Permeability and Effective Stress
JKSSB Junior Engineer (Civil) · Soil Mechanics · 12 practice questions with answers and explanations: 8 Easy, 3 Medium, 1 Hard.
Sample questions
Sample Easy question 1
The quick condition (quicksand) occurs when:
- The soil is very dense and dry
- The effective stress becomes zero because of upward seepage
- The water table is low
- The soil is perfectly dry
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Answer: B. The upward seepage force equals the submerged weight of the soil.
Sample Easy question 2
Capillary rise in soil is greatest in:
- Boulders
- Clean coarse sand
- Fine-grained soil such as silt
- Coarse gravel
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Answer: C. The smaller the pores, the greater the rise.
Sample Easy question 3
In a flow net, the flow lines and the equipotential lines:
- Run parallel to each other
- Intersect each other at right angles
- Never meet
- Meet at 45°
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Answer: B. The flow net is drawn as a set of approximate squares.
Sample Medium question 4
The effective stress at 5 m depth in a saturated clay of unit weight 19 kN/m³ with the water table at the surface is about:
- 144 kPa
- 95 kPa
- 49 kPa
- 46 kPa
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Answer: D. σ = 95 kPa, u = 49.05 kPa, so σ' = 95 − 49.05 = 45.95 kPa.
Sample Medium question 5
A soil with k = 10⁻³ cm/s under a hydraulic gradient of 1 over an area of 100 cm² has a discharge of:
- 0.001 cm³/s
- 0.1 cm³/s
- 10 cm³/s
- 100 cm³/s
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Answer: B. Q = k i A = 10⁻³ × 1 × 100 = 0.1 cm³/s.
Sample Medium question 6
The critical hydraulic gradient for quicksand in a soil with G = 2.7 and e = 0.8 is about:
- 0.94
- 0.5
- 0.1
- 2.0
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Answer: A. ic = (G − 1) ÷ (1 + e) = 1.70 ÷ 1.80 = 0.94.
Sample Hard question 7
A flow net has 4 flow channels and 12 equipotential drops under a head of 6 m in soil with k = 10⁻⁵ m/s. The seepage per metre length is:
- 2 × 10⁻⁵ m³/s
- 6 × 10⁻⁵ m³/s
- 1 × 10⁻⁵ m³/s
- 8 × 10⁻⁵ m³/s
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Answer: A. q = k H Nf ÷ Nd = 10⁻⁵ × 6 × 4 ÷ 12 = 2 × 10⁻⁵ m³/s.