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Shear Strength of Soils MCQs – Part 4

10 questions. Tap an option to check it instantly, then read the solution.

Question 46Easy
The unconfined compressive strength test is a/an:
Full explanation

Answer: A. Undrained test

The UCS test is quick and has no drainage — it is an undrained test (a special UU test with σ₃ = 0).

Question 47Easy
The direct shear test suffers from the following disadvantage:
Full explanation

Answer: D. All of the above

All three are known disadvantages of the direct shear test.

Question 48Easy
The triaxial test is usually carried out as:
Full explanation

Answer: D. All of the above

The triaxial apparatus can run UU, CU and CD tests; all the listed forms are standard.

Question 49Medium
Due to a rise in the water table, the effective stress in soil:
Full explanation

Answer: B. decreases

A rising water table within the soil increases pore pressure (and buoyancy), so effective stress decreases. (If the water table is already above ground, a further rise does not change it.)

Question 50Easy
In a triaxial test, the intermediate principal stress is:
Full explanation

Answer: B. Equal to the confining pressure

In a triaxial compression test σ₂ = σ₃ = cell (confining) pressure.

Question 51Easy
The actual movement of a soil mass is known as:
Full explanation

Answer: C. Slope failure

Downward and outward movement of a soil mass along a slip surface is a slope failure.

Question 52Medium
The total and effective stresses at a depth of 5 m below the top level of water in a pond are respectively:
Full explanation

Answer: B. 0.5 kg/cm² and 0.0

σ = γwh = 10 × 5 = 50 kN/m² ≈ 0.5 kg/cm²; u = 0.5 kg/cm²; σ′ = 0.

Question 53Easy
In a triaxial compression test on a soil specimen, the intermediate principal stress is equal to:
Full explanation

Answer: B. minor principal stress

The two horizontal stresses are both the cell pressure, so σ₂ = σ₃ (the minor principal stress).

Question 54Easy
In a direct shear test, the shear stress and normal stress on a dry sand at failure are 0.6 kg/cm² and 1 kg/cm² respectively. The angle of internal friction of the sand will be nearly:
Full explanation

Answer: B. 31°

Dry sand: c = 0, so \(\tan\phi = \tau/\sigma = 0.6/1 = 0.6\) and \(\phi = \tan^{-1}0.6 \approx 31^\circ\).

Question 55Medium
The static cone penetration test and the standard penetration test are performed on a soil at a certain depth. The value of static cone resistance is 8 MPa and the N value is 20. The soil met at that depth is:
Full explanation

Answer: D. medium dense sand

qc = 8 MPa ≈ 80 kg/cm²; qc/N = 80/20 = 4, which falls in the range for clean sands (≈ 3.5–5). With N = 20, the soil is a medium dense sand.