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A garage jack lets one person lift a car with a few pumps of a handle. Nothing about that breaks physics, but it is easy to set up the numbers wrongly, and examiners know it. This post works a full jack example with force, distance and energy, then adds the height-difference twist that turns a NEET-level question into a JEE one.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "Pascal's law and its applications" next to pressure due to a fluid column. The JEE (Advanced) 2026 brochure lists "Pressure in a fluid; Pascal's law". The textbook is sections 9.2.1 and 9.2.4 of NCERT Class 11 Physics, Mechanical Properties of Fluids (2026-27 reprint). Check the syllabus for your exam year.
What does Pascal's law say?
NCERT gives it two ways. In a fluid at rest, pressure is the same at all points at the same height. And when external pressure is applied on any part of a fluid in a vessel, "it is transmitted undiminished and equally in all directions".
For a hydraulic machine with a small piston of area A₁ and a large one of area A₂ at the same level:
F₁/A₁ = F₂/A₂, so F₂ = (A₂/A₁) × F₁
NCERT calls A₂/A₁ the mechanical advantage. Since area is πd²/4, the ratio of areas is the ratio of diameters squared.
Worked example: a garage jack
These are sample numbers. A car and its lifting platform have a mass of 1200 kg. The large piston has a diameter of 25 cm and the small one 2.5 cm. Take g = 9.8 m/s².
Area ratio: (25 ÷ 2.5)² = 10² = 100.
Load: 1200 × 9.8 = 11,760 N.
Force on the small piston: 11,760 ÷ 100 = 117.6 N.
A handle lever with a 6 : 1 ratio cuts the push further: 117.6 ÷ 6 = 19.6 N, about the weight of a 2 kg bag.
Oil pressure under the large piston: 11,760 ÷ (π × 0.125²) = 11,760 ÷ 0.0491 ≈ 2.40 × 10⁵ Pa, about 2.4 times atmospheric pressure, above atmospheric.
Where does the extra force come from?
Not from nowhere. NCERT treats the liquid as incompressible, so the volume pushed in by the small piston equals the volume pushed out under the large one: L₁A₁ = L₂A₂.
To raise the car 10 cm, the small piston must move 0.10 × 100 = 10 m in total.
With a 5 cm stroke per pump, that is 10 ÷ 0.05 = 200 strokes.
Work done on the small piston: 117.6 N × 10 m = 1176 J.
Work done on the car: mgh = 1200 × 9.8 × 0.10 = 1176 J.
The two match. The jack multiplies force and divides distance; it does not create energy. Real jacks lose a little to friction, so you put in slightly more.
What if the pistons are at different heights?
Pascal's law in its simple form assumes both pistons are at the same level. If not, the pressure also changes with depth, as in our post on hydrostatic pressure and depth.
Sample twist: the large piston sits 0.50 m above the small one and the oil's density is 900 kg/m³. The small piston is deeper, so its pressure must be higher by ρgh:
ρgh = 900 × 9.8 × 0.50 = 4410 Pa.
Pressure at the small piston: 239,573 + 4410 ≈ 243,983 Pa.
Force needed: 243,983 × (π × 0.0125²) ≈ 119.8 N, about 1.8 per cent more than before.
Small here, but JEE questions often make the height large enough to matter.
How do hydraulic brakes use the same law?
NCERT's other example is the car brake. Pressing the pedal moves a piston in the master cylinder, and the pressure is carried by brake oil to pistons of larger area at the wheels, which press the brake shoes against the lining. Because pressure is transmitted equally, the braking effort is equal on all four wheels.
Which mistakes cost marks?
Using the diameter ratio instead of its square. Ten times the diameter is a hundred times the area.
Mixing radius and diameter in π r². Half of 25 cm is 0.125 m.
Leaving areas in cm² when you want pascals.
Adding atmospheric pressure to one side only. NCERT notes that atmospheric pressure acts on both pistons and can be ignored.
Forgetting the distance check and claiming the machine gives free energy.
A practice routine
Redo the jack example with a 3.0 cm small piston and find the new handle force.
Solve NCERT exercise 9.8: a lift for cars up to 3000 kg with a 425 cm² load piston. You should get a maximum pressure of about 6.9 × 10⁵ Pa.
Add a 1.0 m height difference to your answer for step 1 and find the change.
Explain the brake system aloud, naming where Pascal's law acts.
If the height twist trips you up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. For the people behind the idea, read about Blaise Pascal and Joseph Bramah and the hydraulic press.
Next time you see a car on a jack, estimate the area ratio from the piston sizes.
