Question
In a hydraulic lift, the area of the small piston is and the area of the large piston is . A force of N is applied on the small piston. Find the force exerted by the large piston. Also, if the small piston is pushed down by 10 cm, how much does the large piston rise?
(NCERT Class 11, Chapter 10)
Solution — Step by Step
Pascal’s law states: pressure applied at any point in a confined fluid is transmitted equally in all directions. So the pressure on both pistons is equal:
A small 25 N push produces 1000 N of lift — a force multiplication factor of 40.
The fluid is incompressible, so the volume displaced by the small piston equals the volume received by the large piston:
Work done on small piston: J
Work done by large piston: J
Energy is conserved (ignoring friction). The hydraulic lift multiplies force but not energy — what you gain in force, you lose in displacement.
Why This Works
Pascal’s law works because fluids cannot sustain shear stress in static equilibrium. When you push on one piston, the pressure increase propagates uniformly through the entire fluid. The large piston, having a bigger area, experiences the same pressure over a larger surface — producing a proportionally larger force.
The trade-off is displacement. Since fluid volume is conserved, the large piston moves much less than the small one. Force is amplified, but the total work (force times distance) stays the same. This is the hydraulic analogue of a lever — mechanical advantage through geometry.
Alternative Method
Think of it as a “force transformer” analogous to a transformer in electricity. The area ratio is the mechanical advantage, just like the turns ratio is the voltage amplification factor. In both cases, power (or energy per unit time) is conserved.
For NEET and CBSE boards, remember the two key equations: (force amplification) and (displacement reduction). These two always come together. If a question gives three of the four quantities, you can find the fourth.
Common Mistake
Students often forget to convert units consistently. If areas are in cm and displacement in cm, the ratio works out fine. But if you mix cm with m, the answer will be off by a factor of . Either convert everything to SI at the start, or ensure both areas are in the same unit so the ratio is dimensionless.