Hydrostatics – pressure distribution on a vertical wall (single liquid) A vertical wall retains a single, homogeneous liquid on one side only. Which statement best describes the hydrostatic pressure along the wall from the free surface to the bottom?

Introduction:Hydrostatic pressure increases with depth due to the weight of the overlying fluid. Understanding the distribution is essential for computing forces on retaining walls, tanks, and gates.

Given Data / Assumptions:

• Liquid is at rest; density is constant.
• Free surface is exposed to atmosphere.
• Wall is vertical; gravity is uniform.

Concept / Approach:The hydrostatic relation is p = p_atm + w * h, where h is depth below the free surface. Gauge pressure p_g = p − p_atm = w * h increases linearly with h, from zero at the free surface to maximum at the deepest point.

Step-by-Step Solution:1) At h = 0 (free surface), p_g = 0 → pressure equals atmospheric.2) At the bottom (h = H), p_g = w * H → maximum pressure.3) The wall therefore experiences a triangular pressure distribution.4) The resultant hydrostatic force acts at H/3 above the bottom (center of pressure for a vertical plane).

Verification / Alternative check:Tank gauge readings and force integrations on test panels match the linear rise in pressure with depth.

Why Other Options Are Wrong:

• Minimum but not zero: at the free surface, gauge pressure is precisely zero.
• Bottom zero or uniform pressure: contradicts hydrostatic law.

Common Pitfalls:Confusing absolute and gauge pressure; forgetting free-surface reference.

Final Answer:The pressure at the liquid level is zero, and the pressure at the bottom is maximum.