Hydrostatic Pressure and Communicating Vessels
Hydrostatic equilibrium requires the pressure gradient to balance body force; for constant density in uniform gravity, pressure increases linearly with downward vertical distance.
Why this shows up in the exam
Manometers and barometers · Dam and aquarium wall loading · Liquid-level equalisation in connected vessels
Learn the idea
In a fluid at rest, pressure changes with vertical depth and is equal at equal levels in one connected liquid. A deeper point supports a taller column of fluid above it. Container shape is irrelevant to pressure at a point, but the density and vertical height of every layer matter.
🧠 Memory hook: Static pressure counts vertical depth, not the container's shape or sloping path.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- P₂ - P₁ = rho g(h₁-h₂) — pressure difference in one static incompressible fluid, with h measured upward
- P_gauge = rho g d — gauge pressure at depth d below a free surface exposed to the reference pressure
- sum(rho_i g Delta h_i) = net pressure change — piecewise balance through immiscible static columns; signs follow upward or downward travel
How to approach it
- 1Mark one reference pressure such as the atmosphere
- 2Walk through each column with signed rho g Delta h
- 3Check that pressure rises when moving downward
Common slip-ups that cost marks
- •Using path length instead of vertical height
- •Adding atmospheric pressure when gauge pressure is requested
- •Equating pressures at equal heights in disconnected liquids
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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