Density and pressure — OCR A-Level Physics
Test yourself on Density and pressure with OCR A-Level practice questions.
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Density and pressure explained
Density measures how much mass is packed into each unit of volume, defined by ρ = m/V, where ρ is density in kg m⁻³, m is mass in kg and V is volume in m³.
Read the full explanation
To use it, identify the mass and the volume occupied by the same body, convert to SI units, then divide. For a regular solid, find V from its dimensions; for an irregular solid, use displacement; for a liquid or gas, use the container volume. A 2.0 kg block of volume 5.0 × 10⁻⁴ m³ has ρ = 2.0 ÷ (5.0 × 10⁻⁴) = 4.0 × 10³ kg m⁻³. Rearranged, m = ρV and V = m/ρ. Density is a property of the material, so it can identify a substance, but it changes with temperature for gases and slightly for liquids.
(b) pressure; p = F/A for solids, liquids and gases
Pressure is the normal force per unit area, p = F/A, where p is in pascals (Pa), F is the force perpendicular to the surface in newtons and A is the area in m². One pascal equals one newton per square metre. The same equation applies to solids, liquids and gases: a solid block exerts pressure on the floor through its contact area, a liquid exerts pressure on a container base, and a gas exerts pressure on the walls of its container through molecular collisions. For a 60 N force spread over 0.030 m², p = 60 ÷ 0.030 = 2.0 × 10³ Pa. A smaller area gives a larger pressure for the same force, which is why sharp edges cut well and wide foundations reduce pressure on soft ground.
(c) p = ρgh; upthrust on an object in a fluid; Archimedes’ principle.
In a fluid at rest, pressure increases with depth according to p = ρgh, where ρ is the fluid density in kg m⁻³, g is the gravitational field strength in N kg⁻¹ and h is the depth below the surface in m. This pressure acts equally in all directions. A submerged or partly submerged object experiences upthrust because the pressure on its lower surface exceeds the pressure on its upper surface. Archimedes’ principle states that the upthrust equals the weight of fluid displaced. For a 0.20 m³ object fully submerged in water of density 1.0 × 10³ kg m⁻³ with g = 9.81 N kg⁻¹, upthrust = ρVg = 1.0 × 10³ × 0.20 × 9.81 = 1.96 × 10³ N. An object floats when upthrust balances its weight.
Your focus
- State and apply the equation ρ = m/V to find density, mass or volume.
- Convert volumes between cm³, litres and m³ correctly before substitution.
- Explain how density can be used to identify a material and why gas density varies with temperature.
Show all 9 objectives
- State and apply p = F/A to calculate pressure, force or area.
- Use the pascal and convert areas between cm², mm² and m².
- Explain how changing the contact area affects the pressure produced by a given force in solids, liquids and gases.
- Apply p = ρgh to calculate pressure at a given depth in a fluid.
- Explain upthrust using Archimedes’ principle and calculate it from the weight of fluid displaced.
- Use upthrust and weight to explain whether an object floats, sinks or remains in equilibrium.
Density and pressure exam tips
Marking Points
- States ρ = m/V and identifies ρ as density, m as mass and V as volume.
- Uses SI units: kg m⁻³ for density, kg for mass and m³ for volume, converting cm³ by multiplying by 10⁻⁶.
- Calculates density by dividing mass by volume, or rearranges to m = ρV or V = m/ρ.
- Explains that density is a property of the material and can be used to identify a substance.
- Recognises that gas density depends strongly on temperature and pressure because volume changes.
- States p = F/A and identifies p as pressure, F as the normal force and A as the area.
- Uses the pascal as the unit of pressure, with 1 Pa = 1 N m⁻².
- Applies p = F/A to solids, liquids and gases, using the force perpendicular to the relevant surface.
- Calculates pressure, force or area by rearranging the equation correctly.
- Explains that for a fixed force a smaller area produces a larger pressure, and vice versa.
- States p = ρgh and uses it to find pressure at a depth in a fluid of uniform density.
- Identifies upthrust as the resultant upward force on an object in a fluid caused by the pressure difference between its lower and upper surfaces.
- States Archimedes’ principle: the upthrust equals the weight of fluid displaced by the object.
- Calculates upthrust using weight of displaced fluid, ρVg, where V is the displaced volume.
- Explains floating and sinking in terms of the balance between upthrust and the object’s weight.
Examiner Tips
- 💡Write the equation, substitute values with units, then give the answer with the correct unit and a sensible number of significant figures.
- 💡Convert every length to metres first when a volume is given in cm³ or litres; 1 litre = 1 × 10⁻³ m³.
- 💡Check the rearrangement by substituting a simple case, such as doubling the volume while keeping mass fixed, which should halve the density.
- 💡Identify the surface first, then find the force perpendicular to it before dividing by the contact area.
- 💡Convert areas carefully: 1 cm² = 1 × 10⁻⁴ m² and 1 mm² = 1 × 10⁻⁶ m².
- 💡Give pressure in pascals and check that the answer is reasonable, since everyday pressures are often thousands of pascals.
- 💡Sketch the object in the fluid and mark the depths of its top and bottom surfaces to see why a pressure difference produces upthrust.
- 💡For floating objects, equate upthrust to weight and use the submerged volume, not the total volume, in ρVg.
- 💡Keep g as 9.81 N kg⁻¹ unless the question states otherwise, and check that pressure answers are in pascals.
Common Mistakes
- Using volume in cm³ without converting: the error is mixing units, and the correction is to convert cm³ to m³ by multiplying by 10⁻⁶ before dividing.
- Dividing volume by mass: the error is inverting the formula, and the correction is to divide mass by volume, ρ = m/V.
- Treating density as fixed for a gas: the error is ignoring compression, and the correction is to note that gas density rises when the same mass is compressed into a smaller volume.
- Using the total force rather than the component perpendicular to the surface: the error is including a parallel component, and the correction is to resolve the force and use only the normal component in p = F/A.
- Forgetting to convert area from cm² to m²: the error is mixing units, and the correction is to multiply cm² by 10⁻⁴ to obtain m².
- Assuming p = F/A applies only to solids: the error is restricting the equation, and the correction is to recognise that it also describes pressure in liquids and gases.
- Using the total depth to the bottom of the container instead of the depth below the fluid surface: the error is measuring from the wrong reference, and the correction is to measure h from the free surface of the fluid.
- Confusing the object’s volume with the displaced volume for a floating object: the error is using the whole volume, and the correction is to use only the submerged volume, which equals the volume of fluid displaced.
- Treating upthrust as a fixed property of the object: the error is ignoring the fluid, and the correction is to recognise that upthrust depends on the density of the fluid and the displaced volume.