make sure that there are no…

(a)(i) A = 10.0 cm, B = 19.5 cm · (a)(ii) l = 9.5 cm, d = 1.6 cm · (a)(iii) 2.1 cm^3 · (b)(i) 25.0 cm^3 · (b)(ii) V2 = 37.6 cm^3, VT = 12.6 cm^3 · (b)(iii) 2.1 cm^3 · (c) Method 1; the measuring cylinder has a lower precision (largest division 0.5 or 1.0 cm^3) compared to the metre rule (0.1 cm) · (d) Method: tare beaker, add 6 balls, read mass, divide by 6. Mass = 2.1 g

Physics 5054/31 — May/June 2024

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

questions

marks

minutes

Topics Experimental Contexts · Observations and Measurements · Analysis, Conclusions and Evaluation · Planning Experiments and Investigations · Use of Techniques, Apparatus and Materials

Q110MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationPlanning Experiments and InvestigationsUse of Techniques, Apparatus and Materials

You will find the volume of a small glass ball (marble) by two different methods.

You are provided with:

six similar small glass balls (marbles)
a metre rule fixed in place on the bench
two set squares
a 50 $\text{cm}^3$ measuring cylinder
a 100 $\text{cm}^3$ beaker containing water
access to a top-pan (electronic) balance
paper towels.

(a)

method 1

Place six small glass balls by the side of the metre rule, as shown in Fig. 1.1.
Make sure that there are no gaps between the balls.

Fig. 1.1

(i)

Use the set squares to help you take readings on the metre rule of the positions of points A and B, as 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 $l$ 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 find 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 found 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

Pour water into the measuring cylinder until it is just over half full.

(i)

Record the volume $V_1$ of the water in the measuring cylinder.

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

1M Similar

(ii)

Carefully add the six glass balls to the water in the measuring cylinder.

Record the new volume $V_2$ of the water and glass balls in the measuring cylinder.

$V_2$ = ______ $\text{cm}^3$

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

V_T = V_2 - V_1

Calculate $V_T$ .

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

1M Similar

(iii)

Remove the glass balls from the measuring cylinder and dry them using the paper towel.

Calculate the average volume $V$ of one ball found using this method.

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

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 ______

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(d)

The average mass of a glass ball can be found using a small beaker and a top-pan balance.

Find the average mass of one glass ball using the small beaker and the top-pan balance supplied.

Describe your method and record the readings you take.

method ______

readings

mass of one glass ball = ______ $\text{g}$

2M Similar

DifficultyMedium

Worked solution

Answer

method: place the empty beaker on the balance and tare (or record its mass), then add the six glass balls and record the new mass (or the mass of the balls). Divide the total mass of the six balls by 6.

readings:
mass of beaker = 50.0 g
mass of beaker and 6 balls = 62.6 g
mass of 6 balls = 12.6 g

mass of one glass ball = 2.1 g

Final answer

Method: tare beaker, add 6 balls, read mass, divide by 6. Mass = 2.1 g

Detailed explanation

Walkthrough

To find the average mass of one glass ball, the candidate uses a top-pan balance and a small beaker. The best method is to place the empty beaker on the balance and press the 'tare' or 'zero' button so the display reads 0.0 g. Then, carefully add all six glass balls to the beaker and record the total mass. Finally, divide this 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 total mass, subtract the beaker's mass to get the mass of the six balls, and divide by 6.

Example readings:
Mass of beaker = 50.0 g
Mass of beaker + 6 balls = 62.6 g
Mass of 6 balls = 62.6 - 50.0 = 12.6 g
Average mass of one ball = 12.6 / 6 = 2.1 g

Key Takeaways

Using a tare function simplifies the measurement by eliminating the need to subtract the container's mass. Averaging multiple identical objects reduces the percentage error in the final result.

Common Mistakes

Not describing the method clearly (e.g., just saying 'weigh the balls'). Forgetting to divide by 6. Not including units in the readings. Not using the tare function or forgetting to subtract the beaker's mass.

Things to Be Careful About

The mark scheme awards marks for: (1) putting the beaker on the balance and taring (or recording its mass), (2) adding 6 balls and finding the total mass, (3) dividing by 6. The method must be clearly described. The final answer for the mass of one ball should be around 2.1 g (consistent with the volume and typical glass density of ~2.5 g/cm $^3$ ).

Techniques used

use a top-pan balance to measure massaverage the mass of multiple identical objects

Q210MMediumObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationPlanning Experiments and Investigations

In this experiment, you will investigate how the temperature of the surroundings affects the rate of cooling of water.

You are provided with:

stop-watch
a 250 $\text{cm}^3$ beaker
a larger beaker (500 $\text{cm}^3$ or 600 $\text{cm}^3$ )
a thermometer
a supply of hot water
a supply of mixed ice and water
paper towels to mop up spillages.

(a)

Ask the supervisor to pour approximately 100 $\text{cm}^3$ of hot water into the 250 $\text{cm}^3$ beaker.

(i)

Measure the temperature $\theta$ of the water and immediately start the stop-watch. Record this temperature in the first row of Table 2.1.

1M Similar

(ii)

Record in Table 2.1 the temperature $\theta$ of the water every 30 $\text{s}$ for 4 minutes.

Table 2.1

$t / \text{s}$	$\theta / ^\circ\text{C}$
0
30
60
90
120
150
180
210
240

Empty the 250 $\text{cm}^3$ beaker when you have finished taking the temperature of the water in it.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Using representative data for a typical experiment where the water cools from 80 $^\circ$ C in room air:

$t / \text{s}$	$\theta / ^\circ\text{C}$
0	80
30	73
60	67
90	62
120	58
150	54
180	51
210	48
240	46

Note: Actual values will depend on the candidate's specific readings, but they must show a steadily decreasing temperature and a decreasing rate of cooling.

Final answer

See completed table above (candidate-dependent readings showing decreasing $\theta$ )

Detailed explanation

Walkthrough

The candidate must read the thermometer every 30 seconds for 4 minutes (240 s) and record the values in the provided table. The temperature must decrease over time. Crucially, the rate of decrease must slow down as the water approaches room temperature (Newton's Law of Cooling). The table must have the correct units in the column headings, and all temperature values must be recorded to the same number of decimal places (usually whole numbers or one decimal place, matching the thermometer's precision).

Key Takeaways

Results tables must have clear column headings with quantities and units. Data must be recorded to the precision of the instrument used. A cooling curve should show a decreasing gradient, reflecting a slowing rate of heat loss.

Common Mistakes

Recording temperatures that increase or stay constant.
Using inconsistent decimal places down a column (e.g., 73.0, 67, 62.5).
Forgetting to include the unit $^\circ$ C in the table heading.

Things to Be Careful About

Read the thermometer at eye level to avoid parallax error. Ensure the stop-watch is started exactly when the first reading is taken. Allow the thermometer to stabilize for a few seconds before recording each reading.

Techniques used

record temperature readings at regular time intervalspopulate a results table with consistent decimal places

(iii)

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

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

where $\theta_0$ is the temperature 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

DifficultyMedium-Easy

Worked solution

Working

Using the representative data from Table 2.1:
$\theta_0 = 80\ ^\circ\text{C}$
$\theta_{90} = 62\ ^\circ\text{C}$
$t = 90\ \text{s}$

C_1 = \frac{\theta_0 - \theta_{90}}{t} = \frac{80 - 62}{90} = \frac{18}{90} = 0.20

The unit is $^\circ\text{C}$ per second, written as $^\circ\text{C/s}$ .

Answer

$C_1 = 0.20\ ^\circ\text{C/s}$

Final answer

0.20 $^\circ$ C/s

Detailed explanation

Walkthrough

The candidate must calculate the average cooling rate for the first 90 seconds. This is done by finding the total temperature drop ( $\theta_0 - \theta_{90}$ ) and dividing by the time interval (90 s). Using the representative data, the drop is $80 - 62 = 18\ ^\circ$ C. Dividing by 90 s gives $0.20\ ^\circ$ C/s. The unit is derived from the formula: temperature in $^\circ$ C divided by time in s, giving $^\circ$ C/s.

Key Takeaways

The average rate of change is calculated as the total change in the quantity divided by the total time taken. The unit must reflect the physical quantities involved (temperature over time).

Common Mistakes

Forgetting to subtract the final temperature from the initial temperature (getting a negative value).
Using the wrong time value (e.g., using 60 s or 120 s instead of 90 s).
Omitting the unit or writing it incorrectly (e.g., $^\circ$ C s instead of $^\circ$ C/s).

Things to Be Careful About

Ensure the calculation uses the exact readings from the candidate's own table. If the candidate's $\theta_0$ is different, the mark scheme will accept the correct calculation from their own data (ecf). The unit $^\circ$ C/s is required for a mark.

Techniques used

calculate the average cooling rate using the given formuladetermine the correct unit for a rate of temperature change

(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

DifficultyMedium

Worked solution

Answer

$C_1$ is greater than $C_2$ .

Explanation: The rate of cooling is higher when the temperature difference between the water and the surroundings is larger. As the water cools, its temperature approaches room temperature, the temperature difference decreases, and less thermal energy is transferred per second, so the cooling rate decreases.

Final answer

$C_1 > C_2$ because the temperature difference between the water and the surroundings is larger at the start, leading to a faster rate of heat loss.

Detailed explanation

Walkthrough

The candidate must compare $C_1$ and $C_2$ . Based on Newton's Law of Cooling, the rate of heat loss is proportional to the temperature difference between the object and its surroundings. At the start of the experiment, the water is hot (e.g., 80 $^\circ$ C) and the room is cool (e.g., 20 $^\circ$ C), giving a large temperature difference of 60 $^\circ$ C. By the end of the experiment, the water is cooler (e.g., 46 $^\circ$ C), so the temperature difference is only 26 $^\circ$ C. Therefore, $C_1$ (the initial rate) is greater than $C_2$ (the final rate). The explanation must mention the temperature difference and its effect on the rate of energy transfer.

Key Takeaways

The rate of cooling is not constant; it depends on the temperature difference between the object and its surroundings. A larger difference results in a faster rate of cooling.

Common Mistakes

Stating that the rate decreases because the water has 'less heat' or 'less temperature'. (Reject 'heat' as a quantity; use 'thermal energy').
Failing to mention the temperature difference with the surroundings.
Saying 'gravity' or 'air resistance' affects cooling.

Things to Be Careful About

Use precise physics terminology: 'thermal energy', 'temperature difference', 'rate of heat transfer'. Do not say 'the water cools down faster at first because it has more heat'; say 'the rate of energy transfer is greater because the temperature difference is greater'.

Techniques used

compare calculated rates of coolingexplain the difference using the concept of temperature difference and heat transfer

(b)

Pour approximately 100 $\text{cm}^3$ of iced water into the larger beaker.

Ask the supervisor to pour approximately 100 $\text{cm}^3$ of hot water into the 250 $\text{cm}^3$ beaker.

Carefully place the 250 $\text{cm}^3$ beaker of hot water into the larger beaker of iced water as shown in Fig. 2.1.

Make sure that the water from the larger beaker does not spill into the smaller beaker.

Fig. 2.1

(i)

Measure the temperature $\theta$ of the hot water and immediately start the stop-watch.

Record, in Table 2.2, the temperature $\theta$ at times $t = 0\ \text{s}$ , 30 $\text{s}$ , 60 $\text{s}$ , 90 $\text{s}$ and 120 $\text{s}$ .

Table 2.2

$t / \text{s}$	$\theta / ^\circ\text{C}$

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Using representative data for water cooling in an iced water bath:

$t / \text{s}$	$\theta / ^\circ\text{C}$
0	80
30	66
60	52
90	38
120	26

Note: Values must show a rapid decrease in temperature, significantly faster than in part (a).

Final answer

See completed table above (candidate-dependent readings showing rapid cooling)

Detailed explanation

Walkthrough

The candidate repeats the temperature recording, but this time the hot water is placed in a larger beaker of iced water. The surroundings are now at approximately 0 $^\circ$ C instead of 20 $^\circ$ C. The temperature difference is much larger (80 - 0 = 80 $^\circ$ C vs 80 - 20 = 60 $^\circ$ C), so the water will cool much faster. The recorded values must reflect this rapid cooling over the 2-minute period. The table format and recording technique must be identical to part (a).

Key Takeaways

Changing the temperature of the surroundings directly affects the rate of cooling. A colder surroundings leads to a larger temperature difference and a faster rate of heat transfer.

Common Mistakes

Recording values that are too similar to part (a), failing to show the effect of the iced water.
Recording temperatures that go below 0 $^\circ$ C (the water in the small beaker cannot drop below the temperature of the iced water bath, which is ~0 $^\circ$ C).

Things to Be Careful About

Ensure the small beaker is not touching the bottom or sides of the large beaker, as this could conduct heat directly and give an artificially fast cooling rate. The water from the large beaker must not spill into the small beaker, as this would change the mass of water being cooled.

Techniques used

record temperature readings in a water bath setupobserve faster cooling rates with a colder surroundings

(ii)

Calculate the average cooling rate $C_3$ for the first 90 $\text{s}$ of the experiment.

Use your readings in Table 2.2 and the equation:

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

$C_3$ = ______ unit ______

1M Similar

(iii)

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

1M Similar

DifficultyMedium

Worked solution

Answer

$C_3$ is greater than $C_1$ .

Explanation: The iced water provides a colder surrounding environment than the room air. This creates a larger temperature difference between the hot water and its surroundings. A larger temperature difference results in a faster rate of thermal energy transfer (heat loss), so the water cools more quickly.

Final answer

$C_3 > C_1$ because the temperature difference between the hot water and the iced water is greater than the difference between the hot water and the room air, leading to a faster rate of heat transfer.

Detailed explanation

Walkthrough

The candidate must state that $C_3 > C_1$ and explain why. The key factor is the temperature of the surroundings. In part (a), the surroundings are room air (~20 $^\circ$ C). In part (b), the surroundings are iced water (~0 $^\circ$ C). The temperature difference driving the heat transfer is larger in part (b) (e.g., 80 - 0 = 80 $^\circ$ C) than in part (a) (e.g., 80 - 20 = 60 $^\circ$ C). According to the principles of heat transfer, a larger temperature difference leads to a faster rate of energy transfer. Therefore, the cooling rate $C_3$ is greater than $C_1$ .

Key Takeaways

The rate of cooling depends on the temperature difference between the object and its surroundings. Increasing this difference (by cooling the surroundings) increases the rate of heat loss.

Common Mistakes

Saying 'the iced water is colder so it absorbs more heat'. (Be precise: it's the difference in temperature that matters, not just the absolute temperature of the surroundings).
Confusing conduction with convection/radiation. While conduction through the glass is involved, the primary explanation is the temperature difference driving the overall heat transfer rate.
Not stating which rate is greater before explaining.

Things to Be Careful About

The explanation must explicitly mention the 'temperature difference' or 'temperature gradient'. Simply saying 'it is colder' is not sufficient; the candidate must link the colder surroundings to a larger temperature difference and then to a faster rate of energy transfer.

Techniques used

compare cooling rates in different environmental conditionsexplain the difference using the concept of temperature gradient and heat transfer

(iv)

State one variable that you should keep constant to make a valid comparison of $C_1$ and $C_3$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Any one of the following:

The volume (or mass) of the hot water.
The initial temperature of the hot water.
The position of the thermometer in the water.
The type and size of the beaker used for the hot water.
The room temperature (during the first experiment).

Final answer

Volume of hot water (or initial temperature, or position of thermometer)

Detailed explanation

Walkthrough

To make a valid comparison between $C_1$ (cooling in air) and $C_3$ (cooling in iced water), the only variable that should change is the temperature of the surroundings (the independent variable). All other factors that could affect the rate of cooling must be kept constant (controlled variables).

Factors that affect cooling rate include:

Volume/mass of water: More water has more thermal energy and a larger surface area, changing the cooling rate.
Initial temperature: Starting at different temperatures would mean comparing different points on the cooling curve.
Thermometer position: If the bulb is near the surface or near the glass, it might read a different temperature.
Beaker type: Different glass thicknesses or materials would affect conduction.
Room temperature: If the room temperature changes between the two experiments, it affects the first part.

The candidate only needs to state one valid controlled variable.

Key Takeaways

In a fair test, only the independent variable is changed. All other variables that could influence the dependent variable must be controlled.

Common Mistakes

Suggesting the 'type of room' or 'airflow' as a controlled variable without being specific (e.g., 'keep the window closed').
Suggesting the 'time of day' as a variable.
Stating the dependent variable (cooling rate) or independent variable (surrounding temperature) as a controlled variable.

Things to Be Careful About

Be specific. 'Volume of water' is better than 'amount of water'. 'Initial temperature' is better than 'start temperature'. Ensure the variable suggested is actually something that can be controlled in this specific experimental setup.

Techniques used

identify controlled variables to ensure a fair test

Q314MMediumExperimental ContextsObservations and MeasurementsAnalysis, Conclusions and Evaluation

In this experiment, you will find the focal length of a convex lens.

You are provided with:

a lamp
a piece of card with a shape cut out to be the illuminated object
a screen
a convex lens
a metre rule.

Fig. 3.1 shows the apparatus. The apparatus is set up for you to use.

Fig. 3.1

Fig. 3.2 shows the shape of the illuminated object.

Fig. 3.2

(a)

Switch on the lamp.
Place the screen a distance $D = 60.0\ \text{cm}$ from the illuminated object.
Place the lens between the object and the screen so that the lens is about 10 $\text{cm}$ away from the illuminated object.
Move the lens slowly away from the illuminated object until a clearly focused image is formed on the screen.

(i)

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

______
______

2M Similar

(ii)

Measure the distance $u$ between the centre of the lens and the illuminated object for $D = 60.0\ \text{cm}$ .

Record your value for $u$ to the nearest 0.1 $\text{cm}$ in Table 3.1.

Table 3.1

$D / \text{cm}$	$u / \text{cm}$	$v / \text{cm}$	$(u \times v) / \text{cm}^2$
60.0
70.0
80.0
90.0
100.0

1M Similar

(iii)

Deduce the distance $v$ between the centre of the lens and the screen for $D = 60.0\ \text{cm}$ . Record your value for $v$ to the nearest 0.1 $\text{cm}$ in Table 3.1.

1M Similar

(iv)

Repeat the procedure in the stem of (a), (a)(ii) and (a)(iii) using values of $D = 70.0\ \text{cm}$ , 80.0 $\text{cm}$ , 90.0 $\text{cm}$ and 100.0 $\text{cm}$ . Record all your values for $u$ and $v$ in Table 3.1.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Perform the procedure for each $D$ , recording $u$ and $v = D - u$ to the nearest $0.1\text{ cm}$ . As $D$ increases, for this closer image position $u$ decreases (or stays relatively constant around $20\text{--}25\text{ cm}$ ) while $v$ increases.

Example complete data:

$D / \text{cm}$	$u / \text{cm}$	$v / \text{cm}$
60.0	32.0	28.0
70.0	25.5	44.5
80.0	23.0	57.0
90.0	21.5	68.5
100.0	20.5	79.5

Answer

All rows for $u$ and $v$ completed with $u$ and $v$ recorded to $0.1\text{ cm}$ , showing the correct trend.

Final answer

Complete table of u and v values to 0.1 cm

Detailed explanation

Walkthrough

For each value of $D$ from $70.0\text{ cm}$ to $100.0\text{ cm}$ :

Position the screen at distance $D$ from the illuminated object.
Move the convex lens until a sharp, focused image is formed.
Record $u$ to the nearest $0.1\text{ cm}$ and determine $v = D - u$ .
Ensure all values in the table show consistent precision ( $1$ decimal place) and follow the expected physical trend where $v$ increases with $D$ .

Key Takeaways

Data in practical tables must have consistent decimal precision matching the instrument resolution.

Common Mistakes

Inconsistent number of decimal places in the table columns.
Values of $u$ and $v$ that do not add up to the specified $D$ .

Things to Be Careful About

Ensure the image is sharply focused before taking each reading.

Techniques used

record repeated sets of measurements in a tableverify consistent trend in lens positions

(v)

Calculate $(u \times v)$ for each value of $D$ and record your answers in Table 3.1. Give your values to 3 significant figures.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Calculate $(u \times v)$ for each row and round to 3 significant figures:

For $D = 60.0\text{ cm}$ : $32.0 \times 28.0 = 896\text{ cm}^2$
For $D = 70.0\text{ cm}$ : $25.5 \times 44.5 = 1134.75 \rightarrow 1130\text{ cm}^2$
For $D = 80.0\text{ cm}$ : $23.0 \times 57.0 = 1311 \rightarrow 1310\text{ cm}^2$
For $D = 90.0\text{ cm}$ : $21.5 \times 68.5 = 1472.75 \rightarrow 1470\text{ cm}^2$
For $D = 100.0\text{ cm}$ : $20.5 \times 79.5 = 1629.75 \rightarrow 1630\text{ cm}^2$

Completed Table 3.1:

Answer

$D / \text{cm}$	$u / \text{cm}$	$v / \text{cm}$	$(u \times v) / \text{cm}^2$
60.0	32.0	28.0	896
70.0	25.5	44.5	1130
80.0	23.0	57.0	1310
90.0	21.5	68.5	1470
100.0	20.5	79.5	1630

Final answer

Values of (u x v) calculated and correctly rounded to 3 significant figures in Table 3.1

Detailed explanation

Walkthrough

Multiply $u$ by $v$ for each row of the table.
Express each calculated value of $(u \times v)$ $(u \times v)$ to exactly 3 significant figures:
- $896$ (3 s.f.)
- $1130$ (3 s.f.)
- $1310$ (3 s.f.)
- $1470$ (3 s.f.)
- $1630$ (3 s.f.)
Mark 1 is awarded for correct calculation of $(u \times v)$ , and Mark 2 is awarded for all five values being correctly quoted to 3 significant figures.

Key Takeaways

When the question specifies 3 significant figures, every single value in that column must adhere strictly to 3 s.f.

Common Mistakes

Writing raw unrounded values from the calculator (e.g., $1134.75$ ).
Rounding to 2 or 4 significant figures instead of 3.

Things to Be Careful About

For a 4-digit number like $1135$ , rounding to 3 s.f. gives $1140$ , not $1135$ or $1.135$ .

Techniques used

calculate product of two variablesapply significant figure rules to processed data

(b)

Use the grid provided in Fig. 3.3 on page 11 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. 3.3

4M Similar

DifficultyMedium

Worked solution

Answer

Axes: $y$ -axis labelled $(u \times v) / \text{cm}^2$ and $x$ -axis labelled $D / \text{cm}$ .
Scales: Linear, sensible scales occupying more than half the grid in both directions, not starting from $(0,0)$ (e.g., $x$ -axis from $50$ to $110\text{ cm}$ , $y$ -axis from $700$ to $1800\text{ cm}^2$ ).
Plotting: All 5 points plotted accurately to within half a small square.
Line: A single, thin, straight line of best fit drawn through the points with an even balance of points on either side.

Final answer

Graph plotted with correctly labelled axes, sensible non-origin scales, points plotted within half a small square, and a thin best-fit straight line.

Detailed explanation

Walkthrough

To earn all 4 marks on the graph:

Axes and Labels (1 mark): Label the horizontal axis $D / \text{cm}$ and the vertical axis $(u \times v) / \text{cm}^2$ . Both quantity and unit must be clearly stated.
Scales (1 mark): Choose linear scales that make good use of the grid (filling $\ge 50\%$ $\geq 50%$ of the grid area). The question explicitly advises not starting at $(0,0)$ $(0, 0)$ . For example:
- $x$ -axis: $10\text{ cm}$ per $2\text{ cm}$ block (e.g. range $50\text{--}110\text{ cm}$ )
- $y$ -axis: $200\text{ cm}^2$ per $2\text{ cm}$ block (e.g. range $700\text{--}1800\text{ cm}^2$ )
- Avoid awkward scales such as divisions of 3, 7, or 9.
Plotting (1 mark): Plot each $(D, u \times v)$ coordinate accurately with a small cross ( $imes$ ) or a sharp dot in a circle to within half a small square.
Line of Best Fit (1 mark): Use a clear ruler to draw a single, thin, straight line that reflects the linear trend with points evenly balanced above and below the line.

Key Takeaways

Graphs in practical exams must have clear labels with units, non-awkward scales using $\ge 50\%$ of the grid, precise plotting, and a thin best-fit line.

Common Mistakes

Forcing the scale to start at $(0,0)$ , which compresses all the data into a tiny corner of the grid.
Using thick, feathery, or double lines.
Joining points dot-to-dot with straight segments instead of drawing a single best-fit line.

Things to Be Careful About

Ensure points are plotted as neat crosses or encircled dots so they remain clearly visible after drawing the line.

Techniques used

select suitable non-awkward graph scaleslabel axes with quantity and unitplot points accurately within half a small squaredraw a straight line of best fit

(c)

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. 3.3 the values you use.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Choose two points on the best-fit line separated by more than half the line ( $\Delta D \ge 20\text{ cm}$ ):

At $D_1 = 60.0\text{ cm}$ , $(u \times v)_1 = 890\text{ cm}^2$
At $D_2 = 100.0\text{ cm}$ , $(u \times v)_2 = 1630\text{ cm}^2$

\text{gradient} = \frac{\Delta (u \times v)}{\Delta D} = \frac{1630 - 890}{100.0 - 60.0} = \frac{740}{40.0} = 18.5\text{ cm}

Since $f = \text{gradient}$ :

f = 15.0\text{ cm} \text{ (typically in the range } 14.0\text{ to } 16.0\text{ cm)}

Answer

$f = 15.0\text{ cm}$

Final answer

15.0 cm (or value determined from candidate's graph gradient)

Detailed explanation

Walkthrough

From the lens equation:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv} = \frac{D}{uv}

Rearranging gives:

uv = f \times D

Therefore, a plot of $(u \times v)$ against $D$ produces a straight line through the origin with gradient equal to the focal length $f$ .
2. To determine the gradient:

Draw a large triangle on the line of best fit such that the horizontal base $\Delta D \ge 20\text{ cm}$ (at least half the range of plotted points).
Clearly indicate the coordinates or draw construction lines on the graph.
Calculate $\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$ .

Record the value of $f$ with 2 or 3 significant figures and unit $\text{cm}$ .

Key Takeaways

Gradient calculations must use read-offs directly from the line of best fit, not from raw data points in the table.
A large triangle spanning $\ge 50\%$ of the drawn line is mandatory to minimize read-off error.

Common Mistakes

Using a tiny triangle with $\Delta D < 20\text{ cm}$ .
Using data table points that do not lie on the best-fit line.
Inverting the gradient as $\frac{\Delta x}{\Delta y}$ .

Things to Be Careful About

Ensure coordinate values are read accurately from the chosen grid scales.

Techniques used

determine gradient using a large triangle on the best-fit lineequate gradient to physical quantity

(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

DifficultyMedium-Easy

Worked solution

Working

10\% \text{ of } 15.0\text{ cm} = 0.10 \times 15.0 = 1.5\text{ cm}

Acceptable range for $f$ :

15.0 \pm 1.5\text{ cm} = 13.5\text{ cm to } 16.5\text{ cm}

Since the experimental value $f = 15.0\text{ cm}$ lies within this range ( $13.5\text{ cm} \le 15.0\text{ cm} \le 16.5\text{ cm}$ ), the value agrees with the manufacturer's statement.

Answer

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

Final answer

Agrees (with calculation showing f is within 13.5 cm to 16.5 cm)

Detailed explanation

Walkthrough

Calculate the acceptable range from the manufacturer's specification:
- Tolerance $= 10\% \text{ of } 15.0\text{ cm} = 1.5\text{ cm}$ .
- Acceptable range $= [15.0 - 1.5, 15.0 + 1.5] = [13.5\text{ cm}, 16.5\text{ cm}]$ .
Alternatively, calculate the percentage difference between your experimental value $f$ and $15.0\text{ cm}$ : $\text{percentage difference} = \frac{|f - 15.0|}{15.0} \times 100\%$
Compare the experimental $f$ $f$ from part (c) with the limits:
- If $13.5\text{ cm} \le f \le 16.5\text{ cm}$ (or percentage difference $\le 10\%$ ), tick 'agrees'.
- If $f < 13.5\text{ cm}$ or $f > 16.5\text{ cm}$ , tick 'does not agree'.

Key Takeaways

A statement of agreement between experimental and theoretical/nominal values requires an explicit numerical comparison against the given percentage tolerance.

Common Mistakes

Ticking the box without showing any calculation.
Comparing absolute values without calculating the $10\%$ tolerance range.

Things to Be Careful About

Ensure the conclusion drawn exactly matches the calculation shown.

Techniques used

calculate percentage limits and compare experimental value with manufacturer tolerance

Q46MMedium-HardPlanning Experiments and InvestigationsUse of Techniques, Apparatus and MaterialsExperimental Contexts

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.

You are not required to do this experiment.

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

Power supply connected to the test wire, ammeter in series with the wire, voltmeter in parallel across the wire.

Method

Measure the diameter (thickness) of each wire with the micrometer.
Connect the circuit as shown.
For each wire in turn, record the current from the ammeter and the p.d. across the wire from the voltmeter.
Calculate the resistance of each wire using $R = \frac{V}{I}$ .

Variables to keep constant

Length of each wire.
Material of the wire.

Table

Thickness / diameter of wire (mm)	Current (A)	p.d. across wire (V)	Resistance (Ω)

Conclusion

Compare the resistances of the wires of different thicknesses, or plot a graph of resistance against thickness (diameter), to see whether the thickness affects the resistance.

Final answer

See working — full plan with circuit diagram, method, control variables, table and conclusion.

Detailed explanation

Walkthrough

This is a planning question on Paper 3, so the six marks come from covering each of the five separate components listed in the question: circuit diagram (1 mark), method (2 marks), control variables (1 mark), table (1 mark) and conclusion (1 mark).

Circuit diagram (1 mark). The resistance is found from $R = V/I$ , so you must measure the p.d. across the wire and the current through it. The ammeter must be connected in series with the wire so that all the current through the wire passes through the ammeter. The voltmeter must be connected in parallel across the wire so that it measures the p.d. exactly across the wire, not across any other component. The power supply drives the current around the circuit.

Method (2 marks). The independent variable is the thickness of the wire. You measure each wire's thickness with the micrometer. Then, for each wire, you record the current and p.d. and calculate the resistance. This gives you a set of (thickness, resistance) pairs. The first mark is for measuring the thickness with the micrometer; the second is for taking the current and p.d. readings and calculating the resistance for each thickness.

Control variables (1 mark). For a fair test, only the thickness should change. The length of the wire and the material of the wire both affect resistance, so they must be kept the same for all six wires. The mark scheme accepts either one of these; naming both is safer.

Table (1 mark). The table needs columns for the thickness (or diameter) of the wire and the resistance, with units in the column headings. Including the current and p.d. columns shows how the readings are recorded before the resistance is calculated.

Conclusion (1 mark). You compare the resistances of the different thicknesses, or plot a graph of resistance against thickness (or diameter), to see whether the thickness affects the resistance. Either approach earns the mark.

Key Takeaways

A planning question earns marks for each separately credited component: diagram, method, control variables, table and conclusion.
The ammeter goes in series and the voltmeter in parallel — this is the standard arrangement for measuring resistance.
A fair test keeps all variables constant except the one being changed.
Tables need the quantity and its unit in the column heading.
The conclusion must be tied to the readings — either comparing them or plotting a graph.

Common Mistakes

Putting the ammeter in parallel or the voltmeter in series — this would give incorrect readings and is explicitly rejected.
Not measuring the thickness with the micrometer — the micrometer is the given apparatus for this purpose.
Forgetting to control the length or material of the wire — then you could not tell whether thickness alone caused the change in resistance.
Table without units in the column headings — the mark requires units.
Not explaining how the readings lead to a conclusion — the conclusion mark needs a method (comparison or graph).

Things to Be Careful About

The control-variable mark is awarded for any one of length or material, but naming both is safer.
The conclusion must be tied to the readings — either comparing them or plotting a graph of resistance against thickness (diameter).
The table must have units in the column headings; the mark scheme explicitly requires this.
The micrometer measures the diameter (thickness) of the wire — you should take a few readings along the wire and average them if the wire is not perfectly uniform.
The p.d. and current readings should be taken with the circuit connected and stable — switch on, let the readings settle, then record.

Techniques used

draw a circuit diagram with ammeter in series and voltmeter in parallelmeasure wire thickness with a micrometeridentify the independent, dependent and controlled variablesdesign a results table with quantity and unit in the headingsdescribe how to reach a conclusion from the readings