PhysicsICSE Class 9 · Pressure in Fluids

Pascal's Law and Hydraulic Machines: Hydraulic Lift, Press and Brakes (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Pascal's law: pressure applied to an enclosed liquid is transmitted equally and undiminished in all directions. In a hydraulic machine F₁/A₁ = F₂/A₂, so the force is multiplied by A₂/A₁, but the large piston moves less in the same ratio, so work is not gained.

Understand Pascal's Law and Hydraulic Machines

Pascal’s law. Pressure applied at any point of a liquid enclosed in a vessel is transmitted equally and undiminished to every part of the liquid and to the walls of the vessel.

Liquids are nearly incompressible and the pressure is carried through them without loss. Gases are compressible, so they are not used in hydraulic machines.

Hydraulic machine. Two cylinders of unequal area are joined and filled with liquid, each fitted with a piston.

A force $F_{1}$ on the small piston (area $A_{1}$ ) gives the pressure $p = \frac{F _{1}}{A _{1}}$ .
The same pressure acts on the large piston (area $A_{2}$ ), so $F_{2} = p A_{2}$ :

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}, F_{2} = F_{1} \times \frac{A _{2}}{A _{1}} .$

The ratio $\frac{A _{2}}{A _{1}}$ is the force multiplication (mechanical advantage). For circular pistons $\frac{A _{2}}{A _{1}} = (\frac{d _{2}}{d _{1}})^{2}$ : a piston of twice the diameter has four times the area.

What is not multiplied. The liquid does not change in volume, so the volume pushed out of the small cylinder equals the volume entering the large one: $A_{1} d_{1} = A_{2} d_{2}$ , giving

$d_{2} = d_{1} \frac{A _{1}}{A _{2}} .$

The large piston moves less by the same ratio that its force is larger. Work done on the small piston $F_{1} d_{1}$ equals work done by the large piston $F_{2} d_{2}$ (ideal machine): the machine multiplies force, never energy.

Applications.

Hydraulic lift / car jack: a small effort raises a heavy car.
Hydraulic press (squeezing, baling).
Hydraulic brakes: pressing the pedal sends equal pressure to the brake cylinders at all wheels, so each wheel receives the force corresponding to its piston area.

Practical points. The liquid must be free of air bubbles (air compresses and the pressure is not transmitted fully). A non-return valve and a pump allow the small piston to be worked repeatedly, because in one stroke it would need to travel a very long distance.

Pascal's Law and Hydraulic Machines: formulas and key facts

Pascal's law in a machine: $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$; Equal pressure.
Force multiplication: $\frac{F _{2}}{F _{1}} = \frac{A _{2}}{A _{1}} = (\frac{d _{2}}{d _{1}})^{2}$; Circular pistons.
Distances moved: $A_{1} d_{1} = A_{2} d_{2}; F_{1} d_{1} = F_{2} d_{2}$; Ideal machine; no energy gained.

How to solve Pascal's Law and Hydraulic Machines questions

Treat pressure as the common quantity.

1Convert both areas to the same unit (cm² with cm², or both m²).
2Equate $F_{1} / A_{1} = F_{2} / A_{2}$ .
3For circular pistons use the area ratio $(d_{2} / d_{1})^{2}$ .
4For distances use $d_{2} = d_{1} A_{1} / A_{2}$ .
5Check with work: $F_{1} d_{1} = F_{2} d_{2}$ for an ideal machine.

Pascal's Law and Hydraulic Machines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A garage jack

Suppose a garage jack has a small piston of area 4 cm² and a large piston of area 400 cm². A mechanic pushes down with 60 N.

What load can it raise and by how much must the small piston move to raise the load by 1 cm?

Try it first, then show the solutionSolution

$F_{2} = 60 \times \frac{400}{4} = 6000$ N.
$d_{1} = d_{2} \times \frac{A _{2}}{A _{1}} = 1 \times 100 = 100$ cm.

Answer: 6000 N; the small piston moves 100 cm.

Memory hook: Force gain = distance loss.

Tricks to remember Pascal's Law and Hydraulic Machines

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same pressure, different forces
F/A is the same on both pistons.
Use it when: Setting up the equation.
Area ratio is square of diameter ratio
Twice the diameter, four times the area.
Use it when: Pistons given by diameter.
Force gain, distance loss
Multiply force by 80, divide travel by 80.
Use it when: Conceptual questions.

Pascal's Law and Hydraulic Machines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Pascal's law statement · 1 mark · MCQBoard level

According to Pascal’s law, pressure applied to an enclosed liquid is transmitted:
(a) only downward
(b) only to the walls nearest the piston
(c) equally and undiminished in all directions
(d) in proportion to the distance from the piston

Try it first, then show the solutionSolution

That is the statement of the law.

Answer: (c)

Example 2: Hydraulic press · 2 marks · Short answerBoard level

The pistons of a hydraulic press have areas of 5 cm² and 400 cm². A force of 30 N acts on the small piston. Find the pressure in the liquid and the force on the large piston.

Try it first, then show the solutionSolution

$p = \frac{30}{5 \times 1 0 ^{- 4}} = 60000$ Pa. (1)
$F_{2} = p A_{2} = 60000 \times 0.04 = 2400$ N. (1)

Answer: 60 000 Pa; 2400 N

Example 3: Hydraulic lift: effort and travel · 3 marks · Short answerHarder board level

A hydraulic lift raises a load of 12 000 N with an effort of 150 N on the small piston. Find (a) the ratio of piston areas, (b) the small piston area if the large piston area is 2000 cm², (c) the distance the load rises when the effort piston moves 40 cm.

Try it first, then show the solutionSolution

(a) $\frac{A _{2}}{A _{1}} = \frac{12 000}{150} = 80$ . (1)
(b) $A_{1} = \frac{2000}{80} = 25$ cm². (1)
(c) $d_{2} = \frac{40}{80} = 0.5$ cm. (1)

Answer: (a) 80 (b) 25 cm² (c) 0.5 cm

Example 4: Hydraulic brake · 3 marks · Short answerHarder board level

In a hydraulic brake the master piston has diameter 2 cm and each wheel piston has diameter 4 cm. The driver’s pedal exerts 120 N on the master piston. Find the force on each wheel piston and the factor by which the force is multiplied.

Try it first, then show the solutionSolution

Area ratio $= (\frac{4}{2})^{2} = 4$ . (1)
The same pressure acts on every wheel piston, so each gets $120 \times 4 = 480$ N. (1)
Force multiplied by 4 (but each wheel piston moves $\frac{1}{4}$ as far). (1)

Answer: 480 N on each; factor 4

Example 5: A garage lift · 4 marks · Case-basedHOTS / competency

A car of mass 1200 kg is to be raised on a lift ( $g = 10$ m s⁻²). The large piston has radius 20 cm and the small piston radius 2 cm.
(a) Find the ratio of the areas.
(b) Find the least effort on the small piston.
(c) How far must the small piston move to raise the car by 5 cm?
(d) Why does a real lift use a pump and valve?

Try it first, then show the solutionSolution

(a) $(\frac{20}{2})^{2} = 100$ . (1)
(b) $W = 1200 \times 10 = 12000$ N; $F_{1} = \frac{12 000}{100} = 120$ N. (1)
(c) $d_{1} = 5 \times 100 = 500$ cm $= 5$ m. (1)
(d) The small piston would have to move a long way for each small rise; a pump with non-return valves delivers the liquid in repeated short strokes. (1)

Answer: (a) 100 (b) 120 N (c) 5 m (d) repeated strokes

Where students lose marks in Pascal's Law and Hydraulic Machines

The slips examiners see most often, and what to write instead.

Mistake: Areas in different units
cm² and m² cannot be ratioed directly.
Wrong working
F₂ = 30 × (0.04/5)
Right working
F₂ = 30 × (400/5) = 2400 N
Do this: Convert to the same unit first.
Usually costs: The answer.
Mistake: Diameter ratio used as area ratio
Area goes as the square of the diameter.
Wrong working
F₂ = 120 × 2
Right working
F₂ = 120 × 4
Do this: Square the ratio.
Usually costs: The answer.
Mistake: Claiming energy is gained
Work in equals work out.
Wrong working
The hydraulic lift gives more work than we put in.
Right working
It multiplies force; the distance falls by the same factor.
Do this: Quote F₁d₁ = F₂d₂.
Usually costs: The reasoning mark.

Exam tips

State the law in full before using it.
Write both pressures as fractions and equate them.
For distance moved, use A₁d₁ = A₂d₂ and mention volume.
Give the working for "work is not multiplied" with numbers when asked.

Practice questions on Pascal's Law and Hydraulic Machines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these is used in a hydraulic machine instead of a gas because it is nearly incompressible?
(a) a liquid
(b) air
(c) steam
(d) vacuum
Hint
Pressure transmission.
Answer
(a) a liquid
Q2 · 1 mark · Very short answerBoard level
In an ideal hydraulic lift, what is multiplied, force or work?
Hint
Think of the energy.
Answer
Force. Work (energy) is not multiplied.
Q3 · 2 marks · Short answerHarder board level
Pistons of areas 10 cm² and 500 cm² are joined. A force of 40 N is applied on the small one. Find the force on the large piston.
Hint
Area ratio 50.
Answer
$F_{2} = 40 \times \frac{500}{10} = 2000$ N.
Q4 · 3 marks · Short answerHarder board level
A load of 8000 N is supported on the large piston when an effort of 50 N acts on a small piston of area 4 cm². Find the area of the large piston and the mechanical advantage.
Hint
Equal pressure.
Answer
$\frac{A _{2}}{4} = \frac{8000}{50} = 160$ , so $A_{2} = 640$ cm²; mechanical advantage $= 160$ .
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): Hydraulic machines use liquids and not gases.
Reason (R): Liquids are nearly incompressible.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Pressure transmission.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A hydraulic jack lifts a 3600 N load with an effort of 90 N which moves through 24 cm. Assuming no friction, find how far the load rises and show that the work done on and by the jack are equal.
Hint
Use $F_{1} d_{1} = F_{2} d_{2}$ .
Answer
$d_{2} = \frac{90 \times 24}{3600} = 0.6$ cm. Work in $= 90 \times 0.24 = 21.6$ J; work out $= 3600 \times 0.006 = 21.6$ J. Force is multiplied by 40; distance is divided by 40.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Pascal's law.Show answer
Pressure applied to an enclosed liquid is transmitted equally and undiminished in all directions.
2.Force multiplication in a hydraulic lift?Show answer
A₂/A₁.
3.Why not use a gas?Show answer
It is compressible.
4.What changes by the same factor in the opposite sense?Show answer
Distance moved by the piston.
5.Hydraulic brakes work on?Show answer
Pascal's law.

Pascal's Law and Hydraulic Machines: frequently asked questions

What is Pascal's law?

Pressure applied to a liquid enclosed in a vessel is transmitted equally and undiminished to every part of the liquid and the walls of the vessel.

How does a hydraulic lift multiply force?

The same pressure acts on pistons of different areas, so F₂ = F₁ × (A₂/A₁); the large piston moves less in the same ratio.

Does a hydraulic machine violate energy conservation?

No. Work input F₁d₁ equals work output F₂d₂ for an ideal machine; only force is multiplied.