Question 1 of 6: Concept multiple-choice bank

Question 1: Concept multiple-choice bank (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Each item is answered with its reason (no marks without a reason).

1. Item 1 — (D) clay. Optimum moisture content (OMC) rises with specific surface area: fine, plate-shaped clay particles hold far more adsorbed water than coarse gravel, so clay has the highest OMC and gravel the lowest.
2. Item 2 — (B) plastic limit. For most fine-grained soils the compaction OMC falls very close to the plastic limit, where the soil is workable yet not free-draining; this is the classic empirical correlation used to estimate OMC from index tests.
3. Item 3 — (A) shear strength. Compaction packs the grains, raising dry density and interlock, so shear strength increases. Permeability, compressibility and further consolidation all decrease with compaction, so “all of the above” is wrong.
4. Item 4 — (A) gravel. The complement of Item 1: a well-graded granular soil reaches the highest maximum dry density (particles interlock and fill voids at a low OMC), whereas clay reaches the lowest maximum dry density at a high OMC.
5. Item 5 — (C) expulsion of water from the voids. Consolidation settlement of a saturated clay occurs as excess pore water is squeezed out and load transfers to the skeleton; the grains and water are effectively incompressible, so it is the drainage of pore water — not deformation of grains, compression of water or expulsion of grains — that produces the settlement.
6. Item 6 — (A) True. Both a normally consolidated clay and a loose sand are contractive: the stress–strain curve strain-hardens to an ultimate value with no distinct peak, and both tend to decrease in volume (or generate positive excess pore pressure) on shearing.
7. Item 7 — (A) gravel. Even at maximum dry density the pore channels scale with grain size, so the coarsest soil (gravel, largest D₁₀) keeps the highest permeability; by Hazen, $k \approx C\,D_{10}^{2}$.
8. Item 8 — (D) is the incorrect statement. The question asks for the INCORRECT statement(s). (A) is correct: an OC clay carries an effective cohesion intercept $c'>0$. (B) is correct: loaded below its pre-consolidation pressure an OC clay is dilatant, so shearing can generate negative pore pressure. (C) is correct: at 100 kPa, below $\sigma'_p=200$ kPa, the clay is on its stiff recompression line, so its volume change is smaller than that of an NC clay on the virgin line. (D) is incorrect: being denser, with a lower void ratio, an OC clay is typically less permeable than the same clay normally consolidated.
9. Item 9 — (B) and (C). Terzaghi’s theory assumes a homogeneous, fully saturated soil (A), incompressible grains and water (D), one-dimensional flow obeying Darcy’s law, small strains, a unique time-independent void-ratio–effective-stress relationship (E), and $m_v$ and $k$ constant through the process. Statement (B), that $m_v$ and $k$ vary with effective stress, contradicts that, so it is not an assumption. Statement (C), drainage at both top and bottom, is not an assumption of the theory either: it is a boundary condition of a particular problem, and the same solution covers single drainage by taking the drainage path $H_{dr}$ as the full layer thickness.
10. Item 10 — (A) consolidated-undrained triaxial test. An undrained strength needs drainage to be prevented while the specimen is sheared, and a clean sand drains almost instantly unless the drainage is physically closed. The triaxial cell does this: the saturated sand specimen is consolidated, the drainage valve is closed, and it is sheared with pore pressure measured, giving its undrained strength (the basis of liquefaction and steady-state testing). A vane cannot hold a sand undrained, and an unconfined compression test needs a cohesive specimen that stands without confinement, which a sand cannot do. So (D) “none of the above” is wrong.