Pascal's law

Put your foot on a car's brake pedal and, half a metre away and hidden inside a metal drum, a set of pads clamps hard enough to stop a two-tonne vehicle. Nudge the handle of a garage jack a few centimetres and a car rises smoothly off the ground. Neither of these machines has a motor doing the heavy lifting at the point of use — they run on nothing but a sealed tube of oil and a simple fact about how liquids carry pressure. That fact is Pascal's law: squeeze an enclosed fluid anywhere, and the extra pressure you created shows up, completely undiminished, at every other point of that fluid and on every wall that contains it. The idea traces back to the French scientist and philosopher Blaise Pascal in the seventeenth century, who reasoned that a liquid at rest cannot quietly "absorb" a push the way a solid can — it has no internal stiffness to resist a sideways force, so any imbalance in pressure simply drives the fluid to flow until the imbalance disappears. In a sealed container that flow has nowhere to go, so the pressure change must spread through the whole liquid instead. This one idea, applied to two pistons of different sizes joined by oil-filled pipework, is the reason a small force at a garage jack's handle can lift a car, and the reason a light foot on a brake pedal can generate enough clamping force to stop one. In this article you will see why the law holds, derive the force-multiplication relation for a two-piston system, work through both a straightforward board-style numerical and a tougher multi-stroke jack problem, and pin down exactly why "free force" from Pascal's law never means free energy.

Key takeaways

Pascal's law: a pressure change applied anywhere in an enclosed, incompressible fluid at rest is transmitted undiminished to every point of the fluid and to the walls of its container.
It holds because a fluid at rest cannot sustain shear stress — any pressure imbalance drives flow until pressures equalise, and in a sealed container that equalisation spreads the change everywhere.
The law only concerns the change in pressure, $\Delta P$ , added on top of whatever pressure already exists from depth, $P = P_0 + \rho g h$ .
In a two-piston hydraulic system, equal pressure on both pistons gives $F_2 = F_1 \dfrac{A_2}{A_1}$ — a small force on a narrow piston produces a large force on a wide one.
Force multiplication is not energy multiplication: incompressibility forces $A_1 d_1 = A_2 d_2$ , so the wide piston always moves a proportionally smaller distance, and $F_1 d_1 = F_2 d_2$ (ignoring friction).
This is the working principle of hydraulic lifts, hydraulic jacks, and hydraulic brakes — the applications studied right after this topic.
Key idea: because a fluid at rest cannot resist shear, any pressure change applied at one point of an enclosed fluid appears equally everywhere else, so two pistons of different area connected by that fluid must carry the same pressure, giving a force ratio F2/F1 = A2/A1 that trades a smaller piston's large displacement for a larger piston's large force, with total work conserved.

The core insight

An enclosed fluid at rest cannot sustain shear, so a pressure change applied anywhere spreads undiminished to every point and wall of the container, forcing two connected pistons to share the same pressure and giving the force ratio $F_2/F_1 = A_2/A_1$ , with $A_1 d_1 = A_2 d_2$ keeping the work done exactly equal on both sides.

Where it fits in the syllabus

Pascal's law sits in the CBSE Class 11 "Mechanical Properties of Fluids" unit, immediately after the definition of pressure and just before the pressure-versus-depth relation $P = P_0 + \rho g h$ and the standard hydraulic-machine applications. The board expects you to state the law precisely, explain why it follows from a fluid's inability to resist shear, and apply it — mainly through the two-piston force relation — to hydraulic lifts, jacks, and brakes. Everything here stays within that scope: no fluid dynamics, no Bernoulli's theorem, and no viscosity enter this topic at all; those are separate parts of the same chapter.

The physical picture

Pressure at a point in a fluid is defined as force per unit area, $P = F/A$ , acting perpendicular to any surface at that point. Because a liquid has essentially fixed volume — it strongly resists being squeezed smaller — any attempt to raise the pressure at one spot inside a sealed, connected body of liquid cannot be "soaked up" locally; the liquid has nowhere to compress into, so the disturbance must be carried to every other part of the fluid. Pascal's law states this precisely: a pressure change $\Delta P$ applied at any point of an enclosed, incompressible fluid at rest appears, unchanged in size, at every other point of the fluid and at every wall touching it.

\Delta P_1 = \Delta P_2 = \Delta P \quad \text{at every point of the enclosed fluid}

where: $\Delta P_1$ and $\Delta P_2$ are the pressure changes measured at any two points of the same enclosed fluid, and $\Delta P$ is the single applied change common to both. The physical reading is that this equation says nothing about the absolute pressure at each point — that still depends on depth through $P = P_0 + \rho g h$ — it only says that whatever extra pressure you push in at one point turns up, in full, everywhere else, instantly for practical purposes because liquids are so close to incompressible. The most important use of this idea is a two-piston hydraulic system: a small piston of area $A_1$ and a large piston of area $A_2$ , joined by a sealed, oil-filled channel. Push the small piston with force $F_1$ and it creates a pressure $P = F_1/A_1$ in the oil right under it.

P = \frac{F_1}{A_1}

where: $F_1$ is the force applied to the small piston and $A_1$ is its cross-sectional area, so $P$ is the pressure this push produces in the enclosed oil. By Pascal's law that same pressure $P$ acts on the large piston too, so the force it delivers there is $F_2 = P \times A_2$ .

F_2 = P \times A_2 = F_1 \times \frac{A_2}{A_1}

where: $F_2$ is the force delivered at the large piston and $A_2$ is its cross-sectional area. Since $A_2 > A_1$ , this equation guarantees $F_2 > F_1$ : a modest push on the small piston becomes a much larger push on the big one. This is exactly the working principle behind a hydraulic lift raising a car, a hydraulic jack, and a hydraulic brake clamping a wheel — all covered as the direct application right after this topic.

Figure

A small piston on the left pushed down by a small force and a large piston on the right pushed up by a large force, connected underneath by a sealed channel of fluid, with the pressure at the base of both pistons shown as equal

Pascal's law is a statement about what an applied push does on top of the pressure a fluid already has from gravity acting on its own weight. The pressure-versus-depth relation $P = P_0 + \rho g h$ describes how pressure grows naturally from point to point within a static fluid purely because of depth; Pascal's law describes how an externally applied pressure change adds equally to that background everywhere, regardless of depth or the container's shape. Used together, the two ideas let you find the pressure at any point of any enclosed fluid system, whether it is simply sitting still under gravity or being actively driven by a piston.

Derivation AForce multiplication in a two-piston hydraulic system

The goal of this derivation is to turn the bare statement of Pascal's law — that a pressure change is transmitted undiminished through an enclosed fluid — into the working formula every hydraulic-machine numerical needs, $F_2 = F_1 (A_2/A_1)$ , and to show exactly where the assumption of incompressibility enters both the force relation and the companion volume-conservation relation used for piston displacement.

Step 1. Set up the system. Take a sealed, oil-filled vessel connecting a small piston of area $A_1$ to a large piston of area $A_2$ , with the fluid at rest and both pistons free to move against whatever load sits on them. Apply a force $F_1$ to the small piston. By the definition of pressure, this creates a pressure increase right beneath that piston of $\Delta P = F_1/A_1$ .

Step 2. Apply Pascal's law. Because the fluid is enclosed, incompressible, and at rest, this pressure increase $\Delta P$ does not stay local — it is transmitted, unchanged in magnitude, to every other point of the fluid, including the region directly beneath the large piston. So the pressure increase felt at the large piston is the very same $\Delta P = F_1/A_1$ , not some fraction of it and not something reduced by the extra distance travelled through the fluid.

Step 3. Convert the transmitted pressure back into a force. Since pressure is force per unit area, the force this pressure increase produces on the large piston, of area $A_2$ , is $F_2 = \Delta P \times A_2$ . Substituting the expression for $\Delta P$ from Step 1 gives $F_2 = (F_1/A_1) \times A_2$ , which rearranges to the working relation $F_2 = F_1 (A_2/A_1)$ .

Step 4. Read off the multiplication and its limit. Because $A_2$ and $A_1$ are both positive areas, the ratio $A_2/A_1$ is simply the piston-area ratio, and whenever $A_2 > A_1$ the output force exceeds the input force in exact proportion to that ratio — there is no upper limit imposed by the fluid itself, only by how large a piston-area ratio is practical to build.

Step 5. Bring in incompressibility a second time, for displacement. If the small piston moves down a distance $d_1$ , it pushes out a volume of fluid $A_1 d_1$ . Since the fluid cannot compress and has nowhere else to go, this exact volume must appear as a rise of the large piston, $A_2 d_2$ . Equating the two volumes gives $A_1 d_1 = A_2 d_2$ , so $d_2 = d_1 (A_1/A_2)$ — the large piston moves a factor of $A_1/A_2$ less than the small one, which is the reciprocal of the force-multiplication factor.

Step 6. Confirm energy is conserved, not created. Multiplying force by displacement on each side, $F_1 d_1 = \big[F_2 (A_1/A_2)\big] d_1$ and, using $F_2 = F_1(A_2/A_1)$ from Step 3, this becomes $F_1 d_1 = F_1 (A_2/A_1)(A_1/A_2) d_1 = F_1 d_1$ — an identity. So $F_1 d_1 = F_2 d_2$ always, meaning the work put in at the small piston exactly equals the work delivered at the large piston (ignoring friction and any elastic loss in real seals); the hydraulic system multiplies force, never energy.

Worked examples

Board level — force and piston displacement in a hydraulic press

In a hydraulic press, the small piston has area $A_1 = 4\times10^{-4}\ \text{m}^2$ and the large piston has area $A_2 = 0.04\ \text{m}^2$ . A force of $100\ \text{N}$ is applied to the small piston. Find (a) the force exerted on the large piston, and (b) how far the large piston rises if the small piston is pushed down by $20\ \text{cm}$ .

Step 1. Find the pressure produced at the small piston. $P = F_1/A_1 = 100 / (4\times10^{-4}) = 2.5\times10^{5}\ \text{Pa}$ .

Step 2. Apply Pascal's law: this pressure is transmitted, undiminished, to every point of the fluid, including the base of the large piston.

Step 3. Find the force on the large piston. $F_2 = P \times A_2 = 2.5\times10^{5} \times 0.04 = 1.0\times10^{4}\ \text{N}$ . So a $100\ \text{N}$ push produces a $10{,}000\ \text{N}$ output — a mechanical advantage of $100$ , exactly equal to the area ratio $A_2/A_1 = 0.04/(4\times10^{-4}) = 100$ . This is the answer to part (a): $F_2 = 1.0\times10^4\ \text{N}$ .

Step 4. For part (b), use volume conservation for the incompressible fluid: $A_1 d_1 = A_2 d_2$ , with $d_1 = 0.20\ \text{m}$ . Substituting, $4\times10^{-4} \times 0.20 = 0.04 \times d_2$ , so $8\times10^{-5} = 0.04\, d_2$ .

Step 5. Solve for $d_2$ . $d_2 = (8\times10^{-5})/0.04 = 2\times10^{-3}\ \text{m} = 2\ \text{mm}$ . The large piston rises only 2 mm while the small piston moves 200 mm, consistent with work conservation: $F_1 d_1 = 100 \times 0.20 = 20\ \text{J}$ and $F_2 d_2 = 1.0\times10^4 \times 2\times10^{-3} = 20\ \text{J}$ — the two match exactly.

Competitive level — number of strokes to lift a car with a hydraulic jack

A garage hydraulic jack has a small piston of area $A_1 = 4\times10^{-4}\ \text{m}^2$ operated by a hand lever, and a large piston of area $A_2 = 0.04\ \text{m}^2$ supporting a car of mass $1200\ \text{kg}$ . Take $g = 9.8\ \text{m/s}^2$ . Find (a) the minimum force needed on the small piston to lift the car, and (b) if each stroke pushes the small piston down by $15\ \text{cm}$ , how many strokes are needed to raise the car by $6\ \text{cm}$ .

Step 1. Find the weight the large piston must support. $F_2 = mg = 1200 \times 9.8 = 1.176\times10^{4}\ \text{N}$ .

Step 2. Use the force relation from the derivation, rearranged for $F_1$ . Since $F_2 = F_1(A_2/A_1)$ , the minimum force needed is $F_1 = F_2 (A_1/A_2) = 1.176\times10^{4} \times \dfrac{4\times10^{-4}}{0.04} = 1.176\times10^{4} \times 0.01 = 117.6\ \text{N}$ . This is the answer to part (a): about 118 N — comfortably within hand-operated range, which is exactly why a jack this size can lift a car.

Step 3. For part (b), find how far the large piston rises for one stroke of the small piston, using $A_1 d_1 = A_2 d_2$ with $d_1 = 0.15\ \text{m}$ . $d_2 = d_1 (A_1/A_2) = 0.15 \times 0.01 = 1.5\times10^{-3}\ \text{m} = 1.5\ \text{mm}$ per stroke.

Step 4. Divide the total rise needed by the rise per stroke. Number of strokes $n = \dfrac{0.06\ \text{m}}{1.5\times10^{-3}\ \text{m}} = 40$ . The jack needs 40 strokes to raise the car by 6 cm. Timing note: the two-part structure here — force first, then a displacement or repeated-stroke calculation — is the standard competitive pattern for hydraulic-machine questions; solve the force part in under a minute using the direct ratio $F_1 = F_2(A_1/A_2)$ , and do not fall into re-deriving Pascal's law from scratch under time pressure — quote the two working relations, $F_2/F_1 = A_2/A_1$ and $A_1 d_1 = A_2 d_2$ , and substitute directly.

Common traps

Thinking Pascal's law "creates" extra energy — it multiplies force, not energy; the large piston always moves a proportionally smaller distance, so work in equals work out (ignoring friction).
Forgetting that Pascal's law concerns the change in pressure, not the absolute pressure — if the fluid already has pressure from gravity and depth, the applied change adds to it uniformly everywhere, it does not replace it.
Mixing up which piston is $A_1$ and which is $A_2$ when applying $F_2 = F_1(A_2/A_1)$ , which inverts the mechanical advantage and gives an answer smaller than $F_1$ instead of larger.
Using the force ratio $A_2/A_1$ for displacement too, instead of its reciprocal $A_1/A_2$ — the piston with the larger area moves the smaller distance, exactly opposite to the force ratio.
Assuming the law needs a rigid container of a specific shape — it works for any shape of closed, connected fluid, as long as the fluid itself is nearly incompressible and at rest.

Common trapPascal's law never gives you something for nothing

The single biggest misconception is treating a hydraulic lift as a machine that multiplies energy: "a small force in gives a large force out, so I've gained something." It has not. What has actually happened is that the same total work has been redistributed between force and distance. The small piston moves a large distance under a small force; the large piston moves a proportionally small distance under a large force; the product $F \times d$ , the work done, comes out identical on both sides (ignoring friction and seal losses). This is exactly the same trade-off you see in a lever or a pulley system — mechanical advantage in force always costs you an equal factor in displacement. If a hydraulic-machine numerical ever seems to hand you more work out than you put in, recheck the displacement relation $A_1 d_1 = A_2 d_2$ ; you have almost certainly made an arithmetic slip, not discovered a new source of energy.

You should be able to…

State Pascal's law in words and explain why it follows from a fluid's inability to sustain shear.
Derive $F_2 = F_1(A_2/A_1)$ for a two-piston hydraulic system starting from the definition of pressure and Pascal's law.
Derive and use the companion volume-conservation relation $A_1 d_1 = A_2 d_2$ for piston displacement, and recognise it as the reciprocal of the force ratio.
Explain why hydraulic machines multiply force but never energy, using $F_1 d_1 = F_2 d_2$ .
Distinguish Pascal's law (an applied pressure change spreading equally) from the pressure-depth relation $P = P_0 + \rho g h$ (pressure varying naturally with depth), and combine the two when a problem needs both.
Solve multi-part numericals involving force, piston area, displacement, and repeated strokes in hydraulic presses and jacks.

Ask the tutor: Want a guided walkthrough of how Pascal's law scales up into a real hydraulic brake system, or help setting up a trickier multi-piston or multi-stroke numerical? Ask the AI tutor.

Why doesn't a hydraulic lift break the law of conservation of energy?

Because it never multiplies energy, only force. The volume of incompressible fluid pushed out by the small piston must exactly equal the volume received by the large piston, which forces $A_1 d_1 = A_2 d_2$ . Combine that with the force relation $F_2 = F_1(A_2/A_1)$ and you get $F_1 d_1 = F_2 d_2$ — the work done on the small piston always equals the work delivered by the large piston, ignoring friction. You trade a large force for a small distance, or a small force for a large distance; you never get both large force and large distance for free.

Does Pascal's law apply to gases as well as liquids?

The idea that a pressure change spreads through a fluid because it cannot sustain shear applies to gases too, in principle, since gases also flow freely. But the CBSE Class 11 statement of Pascal's law is built specifically around incompressible fluids, because it is the fluid's near-total resistance to volume change that makes the transmitted pressure genuinely undiminished and close to instantaneous. Gases are compressible, so a pressure change can partly go into squeezing the gas itself rather than being passed on unchanged, and the "everywhere-equal" picture is only approximate. Hydraulic machines therefore always use liquids, such as oil, not air.

How is Pascal's law different from the pressure-depth formula P equals P naught plus rho g h?

They describe two different sources of pressure, and a numerical often needs both. The formula $P = P_0 + \rho g h$ tells you how pressure at a point in a static fluid grows purely because of the weight of fluid above it — nothing needs to be "applied" for this to happen. Pascal's law, by contrast, describes what happens when you actively apply an extra pressure change $\Delta P$ at some point, for example by pushing a piston: that extra change is transmitted equally to every point, on top of whatever depth-dependent pressure was already there. Add the two together to get the full pressure at any point in an actively driven, enclosed fluid system.

Why do hydraulic jacks use many small strokes instead of one big push?

Because a hand lever can only move a short, comfortable distance per stroke, and the large piston's rise per stroke is fixed by $A_1 d_1 = A_2 d_2$ — a small fraction of the small piston's stroke, set by the area ratio. To lift a car through several centimetres, the small piston therefore has to repeat its short stroke many times, with a one-way valve system letting fluid through on the push stroke and holding it in place on the return stroke, so each stroke adds a little more height without letting the car sink back down between pushes. The force needed per stroke stays the same throughout, set only by $F_1 = F_2(A_1/A_2)$ , which is why a jack lifts steadily rather than getting harder as the car rises.