Upthrust | CAIE 9702 Paper 1 Notes · Conquer Your A Levels

Paper 1

Physics9702

Upthrust

Anything in a fluid gets an upward push from it. That push is the upthrust from lesson 3.05's force list.

This lesson explains where it comes from and how to calculate it. Both are asked directly.

Where upthrust comes from

Pressure grows with depth ( $\rho g h$ , last lesson). The bottom of a submerged object is deeper than its top.
So the fluid pushes up on the bottom face harder than it pushes down on the top face. The sideways pushes are at equal depths and cancel.
The leftover upward force is the .

The bottom face is deeper, so the water pushes up on it harder than it pushes down on the top. The sideways pushes cancel. The leftover upward force is the upthrust.

The two-mark explanation, in mark-scheme words: pressure is greater at the bottom than the top because the bottom is deeper, so the force on the bottom is greater than the force on the top. The same two lines were the answer in both October 2025 papers (9702/22/O/N/25 Q1(b)(ii)).

The formula

For a cuboid with horizontal face area $A$ and height $\Delta h$ , the bottom pressure exceeds the top pressure by $\rho g\Delta h$ . Multiplying this pressure difference by the face area gives the resultant upward force:

F = (\rho g\Delta h)A = \rho gV

Archimedes' principle extends this result to any shape: the upthrust equals the weight of the displaced fluid.

F = \rho g V

Symbols

$F$ = upthrust (N)
$\rho$ = density of the fluid (not the object) (kg m⁻³)
$g$ = gravitational field strength (N kg⁻¹)
$V$ = volume of fluid pushed aside (the submerged volume) (m³)

Two inputs cause most lost marks. $\rho$ is the fluid's density. $V$ is only the volume below the surface.

Worked example

Smallest case: a submerged block

A block pushes aside 2.0 × 10⁻³ m³ of water. Find the upthrust on it.

$F = \rho g V = 1000 \times 9.81 \times 2.0 \times 10^{-3} = 20 \text{ N}$ .

Answer

F \approx 20 \text{ N}

Worked example

Exam version: a weather balloon

A weather balloon is a sphere of radius 0.90 m in air of density 1.1 kg m⁻³. Its total weight is 19 N. Find the upthrust on it, then its initial acceleration when released. (9702/22/O/N/25 Q1(b))

Volume: $V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (0.90)^3 = 3.1 \text{ m}^3$ .
Upthrust: $F = \rho g V = 1.1 \times 9.81 \times 3.1 = 33 \text{ N}$ .
Resultant up: $33 - 19 = 14 \text{ N}$ . Mass: $m = 19/9.81 = 1.9 \text{ kg}$ .
$a = \dfrac{14}{1.9} = 7.2 \text{ m s}^{-2}$ upwards.

Answer

F = 33 \text{ N}, \; a = 7.2 \text{ m s}^{-2}

Your turnExam-style

A fully submerged object displaces 0.015 m³ of oil of density 800 kg m⁻³. Calculate the upthrust.

Show worked answer

Use the fluid density and displaced volume: $F = \rho gV$ .
$F = 800 \times 9.81 \times 0.015 = 118 \text{ N}$ .

Answer: 118 N upward.

Common mistake

For a fully submerged object,

V

is fixed, so the upthrust does not grow as the object goes deeper when the fluid density and g are constant. It does change in a denser fluid, or when the submerged volume changes.

Upthrust

Anything in a fluid gets an upward push from it. That push is the upthrust from lesson 3.05's force list.

This lesson explains where it comes from and how to calculate it. Both are asked directly.

Where upthrust comes from

Pressure grows with depth ( $\rho g h$ , last lesson). The bottom of a submerged object is deeper than its top.
So the fluid pushes up on the bottom face harder than it pushes down on the top face. The sideways pushes are at equal depths and cancel.
The leftover upward force is the .

The bottom face is deeper, so the water pushes up on it harder than it pushes down on the top. The sideways pushes cancel. The leftover upward force is the upthrust.

The formula

F = (\rho g\Delta h)A = \rho gV

Archimedes' principle extends this result to any shape: the upthrust equals the weight of the displaced fluid.

F = \rho g V

Symbols

$F$ = upthrust (N)
$\rho$ = density of the fluid (not the object) (kg m⁻³)
$g$ = gravitational field strength (N kg⁻¹)
$V$ = volume of fluid pushed aside (the submerged volume) (m³)

Two inputs cause most lost marks. $\rho$ is the fluid's density. $V$ is only the volume below the surface.

Worked example

Smallest case: a submerged block

A block pushes aside 2.0 × 10⁻³ m³ of water. Find the upthrust on it.

$F = \rho g V = 1000 \times 9.81 \times 2.0 \times 10^{-3} = 20 \text{ N}$ .

Answer

F \approx 20 \text{ N}

Worked example

Exam version: a weather balloon

A weather balloon is a sphere of radius 0.90 m in air of density 1.1 kg m⁻³. Its total weight is 19 N. Find the upthrust on it, then its initial acceleration when released. (9702/22/O/N/25 Q1(b))

Volume: $V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (0.90)^3 = 3.1 \text{ m}^3$ .
Upthrust: $F = \rho g V = 1.1 \times 9.81 \times 3.1 = 33 \text{ N}$ .
Resultant up: $33 - 19 = 14 \text{ N}$ . Mass: $m = 19/9.81 = 1.9 \text{ kg}$ .
$a = \dfrac{14}{1.9} = 7.2 \text{ m s}^{-2}$ upwards.

Answer

F = 33 \text{ N}, \; a = 7.2 \text{ m s}^{-2}

Your turnExam-style

A fully submerged object displaces 0.015 m³ of oil of density 800 kg m⁻³. Calculate the upthrust.

Show worked answer

Use the fluid density and displaced volume: $F = \rho gV$ .
$F = 800 \times 9.81 \times 0.015 = 118 \text{ N}$ .

Answer: 118 N upward.

Common mistake

For a fully submerged object,

V

is fixed, so the upthrust does not grow as the object goes deeper when the fluid density and g are constant. It does change in a denser fluid, or when the submerged volume changes.