Pressure in a Fluid | CAIE 9702 Paper 2 Notes

Paper 2

Physics9702

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 change in pressurebetween two depths. The pressure at the surface is added only when total pressure is required.

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 increase across depth h is Δp = 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: $\Delta p = \dfrac{\rho A\Delta h g}{A} = \rho g\Delta h$ .
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 pressure at the surface. For a liquid open to the atmosphere, $p = p_{\text{atm}} + \rho g h$ .

A fish near the surface pushed by a few arrows and a deeper fish pushed by more arrows from all sides. — Pressure acts from every direction, and its value is greater at the deeper fish.

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. Always draw the surface and mark the vertical distance down to the point.

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: $\Delta p = \rho g h$ .
The side with the lower liquid level has the higher pressure. If the gas pressure is below atmospheric pressure, the gas side stands higher instead.

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 turnExam-style

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.

Show worked answer

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 pressure increase; stacked liquids require one $\rho g h$ term for each layer; and a U-tube can contain two liquids balanced at their boundary.

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