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Nepal Engineering Council Registration Exam Soil Mechanics: practice questions

Nepal Engineering Council Registration Exam — Agricultural Engineering · Nepal · Nepal Engineering Council

Original practice questions written to the published syllabus · not past papers
10worked here, free
75written to this exam in this subject
11syllabus units

What this subject covers

Soil as an engineering material: its phases, flow of water through it, compaction, consolidation, shear strength and stability.

In the exams: GATE AG Section 4 (soil mechanics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit IV; NCEES PE soils and foundations

Where the 75 questions fall

How our questions for Nepal Engineering Council Registration Exam spread across the units of Soil Mechanics. Difficulty: 38 easy, 29 medium, 8 hard.

Phase relations9
Terzaghi one-dimensional consolidation9
Bearing capacity basics9
Index properties and soil classification8
Compaction and the Proctor test8
Shear strength and the Mohr-Coulomb criterion8
Permeability7
Earth pressure: active and passive (Rankine)6
Slope stability, including infinite slopes5
Effective stress4
Seepage and flow nets2

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Phase relations · medium

A soil has a bulk unit weight of 19.0 kN/m³, a water content of 12% and G = 2.65. Taking γw = 9.81 kN/m³, what is its degree of saturation?

  1. 0.532
  2. 0.597
  3. 0.318
  4. 0.864
Answer and working

B. 0.597

γd = γ/(1 + w) = 19.0/1.12 = 16.96 kN/m³. e = Gγw/γd − 1 = 2.65 × 9.81/16.96 − 1 = 0.532. S = wG/e = 0.12 × 2.65/0.532 = 0.597. Using the bulk instead of the dry unit weight to find e gives e = 0.368 and S = 0.864; 0.318 is just wG with no division by e.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

2 · Phase relations · easy

A sand deposit has an in-situ void ratio of 0.62. Laboratory tests give a maximum void ratio of 0.85 and a minimum void ratio of 0.45. What is its relative density?

  1. 42.5%
  2. 72.9%
  3. 27.1%
  4. 57.5%
Answer and working

D. 57.5%

Dr = (emax − e)/(emax − emin) = (0.85 − 0.62)/(0.85 − 0.45) = 0.23/0.40 = 57.5%, a medium-dense sand. 42.5% comes from putting emin in the numerator, which measures looseness instead of density. 27.1% divides by emax alone (0.23/0.85), and 72.9% is e/emax (0.62/0.85).

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

3 · Index properties and soil classification · medium

A fine-grained inorganic soil has a liquid limit of 45% and a plasticity index of 22%. What is its group symbol in the Indian Standard system (IS 1498)?

  1. CL
  2. MI
  3. CH
  4. CI
Answer and working

D. CI

The A-line is Ip = 0.73(wL − 20) = 0.73 × 25 = 18.25. Ip = 22 lies above it, so the soil is a clay. IS 1498 uses three plasticity bands (L below 35, I from 35 to 50, H above 50), so wL = 45 gives CI. CL is the USCS symbol, which uses only L and H with the boundary at 50.

Further reading: IS 1498

4 · Permeability · medium

In a falling-head test, the standpipe area is 1.0 cm², the specimen is 10 cm long with an area of 50 cm², and the head falls from 100 cm to 50 cm in 600 s. What is k?

  1. 1.00e-04 cm/s
  2. 3.33e-04 cm/s
  3. 5.78e-01 cm/s
  4. 2.31e-04 cm/s
Answer and working

D. 2.31e-04 cm/s

k = (aL/(At))·ln(h1/h2) = (1.0 × 10)/(50 × 600) × ln 2 = 3.333e-04 × 0.693 = 2.31e-04 cm/s. Using log10 without the 2.303 factor gives 1.00e-04; swapping the standpipe and specimen areas gives an absurdly high 5.78e-01.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

5 · Effective stress · medium

A sand deposit has a water table 3 m below ground. Above it the sand's unit weight is 17 kN/m³; below it the saturated unit weight is 19.5 kN/m³. With γw = 9.81 kN/m³, what is the vertical effective stress at 8 m depth?

  1. 148.50 kPa
  2. 106.95 kPa
  3. 70.02 kPa
  4. 99.45 kPa
Answer and working

D. 99.45 kPa

σ = 17 × 3 + 19.5 × 5 = 148.5 kPa. u = 9.81 × 5 = 49.05 kPa (water depth is 5 m, not 8 m). σ′ = 99.45 kPa. Measuring u from the ground surface gives 70.02; using γsat above the water table gives 106.95.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

6 · Compaction and the Proctor test · easy

If the compactive effort applied to a soil is increased, how do its maximum dry density and optimum moisture content change?

  1. Both rise
  2. Both fall
  3. Maximum dry density falls; optimum moisture content rises
  4. Maximum dry density rises; optimum moisture content falls
Answer and working

D. Maximum dry density rises; optimum moisture content falls

More energy packs the grains more closely with less water needed for lubrication, so the curve moves up and to the left. This is why the modified Proctor test gives a higher maximum dry density at a lower optimum than the standard test.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

7 · Earth pressure: active and passive (Rankine) · easy

Which assumption underlies Rankine's earth pressure theory?

  1. Wall friction equals the soil friction angle
  2. The failure surface is a logarithmic spiral
  3. The backfill is always cohesive
  4. The back of the wall is vertical and perfectly smooth
Answer and working

D. The back of the wall is vertical and perfectly smooth

Rankine assumed a smooth vertical wall so that the stresses in a semi-infinite soil mass in plastic equilibrium apply directly. Wall friction is included in Coulomb's wedge theory; the log-spiral surface is used in more refined passive calculations.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

8 · Slope stability, including infinite slopes · easy

What is the steepest slope angle for a long, dry cohesionless embankment face (φ′ = 35°) if a factor of safety of 1.5 is required?

  1. 23.3°
  2. 35.0°
  3. 25.0°
  4. 46.4°
Answer and working

C. 25.0°

F = tanφ′/tanβ, so tanβ = tan35°/1.5 = 0.4668 and β = 25.0°. Dividing the angle directly (35/1.5 = 23.3°) is not the same because tan is not linear. 35° gives F = 1. 46.4° multiplies tanφ′ by 1.5 instead of dividing.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

9 · Bearing capacity basics · medium

A plate load test with a 0.3 m square plate on a deep, uniform dense sand gives an ultimate bearing pressure of 400 kPa. By the usual plate-load relation for sand, what is the ultimate bearing capacity of a 1.5 m square footing at the same level?

  1. 2000 kPa
  2. 400 kPa
  3. 80 kPa
  4. 10000 kPa
Answer and working

A. 2000 kPa

For sand the ultimate capacity from the γBNγ term is proportional to width: qf = qp × Bf/Bp = 400 × 1.5/0.3 = 2000 kPa. 400 kPa (no scaling) applies to clays, where capacity does not depend on width. The test reflects only soil within about twice the plate width, so deeper weak layers must be checked separately. 80 kPa inverts the width ratio (400 × 0.3/1.5), and 10000 kPa squares it (400 × 25).

Further reading: IS 1888; Punmia, Jain and Jain, Soil Mechanics and Foundations

10 · Compaction and the Proctor test · easy

How does compaction differ from consolidation?

  1. Both expel water only
  2. Compaction slowly expels pore water under a sustained static load; consolidation expels air quickly by rolling or ramming the soil
  3. They are two names for the same process
  4. Compaction expels air quickly by mechanical energy; consolidation slowly expels water from saturated soil under sustained load
Answer and working

D. Compaction expels air quickly by mechanical energy; consolidation slowly expels water from saturated soil under sustained load

Compaction is a rapid densification of partly saturated soil by rolling, tamping or vibration that reduces air voids. Consolidation is the gradual volume decrease of a saturated soil as pore water drains under a sustained load, governed by permeability.

Further reading: Das and Sobhan, Principles of Geotechnical Engineering

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Soil MechanicsThe full study manual, free Nepal Engineering Council Registration ExamFormat, sections and how it has changed

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