Q21GATE 2017MCQ

Group I lists the type of gain or loss of strength in soils. Group II lists the property or process responsible for the loss or gain of strength in soils. The correct match between Group I and Group II is

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Q23GATE 2017MCQ

The laboratory test on a soil sample yields the following results: natural moisture content = 18%, liquid limit = 60%, plastic limit = 25%, percentage of clay sized fraction = 25%. The liquidity index and activity (as per the expression proposed by skempton) of the soil, respectively, are

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📖 Explanation

Given data: $w_{n}=18 \%, w_{L}=60 \%, w_{p}=25 \%$ % of clay size particle =25 % Liquidity index $I_{L}=\frac{w_{n}-w_{p}}{w_{L}-w_{p}}=\frac{0.18-0.25}{0.60-0.25}=-0.2$ Activity $=\frac{I_{p}}{\% C}=\frac{w_{L}-w_{p}}{\% C}=\frac{60-25}{25}=1.4$

Q24GATE 2016NAT

The porosity (n) and the degree of saturation (S) of a soil sample are 0.7 and 40%, respectively. In a 100 $m^{3}$ volume of the soil, the volume (expressed in $m^{3}$ ) of air is ________ .

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📖 Explanation

\begin{aligned} n=0.7 .5 &=40 \%, V=100 \mathrm{m}^{3}, V_{a}=? \\ n &=\frac{V_{v}}{V} \\ \Rightarrow \quad 0.7 &=\frac{V_{v}}{100} \\ \Rightarrow \quad V_{v} &=70 \mathrm{m}^{3} \\ \text { Now, } \quad S &=\frac{V_{W}}{V_{v}}=\frac{V_{v}-V_{a}}{V_{v}}=1-\frac{V_{a}}{V_{v}}\\ \Rightarrow \quad 0.40&=1-\frac{V_{a}}{V_{v}} \\ \Rightarrow \quad \frac{V_{a}}{V_{v}}&=0.60 \Rightarrow V_{a}=0.6 \times 70=42 \mathrm{m}^{3} \end{aligned}

Q25GATE 2015NAT

A 588 $cm^{3}$ volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G is 2.67. Assuming density of water as 1 gm/ $cm^{3}$ , the void ratio is __________.

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📖 Explanation

\begin{aligned} \gamma_{\mathrm{d}} &=\frac{918}{588}=1.561 \mathrm{g} / \mathrm{cc} \\ \gamma_{\mathrm{d}} &=\frac{G \gamma_{\mathrm{w}}}{1+e} \\ 1.561 &=\frac{2.67 \times 1}{1+\mathrm{e}} \\ e &=0.71 \end{aligned}

Q26GATE 2015MCQ

A fine-grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when water content is between 15% and 28%. The soil behaves fluid-like when the water content is more than 40%. The 'Activity' of the soil is

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📖 Explanation

\begin{aligned} I_{p} &=W_{L}-W_{p}=40-28=12 \% \\ \text { Activity } &=\frac{I p}{C(\% \text { clay fraction finer than } 2 \mu)} \\ &=\frac{12}{40}=0.3 \end{aligned}

Q27GATE 2015MCQ

In the water content of a fully saturated soil mass is 100% the void ratio of the sample is:

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📖 Explanation

\begin{aligned} S.e&=w.G_{s}\\ \text{Given }S&=1\text{ (Fully saturated soil)}\\ w&=100 \%=1 \\ e&=G_{s} \end{aligned}

Q28GATE 2015MCQ

An earth embankment is to be constructed with compacted cohesionless soil. The volume of the embankment is 5000 $m^{3}$ and the target dry unit weight is 16.2 $kN/m^{3}$ . Three nearby sites (see figure blwo) have been identified from where the required soil can be transported to the construction site. The void ratios (e) of different sites are shown in the figure. Assume the specific gravity of soil to be 2.7 for all three sites. If the cost of transportation per km is twice the cost of excavation per $m^{3}$ of borrow pits, which site would you choose as the most economic solution? (Use unit weight of water = 10 $kN/m^{3}$ )

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📖 Explanation

Volume of solids at embankment,

\begin{aligned} V_{s} &=\frac{W_{s}}{G_{s} \gamma_{w}} \\ V_{s} &=\frac{16.2 \times 5000}{27 \times 10} \mathrm{m}^{3} \\ &=3000 \mathrm{m}^{3} \end{aligned}

Volume of solids will remain constant. also,

\begin{aligned} V &=V_{V}+V_{s} \\ &=e V_{s}+V_{s}=V_{s}(1+e) \\ V &=3000(1+e) \\ V_{X} &=3000 \times 1.6=4800 \mathrm{m}^{3} \\ V_{Y} &=3000 \times 1.7=5100 \mathrm{m}^{3} \\ v_{z} &=3000 \times 1.64=4920 \mathrm{m}^{3} \end{aligned}

Let cost of excavation be $\alpha / \mathrm{m}^{3}$

\begin{aligned} P_{X} &=4800 \alpha+140 \times \alpha \times 2 \\ &=5080 \alpha \\ P_{Y} &=5100 \alpha+80 \times \alpha \times 2 \\ &=5260 \alpha \\ P_{Z} &=4920 \alpha+100 \times \alpha \times 2 \\ &=5120 \alpha \end{aligned}

Hence $P_{X}$ is least. Hence total cost for site X is least.

Q29GATE 2014NAT

A certain soil has the following properties: $G_{s}$ =2.71, n=40% and w=20%. The degree of saturation of soil (rounded off to the nearest percent) is _____

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📖 Explanation

\begin{aligned} Se&=wG \\ \text{and }e&=\frac{n}{1-n}=0.67 \\ \Rightarrow \quad S&=\frac{2.71 \times 20}{0.67}=81.3 \% \end{aligned}

Q30GATE 2014MCQ

A given cohesionless soil has $e_{max}=0.85$ and $e_{min}=0.50$ . In the field, the soil is compacted to a mass density of 1800 $kg/m^{3}$ at a water content of 8%. Take the mass density of water as 1000 $kg/m^{3}$ and $G_{s}$ as 2.7. The relative density (in %) of the soil is

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📖 Explanation

\begin{aligned} e_{max} &=0.85 \\ e_{min} &=0.5 \\ \rho_{\text {field }} &=1800 \mathrm{kg} / \mathrm{m}^{3}\\ &\text{at water content }=8\% \\ \rho_{w}&=1000 \mathrm{kg} / \mathrm{m}^{3} \\ G_{S}&=2.7 \end{aligned}

Relative density, $\mathrm{I}_{D}=?$

\begin{aligned} \rho=\frac{G_{p_{W}}(1+w)}{1+e}&=\frac{2.7 \times 1000(1.08)}{1+e}\\ \Rightarrow\quad 1+e &=\frac{2.7 \times 1000 \times 1.08}{1800} \\ \Rightarrow\quad e &=0.62\\ \Rightarrow\quad I_{D} &=\frac{e_{\max }-e}{e_{\max }-e_{\min }} \times 100 \\ &=\frac{0.85-0.62}{0.85-0.5} \times 100 \\ &=65.714 \% \end{aligned}

Q31GATE 2013MCQ

In its natural condition, a soil sample has a mass of 1.980 kg and a volume of 0.001 $m^{3}$ . After being completely dried in an oven, the mass of the sample is 1.800 kg. Specific gravity G is 2.7. Unit weight of water is 10 kN/ $m^{3}$ . The degree of saturation of the soil is:

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📖 Explanation

\begin{aligned} \rho &=\frac{M}{V} \\ &=\frac{1.980}{0.001}=1980 \mathrm{kg} / \mathrm{m}^{3} \\ \rho_{d} &=\frac{M_{d}}{V}\\ &=\frac{1.800}{0.001}=1800 \mathrm{kg} / \mathrm{m}^{3} \\ \rho_{d} &=\frac{\rho}{1+\mathrm{w}} \\ \Rightarrow \quad 1800 &=\frac{1980}{1+\mathrm{w}} \\ \Rightarrow \quad & \mathrm{w}=0.1 \\ \because \quad & \rho_{\mathrm{d}}=\frac{\mathrm{G} \cdot \rho_{\mathrm{w}}}{1+\mathrm{e}}\\ \Rightarrow \quad 1800&=\frac{2.7 \times 1000}{1+e} \\ \Rightarrow \quad e&=0.5 \\ \because wG&=e.s\\ \Rightarrow 0.1 \times 2.7&=0.5 S \\ \Rightarrow \quad S&=0.54 \end{aligned}

Q32GATE 2008MCQ

The liquid limit (LL), plastic limit (PL) and shrinkage limit (SL) of a cohesive soil satisfy the relation

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Q33GATE 2007MCQ

The water content of a saturated soil and the specific gravity of soil solids were found to be 30% and 2.70, respectively. Assuming the unit weight of water to be 10 kN/ $m^{3}$ , the saturated unit weight kN/ $m^{3}$ and the void ratio of the soil are

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📖 Explanation

We know that,

\begin{aligned} S \times e&=G \times w \\ \Rightarrow\quad 1 \times e&=2.70 \times 0.30 \\ \Rightarrow\quad e&=0.81 \\ \text{and }\quad \gamma_{\text {sat }}&=\left(\frac{G+e}{1+e}\right) \gamma_{w} \\ &=\left(\frac{2.7+0.81}{1+0.81}\right) \times 10 \\ &=19.4 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Q34GATE 2005MCQ

A saturated soil mass has a total density 22 kN/ $m^{3}$ and a water content of 10%. The bulk density and dry density of this soil are

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📖 Explanation

Total density is bulk density, $\gamma_{t}=22 \mathrm{kN} / \mathrm{m}^{3}$

\begin{aligned} \text { Dry density, } \gamma_{d} &=\frac{\gamma_{1}}{1+w}=\frac{22}{1+\frac{10}{100}}=\frac{22}{1.1} \\ &=20 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Q35GATE 2004MCQ

The undrained cohesion of a remoulded clay soil is 10 kN/ $m^{2}$ . If the sensitivity of the clay is 20, the corresponding remoulded compressive strength is

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📖 Explanation

We know that,

\begin{aligned} c_{u} &=\frac{q_{u}}{2} \\ \Rightarrow \quad q_{u} &=2 c_{u}=2 \times 10 \\ &=20 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Note: Sensitivity of the clay is of no use

Q36GATE 2004MCQ

The ratio of saturated unit weight to dry unit weight of soil is 1.25. If the specific gravity of solids ( $G_{s}$ ) is 2.65, the void ratio of the soil is

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📖 Explanation

\begin{aligned} \gamma_{\text {Sat }} &=\left(\frac{G+e}{1+e}\right) \gamma_{w} &\dots(i)\\ \gamma_{d} &=\frac{G \gamma_{w}}{1+e} &\dots(ii)\\ \Rightarrow \quad \frac{\gamma_{\text {sat }}}{\gamma_{d}} &=\frac{G+e}{G} \\ \Rightarrow \quad 1.25 &=\frac{2.65+e}{2.65} \\ e &=0.663 \end{aligned}

Q37GATE 2002MCQ

The void ratios at the densest, loosest and the natural states of a sand deposit are 0.2, 0.6 and 0.4, respectively. The relative density of the deposit is

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📖 Explanation

Relative density

\begin{aligned} &=\frac{e_{\max }-e_{\text {natuat }}}{e_{\max }-e_{\min }} \times 100 \\ &=\frac{0.6-0.4}{0.6-0.2} \times 100=50 \% \end{aligned}

Q38GATE 2002MCQ

The following data were obtained from a liquid limit test conducted on a soil sample. The liquid limit of the soli is

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📖 Explanation

The liquid limit is the water content corresponding to 25 number of blows $\therefore w_{L}=61.9\%$

Q39GATE 2001MCQ

The void ratio and specific gravity of a soil are 0.65 and 2.72 respectively. The degree of saturation (in percent) corresponding to water content of 20% is

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📖 Explanation

\begin{array}{ll} \text { Given } & e=0.65 \\ & \mathrm{G}=2.72 \\ & \mathrm{w}=20 \% \\ \text { From } & \mathrm{e}=\frac{\mathrm{wG}}{\mathrm{S}} \\ \therefore \quad & \mathrm{S}=\frac{\text { wG }}{\mathrm{e}}=\frac{0.2 \times 2.72}{0.65} \\ & =0.8369 \approx 83.7 \% \end{array}