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IMAT Physics · Fluids

Pressure, Depth and the Hydraulic Press

Pressure is force per area; in a liquid at rest it grows with depth, and any extra pressure you add is passed on to every point.

Why this matters

Pressure appears in the 2012, 2021, 2025 and 2026 papers. The two ministry questions asked for a fact: which way a still liquid pushes on its container, and how deep a diver goes before the pressure doubles.

Concept 1 of 3: Pressure as force per area, and its units

The same push hurts more through a needle than through a palm. Pressure measures how concentrated a force is: the force divided by the area it is spread over. Halve the area and the pressure doubles, though the force has not changed.

Definition

Pressure is the perpendicular force on a surface divided by the area of that surface.

  • SI unit: the pascal, 1 Pa=1 N/m21\ \text{Pa} = 1\ \text{N/m}^2.
  • IMAT also uses units outside SI: 1 atm≈1.013×105 Pa=760 mmHg1\ \text{atm} \approx 1.013 \times 10^5\ \text{Pa} = 760\ \text{mmHg}, and 1 bar=105 Pa1\ \text{bar} = 10^5\ \text{Pa}.
  • Area conversions are the usual slip: 1 cm2=10−4 m21\ \text{cm}^2 = 10^{-4}\ \text{m}^2 and 1 mm2=10−6 m21\ \text{mm}^2 = 10^{-6}\ \text{m}^2.

Pressure

p=FAp = \frac{F}{A}
  • pppressure, in Pa
  • FFforce perpendicular to the surface, in N
  • AAarea, in m²

Worked example

A crate weighing 600 N rests on the floor on a face measuring 0.20 m by 0.30 m. It is then turned over onto a face measuring 0.10 m by 0.20 m. Find the pressure on the floor in each position.
Practice this conceptself-check · 4 quick reps

Square millimetres are not a thousandth of a square metre

A millimetre is 10−310^{-3} m, so a square millimetre is (10−3)2=10−6 m2(10^{-3})^2 = 10^{-6}\ \text{m}^2. Converting the length and forgetting to square it gives an answer a thousand times too small, and IMAT puts that wrong answer among the options.

Concept 2 of 3: Pressure at a depth in a liquid (Stevin's law)

At any depth, the liquid is holding up the column of liquid above it. A taller column weighs more, so the pressure rises steadily as you go down. Only the depth matters, not the width or shape of the container. A still liquid can only push straight against a surface, never along it, so its force on every wall is perpendicular to that wall.

Definition

In a liquid at rest:

  • The pressure at depth hh is the pressure at the surface plus ρgh\rho g h (Stevin's law).
  • Points at the same depth in the same connected liquid have the same pressure, whatever the shape of the vessel.
  • The force a still liquid exerts on a wall is perpendicular to the wall at every point.
  • For water, ρgh\rho g h reaches about 1 atm1\ \text{atm} at h≈10 mh \approx 10\ \text{m}. So at about 10 m down the total pressure is twice the surface pressure.
  • Gauge pressure is ρgh\rho g h alone; absolute pressure adds the atmosphere on top.

Pressure at depth h

p=p0+ρghp = p_0 + \rho g h
  • p0p_0pressure at the surface (often atmospheric)
  • ρ\rhodensity of the liquid, in kg/m³
  • gggravitational field strength, about 9.8 N/kg (IMAT often says 10)
  • hhdepth below the surface, in m

Worked example

A diver is 25 m below the surface of a freshwater lake. Take ρ=1000 kg/m3\rho = 1000\ \text{kg/m}^3, g=10 N/kgg = 10\ \text{N/kg} and atmospheric pressure 1.0×105 Pa1.0 \times 10^5\ \text{Pa}. Find the gauge and the absolute pressure on the diver.
Practice this conceptself-check · 4 quick reps

Doubling the pressure means doubling the absolute pressure

At the surface the pressure is already about 1 atm from the air. It doubles when the water adds another 1 atm, at about 10 m. Answering with the depth where the water's own pressure doubles has no meaning, because that pressure starts at zero.

A still liquid pushes perpendicular to the wall

A liquid at rest cannot push along a surface; any sideways component would make it flow. So the force on each part of a wall is at right angles to that part, at every point. Options saying the force can be in any direction, or parallel to the wall, are wrong.

Concept 3 of 3: Pascal's principle and the hydraulic press

Squeeze a closed bag of liquid at one spot and it bulges everywhere: the extra pressure reaches every point. A hydraulic press uses this. A small force on a small piston makes a pressure, and that same pressure acting on a large piston makes a large force. Nothing is free, though: the large piston moves only a short way.

Definition

Pascal's principle: a change in pressure applied to an enclosed liquid is passed on undiminished to every point of the liquid and to the walls.

  • In a hydraulic press the two pistons are at the same level, so the pressure under each is the same: F1/A1=F2/A2F_1/A_1 = F_2/A_2.
  • The force is multiplied by the area ratio A2/A1A_2/A_1. If you are given diameters, the area ratio is the diameter ratio squared.
  • The liquid's volume is fixed, so A1d1=A2d2A_1 d_1 = A_2 d_2: the large piston moves a shorter distance, and the work done on each piston is the same.

Hydraulic press

F1A1=F2A2A1d1=A2d2\frac{F_1}{A_1} = \frac{F_2}{A_2} \qquad A_1 d_1 = A_2 d_2
  • F1,A1,d1F_1, A_1, d_1force on, area of, and distance moved by the small piston
  • F2,A2,d2F_2, A_2, d_2the same for the large piston

Worked example

A hydraulic jack has a small piston of area 4.0 cm² and a large piston of area 200 cm². A force of 30 N pushes the small piston down by 25 cm. Find the force on the large piston and how far it rises.
Practice this conceptself-check · 3 quick reps

A hydraulic press multiplies force, not energy

The output force is larger by the area ratio, but the output piston moves a distance smaller by the same ratio. Work in equals work out (ignoring friction). An option claiming the press gives out more energy than it takes in is always wrong.

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