Your Answer : (Not Answered)

Correct Answer : B

The maximum shear stress is \((\sigma_{1} - \sigma_{3})/2\).

On Mohr's circle the major and minor principal stresses \(\sigma_1\) and \(\sigma_3\) mark the two ends of a diameter. The greatest shear stress is the radius of the circle, acting on planes at 45° to the principal planes:

\[ \tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \]

\((\sigma_1 + \sigma_3)/2\) is the centre of the circle, the normal stress on the plane of maximum shear. \((\sigma_3 - \sigma_1)/2\) has the right size but is negative, and the product of the stresses has the wrong units.

Your Answer : (Not Answered)

Correct Answer : A

The correct matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

• The hydrometer measures the settling of fine particles in suspension and gives the grain size distribution of clays and silts (a-iv).
• The plate load test loads a steel plate at foundation level to find the bearing capacity and settlement of the soil (b-iii).
• The pycnometer, a glass jar with a conical cap, is used to find the specific gravity of soil solids (c-ii).
• The direct shear apparatus shears a sample along a fixed plane to give the shear parameters c and φ (d-i).

The other options either swap the hydrometer and pycnometer or swap the plate load and direct shear tests.

Your Answer : (Not Answered)

Correct Answer : D

Clay has the highest plasticity index.

The plasticity index is the range of water content over which a soil behaves plastically, the liquid limit minus the plastic limit. Clay minerals are fine, plate-shaped and electrically charged, so they hold water films and stay mouldable over a wide range of moisture; clays typically have plasticity indices well above 17.

Silt has low plasticity. Sand and gravel are coarse and non-plastic, with a plasticity index of practically zero.

Your Answer : (Not Answered)

Correct Answer : C

At the liquid limit the shearing strength is small but definite.

The liquid limit is the water content at the boundary between the plastic and liquid states. The soil still has a small, measurable shear strength there, of the order of 2 kN/m², and the standard cup test is in effect a way of finding the water content at which the soil has that strength.

The soil is fully saturated at the liquid limit, so A is wrong. Zero strength would describe a true liquid, beyond the limit. Easy moulding describes the plastic state between the plastic and liquid limits rather than the limit itself.

Your Answer : (Not Answered)

Correct Answer : C

A soil with a plasticity index of zero is Non Plastic.

The plasticity index is the liquid limit minus the plastic limit, the range of water content over which the soil behaves plastically. If it is zero, there is no such range, as with sands and non-plastic silts, so the soil is called non-plastic.

In the usual classification, a plasticity index below 7 is low plasticity, 7 to 17 medium, and above 17 high. All three describe soils that do have some plastic range, unlike a soil with an index of zero.

Your Answer : (Not Answered)

Correct Answer : D

The coefficients are 8 and 1.125.

Use the standard definitions:

\[ \begin{aligned} C_u &= \frac{D_{60}}{D_{10}} = \frac{0.8}{0.1} = 8 \\ C_c &= \frac{D_{30}^{2}}{D_{10} \times D_{60}} = \frac{0.09}{0.08} = 1.125 \end{aligned} \]

Since \(C_u > 6\) and \(C_c\) lies between 1 and 3, a sand with these values is well graded. 2.67 is \(D_{60}/D_{30}\), the wrong pair of sizes for \(C_u\), and 3 is \(D_{30}/D_{10}\), not the curvature coefficient.

Your Answer : (Not Answered)

Correct Answer : A

Permeability is directly proportional to the square of grain size.

Allen Hazen's empirical relation for clean sands is \(k = C\,D_{10}^{2}\), where \(D_{10}\) is the effective size. The reason is physical: the size of the pore channels grows with the grain size, and flow through a channel grows with the square of its size, as in Poiseuille's law.

So doubling the effective size makes the soil about four times as permeable. That rules out the inverse relations, which would make clay more permeable than gravel, and the simple direct proportion, which understates the effect.

Your Answer : (Not Answered)

Correct Answer : D

Darcy's law gives Q = kiA.

For laminar flow through saturated soil, the discharge velocity is proportional to the hydraulic gradient, v = ki, where k is the coefficient of permeability. Multiplying by the total cross-sectional area of the soil perpendicular to the flow gives the rate of flow Q = kiA.

The other options square one of the three quantities, which breaks the linear proportionality on which Darcy's law rests and gives the wrong dimensions for a flow rate.

Your Answer : (Not Answered)

Correct Answer : C

The coefficient of permeability is \(3.75 \times 10^{-1}\) cm/min.

For a constant head test, Darcy's law gives \(K = \frac{QL}{Aht}\).

\[ \begin{aligned} K &= \frac{30 \times 20}{10 \times 40 \times 4} \\ &= \frac{600}{1600} = 0.375\ \text{cm/min} \end{aligned} \]

That is \(3.75 \times 10^{-1}\) cm/min. The other options differ only in the leading digit and come from arithmetic slips in the same formula.

Your Answer : (Not Answered)

Correct Answer : D

The A-line is Ip = 0.73 (\(W_{L}\) - 20).

On Casagrande's plasticity chart, plasticity index is plotted against liquid limit. The empirical A-line, \(I_p = 0.73(w_L - 20)\), separates inorganic clays above it from silts and organic soils below it. The U-line, \(I_p = 0.9(w_L - 8)\), marks the upper limit found for natural soils.

The other three equations, with 0.93, 0.83 or 0.65 and an intercept of 22, are not lines of the plasticity chart.

Your Answer : (Not Answered)

Correct Answer : D

The saturated density is \(\frac{G+e}{1+e}\rho w\).

Take a soil element with unit volume of solids. Its total volume is \(1 + e\). The solids have mass \(G\rho_w\), and when saturated the voids hold water of mass \(e\rho_w\):

\[ \begin{aligned} \rho_{sat} &= \frac{G\rho_w + e\rho_w}{1 + e} \\ &= \frac{G + e}{1 + e}\rho_w \end{aligned} \]

This is the general form \(\frac{G + Se}{1 + e}\rho_w\) with the degree of saturation S = 1. The option with \(G + eS\) over \(1 + w\) uses the wrong denominator, \(G + S\) drops the void ratio from the water term, and \(G + w\) adds a water content to a specific gravity.

Your Answer : (Not Answered)

Correct Answer : C

The saturated and submerged densities are 2.25 g/cc and 1.25 g/cc.

At saturation the soil holds water equal to 50% of the weight of its solids, so its bulk density is the dry density times \((1 + w)\):

\[ \begin{aligned} \rho_{sat} &= \rho_d (1 + w) = 1.5 \times 1.5 = 2.25 \\ \rho' &= \rho_{sat} - \rho_w = 2.25 - 1 = 1.25 \end{aligned} \]

Both in g/cc. 1.5 and 1.0 repeats the dry density, 2.0 and 1.0 adds the water content as if it were a density, and 2.50 and 1.50 overstates the saturated density.

Your Answer : (Not Answered)

Correct Answer : A

The void ratio is 0.667.

Porosity n and void ratio e are directly related, so only n is needed; the densities and G given are not used.

\[ \begin{aligned} e &= \frac{n}{1 - n} = \frac{0.40}{1 - 0.40} \\ &= \frac{0.40}{0.60} \approx 0.667 \end{aligned} \]

A common slip is to take e = n, or to use the reverse relation \(n = e/(1+e)\), which gives the other options.

Your Answer : (Not Answered)

Correct Answer : C

The critical gradient is about 0.94.

A quick condition occurs when the upward seepage force equals the submerged weight of the soil:

\[ \begin{aligned} i_c &= \frac{G - 1}{1 + e} = \frac{2.67 - 1}{1 + 0.78} \\ &= \frac{1.67}{1.78} = 0.94 \end{aligned} \]

The rule of thumb is that \(i_c\) is close to 1 for most sands. 0.62 is \((G - 1)/G\), dividing by the specific gravity instead of \(1 + e\); 0.74 and 0.82 do not follow from these values.

Your Answer : (Not Answered)

Correct Answer : D

The critical hydraulic gradient is 1.00.

The critical (quick sand) gradient is the gradient at which upward seepage pressure just balances the submerged weight of the soil.

\[ \begin{aligned} i_c &= \frac{G - 1}{1 + e} = \frac{2.71 - 1}{1 + 0.7} \\ &= \frac{1.71}{1.7} \approx 1.00 \end{aligned} \]

For most soils \(i_c\) is close to 1. The values 0.85 to 0.95 come from slips such as using G instead of G - 1 or adding e wrongly.

Your Answer : (Not Answered)

Correct Answer : B

The bulk unit weight is 18639.00 N/m³.

For a soil with specific gravity \(G\), void ratio \(e\) and degree of saturation \(S\), the bulk unit weight is

\[ \begin{aligned} \gamma &= \frac{(G + Se)\gamma_{w}}{1 + e} \\ &= \frac{(2.7 + 0.9 \times 0.8) \times 9810}{1.8} \\ &= \frac{3.42 \times 9810}{1.8} = 18639\ \text{N/m}^{3} \end{aligned} \]

18.64 is the same figure in kN/m³, but the question asks for N/m³. 10594.80 is far too low: even the dry unit weight, \(G\gamma_{w}/(1 + e) = 14715\ \text{N/m}^{3}\), is larger, and adding the water in the voids can only raise it. 10.60 is that low value written in kN/m³.

Your Answer : (Not Answered)

Correct Answer : A

The porosity is 0.30.

Void ratio \(e\) is void volume over solid volume, while porosity \(n\) is void volume over total volume, so

\[ n = \frac{e}{1 + e} = \frac{0.43}{1.43} = 0.30 \]

0.70 is \(1 - n\), the solid fraction. 0.40 and 0.75 do not follow from the relation; porosity must always be smaller than the void ratio.

Your Answer : (Not Answered)

Correct Answer : B

The porosity is 0.44 and the void ratio 0.79.

Porosity follows from the two densities, and void ratio from porosity:

\[ \begin{aligned} n &= 1 - \frac{\rho_{b}}{\rho_{s}} = 1 - \frac{1.48}{2.64} \approx 0.44 \\ e &= \frac{n}{1 - n} = \frac{0.44}{0.56} \approx 0.79 \end{aligned} \]

The saturated volumetric moisture content of 36% is not needed. The option 0.79 and 0.44 swaps the two quantities. 0.36 and 0.56 takes the moisture figure as the porosity, and 0.56 and 0.36 takes the solid fraction \(1 - n\) as the porosity.

Your Answer : (Not Answered)

Correct Answer : D

A flow net is a graphical solution of the Laplace equation for the seepage of water through soil, so it is used to find the seepage quantity (for example beneath a dam or through an earthen dam), the exit gradient, the uplift (hydrostatic) pressure at points and the pore pressure. Passive earth pressure is the resistance of soil to a wall moving against it and is found from Rankine's or Coulomb's theories, not from a flow net, so it is not an application.

Your Answer : (Not Answered)

Correct Answer : B

Under the Indian Standard soil classification (IS 1498), a clean sand is well graded (SW) when the uniformity coefficient Cu is greater than 6 and the coefficient of curvature Cc lies between 1 and 3. Both conditions are met here. If either of them fails, the sand is poorly graded (SP). Silty sand and clayey sand are sands with more than 12 percent fines, and are classified by plasticity and fines, not by Cu and Cc, which apply to clean sand.