The average volume V of one glass ball found using method 1 is given by V = 3.14d³ ÷ 6. Using d = 1.5 cm from (a)(ii), calculate V.
Show model answer
V ≈ 1.77 cm³
V = 3.14 × 1.5³ ÷ 6 = 3.14 × 3.375 ÷ 6 = 10.5975 ÷ 6 = 1.77 cm³ (3 s.f.).
O Level Physics: Practical, questions with answers
Practise 65 real questions on Practical from 2021 to 2026. See the answer and explanation as soon as you answer.
65real questions
2papers covered
11exam sessions
About Practical in O Level Physics
Practical is one of the 14 main topics in our O Level Physics (Cambridge (CAIE) 5054) question bank, and about 9% of its real questions come from it. Tayyari Karo has 65 real questions on this topic from 2021 to 2026, from Paper 3, Paper 4. By difficulty: 44 medium, 16 hard, 5 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 8 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
In method 2, water is poured into the 50 cm³ measuring cylinder until just over half full (volume V1 recorded), then the six glass balls are carefully added and the new volume V2 recorded; VT = V2 − V1 is the volume of the six balls together. Using specimen readings V1 = 25.0 cm³ and V2 = 35.5 cm³, calculate VT and the average volume V of one ball by this method.
Show model answer
VT = 10.5 cm³; V (one ball) = 1.75 cm³
VT = V2 − V1 = 35.5 − 25.0 = 10.5 cm³; average volume of one ball = VT ÷ 6 = 10.5 ÷ 6 = 1.75 cm³ — closely consistent with the 1.77 cm³ found by method 1.
Suggest whether method 1 or method 2 gives the more accurate value for the volume of the ball. Explain your answer.
Show model answer
Method 1 (the line-up method) gives the more accurate value.
Method 1 measures a much larger length (six balls, about 9 cm) on a metre rule that can be read precisely with set squares, so the percentage/proportional uncertainty in the reading is small. Method 2 relies on reading a fairly small volume change (about 10 cm³ out of 50 cm³) on a measuring cylinder, which typically can only be read to the nearest 0.5–1 cm³, giving a much larger percentage uncertainty in the small volume difference, and is also affected by the meniscus and any trapped air bubbles between the balls.
Find the average mass of one glass ball using a small beaker and the top-pan balance supplied. Describe your method and record the readings you take.
Show model answer
Method: place the empty small beaker on the top-pan balance and record its mass; add all six glass balls to the beaker and record the new mass; subtract to find the mass of the six balls, then divide by 6. Specimen readings: mass of empty beaker = 30.0 g; mass of beaker + 6 balls = 56.5 g; mass of six balls = 26.5 g; average mass of one glass ball = 26.5 ÷ 6 ≈ 4.4 g.
Weighing all six balls together and dividing by 6 (rather than weighing one ball alone) reduces the percentage error from the balance's reading precision.
Compare the specimen values of C1 (0.20 °C/s) and C2 (0.089 °C/s). Explain any difference.
Show model answer
C1 is greater than C2 — the cooling rate decreases as time passes.
By Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the water and its surroundings; as the water cools towards room temperature this difference decreases, so the rate of cooling (and hence C2 compared with C1) becomes smaller.
Describe how C3 differs from C1. Explain your answer.
Show model answer
C3 (≈0.39 °C/s) is much larger than C1 (0.20 °C/s) — the water surrounded by iced water cools faster.
By Newton's law of cooling, the rate of cooling depends on the temperature difference between the water and its surroundings; the iced-water surroundings (≈0 °C) give a much greater temperature difference than room-temperature air surroundings, so the cooling rate is greater.
State one variable that you should keep constant to make a valid comparison of C1 and C3.
Show model answer
The initial temperature (and volume/mass) of the hot water used in the 250 cm³ beaker, and the type and size of beaker, should be kept the same in both parts of the experiment.
A fair comparison requires only the surroundings (air vs iced water) to differ between the two trials.
The focal length f of the lens is numerically equal to the gradient of the line. Calculate the gradient, showing the values used.
Show model answer
f = 15.0 cm
Using two well-separated points on the line, e.g. (60, 900) and (100, 1500): gradient = (1500 − 900) ÷ (100 − 60) = 600 ÷ 40 = 15.0 cm.
Sessions with Practical questions
| Session | Paper 3 | Paper 4 |
|---|---|---|
| May/June 2026 | 4 | – |
| May/June 2025 | 4 | 4 |
| Oct/Nov 2025 | 4 | – |
| May/June 2024 | 17 | – |
| Oct/Nov 2024 | 4 | – |
| May/June 2023 | 4 | – |
| Oct/Nov 2023 | 4 | – |
| May/June 2022 | 4 | – |
| Oct/Nov 2022 | 4 | 4 |
| May/June 2021 | 4 | – |
| Oct/Nov 2021 | 4 | – |
Common questions
What does the Practical topic cover on this page?
65 real O Level Physics (5054) questions on Practical from 2021 to 2026, from Paper 3, Paper 4. Each one shows the year, session and paper it came from.
Do I get the answers and explanations?
Yes. Every question shows the correct answer and a short explanation the moment you answer.
Is it free?
You can start free with no signup. A small set of papers per subject is free, and Premium unlocks every paper, the full question bank and the paper PDFs.