Hydrostatics — Pressure at a Depth in Water Calculate the gauge pressure at a point located 4 m below the free surface of water (assume rho ≈ 1000 kg/m^3 and g ≈ 9.81 m/s^2).

Introduction:Hydrostatic pressure in a liquid at rest increases linearly with vertical depth below the free surface. This simple but powerful relation is fundamental for manometer readings, dam design, and underwater equipment ratings.

Given Data / Assumptions:

• Depth h = 4 m below a water free surface.
• Density of water rho ≈ 1000 kg/m^3.
• Acceleration due to gravity g ≈ 9.81 m/s^2.
• Gauge reference at the free surface (p = 0 gauge at surface).

Concept / Approach:Use the hydrostatic relation p = rho * g * h. This gives pressure directly in pascals when SI units are used. Convert to kilopascals by dividing by 1000.

Step-by-Step Solution:Write p = rho * g * h.Substitute: p = 1000 * 9.81 * 4.Compute: p = 39,240 Pa.Convert to kPa: 39,240 Pa / 1000 = 39.24 kPa.

Verification / Alternative check:A quick estimate using g ≈ 10 m/s^2 gives p ≈ 40 kPa, which is close and confirms the detailed result 39.24 kPa.

Why Other Options Are Wrong:19.24 kPa and 29.24 kPa correspond to shallower depths; 49.24 kPa would imply a depth near 5 m, not 4 m.

Common Pitfalls:Forgetting the gauge vs absolute reference; mixing units (using g in cm/s^2 or rho in g/cm^3 without converting); using slanted distance instead of vertical depth.

Final Answer:39.24 kPa