Q.The given figure shows the principle behind the hydraulic lift.
The radius of small piston is 5.0 cm and that of larger piston is 15 cm. Calculate the force F, if the mass of the car to be lifted is 1350 kg (g = 9.8 ms^-2).

Pascal's Principle: The Intuition

Imagine you have a sealed plastic bottle completely filled with water, with no air inside. If you squeeze the bottle at one spot, what happens? The water doesn't just compress at that spot — the entire bottle bulges outward everywhere. The pressure you applied at one point travels instantly through the entire liquid.

That's the core idea: in a confined fluid, pressure applied anywhere is transmitted undiminished to every part of the fluid and to the walls of the container.

Now, why does this happen? Fluids (liquids and gases) are made of particles that can slide past each other. When you push on one part of a confined fluid, those particles bump into their neighbours, which bump into their neighbours, and so on — like a chain of dominoes. Because the fluid is nearly incompressible (especially liquids), the push doesn't get absorbed by squishing the fluid itself. It gets passed along perfectly.

The Precise Statement

In simpler words: if you increase the pressure at one point by $\Delta P$, every other point in the fluid experiences exactly the same increase $\Delta P$.

The Key Equation

The pressure $P$ is defined as force per unit area:

So if you apply a force $F_1$ over a small area $A_1$, you create a pressure $P = F_1/A_1$. That same pressure pushes on a larger area $A_2$ elsewhere, producing a larger force:

This is the force multiplication that makes hydraulic systems work.

The Classic Example: Hydraulic Lift

A car repair shop uses a hydraulic jack. Here's how it works:

1. You push down on a small piston (area $A_1 = 10\ \text{cm}^2$) with a force of $100\ \text{N}$.
2. This creates a pressure $P = 100\ \text{N} / 0.001\ \text{m}^2 = 100,000\ \text{Pa}$.
3. That same pressure pushes up on a large piston (area $A_2 = 1000\ \text{cm}^2 = 0.1\ \text{m}^2$).
4. The upward force is $F_2 = P \times A_2 = 100,000 \times 0.1 = 10,000\ \text{N}$.

You applied $100\ \text{N}$ and got $10,000\ \text{N}$ — a 100× force multiplication. But notice: the small piston moves down 100 cm while the large piston moves up only 1 cm. Work done ($F \times d$) is the same on both sides — you don't get energy for free.

Where It Applies (and Where It Doesn't)