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S4 Physics: Pressure and Upthrust Paper 1 (Solid and Liquid Pressure)
S4 Physics: Pressure and Upthrust Paper 1 (Solid and Liquid Pressure)
This paper drills solid pressure P = F/A and liquid pressure P = pgh, using g = 10 N/kg and water density 1,000 kg/m3 throughout.
Quick answer 1. 400 Pa | 2. 600 N and 20,000 Pa | 3. 20,000 Pa | 4. 1,000 N | 5. 30,000 Pa (30 kPa) | 6. 20 N | 7. 8 N and 800 cm3 | 8. 6 N, 0.06 m immersed | 9. 0.4 m3 | 10. 15,000 Pa
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Physics – Pressure and Upthrust |
| Level | Foundation |
02 Questions (10 items)
- FoundationA box weighs 200 N and its base area is 0.5 m2. What pressure does it exert on the floor?
- FoundationA 60 kg student stands on both feet covering 0.03 m2 (g = 10 N/kg). What is the pressure?
- FoundationAt a depth of 2 m in water of density 1,000 kg/m3 (g = 10 N/kg), what is the water pressure?
- FoundationA pressure of 5,000 Pa acts on 0.2 m2. What is the force?
- IntermediateWhat is the water pressure at 3 m depth in kPa?
- IntermediateAn object of volume 0.002 m3 is fully immersed in water. What upthrust acts on it?
- IntermediateA metal block weighs 20 N in air and 12 N when immersed in water. Find the upthrust and volume.
- ChallengeA cube of side 0.1 m and density 600 kg/m3 floats in water. Find the upthrust and immersed depth.
- ChallengeA 0.5 m3 block of density 800 kg/m3 floats in water. What volume of water does it displace?
- Error redoA student gets 0.15 Pa for 300 N on 0.02 m2 by dividing area by force. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | P = F / A = 200 / 0.5 = 400 Pa. | Pressure is force divided by area, in pascals. |
| 2 | F = 60 x 10 = 600 N; P = 600 / 0.03 = 20000 Pa. | Mass must become a force by multiplying g before dividing by area. |
| 3 | P = pgh = 1,000 x 10 x 2 = 20000 Pa. | Liquid pressure depends on depth, density and g, not on container shape. |
| 4 | F = P x A = 5,000 x 0.2 = 1000 N. | Pressure times area gives the total force, acting perpendicular to the surface. |
| 5 | P = 1,000 x 10 x 3 = 30000 Pa = 30 kPa. | One kPa is 1,000 Pa, so the answer is 30. |
| 6 | U = pVg = 1,000 x 0.002 x 10 = 20 N. | When fully immersed the upthrust uses the object volume, not its weight. |
| 7 | Upthrust = 20 – 12 = 8 N; V = 8 / (1,000 x 10) = 0.0008 m3 = 800 cm3. | The loss in apparent weight is the upthrust, which gives the volume. |
| 8 | Weight = 600 x 0.001 x 10 = 6 N, so upthrust is 6 N; volume displaced 0.0006 m3, depth 0.05999999999999999 m. | For a floating body upthrust equals weight; divide by base area for depth. |
| 9 | Mass 400 kg, weight 4000 N; displaced volume 4000 / (1,000 x 10) = 0.4 m3. | Displaced volume is weight / (pg) and is less than the total volume. |
| 10 | P = 300 / 0.02 = 15000 Pa. | Pressure is force over area; the inverted ratio is off by orders of magnitude. |
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.
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