Hydrostatics – Pressure variation with depth in a liquid at rest At any point in a static liquid, the intensity of pressure is __________ to the depth measured below the free surface.

Introduction:
Hydrostatic pressure is foundational for manometers, dam design, and submerged structural analysis. In incompressible fluids at rest, pressure increases with depth due to the weight of the overlying fluid column, a relationship sometimes called Pascal's hydrostatic law.

Given Data / Assumptions:

• Liquid at rest; density rho approximately constant.
• Uniform gravitational field g.
• Depth h is measured vertically below the free surface.

Concept / Approach:

The hydrostatic equilibrium equation is dp/dz = − rho * g (z upward). Integrating between the free surface (pressure P0) and a point at depth h gives p = P0 + rho * g * h. Therefore, gauge pressure p − P0 is directly proportional to depth h, independent of the vessel shape (hydrostatic paradox).

Step-by-Step Solution:

Verification / Alternative check:

Manometer readings confirm that equal increments in depth produce equal increments in pressure; changing vessel shape does not alter the local pressure at a given depth.

Why Other Options Are Wrong:

Equal / inverse / independent: Contradict the derived linear relation.Proportional to container volume: Pressure is intensive and does not depend on total volume.

Common Pitfalls:

Assuming wider containers give higher pressure; only depth and density matter (for given P0 and g).

Final Answer:

directly proportional