Hydrostatics – Pressure variation with depth At any point within a static liquid, the intensity of pressure varies how with depth measured below the free surface?

Introduction:
Hydrostatic pressure distribution in an incompressible fluid at rest follows a simple linear law with depth under a uniform gravitational field. Recognizing this relationship is foundational for manometry, buoyancy, and design of storage tanks and dams.

Given Data / Assumptions:

• Liquid at rest (no shear stresses) and constant density rho.
• Uniform gravitational acceleration g.
• Reference free surface at pressure P0 (often atmospheric).

Concept / Approach:

From hydrostatic equilibrium, dp/dz = − rho * g with z positive upward. Integrating between the free surface and a depth h gives p = P0 + rho * g * h. Therefore, the gauge pressure increases linearly and is directly proportional to depth h beneath the surface, independent of the container’s shape or plan area (Pascal’s hydrostatic paradox).

Step-by-Step Solution:

Verification / Alternative check:

Experimental manometer readings show equal increments of depth produce equal increments of pressure, regardless of vessel shape, confirming the linear law.

Why Other Options Are Wrong:

Proportional to area or length of the vessel: Pressure is an intensive property and does not depend on container dimensions.Inversely proportional to depth: Opposite to the hydrostatic law.Independent of depth: Only true at the free surface or in microgravity; not in normal hydrostatics.

Common Pitfalls:

Assuming a wide container has higher pressure because it holds more fluid; pressure depends only on vertical depth and density, not on total volume or shape.

Final Answer:

directly proportional to the depth of liquid from the surface