Hydrostatics — Converting Pressure to Water Column Height What is the height of a static water column (in metres) that is equivalent to a pressure of 0.15 MPa? Use water's specific weight w ≈ 9.81 kN/m^3 and the relation p = w * h.

Introduction:
Converting a given pressure into an equivalent head of liquid is a standard hydrostatics task used in civil engineering (e.g., water supply, reservoir levels, manometer calibration). The key relation is p = w * h, where p is pressure, w is the liquid's specific weight, and h is the vertical depth (or head).

Given Data / Assumptions:

• Pressure p = 0.15 MPa = 150 kPa = 150 kN/m^2.
• Specific weight of water w ≈ 9.81 kN/m^3.
• Hydrostatic condition (fluid at rest), uniform w, gravity standard.

Concept / Approach:
The hydrostatic relation p = w * h directly connects pressure to fluid column height. Rearranging gives h = p / w. With consistent units (kN/m^2 divided by kN/m^3), the result is in metres. This linear relationship is foundational for pressure head, piezometric head, and energy grade line concepts.

Step-by-Step Solution:
Convert pressure: 0.15 MPa = 150 kN/m^2.Use h = p / w = 150 / 9.81 (m).Compute: 150 / 9.81 ≈ 15.29 m, typically rounded to 15.3 m.

Verification / Alternative check:
Sanity check using 10 kN/m^3 as an easy round value for w gives 150/10 = 15 m. The precise value with 9.81 kN/m^3 is slightly larger (≈15.29 m), which matches the detailed calculation.

Why Other Options Are Wrong:
25.3 m, 35.3 m, and 45.3 m would require much larger pressures (≈0.25–0.45 MPa) for water; they do not satisfy p = w * h for 0.15 MPa.

Common Pitfalls:
Mixing density (kg/m^3) with specific weight (kN/m^3); forgetting unit conversions between MPa, kPa, and kN/m^2; using oblique depth instead of vertical depth.

Final Answer:
15.3 m