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

This guide covers statements P1.3h, P1.3i and P1.3j of OCR GCSE (9-1) Physics A (Gateway Science), as set out in the specification Version 5.0 (July 2026), first assessment 2018. All three statements are marked Higher Tier only. They sit in the pressure section of Topic P1 (Matter). Gas pressure and the atmosphere are covered separately in the OCR GCSE Physics Matter guide and are not repeated here.

Use this page with the revision notes and the practice questions. Course pages: OCR GCSE Physics hub and the printable checklist. To find weak spots first, try the free 10-minute diagnostics.

What this guide covers

Statement	What you must be able to do	Tier
P1.3h	Describe the factors which influence floating and sinking	Higher Tier only
P1.3i	Explain why pressure in a liquid varies with depth and density, and how this leads to an upwards force on a partially submerged object	Higher Tier only
P1.3j	Calculate the differences in pressure at different depths in a liquid. Know that g is the strength of the gravitational field and has a value of 10 N/kg near the Earth’s surface	Higher Tier only

The maths outcome is: pressure due to a column of liquid (Pa) = height of column (m) x density of liquid (kg/m³) x gravitational field strength (N/kg).

Throughout, use g = 10 N/kg unless a question gives a different value.

Why pressure changes with depth

Pressure is the force acting on each unit of area, measured at right angles to the surface:

pressure = force / area        p = F / A

Pressure is measured in pascals (Pa). One pascal is one newton per square metre.

Picture a point under the surface of a liquid. Above it sits a column of liquid. That column has weight, and the weight presses down on the area beneath it. A deeper point has a taller column above it, so more weight on the same area. That is why pressure increases with depth.

A denser liquid packs more mass into the same volume. The same height of column is heavier, so pressure also increases with density.

You can turn this reasoning into the equation. Take a column with base area A and height h in a liquid of density ρ:

volume of column  = A x h
mass of column    = ρ x A x h
weight of column  = ρ x A x h x g
pressure          = weight / A = h x ρ x g

The area cancels. Pressure from the liquid depends only on the depth, the density and g. It does not depend on how wide the container is.

The equation p = h x ρ x g

p = h x ρ x g

Symbol	Quantity	Unit
p	pressure due to the liquid	Pa
h	depth below the surface (height of column)	m
ρ	density of the liquid	kg/m³
g	gravitational field strength (10 N/kg near the Earth’s surface)	N/kg

The pressure at the surface from the liquid is zero, because there is no column above it. The equation gives the pressure caused by the liquid alone. Air above the liquid adds its own pressure on top, so a question must say if it wants that included. If it does not say, give the liquid’s contribution.

Check units before you calculate. Depths in cm or densities in g/cm³ must be converted: 1 g/cm³ = 1000 kg/m³, and 100 cm = 1 m.

Worked example 1: pressure at a depth

A diver is 7.5 m below the surface of a lake. The density of the water is 1000 kg/m³. Calculate the pressure due to the water.

p = h x ρ x g
p = 7.5 x 1000 x 10
p = 75 000 Pa

Worked example 2: rearranging

The pressure due to a liquid at the bottom of a tank is 54 000 Pa. The tank is 6.0 m deep. Calculate the density of the liquid.

ρ = p / (h x g)
ρ = 54 000 / (6.0 x 10)
ρ = 54 000 / 60
ρ = 900 kg/m³

P1.3j: pressure differences between depths

The specification asks you to calculate differences in pressure at different depths. There are two equivalent methods.

Method 1. Work out the pressure at each depth and subtract.

Method 2. Find the difference in depth and use it in the same equation:

Δp = Δh x ρ x g

Method 2 is quicker. It works because the equation is a straight multiplication, so a change in h changes p by the same factor.

Worked example 3: difference between two sensors

Two sensors are fixed in a tank of oil of density 850 kg/m³. One is 4.5 m below the surface and the other is 12 m below the surface. Calculate the difference in pressure.

Method 1:

p at 12 m   = 12 x 850 x 10  = 102 000 Pa
p at 4.5 m  = 4.5 x 850 x 10 =  38 250 Pa
difference  = 102 000 - 38 250 = 63 750 Pa

Method 2:

Δh = 12 - 4.5 = 7.5 m
Δp = 7.5 x 850 x 10 = 63 750 Pa

Both methods give 63 750 Pa. Use one as a check on the other when you have time.

A useful pattern: in water (1000 kg/m³) with g = 10 N/kg, every extra metre of depth adds 10 000 Pa.

P1.3i: why liquid pressure gives an upward force

The force from liquid pressure acts at right angles to any surface it touches. Now lower a block into water. Its bottom face is deeper than its top face, so the pressure on the bottom is greater. The upward force on the bottom (pressure x area) is bigger than the downward force on the top. The sides are at the same depths in pairs and their sideways forces cancel.

The difference is a net force pointing up. It is called upthrust.

For a block with horizontal faces of area A, the faces are a height h apart:

Δp = h x ρ x g                    (bottom face minus top face)
upthrust = Δp x A = h x A x ρ x g

Here h x A is the volume V of the block, so upthrust = V x ρ x g. For a block, this is the weight of the liquid that the block pushes out of the way. You do not need this as a named law for the specification statements, but it is a good check on your working.

Worked example 4: upthrust on a submerged block

A solid block has a square top face of side 0.20 m and a height of 0.50 m. It is held completely under water with its top face 0.30 m below the surface. The density of water is 1000 kg/m³.

depth of bottom face = 0.30 + 0.50 = 0.80 m
pressure on top      = 0.30 x 1000 x 10 = 3 000 Pa
pressure on bottom   = 0.80 x 1000 x 10 = 8 000 Pa
pressure difference  = 5 000 Pa
area of face         = 0.20 x 0.20 = 0.040 m²
upthrust             = 5 000 x 0.040 = 200 N

Check with volume: V = 0.040 x 0.50 = 0.020 m³, and 0.020 x 1000 x 10 = 200 N. The two agree.

If the block has mass 15 kg, its weight is 15 x 10 = 150 N. The upthrust of 200 N is bigger, so the resultant force is 200 - 150 = 50 N upwards. When released, the block accelerates upwards.

Partially submerged objects

When an object floats, only the part below the surface is in the liquid. The top is in air and the bottom is in the liquid. The pressure on the underside is larger than on the top. The upthrust depends on the volume below the surface, because only that part has liquid pressure acting on it. As an object sinks deeper into the liquid, more of it is submerged and the upthrust grows, until the object is fully under or the upthrust equals the weight.

P1.3h: factors that decide floating and sinking

Two forces compare: the weight of the object (down) and the upthrust (up).

Situation	What happens
Weight > largest possible upthrust	The object sinks
Weight = upthrust	The object floats (partly out) or hangs at a steady depth (fully under)
Weight < upthrust when fully under	The object rises until it floats at the surface

The factors you should be able to describe:

1. Density of the object compared with the density of the liquid. An object less dense than the liquid floats. An object denser than the liquid sinks. An object with the same density hangs in the liquid.
2. Density of the liquid. A denser liquid gives more upthrust for the same submerged volume, so objects float higher in it, and some objects that sink in fresh water float in salt water.
3. Volume of liquid pushed aside (the submerged volume). More submerged volume means more upthrust. Spreading the same mass into a larger shape raises the displaced volume.
4. Weight (mass) of the object. A heavier object needs more upthrust to float.

Size alone does not decide the result. A huge steel ship floats and a small steel nail sinks. A tiny cork floats and a large rock sinks. What matters is the balance between weight and upthrust, and that depends on density and shape.

Worked example 5: density and fraction submerged

A plastic block has mass 2.4 kg and volume 0.0030 m³. Will it float in water (1000 kg/m³)? How much of it is under water?

density = mass / volume = 2.4 / 0.0030 = 800 kg/m³

800 kg/m³ is less than 1000 kg/m³, so it floats. For a floating object of uniform density, upthrust equals weight. The submerged fraction is the ratio of densities:

fraction under water = 800 / 1000 = 0.80   (80%)

In oil of density 900 kg/m³ it still floats, but sits lower: 800 / 900 = 0.89 under the surface. In seawater of density 1030 kg/m³: 800 / 1030 = 0.78. In a liquid of density 700 kg/m³ the block is denser than the liquid, so it sinks.

Worked example 6: why a steel boat floats

A steel boat has mass 5000 kg. Its weight is 5000 x 10 = 50 000 N. To float in water it must displace enough water to give an upthrust of 50 000 N:

V = upthrust / (ρ x g) = 50 000 / (1000 x 10) = 5.0 m³

Because the hull is hollow and wide, it can push aside 5.0 m³ of water while still having most of its volume above the waterline. A solid lump of steel of the same mass has a volume of only about 0.64 m³ (using 7800 kg/m³), so it displaces far less water and sinks.

Common errors

• Writing that heavy things sink and light things float. Compare density, or weight against upthrust.
• Saying the pressure at a depth depends on the width or shape of the container. It depends on depth, density and g.
• Mixing up depth and total height of the liquid. Measure h from the surface down to the point.
• Forgetting that the upthrust comes from a pressure difference between the top and bottom of the object.
• Leaving the answer in kPa or MPa without converting, or leaving the unit off completely.
• Using g = 9.8 when the question says 10 N/kg. Use the value you are given.
• Saying the object floats because it “has air in it” without linking this to a larger displaced volume and upthrust.
• Treating a pressure answer as a force. Pressure x area gives force.

Official syllabus

This guide follows statements P1.3h, P1.3i and P1.3j of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, published by OCR. Practical suggestions in the specification for this section include a discussion of the buoyancy of a ping pong ball in water and a demonstration of differences in water pressure using a pressure can with holes.

Next: the revision notes, then the practice questions.