This set covers the skills in pressure in solids and fluids: p = F ÷ A, p = ρgh, explanations with depth, manometers and converting mass to weight. All questions are original and all data are invented for practice.
Attempt each question on paper first, with the equation, the substitution and the unit. Use g = 10 N/kg and atmospheric pressure 1.0 × 10⁵ Pa unless told otherwise. Questions run from easier to harder.
Questions
1. A force of 360 N acts on an area of 0.12 m². Find the pressure.
Show answer
p = F ÷ A = 360 ÷ 0.12 = 3000 Pa
2. A force of 50 N acts on an area of 25 cm². Find the pressure in Pa.
Show answer
Convert: 25 cm² = 25 ÷ 10 000 = 0.0025 m². p = 50 ÷ 0.0025 = 20 000 Pa
3. A crate of mass 80 kg rests on a base of area 0.25 m². Find the pressure under the crate.
Show answer
Weight: W = mg = 80 × 10 = 800 N. p = 800 ÷ 0.25 = 3200 Pa. Using 80 in place of 800 would give 320, which is the mass-for-force mistake.
4. The fluid in a hydraulic piston is at a pressure of 2.4 × 10⁵ Pa. The piston has an area of 0.0050 m². What force does the fluid exert on the piston?
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F = pA = 240 000 × 0.0050 = 1200 N
5. Find the pressure caused by fresh water (1000 kg/m³) at a depth of 3.5 m.
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p = ρgh = 1000 × 10 × 3.5 = 35 000 Pa
6. Oil of density 850 kg/m³ causes a pressure of 6800 Pa at the bottom of a tank. How deep is the oil?
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h = p ÷ (ρg) = 6800 ÷ (850 × 10) = 6800 ÷ 8500 = 0.80 m
7. A dam wall is built much thicker near the base than near the water surface. Explain why.
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The pressure caused by the water increases with depth, because there is a greater height of water above a deeper point and its weight presses on each square metre. The pressure, and so the force on the wall, is greatest at the bottom. A thicker base is needed to withstand it.
8. The water in a tank is 2.0 m deep. A hatch of area 0.30 m² is fixed in the floor. Find the pressure from the water and the force on the hatch.
Show answer
p = ρgh = 1000 × 10 × 2.0 = 20 000 Pa. F = pA = 20 000 × 0.30 = 6000 N. The pressure is 20 000 Pa.
9. A diver is 12 m below the surface of the sea. Sea water has density 1030 kg/m³. Find the total pressure on the diver, including the atmosphere.
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Pressure from the water: 1030 × 10 × 12 = 123 600 Pa. Add the atmosphere: 123 600 + 100 000 = 223 600 Pa. To 2 significant figures, 2.2 × 10⁵ Pa.
10. A mercury manometer (13 600 kg/m³) is joined to a gas supply. The mercury surface in the open arm is 30 mm lower than in the gas arm. Find the gas pressure.
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Convert: 30 mm = 0.030 m. ρgh = 13 600 × 10 × 0.030 = 4080 Pa. The open arm is lower, so the gas pressure is below atmospheric: 100 000 − 4080 = 95 920 Pa.
11. A person of mass 50 kg stands on both feet (total area 0.030 m²), then lifts one foot so the contact area is 0.015 m². Find the pressure in each case and compare them.
Show answer
Weight: 50 × 10 = 500 N. Two feet: 500 ÷ 0.030 = 16 667 Pa, or 1.7 × 10⁴ Pa. One foot: 500 ÷ 0.015 = 33 333 Pa, or 3.3 × 10⁴ Pa. The pressure doubles, because the area halves and the weight is the same.
If you got these wrong
| What went wrong | Lesson to revisit |
|---|---|
| Wrong answer with pressure = force ÷ area, or the unit of area | Relate pressure to force and area |
| Wrong p = ρgh substitution, or the wrong depth | Calculate liquid pressure under stated conditions |
| Lost marks on the written reason in question 7 | Explain a pressure difference with depth |
| Added when you should subtract in question 10 | Interpret a manometer-style diagram |
| Used kilograms in place of newtons in questions 3 or 11 | Avoid using mass where a force is required |
After marking, write one line for each error in the mistake log and retest queue. Attempt a similar question a few days later to see whether the fix has held.
Working through a set like this alone shows you the wrong answers, but not always the reason. A teacher who reads your working line by line, as in online one-to-one Physics tuition, can find that reason quickly.