The total and effective stresses at a depth of 5 m below the top level of water in a pond are respectively:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2004
Detailed Explanation
At a point under water (with no soil above it), the total vertical stress comes only from the weight of the water column, and the pore water pressure is the same hydrostatic pressure.
\[ \sigma = \gamma_w h = 10 \times 5 = 50\ \text{kN/m}^2 \approx 0.5\ \text{kg/cm}^2 \]
\[ u = \gamma_w h = 0.5\ \text{kg/cm}^2 \qquad \sigma' = \sigma - u = 0 \]
Why “0.5 kg/cm² and 0.0” Is Correct
Total stress is 0.5 kg/cm² and effective stress is zero, because the water above carries no load through grain contacts. (Effective stress in the pond bed soil then builds up only from the submerged weight of the soil itself.)
Why the Other Options Are Wrong
- A. 0.0 and 0.0: Total stress is not zero; the 5 m of water has weight.
- C. 0.5 kg/cm² and 0.6 kg/cm²: Effective stress cannot exceed total stress here; it is zero.
- D. none of these: Option (b) is correct.
Key Point
Under standing water: σ = u = γw·h ⇒ σ′ = 0 at the bed surface.
Key Concepts for Students
- 1 kg/cm² ≈ 98 kN/m² ≈ 100 kN/m².
- Raising the pond level does not change effective stress in the bed soil.
- That is why the effective stress in seabed sediments depends only on γ′ × depth into the soil.
Practice more: this is question 52 in Shear Strength of Soils MCQs – Part 4. Try the whole set and check every answer instantly.