Hydrostatics — Pressure at Depth in a Liquid For a static liquid, the pressure intensity p (in kN/m^2 or kPa) at a depth h below the free surface is given by which expression in terms of specific weight w?

Introduction:Hydrostatic pressure increases linearly with depth due to the weight of the overlying fluid. This principle underlies manometers, dams, submarine hulls, and pressure vessels in contact with liquids.

Given Data / Assumptions:

• Liquid at rest (no shear stresses).
• Constant specific weight w = rho * g (in kN/m^3).
• Depth measured vertically from the free surface by h (in m).

Concept / Approach:The hydrostatic equation is dp/dz = -rho * g. Integrating from the surface (p = 0 gauge) to depth h gives p = rho * g * h = w * h (in consistent units).

Step-by-Step Solution:Define w = rho * g.Write pressure-depth relation: p = w * h.Interpretation: pressure increases linearly with h; slope equals w.

Verification / Alternative check:Units check: w (kN/m^3) times h (m) yields kN/m^2, which equals kPa. This confirms dimensional consistency and correctness.

Why Other Options Are Wrong:Using w alone ignores depth; w/h or h/w have incorrect dimensions and do not represent hydrostatic pressure.

Common Pitfalls:Confusing specific weight with specific gravity; forgetting to convert units; using density without multiplying by g; measuring oblique distance rather than vertical depth.

Final Answer:w * h