Lesson 2 of 12 · 7 min
Pascal's law and pressure with depth
NCERT §9.2.1, §9.2.2
On Sunday Uncle Joseph takes Kabir swimming in the backwaters. Ten metres down, Kabir's ears ache. He wonders whether the pain comes from all the water in the lake or only the water directly above him.
The lesson in notes
In short
Pascal's observation: in a fluid at rest, any two points at one common height share one common pressure.
Proof with a tiny right-angled prism of fluid: the normal forces balance (F_b sin θ = F_c, F_b cos θ = F_a) and the face areas obey the same relations (A_b sin θ = A_c, A_b cos θ = A_a), so P_a = P_b = P_c.
So pressure at a point acts equally in every direction; like other stresses it has no direction. The force on any surface inside a resting fluid is normal to that surface, whatever its orientation.
A horizontal bar of fluid in equilibrium must have equal pressures at its two ends. Unequal pressures in a horizontal plane would drive a flow, so without flow the pressure is uniform across any horizontal plane.
A vertical cylinder of fluid, base area A and height h: (P₂ − P₁)A = mg and m = ρhA, so P₂ − P₁ = ρgh.
With point 1 at an open surface, P₁ = P_a (atmospheric) and P = P_a + ρgh. The excess P − P_a = ρgh at depth h is the gauge pressure.
Neither the area nor the shape of the container appears in P = P_a + ρgh: only the height of the fluid column matters.
Hydrostatic paradox: vessels A, B and C of different shapes, joined at the bottom by a horizontal pipe, fill to the same level although they hold different amounts of water, because the pressure at the bottom is the same under each.
Swimmer 10 m below a lake surface (ρ = 1000 kg m⁻³, g = 10 m s⁻²): P = 1.01 × 10⁵ + 1000 × 10 × 10 = 2.01 × 10⁵ Pa ≈ 2 atm, double the surface value. At 1 km depth the rise is about 100 atm, which submarines must be built to withstand.