Q.A swimmer is swimming in a swimming pool at 6 m below the surface of the water. Calculate the pressure on the swimmer due to water above. (Density of water = 1000 kg/m³, g = 9.8 m/s²)

Pressure is defined as the normal force a fluid at rest exerts per unit area of any surface it touches, $p = F/A$, with SI unit N/m² (pascal) and dimensional formula $[L^{-1}M^1T^{-2}]$ — the same dimensions as stress, since both are a force per unit area, though pressure, unlike stress, is a scalar with no direction of its own. A small force concentrated on a tiny area can produce an enormous pressure (a knife's edge, a needle's tip), while the same force spread over a large area produces only a small one.

For a liquid of density ρ at rest, the pressure at a depth h below its free surface is $p = h\rho g$, derived by balancing the weight of an imaginary vertical column of the liquid against the force on its own base — a result that turns out to be entirely independent of the shape or cross-sectional area of the container, depending only on the depth and the liquid's density. This is why liquid poured into a set of interconnected vessels of very different shapes always settles to the same level in every vessel (the so-called hydrostatic paradox, resolved once the sideways components of a tapering wall's reaction force on the liquid are properly included) — and it is …