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S3 Physics: Pressure and Buoyancy (Paper 4: mistake clinic)
S3 Physics: Pressure and Buoyancy (Paper 4: mistake clinic)
The aim of the mistake clinic is to name the cause, not to redo the question: wrong formula, wrong area, or wrong unit. If a cause appears twice, write it on the first page of your notes.
Direct answers Spend three seconds per question checking units, whether you used the total area, and whether buoyancy used the displaced volume.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Physics: Pressure and Buoyancy |
| Level | S3 Physics |
02 Questions (10 items)
- FoundationWith a pressure of 6 000 Pa and a force of 300 N, what is the contact area?
- IntermediateWhat is the pressure at 1.5 m depth in water (1 000 kg/m3, g = 10 N/kg)?
- IntermediateAn object displaces 0.004 m3 of water. With g = 10 N/kg, what buoyant force acts on it?
- FoundationA stone sinks in water. Does it still feel a buoyant force?
- FoundationA force of 200 N acts on 0.4 m2. What is the pressure?
- ChallengeThe same object floats first in fresh water and then in salt water. In which case is the buoyant force larger?
- IntermediateWhat is the pressure at 4 m depth (1 000 kg/m3, g = 10 N/kg)?
- ChallengeAn object of weight 30 N floats at rest. What is the buoyant force?
- ChallengeWith the force unchanged, the contact area falls from 0.05 m2 to 0.01 m2. If the old pressure was 2 000 Pa, what is the new one?
- Error redoWhat is wrong with saying liquid pressure depends on the container shape and the amount of water?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.05 m2 | A = F / p = 300 / 6 000 = 0.05 m2; divide by the pressure rather than multiply. |
| 2 | 15 000 Pa | p = 1 000 x 10 x 1.5 = 15 000 Pa. |
| 3 | 40 N | F = 1 000 x 10 x 0.004 = 40 N. |
| 4 | Yes, but the buoyancy is smaller than its weight. | Sinking does not mean there is no buoyancy, only that the buoyancy cannot support the weight. |
| 5 | 500 Pa | p = 200 / 0.4 = 500 Pa. |
| 6 | Neither differs: both equal its weight; the salt water case displaces less volume. | Both cases must balance the same weight, so the denser salt water only needs a smaller displaced volume. |
| 7 | 40 000 Pa | p = 1 000 x 10 x 4 = 40 000 Pa. |
| 8 | 30 N | Floating at rest means buoyancy equals weight, equal in size and opposite in direction. |
| 9 | 10 000 Pa | The area becomes one fifth, so the pressure becomes five times larger: 10 000 Pa. |
| 10 | It depends only on density and depth; at the same depth in the same liquid the pressure is equal. | Wide and narrow containers give the same pressure at the same depth, though the total force differs. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Keep an error log with three columns: question, cause, corrected working. The usual causes are the wrong formula, the wrong area, and forgetting that buoyancy stops increasing once an object is fully submerged.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.