Density, Upthrust and Pressure

This lesson covers AQA Spec 3.4.2.1 (Density) and related concepts including Archimedes' principle, atmospheric pressure, pressure in fluids, and hydraulic systems.

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1 Density

Key Definition: Density is the mass per unit volume of a substance.

ρ = m / V

SI unit: kg m⁻³

Material	Density (kg m⁻³)
Air (at STP)	1.2
Water	1000
Ice	917
Aluminium	2700
Iron/Steel	7800
Copper	8900
Lead	11 340
Gold	19 300
Mercury	13 600

Worked Example 1 — Finding Density

A block of metal has dimensions 5.0 cm × 4.0 cm × 3.0 cm and a mass of 0.47 kg. Find its density and identify the metal.

Solution:

V = 0.050 × 0.040 × 0.030 = 6.0 × 10⁻⁵ m³

ρ = m/V = 0.47 / (6.0 × 10⁻⁵) = 7833 kg m⁻³ ≈ 7800 kg m⁻³

This is closest to iron/steel.

Worked Example 2 — Mass of Air in a Room

A room measures 5.0 m × 4.0 m × 3.0 m. The density of air is 1.2 kg m⁻³. Find the mass of air in the room.

Solution:

V = 5.0 × 4.0 × 3.0 = 60 m³

m = ρV = 1.2 × 60 = 72 kg

Exam Tip: Always convert volumes to m³ before using the density formula. 1 cm³ = 1 × 10⁻⁶ m³. 1 litre = 1 × 10⁻³ m³.

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2 Pressure

Key Definition: Pressure is the force per unit area acting perpendicular to a surface.

p = F / A

SI unit: pascal (Pa). 1 Pa = 1 N m⁻².

Worked Example 3

A person of mass 70 kg stands on the ground. Each shoe has a contact area of 200 cm². Find the pressure on the ground.

Solution:

Total area = 2 × 200 cm² = 400 cm² = 400 × 10⁻⁴ m² = 0.040 m²

Weight = mg = 70 × 9.81 = 686.7 N

p = F/A = 686.7 / 0.040 = 17 200 Pa ≈ 17.2 kPa

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3 Pressure in a Fluid Column

The pressure at depth h in a fluid of density ρ:

p = ρgh

This is the pressure due to the fluid column only (gauge pressure). The absolute pressure at depth h is:

p_absolute = p_atm + ρgh

where p_atm ≈ 101 325 Pa ≈ 101 kPa.

Derivation

Consider a column of fluid of height h, cross-sectional area A, and density ρ.

Weight of fluid column: W = mg = ρVg = ρAhg

Pressure at the base: p = F/A = W/A = ρAhg/A = ρgh

Worked Example 4 — Pressure at Depth

Find the total pressure at a depth of 25 m in a lake. Density of water = 1000 kg m⁻³, atmospheric pressure = 101 kPa.

Solution:

Pressure due to water: p_water = ρgh = 1000 × 9.81 × 25 = 245 250 Pa = 245.3 kPa

Total pressure = p_atm + p_water = 101 + 245.3 = 346 kPa

This is about 3.4 atmospheres.

Worked Example 5 — Mercury Barometer

A mercury barometer reads a column height of 760 mm. Density of mercury = 13 600 kg m⁻³. Calculate atmospheric pressure.

Solution:

p_atm = ρgh = 13 600 × 9.81 × 0.760 = 101 400 Pa ≈ 101 kPa ✓

Exam Tip: The pressure at a given depth depends only on the depth, the fluid density, and g — it does NOT depend on the shape of the container. This is sometimes called the "hydrostatic paradox."

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4 Archimedes' Principle

Archimedes' Principle: When an object is wholly or partially immersed in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced.

Upthrust (buoyancy force) = ρ_fluid × V_displaced × g

where V_displaced is the volume of the object that is submerged.

Floating Condition

An object floats when the upthrust equals its weight:

ρ_fluid × V_displaced × g = m_object × g

For a fully submerged object: V_displaced = V_object

If ρ_object < ρ_fluid, the object floats (only partially submerged). If ρ_object > ρ_fluid, the object sinks. If ρ_object = ρ_fluid, the object is neutrally buoyant.

Worked Example 6 — Floating Object