A vertical retaining wall holds back water to a depth H (free surface at the top, vertical plane surface). The height of the centre of pressure above the bottom of the wall is:

Introduction / Context:
The centre of pressure locates the line of action of the resultant hydrostatic force on a submerged plane surface. For design of retaining walls and gates, its position determines overturning moments and required stability.

Given Data / Assumptions:

• Vertical plane surface with its top at the free surface, depth H.
• Hydrostatic pressure distribution increases linearly with depth.
• Fluid is at rest; density constant.

Concept / Approach:
For a vertical rectangle with top at the free surface, the resultant force acts at depth 2H/3 below the free surface. Therefore, the distance above the bottom (which is at depth H) is H − 2H/3 = H/3.

Step-by-Step Solution:
Pressure at depth y: p = ρ g y.Resultant force acts at y_cp = (∫ y p dA)/(∫ p dA) = 2H/3 below free surface for a vertical rectangle starting at the surface.Height above bottom = H − 2H/3 = H/3.

Verification / Alternative check:
Using the second-moment formula: y_cp = ȳ + I_G/(ȳ A), for this case ȳ = H/2 and I_G = b H^3/12; substituting yields y_cp = 2H/3, consistent with the simple result.

Why Other Options Are Wrong:
H/2 and 3H/4 contradict the linear pressure centroid location.2H/3 is the depth below the free surface, not height above bottom.

Common Pitfalls:

• Reporting the depth below the surface when the question asks distance above the bottom.
• Confusing centroid of area with centre of pressure (they differ due to pressure variation).

Final Answer:
H/3