Pascal's law states that a change in pressure applied to an enclosed, incompressible fluid at rest is transmitted undiminished to every point of the fluid and to the walls of the container. This is documented in NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), page 181.
The law follows directly from the fact that pressure in a static fluid acts equally in all directions at a given depth. If you increase the pressure at one point — say by pushing a piston — that increase propagates throughout the fluid without loss. The underlying formula for pressure in a static fluid, P = P₀ + ρgh, already encodes this: the surface pressure P₀ appears as an additive constant at every depth.
Where NEET tests this: Questions on Pascal's law typically ask you to apply the hydraulic-machine principle. A hydraulic lift has two pistons of different cross-sectional areas A₁ and A₂. Because pressure is transmitted equally, F₁/A₁ = F₂/A₂, so the output force is amplified by the area ratio: F₂ = F₁ × (A₂/A₁). The trade-off is displacement — the smaller piston must travel a proportionally greater distance (work in = work out, assuming an ideal system).
Common confusion: Students sometimes think Pascal's law means pressure is the same everywhere in a fluid. It does not. Pressure still varies with depth (ρgh term). Pascal's law says a change in pressure is transmitted uniformly. A question that gives two points at different depths and asks whether the pressure change is the same at both — the answer is yes. A question that asks whether the absolute pressure is the same — the answer is no.
Watch-out for NEET: Hydraulic-lift problems sometimes give diameters instead of radii. Area scales as the square of radius (or diameter), so a diameter ratio of 1:10 gives an area ratio of 1:100, not 1:10.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Pascal's law applies to which of the following?
Answer: B. Pascal's law applies to an enclosed, incompressible fluid at rest. This is directly stated in NCERT Class 11 Physics Chapter 9, page 181.
Why A is wrong: A is wrong because Pascal's law requires the fluid to be at rest, not in turbulent motion.
Why C is wrong: C is wrong because Pascal's law requires an enclosed, incompressible fluid. A compressible gas in an open container violates both conditions.
Why D is wrong: D is wrong because a fluid flowing through a pipe is not at rest. Pascal's law applies to static fluids.
Pascal's law states that a pressure change applied to an enclosed fluid at rest is transmitted:
Answer: C. Pascal's law explicitly states the pressure change is transmitted undiminished to every point of the fluid and to the walls of the container (NCERT Class 11 Physics Chapter 9, page 181).
Why A is wrong: A is wrong because the pressure change is not confined to the bottom — it transmits to every point in the fluid.
Why B is wrong: B is wrong because pressure in a fluid acts equally in all directions at a point. The change is not directional.
Why D is wrong: D is wrong because there is no attenuation. The entire pressure increment appears at every point — that is the core of Pascal's law.
Which device directly uses Pascal's law as its working principle?
Answer: D. A hydraulic lift uses Pascal's law: pressure applied at one piston is transmitted undiminished to the other piston, producing a magnified force proportional to the area ratio (NCERT Class 11 Physics Chapter 9, page 181).
Why A is wrong: A is wrong because a mercury barometer measures atmospheric pressure using the height of a mercury column. It relies on the pressure-depth relation, not Pascal's law of pressure transmission.
Why B is wrong: B is wrong because a U-tube manometer measures pressure difference using the height difference of liquid columns. It uses the hydrostatic pressure formula, not Pascal's law of transmission.
Why C is wrong: C is wrong because a Venturi meter measures flow rate using Bernoulli's principle, not Pascal's law.
In a hydraulic lift, the cross-sectional area of the smaller piston is 5.0 cm² and that of the larger piston is 250 cm². A force of 20 N is applied on the smaller piston. What is the maximum force that can be lifted by the larger piston? (Assume an ideal system.)
Answer: D. By Pascal's law, F₁/A₁ = F₂/A₂. So F₂ = F₁ × (A₂/A₁) = 20 × (250/5.0) = 20 × 50 = 1000 N. This is a direct application of the hydraulic-machine principle from NCERT Class 11 Physics Chapter 9, page 181.
Why A is wrong: A (100 N) results from dividing the area ratio by 10 instead of computing it correctly — a careless arithmetic error.
Why B is wrong: B (500 N) is half the correct value: it uses an area ratio of 25 instead of 250/5.0 = 50.
Why C is wrong: C (2500 N) results from multiplying force by the area of the larger piston directly (20 × 250/2), ignoring the ratio. The correct calculation requires dividing by the smaller piston's area.
A hydraulic brake system has a master cylinder of diameter 2.0 cm and a wheel cylinder of diameter 6.0 cm. If the driver applies a force of 50 N on the master cylinder, what force acts on the brake pad at the wheel cylinder?
Answer: A. Area of master cylinder = π(1.0)² = π cm². Area of wheel cylinder = π(3.0)² = 9π cm². Area ratio = 9. Force at wheel cylinder = 50 × 9 = 450 N. Note: the problem gives diameters, so radii are 1.0 cm and 3.0 cm respectively.
Why B is wrong: B (150 N) results from using the diameter ratio (6.0/2.0 = 3) instead of the area ratio. Force scales with area (which goes as diameter squared), not diameter. This is a common confusion in hydraulic-lift problems.
Why C is wrong: C (300 N) results from using the diameter ratio of 3 and then doubling — an arithmetic confusion that doesn't follow from the physics.
Why D is wrong: D (600 N) likely results from squaring the diameter ratio but then multiplying an extra factor. The correct area ratio is (3.0/1.0)² = 9, giving 50 × 9 = 450 N.
Two points X and Y are at depths h and 2h respectively in an enclosed static liquid of density ρ. If the pressure at the surface is increased by ΔP, the new pressure at point Y minus the new pressure at point X equals:
Answer: C. Before the change: P_X = P₀ + ρgh, P_Y = P₀ + 2ρgh, so P_Y − P_X = ρgh. After the surface pressure increases by ΔP (transmitted undiminished by Pascal's law): P_X' = (P₀ + ΔP) + ρgh, P_Y' = (P₀ + ΔP) + 2ρgh. The difference P_Y' − P_X' = ρgh — unchanged. The ΔP cancels because it adds equally to both points.
Why A is wrong: A (ΔP) confuses the pressure change at the surface with the pressure difference between two depths. Pascal's law says the change ΔP is the same at both points, so it cancels in the difference.
Why B is wrong: B (ρgh + ΔP) incorrectly assumes ΔP adds to the difference. Since ΔP is transmitted equally to both X and Y, it cancels when you subtract.
Why D is wrong: D (2ΔP) has no physical basis. The pressure change ΔP appears once at each point; it does not double.
A student claims: "Pascal's law means the pressure is the same at all points in a static fluid." Which of the following is the correct assessment?
Answer: B. Pascal's law states that a change in pressure applied to an enclosed fluid at rest is transmitted undiminished. Absolute pressure still depends on depth (P = P₀ + ρgh). The student's claim conflates uniform transmission of a pressure change with uniform absolute pressure — a common conceptual error.
Why A is wrong: A is wrong because absolute pressure varies with depth due to the ρgh term. Pressure is NOT the same everywhere in a static fluid.
Why C is wrong: C is wrong because even for an incompressible fluid, pressure varies with depth. Incompressibility is a condition for Pascal's law to hold, but it does not make absolute pressure uniform.
Why D is wrong: D is wrong because Pascal's law says the pressure change transmits to every point — in all directions, not just horizontally. The error in the student's claim is about confusing pressure change with absolute pressure, not about direction.
A hydraulic press has pistons of cross-sectional areas 10 cm² and 500 cm². The smaller piston is pushed down by 20 cm. Assuming the fluid is incompressible, how far does the larger piston rise?
Answer: A. By conservation of volume (incompressible fluid): A₁ × d₁ = A₂ × d₂. So d₂ = (A₁/A₂) × d₁ = (10/500) × 20 = 0.40 cm. This is a two-step problem: first apply Pascal's law to recognize the system, then apply volume conservation to find displacement.
Why B is wrong: B (2.8 cm) results from using the square root of the area ratio instead of the area ratio itself: √(10/500) × 20 ≈ 2.8 cm. Volume conservation needs the area ratio itself.
Why C is wrong: C (4.0 cm) results from a decimal slip in the area ratio: reading 10/500 as 0.2 instead of 0.02, giving 20 × 0.2 = 4.0 cm.
Why D is wrong: D (0.20 cm) results from dividing 20 by 100 instead of by 50. The correct area ratio is 500/10 = 50, giving d₂ = 20/50 = 0.40 cm.
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Given
A hydraulic lift has a small piston of diameter 4.0 cm and a large piston of diameter 40 cm. A car of mass 3000 kg is to be lifted. Take g = 10 m/s² (exact, problem-defined).
Required
Find the minimum force F₁ that must be applied on the smaller piston.
Concept
By Pascal's law, pressure applied at the small piston is transmitted undiminished to the large piston. Therefore F₁/A₁ = F₂/A₂, where F₂ is the weight of the car.
Formula
F₁ = F₂ × (A₁/A₂)
Since area = π(d/2)², the area ratio simplifies to (d₁/d₂)².
Substitution
F₂ = mg = 3000 × 10 = 30000 N (exact, since g = 10 m/s² is problem-defined exact).
d₁ = 4.0 cm, d₂ = 40 cm.
Area ratio = (4.0/40)² = (0.10)² = 0.010.
F₁ = 30000 × 0.010 = 300 N.
Calculation
F₁ = 30000 × 0.010 = 300 N.
Note on exact values: g = 10 m/s² is an explicitly exact value given in the problem. The mass 3000 kg and diameters 4.0 cm, 40 cm carry 2 significant figures. The final answer is reported to 2 significant figures.
Final answer
F₁ = 3.0 × 10² N.
The diameter ratio is 1:10, giving an area ratio of 1:100. A 300 N push (roughly the weight of a 30 kg object) lifts a 3000 kg car — that is the mechanical advantage of a hydraulic lift.
Common trap
Using the diameter ratio (1:10) directly as the force ratio instead of squaring it to get the area ratio (1:100). This would give F₁ = 3000 N — ten times the correct answer. Always convert diameters to areas before applying Pascal's law.
Similar NEET-style question
A hydraulic jack has cylinders of diameters 2.0 cm and 20 cm. What force on the smaller cylinder is needed to support a 1500 kg load? (Answer: F₁ = 1500 × 10 × (2.0/20)² = 150 N.)
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A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. Basis of hydraulic press, brakes, lift.
-- NCERT Class 11 Physics, Ch. 9, p. 185Pressure at depth h below free surface of fluid of density rho.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | total pressure | Pa |
| P0 | atmospheric/surface pressure | Pa |
| rho | density | kg/m^3 |
| h | depth | m |
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