OCR GCSE Physics A: Liquid pressure, upthrust, floating and sinking (J249) – Revision Notes

These notes condense statements P1.3h, P1.3i and P1.3j of OCR GCSE (9-1) Physics A (Gateway Science), specification Version 5.0 (July 2026). All three are Higher Tier only. For full explanations and longer worked examples, read the study guide. Then test yourself with the practice questions.

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Key ideas at a glance

Statement	Tier	One-line summary
P1.3h	Higher Tier only	Floating or sinking depends on weight against upthrust, which depends on density and displaced volume
P1.3i	Higher Tier only	Deeper liquid is under more weight above it, so pressure rises with depth and density; the pressure difference gives upthrust
P1.3j	Higher Tier only	Use p = h x ρ x g and Δp = Δh x ρ x g with g = 10 N/kg

Definitions

• Pressure: force per unit area, acting at right angles to the surface. p = F / A. Unit: pascal (Pa), which is N/m².
• Gravitational field strength, g: the force per kilogram on a mass in a gravitational field. Near the Earth’s surface it is 10 N/kg.
• Density, ρ: mass per unit volume. ρ = m / V. Unit: kg/m³.
• Upthrust: the upward resultant force from a liquid on an object in it, caused by greater pressure on the lower surface than the upper surface.
• Submerged: fully under the liquid. Partially submerged: floating with part above the surface.

Equations

Equation	Use	Rearrangements
p = h x ρ x g	Pressure due to a column of liquid	h = p / (ρ x g), ρ = p / (h x g)
Δp = Δh x ρ x g	Pressure difference between two depths	Δh = Δp / (ρ x g)
p = F / A	Pressure, force and area	F = p x A, A = F / p
ρ = m / V	Density	m = ρ x V, V = m / ρ
W = m x g	Weight	m = W / g

Why pressure rises with depth

1. A column of liquid sits above any point.
2. The column has weight (mass x g).
3. More depth means more liquid, so more weight on the same area.
4. A denser liquid has more mass in the same volume, so more weight.
5. So pressure rises with both depth and density.

Algebra check: p = weight / area = (ρ x A x h x g) / A = h x ρ x g. The base area cancels, so the width of the container does not matter.

Why upthrust happens

1. The bottom of an immersed object is deeper than the top.
2. Pressure at the bottom is larger.
3. The upward force on the bottom is bigger than the downward force on the top.
4. The sideways forces cancel in pairs.
5. The resultant is an upward force: upthrust.

For a block with horizontal faces, upthrust = (pressure difference) x (face area).

Floating and sinking: decision table

Comparison	Result
Object’s density < liquid’s density	Floats, partly out of the liquid
Object’s density = liquid’s density	Hangs at any depth when fully under
Object’s density > liquid’s density	Sinks
Weight > upthrust at full submersion	Sinks
Weight = upthrust	Steady: floats or hangs

Factors: density of the object, density of the liquid, volume of liquid displaced (shape), and weight of the object.

For a floating object of uniform density: fraction under the surface = density of object / density of liquid.

Method in steps: pressure difference

1. Write down the two depths (or the depth difference).
2. Subtract to get Δh in metres.
3. Convert the density to kg/m³ if needed.
4. Substitute into Δp = Δh x ρ x g with g = 10 N/kg.
5. Give the unit (Pa) and round sensibly.

Method in steps: upthrust on a submerged block

1. Find the depth of the top face and bottom face.
2. Calculate p = h x ρ x g for each.
3. Subtract: Δp = p(bottom) - p(top).
4. Multiply by the face area: upthrust = Δp x A.
5. Compare with weight (m x g): the larger force decides the motion.

Small worked reminders

Pressure at depth. A sensor in a lake is 35 m down (water 1000 kg/m³): 35 x 1000 x 10 = 350 000 Pa.

Seawater. At 60 m in seawater of density 1030 kg/m³: 60 x 1030 x 10 = 618 000 Pa.

Difference. Between 3 m and 11 m in water: Δh = 8 m, so Δp = 8 x 1000 x 10 = 80 000 Pa.

Finding depth. A pressure of 96 000 Pa in oil of density 800 kg/m³: h = 96 000 / (800 x 10) = 12 m.

Floating fraction. A wooden block of density 600 kg/m³ in water: 600 / 1000 = 0.60, so 60% of it is under the surface.

Depth of a floating block. A flat-bottomed block of mass 1.5 kg has a base 0.30 m by 0.25 m. It floats in water. Weight = 1.5 x 10 = 15 N. Base area = 0.075 m². Upthrust = weight, so pressure on the base = 15 / 0.075 = 200 Pa. Depth of the base = 200 / (1000 x 10) = 0.020 m.

Model explanations to learn

Pressure and depth (P1.3i). Write three linked points: the liquid above a point has weight; a greater depth means a greater weight of liquid pressing on each unit of area; so the pressure is greater. For density, say that a denser liquid has more mass in each cubic metre, so the same column is heavier.

Upthrust (P1.3i). Write: the bottom of the object is deeper than the top; the pressure is greater at the bottom; this gives a bigger upward force on the bottom than the downward force on the top; the resultant force is upwards and is the upthrust.

Floating (P1.3h). Write: an object floats when the upthrust equals its weight. For an object that is fully under, compare its density with the liquid’s density.

Must-know distinctions

• Pressure vs force. Pressure is spread over area. Force = pressure x area.
• Depth vs height of container. h is the distance below the surface to the point.
• Density vs weight. A big heavy object can float if its average density is lower than the liquid’s. A small light object can sink if its density is higher.
• Weight vs upthrust. Weight acts down and is fixed by mass. Upthrust acts up and depends on the liquid and the volume under the surface.
• Liquid pressure vs atmospheric pressure. The equation here gives the pressure caused by the liquid. Atmospheric pressure is a separate topic.

Quick self-test

1. State the equation linking pressure due to a liquid, depth, density and g.
2. Calculate the pressure due to water 9.0 m below the surface (1000 kg/m³).
3. Calculate the difference in pressure between depths of 2.0 m and 7.0 m in water.
4. A liquid causes a pressure of 40 500 Pa at a depth of 4.5 m. Find its density.
5. Why does pressure increase with depth in a liquid?
6. Why is there an upthrust on a block placed in water?
7. An object has density 1200 kg/m³. State whether it floats in water and give a reason.
8. A block of density 750 kg/m³ floats in water. What fraction is under the surface?
9. Why does a steel ship float but a steel bolt sink?
10. Pressure difference across a submerged block is 3 000 Pa and each horizontal face has area 0.050 m². Find the upthrust.

Answers

1. p = h x ρ x g.
2. 9.0 x 1000 x 10 = 90 000 Pa.
3. Δh = 5.0 m, so Δp = 5.0 x 1000 x 10 = 50 000 Pa.
4. ρ = 40 500 / (4.5 x 10) = 900 kg/m³.
5. A deeper point has a taller column of liquid above it, so more weight acts on each unit of area.
6. The bottom face is deeper than the top face, so the pressure and the upward force on it are greater. The resultant force is upward.
7. It sinks, because its density is greater than that of water (1000 kg/m³).
8. 750 / 1000 = 0.75, so three quarters.
9. The ship’s hollow hull displaces a large volume of water, giving upthrust equal to its weight. The bolt displaces little water, so the upthrust is less than its weight.
10. 3 000 x 0.050 = 150 N upwards.

Where marks are usually lost

• Substituting the density in g/cm³ without converting to kg/m³.
• Using the total depth of the tank when the question asks about a point partway down.
• Subtracting the pressures but then also multiplying the answer by something extra.
• Giving the answer in pascals when the question asks for kPa, or the reverse.
• Explaining floating with “light things float” instead of comparing densities or weight and upthrust.
• Saying upthrust is caused by the water “pushing up” with no mention of a pressure difference between top and bottom.
• Forgetting that a floating object has upthrust equal to its weight, not larger.
• Rounding too early, so the final digits drift.

Official syllabus

Source: OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, published by OCR. Statements P1.3h to P1.3j.