Find pressure intensity in the depth of 50 metre below the free water level. Assume that specific weight of water is equal to 9.81 kN/m³.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 25-09-2019 (Morning)
Detailed Explanation
The intensity of hydrostatic pressure increases linearly with depth below the free surface of an incompressible liquid, governed directly by the fluid's specific weight.
1. Identify given values:
- Depth below free surface \(h = 50\text{ m}\)
- Specific weight of water \(\gamma_w = 9.81\text{ kN/m}^3\)
2. Calculate the hydrostatic pressure intensity:
\[ p = \gamma_w \times h \]
\[ p = 9.81\text{ kN/m}^3 \times 50\text{ m} = 490.50\text{ kN/m}^2 \]
In standard SI units: \(490.50\text{ kPa} = 0.4905\text{ MPa}\).
Why “490.50 kN/m²” Is Correct
Applying the hydrostatic law \(p = \gamma h\), multiplying the specific weight of water (\(9.81\text{ kN/m}^3\)) by the depth of \(50\text{ m}\) yields exactly \(490.50\text{ kN/m}^2\).
Why the Other Options Are Wrong
- A. 550.50 kN/m²: 550.50 kN/m² results from adding 60 to the correct pressure.
- B. 390.50 kN/m²: 390.50 kN/m² results from mistakenly using specific weight γ = 7.81 kN/m².
- C. 450.50 kN/m²: 450.50 kN/m² results from using γ = 9.01 kN/m².
Key Point
Hydrostatic pressure intensity: p = γ · h = ρ · g · h.
Key Concepts for Students
- Pressure intensity increases by 9.81 kPa for every 1 metre of water depth.
- 10 metres of water depth corresponds approximately to 1 atmosphere (98.1 kPa ≈ 100 kPa = 1 bar).
- Pressure intensity is independent of surface area and depends exclusively on vertical depth.
Practice more: this is question 15 in Hydrostatic Pressure MCQs – Part 1. Try the whole set and check every answer instantly.