Fluid Pressure and Pascal's Law

A static fluid exerts pressure that depends only on depth, density and the value at the surface. Pascal's law converts this idea into the hydraulic press, a recurring NEET application.

Concept

In a fluid at rest pressure acts equally in all directions on every surface. With increasing depth the weight of the column above raises the pressure linearly. Pascal's law states that any change in pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

The shape of the vessel does not affect the pressure at a given depth — this is the hydrostatic paradox.

Formula Derivation

Consider a thin horizontal slab of area $A$ and thickness $dh$ inside a fluid of density $\rho$. Vertical equilibrium gives

$(P + dP) A - P A = \rho A\,dh\,g \implies \frac{dP}{dh} = \rho g.$

Integrating from the free surface (pressure $P_{0}$) to depth $h$,

$P = P_{0} + \rho g h.$

For a hydraulic press with pistons of areas $A_{1}$ and $A_{2}$, Pascal's law equates the transmitted pressures:

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \implies F_{2} = F_{1}\,\frac{A_{2}}{A_{1}}.$

The mechanical advantage is the area ratio $A_{2}/A_{1}$.

NEET-style Worked Example

A diver is at $20\ \text{m}$ below the surface of a lake. Take $\rho_{\text{water}} = 10^{3}\ \text{kg/m}^{3}$, $g = 10\ \text{m/s}^{2}$, $P_{0} = 10^{5}\ \text{Pa}$. Total pressure on the diver?

$P = P_{0} + \rho g h = 10^{5} + 10^{3} \times 10 \times 20 = 3 \times 10^{5}\ \text{Pa} = 3\ \text{atm}.$

In a hydraulic press, if $A_{1} = 2\ \text{cm}^{2}$ and $A_{2} = 200\ \text{cm}^{2}$, a force of $50\ \text{N}$ on the small piston produces $F_{2} = 50 \times 100 = 5000\ \text{N}$ on the load.

Common Confusions

• The pressure depends only on depth, not on the shape or amount of fluid above.
• $P = P_{0} + \rho g h$ is gauge pressure plus atmospheric: drop $P_{0}$ to get gauge pressure alone.
• The hydraulic press multiplies force but not work — the small piston moves a longer distance.
• In a U-tube with two immiscible liquids, pressure equality at the lowest point determines the height ratio.

Key Takeaways

• $\dfrac{dP}{dh} = \rho g$ gives $P = P_{0} + \rho g h$.
• Pressure acts equally in all directions in a static fluid.
• Hydraulic press: $F_{2}/F_{1} = A_{2}/A_{1}$.
• Atmospheric pressure $\approx 1.013 \times 10^{5}\ \text{Pa} = 760\ \text{mm Hg}$.
• Energy conservation forbids work-multiplication in a press.