Hydrostatics of plane areas: For a plane surface of area A immersed in water with its centroid at depth x below the free surface (water density ω), which statement is correct about total pressure and the depth of the centre of pressure?

Introduction / Context:
The hydrostatic pressure on a plane surface varies linearly with depth. Designers need both the resultant (total) force and the location (centre of pressure) to design gates, slabs, and tank walls.

Given Data / Assumptions:

• Plane surface of area A; centroid at depth x below the free surface.
• Water density (specific weight) is ω.

Concept / Approach:
Total hydrostatic pressure on a plane area is P = ω A x (for plane surface of uniform density fluid). The centre of pressure is deeper than the centroid; its depth y_cp from the free surface is y_cp = I_o / (A x), where I_o is the second moment of area of the surface about the free surface. Equivalently, using the parallel-axis theorem, y_cp = x + (I_G / (A x)).

Step-by-Step Solution:
Compute total force: P = ω A x (depends on A and x).Locate centre: y_cp = I_o / (A x) = x + I_G / (A x).Recognize the correct statement corresponds to the I/(A x) relation.

Verification / Alternative check:
Derivation integrates pressure p = ω y over the surface and equates first moments to obtain the I/(A x) formula; standard in fluid mechanics texts.

Why Other Options Are Wrong:

• (a) Omits area A; incomplete.
• (c) 2/3 H applies only to a triangular pressure diagram on a vertical rectangle with the free surface at the top; not general.
• (d) Cannot be true because (a) and (c) are not universally valid.

Common Pitfalls:
Confusing I_o and I_G; forgetting that centre of pressure lies below centroid for vertical planes.

Final Answer:
Depth of the point of action (centre of pressure) equals its second moment of area about the free surface divided by A x