IGCSE Physics Pressure Questions and Worked Solutions (1.8) | 0625

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IGCSE Physics 0625 · Topic 1.8

IGCSE Physics Pressure Questions and Worked Solutions

50 topic-wise practice questions on Pressure (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions25 Core15 Extended10 Challenge134 marksTextbook §1.8 · pp. 78–84
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Pressure

Pressure is force per unit area. A small area gives a large pressure (knives, pins) and a large area gives a small pressure (snowshoes, tractor tyres). In liquids, pressure increases with depth and density.

Key formulas

p = F ÷ Apressure p (Pa = N/m²), force F (N), area A (m²)Core
Δp = ρ g Δhchange in liquid pressure, density ρ (kg/m³), g (N/kg), depth Δh (m)Extended
total pressure = atmospheric pressure + ρ g hat depth h in a liquid open to the airExtended

Prerequisite knowledge

  • Weight W = mg (1.3)
  • Density (1.4)
  • Area and unit conversion (cm² to m²: ÷ 10 000)

Exam tips & common mistakes

  • Convert cm² to m² by dividing by 10 000
  • Use total contact area (e.g. four tyres, two feet)
  • Use the vertical depth for liquid pressure, not the length of a sloping pipe

Key facts from the syllabus (1.8)

  • Pressure is force per unit area, measured in pascals (Pa) or N/m²
  • Increasing area reduces pressure; reducing area increases pressure
  • Pressure in a liquid increases with depth and with the density of the liquid
  • Pressure in a liquid acts in all directions and does not depend on the shape of the container
  • Examples: dams thicker at the base, divers' ears hurting at depth

Textbook: §1.8 Pressure, pp. 78–84

1
Core[1 mark]

Define pressure.

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Method: p = F/A.

Force per unit area
2
Core[2 marks]

A box of weight 600 N rests on an area of 0.20 m². Calculate the pressure.

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Method: p = F ÷ A.

p = 600 ÷ 0.20 = 3000 N/m² (Pa)
3
Core[2 marks]

A force of 50 N acts on 2.5 cm². Calculate the pressure in N/cm².

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Method: p = F ÷ A.

p = 50 ÷ 2.5 = 20.0 N/cm²
4
Core[2 marks]

A pressure of 2.0 × 10⁵ Pa acts on 0.030 m². Calculate the force.

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Method: F = pA.

F = 2.0×10⁵ × 0.030 = 6000 N
5
Core[2 marks]

Explain why a drawing pin has a sharp point.

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Method: Small area → high pressure.

Small area at the tip gives a high pressure for the same force, so it pierces easily
6
Core[2 marks]

Explain why snowshoes stop a person sinking into snow.

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Method: Large area → low pressure.

They spread the weight over a large area, reducing pressure on the snow
7
Core[2 marks]

Explain why tractors have wide tyres.

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Method: Lower pressure on soft ground.

Large area reduces the pressure on the ground, so they do not sink
8
Core[2 marks]

State how pressure in a liquid changes with depth and with the density of the liquid.

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Method: Increases with both.

Pressure increases with depth and increases with density
9
Core[2 marks]

Explain why a dam is thicker at the bottom than at the top.

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Method: Water pressure is greater at depth.

Pressure increases with depth, so the base must withstand greater pressure
10
Core[3 marks]

A 70 kg person (g = 9.8 N/kg) stands on two feet of total area 0.035 m². Calculate the pressure on the floor.

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Method: p = mg ÷ A.

p = (70 × 9.8) ÷ 0.035 = 19600 Pa
11
Core[2 marks]

The person in the previous question stands on one foot. State and explain the change in pressure.

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Method: Area halves.

Pressure doubles because the same force acts on half the area
12
Core[3 marks]

A brick of weight 24 N measures 0.20 m × 0.10 m × 0.060 m. Calculate the maximum pressure it can exert on the ground.

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Method: Smallest face = 0.10 × 0.060.

A = 0.00600 m²; p = 24 ÷ 0.006 = 4000 Pa
13
Core[2 marks]

Explain why knives are sharpened.

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Method: Smaller area, higher pressure.

Sharpening reduces the edge area, increasing pressure so it cuts more easily
14
Core[2 marks]

A force of 120 N acts on an area of 0.040 m². Calculate the pressure.

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Method: p = F/A.

p = 3000 Pa
15
Core[2 marks]

A pressure of 250 kPa acts on an area of 0.020 m². Calculate the force.

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Method: F = pA.

F = 250 000 × 0.020 = 5000 N
16
Core[2 marks]

A force of 600 N produces a pressure of 15 000 Pa. Calculate the area.

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Method: A = F/p.

A = 600 ÷ 15000 = 0.0400 m²
17
Core[2 marks]

State the unit of pressure and its equivalent in N and m.

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Method: Pa.

pascal (Pa); 1 Pa = 1 N/m²
18
Core[2 marks]

Explain why camels have large flat feet.

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Method: Low pressure on sand.

Large area reduces pressure, so they do not sink into sand
19
Core[2 marks]

Explain why high-heeled shoes can damage some floors.

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Method: Small area.

The small heel area gives a very high pressure on the floor
20
Core[2 marks]

Explain why lorries carrying heavy loads have many wheels.

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Method: Spread weight.

More wheels give a larger contact area, reducing the pressure on the road
21
Core[2 marks]

State why water comes out faster from a hole near the bottom of a tall container.

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Method: Pressure greater at depth.

The pressure is greater at greater depth
22
Core[3 marks]

An elephant of weight 60 000 N stands on four feet each of area 0.10 m². Calculate the pressure on the ground.

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Method: Total area 0.40 m².

p = 60000 ÷ 0.40 = 150000 Pa
23
Core[3 marks]

A 50 kg woman stands on one stiletto heel of area 1.0 cm² (g = 9.8 N/kg). Calculate the pressure in N/cm² and Pa.

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Method: p = F/A.

p = 490 N/cm² = 4.9 × 10⁶ Pa
24
Core[2 marks]

Explain why a deep-sea submarine must have very strong walls.

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Method: Huge pressure at depth.

Water pressure increases with depth, so at great depth the forces on the walls are very large
25
Core[2 marks]

State whether pressure in a liquid acts only downwards.

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Method: All directions.

No, it acts in all directions
26
Extended[2 marks]

Calculate the pressure due to 12 m of water (density 1000 kg/m³, g = 9.8 N/kg).

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Method: Δp = ρgΔh.

Δp = 1000 × 9.8 × 12 = 118000 Pa
27
Extended[3 marks]

A diver descends from 5.0 m to 25 m in sea water (density 1030 kg/m³). Calculate the increase in pressure.

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Method: Δh = 20 m.

Δp = 1030 × 9.8 × 20 = 202000 Pa
28
Extended[3 marks]

The pressure due to a liquid at a depth of 0.40 m is 5300 Pa. Calculate the density of the liquid.

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Method: ρ = Δp ÷ (gΔh).

ρ = 5300 ÷ (9.8 × 0.40) = 1350 kg/m³
29
Extended[3 marks]

Calculate the depth of oil (density 800 kg/m³) that produces a pressure of 3920 Pa.

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Method: Δh = Δp ÷ (ρg).

Δh = 3920 ÷ (800 × 9.8) = 0.500 m
30
Extended[3 marks]

Atmospheric pressure is 1.0 × 10⁵ Pa. Calculate the total pressure on a diver 10 m below the surface of fresh water.

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Method: Total = atmospheric + ρgh.

p = 1.0×10⁵ + 1000 × 9.8 × 10 = 198000 Pa
31
Extended[3 marks]

A mercury column (density 13 600 kg/m³) is 0.76 m high. Calculate the pressure it exerts at its base.

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Method: p = ρgh.

p = 13600 × 9.8 × 0.76 = 101000 Pa
32
Extended[2 marks]

Two tanks contain water to the same depth; one is wider. Compare the pressures at the bottom.

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Method: Depends on depth and density only.

Equal, because pressure depends only on depth and density, not width
33
Extended[2 marks]

Calculate the pressure due to 5.0 m of sea water (density 1030 kg/m³, g = 9.8 N/kg).

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Method: ρgh.

Δp = 1030 × 9.8 × 5.0 = 50500 Pa
34
Extended[3 marks]

Calculate the depth of fresh water at which the water pressure equals atmospheric pressure (1.0 × 10⁵ Pa).

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Method: h = p/(ρg).

h = 1.0×10⁵ ÷ (1000 × 9.8) = 10.2 m
35
Extended[3 marks]

A water tank on a roof is 15 m above a tap. Calculate the pressure due to the water at the tap.

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Method: ρgh.

Δp = 1000 × 9.8 × 15 = 147000 Pa
36
Extended[3 marks]

A manometer contains water. The difference in levels is 12 cm. Calculate the pressure difference.

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Method: ρgh.

Δp = 1000 × 9.8 × 0.12 = 1180 Pa
37
Extended[3 marks]

A dam holds water 40 m deep. Calculate the water pressure at the base and the force on a 2.0 m² panel there.

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Method: ρgh then F = pA.

p = 392000 Pa; F = 784000 N
38
Extended[3 marks]

A liquid produces a pressure of 9408 Pa at a depth of 1.2 m. Calculate its density.

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Method: ρ = p/(gh).

ρ = 9408 ÷ (9.8 × 1.2) = 800 kg/m³
39
Extended[3 marks]

A swimmer descends from 2.0 m to 6.0 m in a pool. Calculate the change in pressure.

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Method: Δh = 4.0 m.

Δp = 1000 × 9.8 × 4.0 = 39200 Pa
40
Extended[2 marks]

Explain why a diver's ears hurt as they swim deeper.

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Method: Pressure increases with depth.

Water pressure increases with depth, pushing on the eardrum

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A 1200 kg car rests on four tyres, each with a contact area of 0.015 m² (g = 9.8 N/kg). Calculate the pressure on the road. Explain the effect of under-inflating the tyres.

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Method: p = F/A.

p = 11 760 ÷ 0.060 = 196000 Pa; under-inflated tyres flatten, increasing area and reducing pressure (but increasing friction and wear)
42
ChallengeCore[4 marks]

A brick 20 cm × 10 cm × 5.0 cm weighs 22 N. Calculate the maximum and minimum pressures it can exert on the ground.

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Method: Smallest and largest faces.

Max: 22 ÷ 0.0050 = 4400 Pa; min: 22 ÷ 0.020 = 1100 Pa
43
ChallengeCore[4 marks]

A drawing pin is pushed with 20 N. The head has area 1.0 cm² and the point 0.010 mm². Calculate the pressure on the thumb and at the point in N/m².

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Method: Convert areas.

Thumb: 20 ÷ 1.0×10⁻⁴ = 2.0 × 10⁵ Pa; point: 20 ÷ 1.0×10⁻⁸ = 2.0 × 10⁹ Pa
44
ChallengeCore[3 marks]

Explain why a person lying flat on thin ice is less likely to fall through than one standing.

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Method: Larger area.

The same weight is spread over a larger area, so the pressure on the ice is lower
45
ChallengeExtended[5 marks]

A diver is 15 m below the surface of the sea (density 1030 kg/m³). Atmospheric pressure is 1.0 × 10⁵ Pa. Calculate (a) the water pressure, (b) the total pressure, (c) the force on a mask of area 0.020 m².

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Method: Δp = ρgh.

(a) 151000 Pa (b) 251000 Pa (c) 5030 N
46
ChallengeExtended[4 marks]

A tank contains 0.50 m of oil (density 800 kg/m³) floating on 1.2 m of water. Calculate the pressure due to the liquids at the bottom of the tank.

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Method: Add both layers.

p = 800 × 9.8 × 0.50 + 1000 × 9.8 × 1.2 = 15700 Pa
47
ChallengeExtended[4 marks]

A submarine window has an area of 0.30 m². The submarine is 200 m deep in sea water (1030 kg/m³). Calculate the force on the window due to the water.

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Method: F = ρghA.

F = 1030 × 9.8 × 200 × 0.30 = 6.06 × 10⁵ N
48
ChallengeExtended[4 marks]

In a water manometer connected to a gas supply, the levels differ by 25 cm. Atmospheric pressure is 101 kPa. Calculate the gas pressure.

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Method: Gas = atmospheric + ρgh.

p = 101 000 + 1000 × 9.8 × 0.25 = 103000 Pa
49
ChallengeExtended[4 marks]

A water tower must provide a pressure of 2.0 × 10⁵ Pa (above atmospheric) at taps. Calculate the minimum height of water above the taps.

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Method: h = p/(ρg).

h = 2.0×10⁵ ÷ 9800 = 20.4 m
50
ChallengeExtended[4 marks]

Explain why a dam must be designed for the depth of water behind it, not the length of the lake, and calculate the pressure at the base of a 60 m deep reservoir.

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Method: p depends on depth only.

Pressure depends only on depth and density; p = 1000 × 9.8 × 60 = 588000 Pa
All IGCSE Physics topics← 1.7.4 Power2.1.1 States of Matter →

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