Pascal's law — hydraulic lift calculation and applications

Question

In a hydraulic lift, the area of the small piston is $A_1 = 5\text{ cm}^2$ and the area of the large piston is $A_2 = 200\text{ cm}^2$ . A force of $F_1 = 25$ 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

Apply Pascal's law for force

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:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

Calculate the output force

$F_2 = F_1 \times \frac{A_2}{A_1} = 25 \times \frac{200}{5} = 25 \times 40$

$\boxed{F_2 = 1000 \text{ N}}$

A small 25 N push produces 1000 N of lift — a force multiplication factor of 40.

Find displacement using volume conservation

The fluid is incompressible, so the volume displaced by the small piston equals the volume received by the large piston:

$A_1 \times d_1 = A_2 \times d_2$

$d_2 = d_1 \times \frac{A_1}{A_2} = 10 \times \frac{5}{200} = \mathbf{0.25 \text{ cm}}$

Verify with energy conservation

Work done on small piston: $W_1 = F_1 \times d_1 = 25 \times 0.10 = 2.5$ J

Work done by large piston: $W_2 = F_2 \times d_2 = 1000 \times 0.0025 = 2.5$ 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 $A_2/A_1$ is the mechanical advantage, just like the turns ratio $N_2/N_1$ is the voltage amplification factor. In both cases, power (or energy per unit time) is conserved.

💡 Expert Tip

For NEET and CBSE boards, remember the two key equations: $F_2/F_1 = A_2/A_1$ (force amplification) and $d_1/d_2 = A_2/A_1$ (displacement reduction). These two always come together. If a question gives three of the four quantities, you can find the fourth.

Common Mistake

⚠️ Common Mistake

Students often forget to convert units consistently. If areas are in cm $^2$ and displacement in cm, the ratio works out fine. But if you mix cm $^2$ with m $^2$ , the answer will be off by a factor of $10^4$ . Either convert everything to SI at the start, or ensure both areas are in the same unit so the ratio is dimensionless.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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