Physics 9702 · Paper 2

Pressure in a Fluid

Divers feel more pressure the deeper they go. The extra pressure comes from the weight of the fluid above them.
One short formula gives the pressure at any depth, and one derivation explains it.

The formula and where it comes from

\Delta p = \rho g \Delta h

Symbols

$\Delta p$ = change in pressure (Pa)
$\rho$ = density of the fluid (kg m⁻³)
$g$ = gravitational field strength (N kg⁻¹)
$\Delta h$ = change in depth (m)

The column of liquid above area A has volume Ah, mass ρAh and weight ρAhg. Divide by A: the pressure at depth h is p = ρgh. The area cancels, so only depth matters.

The derivation is three lines. A column of fluid of height $h$ over area $A$ has volume $Ah$ , so mass $\rho A h$ and weight $\rho A h g$ .
Pressure = force ÷ area: $p = \dfrac{\rho A h g}{A} = \rho g h$ .
The area cancels. So the pressure depends only on depth, density and g, never on the width or shape of the container.
For the total pressure at a depth, add the atmosphere pressing on the surface: $p = p_{\text{atm}} + \rho g h$ .

A fish near the surface pushed by a few arrows; a much deeper fish pushed by many arrows from all sides. — Deeper means more pressure, from every direction at once.

Worked example

Smallest case: 2.0 m under water

Find the extra pressure 2.0 m below the surface of water (density 1000 kg m⁻³), then the total pressure if the atmosphere is 1.0 × 10⁵ Pa.

$\rho g h = 1000 \times 9.81 \times 2.0 = 2.0 \times 10^{4} \text{ Pa}$ .
Total: $p = 1.0 \times 10^{5} + 2.0 \times 10^{4} = 1.2 \times 10^{5} \text{ Pa}$ .

Answer

\Delta p = 2.0 \times 10^{4} \text{ Pa}, \; p_{\text{total}} = 1.2 \times 10^{5} \text{ Pa}

Common mistake

h

is the depth below the fluid surface. A mark on a measuring cylinder or the length of a tube is not the depth. Questions place the point of interest by its scale reading, and the wrong options use that reading directly (9702/13/M/J/25 Q15).

The U-tube: comparing two pressures

A manometer is a U-shaped tube of liquid. One side connects to a gas; the other is open to the air.
The gas pushes its side of the liquid down, so the open side stands higher. In equilibrium the level difference h measures the pressure difference: $\Delta p = \rho g h$ .

A manometer. The gas pushes the liquid down on its side, so the open side stands higher. The pressure difference between gas and atmosphere is ρgh, using the difference h between the two levels.

Your turn— tap to reveal the worked answer (9702/12/M/J/23 Q14)

In a U-tube, a liquid of density 2000 kg m⁻³ stands at 10.0 cm on one side and 6.0 cm on the other. Find the pressure difference between the two sides. (9702/12/M/J/23 Q14)

Use the difference in levels: $h = 0.100 - 0.060 = 0.040 \text{ m}$ .
$\Delta p = 2000 \times 9.81 \times 0.040 = 780 \text{ Pa}$ .

Answer: about 780 Pa.

Ways the exam changes it: doubling the depth doubles the extra pressure (a submarine going from 130 m to 260 m) (9702/11/M/J/25 Q16); two different liquids stacked in layers, where you add $\rho g h$ layer by layer (9702/12/F/M/24 Q12); two liquids in a U-tube balanced at the boundary between them (9702/13/O/N/24 Q16).

Only vertical depth matters. Wide tank, thin tube, tilted pipe: at the same depth in the same liquid, the pressure is the same.

PreviousPressure NextUpthrust