Statement of Pascal's law
Pascal's law has two faces in NCERT, and NEET expects both. The first is about an undisturbed fluid; the second, more famous one, is about an applied pressure change.
| Form | Statement | Where NEET uses it |
|---|---|---|
| Pressure at a level | The pressure in a fluid at rest is the same at all points that lie at the same height. | Manometers, U-tubes, connected vessels |
| Transmission of pressure | A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. | Hydraulic lift, hydraulic brakes, hydraulic press |
The operative word in the second form is undiminished. Add an extra pressure $\Delta P$ at one point of a confined fluid, and exactly that $\Delta P$ — not a weakened fraction of it — appears at every other point. NCERT states the equivalent everyday version verbatim: "whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions." NIOS adds the alternative name, the law of transmission of liquid pressure.
Why it holds — justification
Two physical facts combine to give Pascal's law. The first concerns direction, the second concerns transmission.
Pressure has no direction. Consider a tiny right-angled prismatic element ABC-DEF deep inside a fluid at rest (NCERT Fig. 9.2). It is small enough that gravity acts equally on all of it. The surrounding fluid pushes on each face along the normal to that face. Writing force balance and using the geometry of the prism gives equal pressures on all three faces:
$$ P_a = P_b = P_c. $$
So the pressure at a point is the same regardless of the orientation of the surface you measure it on — pressure is a scalar, not a vector. The force on any area within a fluid at rest is normal to that area, whatever way the area faces.
An applied pressure cannot stay local. Take a fluid element shaped like a horizontal bar of uniform cross-section. In equilibrium the horizontal forces on its two ends must balance, so the pressure at the two ends is equal — pressure is the same everywhere on one horizontal level. If it were not, the unbalanced force would set the fluid flowing. Now push a piston and add $\Delta P$ at one wall. Because every element must stay in equilibrium, the only consistent outcome is that $\Delta P$ propagates to every point; otherwise some element feels a net force and the "at rest" condition breaks. That is Pascal's law of transmission.
Pascal's law vs pressure variation with depth
NEET routinely blurs these two ideas, so separate them firmly. The pressure inside a fluid already varies with depth because of gravity, $P = P_a + \rho g h$. Pascal's law is about something added on top of that — an external pressure change applied through a piston.
| Variation with depth | Pascal's law (transmission) | |
|---|---|---|
| Question answered | How much pressure does gravity build up? | What happens to an externally applied pressure? |
| Relation | $P = P_a + \rho g h$ | $\Delta P$ is the same at every point |
| Depends on depth? | Yes — grows with $h$ | No — added equally everywhere |
| Cause | Weight of the fluid above | External force on a confined fluid |
In the hydraulic lift the two pistons sit at nearly the same level, so the $\rho g h$ difference between them is tiny next to the applied pressure and NCERT ignores it. The depth-dependence is the subject of the companion note on pressure in fluids; the present law is what acts on top of it.
Pascal's law assumes you already know how pressure grows with depth — revisit pressure in fluids for $P = P_a + \rho g h$ and gauge pressure.
The hydraulic lift
A hydraulic lift is two pistons of different cross-sections joined by a body of enclosed liquid (NCERT Fig. 9.6b). A small force $F_1$ is applied on the small piston of area $A_1$. The pressure it creates,
$$ P = \frac{F_1}{A_1}, $$
is transmitted undiminished through the liquid to the large piston of area $A_2$. That same pressure acting over the larger area produces an upward force
$$ F_2 = P A_2 = F_1 \frac{A_2}{A_1}. $$
Since $A_2 > A_1$, the output force $F_2$ exceeds the input force $F_1$. The ratio $A_2/A_1$ is the mechanical advantage of the device. By varying $F_1$ the platform is raised or lowered. The same arrangement, with minor changes, becomes a hydraulic press, a hydraulic balance and a hydraulic jack.
Force multiplication and the conservation of work
The factor $A_2/A_1$ tempts students to imagine free energy. It is not. The fluid is essentially incompressible, so the volume of liquid pushed out from under the small piston equals the volume that lifts the large piston:
$$ A_1 d_1 = A_2 d_2 \quad\Rightarrow\quad d_2 = d_1 \frac{A_1}{A_2}. $$
The large piston rises through a smaller distance $d_2$ in exactly the proportion by which its area is larger. Multiply force and distance together for each piston:
$$ W_{\text{out}} = F_2 d_2 = \left(F_1 \frac{A_2}{A_1}\right)\left(d_1 \frac{A_1}{A_2}\right) = F_1 d_1 = W_{\text{in}}. $$
The work done on the system equals the work it delivers. The device trades distance for force; it does not create energy. This is the single most examined idea on the topic.
Hydraulic brakes
Hydraulic brakes apply the transmission form of Pascal's law to stopping a vehicle. Pressing the pedal pushes a master piston into a master cylinder filled with brake oil. The pressure this creates is transmitted, undiminished, through the oil to slave pistons of larger area at the wheels, which force the brake shoes against the drum or disc. A gentle push on the pedal becomes a large retarding force at the wheel.
Because the same pressure reaches every wheel cylinder, the braking effort is shared equally across all four wheels. That uniform distribution is the safety advantage: the vehicle does not slew when the brakes bite.
Worked relations
Two syringes (no needles) filled with water are joined by a tube. The smaller and larger piston diameters are 1.0 cm and 3.0 cm. (a) Find the force on the larger piston when 10 N is applied to the smaller. (b) If the smaller piston is pushed in 6.0 cm, how far does the larger piston move out?
(a) Force. Pressure is transmitted undiminished, so $F_2 = F_1 (A_2/A_1) = F_1 (r_2/r_1)^2 = 10 \times (1.5/0.5)^2 = 10 \times 9 = 90~\text{N}$.
(b) Displacement. Water is incompressible, so $A_1 L_1 = A_2 L_2$, giving $L_2 = L_1 (r_1/r_2)^2 = 6.0 \times (0.5/1.5)^2 = 6.0/9 \approx 0.67~\text{cm}$. The larger piston moves only 0.67 cm — force up by 9, distance down by 9.
In a car lift, compressed air exerts force $F_1$ on a small piston of radius 5.0 cm. The pressure is transmitted to a second piston of radius 15 cm. To lift a car of mass 1350 kg, find $F_1$ and the pressure needed. ($g = 9.8~\text{m s}^{-2}$.)
Force. The big piston must support $F_2 = mg = 1350 \times 9.8 = 13230~\text{N}$. Since $F_1 = F_2 (A_1/A_2) = F_2 (r_1/r_2)^2 = 13230 \times (5/15)^2 = 13230/9 \approx 1470~\text{N} \approx 1.5\times10^3~\text{N}$.
Pressure. $P = F_1/A_1 = 1470 / (\pi \times 0.05^2) \approx 1.9\times10^5~\text{Pa}$ — almost double atmospheric pressure, which is common to both pistons and cancels.
A hydraulic automobile lift is designed to lift cars of maximum mass 3000 kg. The load-bearing piston has cross-section 425 cm². What maximum pressure must the smaller piston bear?
Solution. The same pressure acts on both pistons. The maximum load force on the big piston is $F_2 = mg = 3000 \times 9.8 = 29400~\text{N}$ over $A_2 = 425\times10^{-4}~\text{m}^2$. So $P = F_2/A_2 = 29400 / (425\times10^{-4}) \approx 6.92\times10^5~\text{Pa}$. The small piston must bear this same pressure.
Other applications
The transmission of pressure powers a family of machines. They differ only in what the output force does.
| Device | Input | Output use |
|---|---|---|
| Hydraulic lift / jack | Small force on small piston | Raises a car or truck on the platform |
| Hydraulic press | Force on plunger | Compacts cotton bales, presses metal |
| Hydraulic brakes | Foot on pedal → master cylinder | Equal braking force on all four wheels |
| Hydraulic balance | Pressure from a known load | Compares large weights |
In every case the same physics applies: an enclosed fluid carries an applied pressure unchanged to a larger area, and the force grows by the area ratio while the displacement shrinks by it. The fluid chosen is a liquid, not a gas, because a liquid's near-incompressibility makes the response immediate. This is where the topic hands off to fluids in motion — once a fluid flows, the rest comes from Bernoulli's principle and the equation of continuity.
Pascal's law in one breath
- A pressure change applied to an enclosed fluid at rest is transmitted undiminished to every point and to the walls.
- Pressure at a point acts equally in all directions — it is a scalar; force on any area is normal to it.
- Hydraulic lift: $P = F_1/A_1$ reaches the big piston, so $F_2 = F_1(A_2/A_1)$. The ratio $A_2/A_1$ is the mechanical advantage.
- Incompressibility gives $A_1 d_1 = A_2 d_2$, so $F_1 d_1 = F_2 d_2$ — force is multiplied, work is not.
- Small piston moves the large distance; large piston moves the small distance.
- Hydraulic brakes transmit the master-cylinder pressure equally to all four wheels.
- Do not confuse this with $P = P_a + \rho g h$, which is gravity-driven variation with depth.