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S4 Physics pressure and fluids paper 4: ten common slips
S4 Physics pressure and fluids paper 4: ten common slips
All ten items are rewritten from the most common slips; redo each one before checking.
Direct answers Items 01 to 03 retry the pressure formula and units, 04 to 06 hydraulics, upthrust and density, 07 to 09 deep water and floating ratios, and 10 the cm2 conversion.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Physics Pressure and Fluids |
| Level | S4 |
02 Questions (10 items)
- FoundationA force of 60 N acts perpendicular to an area of 0.03 m2. Find the pressure.
- FoundationA force of 200 N acts on 50 cm2. Find the pressure.
- FoundationWhat is the water pressure at a depth of 3 m? (rho = 1000 kg/m3, g = 10 N/kg)
- IntermediateA hydraulic press has a small piston of 0.002 m2 with 100 N applied and a large piston of 0.05 m2. Find the force lifted.
- IntermediateA stone displaces 0.002 m3 of water. Find the upthrust, and its weight if it floats.
- IntermediateA metal block weighs 8 N in air and 6 N when fully in water. Find the upthrust, its volume and its density.
- ChallengeA diver is 20 m deep where the water pressure is 200000 Pa. The mask area is 0.02 m2. Find the force on it.
- ChallengeAn iceberg of volume 0.01 m3 floats in sea water. Ice is 900 kg/m3 and sea water 1025 kg/m3. Find the exposed volume.
- ChallengeA cube of side 0.2 m with density 600 kg/m3 floats in water. Find the submerged depth.
- Error redoA force of 400 N acts on 200 cm2. A student answers 2 Pa. What is the correct pressure?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2000 Pa | P = F / A = 60 / 0.03 = 2000 Pa; the area must be in m2. |
| 2 | 40000 Pa | 50 cm2 = 0.005 m2, so P = 200 / 0.005 = 40000 Pa; this conversion is the usual slip. |
| 3 | 30000 Pa | P = rho g h = 1000 x 10 x 3 = 30000 Pa, with h measured from the surface. |
| 4 | 2500 N | Pressure is equal throughout, so F2 = F1 x A2 / A1 = 100 x 0.05 / 0.002 = 2500 N. |
| 5 | 20 N; 20 N | Upthrust = 1000 x 10 x 0.002 = 20 N; a floating body has upthrust equal to its weight. |
| 6 | 2 N; 0.0002 m3; 4000 kg/m3 | Upthrust = 8 – 6 = 2 N, V = 2 / 10000 = 0.0002 m3, mass 0.8 kg, so density is 4000 kg/m3. |
| 7 | 4000 N | F = PA = 200000 x 0.02 = 4000 N; the water pressure alone is enough here. |
| 8 | 0.0012 m3 (about 12%) | Upthrust equals weight, so displaced volume = 0.01 x 900 / 1025 = 0.00878 m3, leaving 0.0012 m3 above water. |
| 9 | 0.12 m | The submerged fraction equals the density ratio 600 / 1000 = 0.6, so depth = 0.2 x 0.6 = 0.12 m. |
| 10 | 20000 Pa | 200 cm2 = 0.02 m2, so P = 400 / 0.02 = 20000 Pa; treating cm2 as m2 is out by a factor of ten thousand. |
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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