Q1GATE 2024NAT

An embankment is constructed with soil by maintaining the degree of saturation as $75 \%$ during compaction. The specific gravity of soil is 2.68 and the moisture content is $17 \%$ during compaction. Consider the unit weight of water as $10 \mathrm{kN} / \mathrm{m}^{3}$ . The dry unit weight (in $\mathrm{kN} / \mathrm{m}^{3}$ ) of the compacted soil is (rounded off to 2 decimal places)

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📖 Explanation

Dry density,

\begin{aligned} \gamma_{d}&=\frac{G \gamma_{w}}{1+e}\\ e & =\frac{w G}{S}=\frac{0.17 \times 2.68}{0.75}=0.6 \\ \therefore \quad \gamma_{d} & =\frac{2.68 \times 10}{1+0.6} \\ & =16.75 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Q2GATE 2023MCQ

With reference to the compaction test conducted on soils, which of the following is INCORRECT ?

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Q3GATE 2022NAT

An uncompacted heap of soil has a volume of 10000 $m^3$ and void ratio of 1. If the soil is compacted to a volume of 7500 $m^3$ , then the corresponding void ratio of the compacted soil is __________. (round off to one decimal place)

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📖 Explanation

\begin{aligned} (V_S)_{\text{heap}}&=(V_S)_{\text{compacted soil}}\\ \left ( \frac{V_T}{1+e} \right )_{\text{heap}}&=\left ( \frac{V_T}{1+e} \right )_{\text{compacted soil}}\\ \frac{10000}{1+1}&=\frac{7500}{1+e}\\ e&=0.5 \end{aligned}

Q4GATE 2016MCQ

OMC-SP and MDD-SP denote the optimum moisture content and maximum dry density obtained from standard Proctor compaction test, respectively. OMC-MP and MDD-MP denote the optimum moisture content and maximum dry density obtained from the modified Proctor compaction test, respectively. Which one of the following is correct?

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📖 Explanation

\begin{array}{r} \mathrm{OMC}-\mathrm{SP} \gt \mathrm{OMC}-\mathrm{MP} \\ \mathrm{MDD}-\mathrm{SP} \lt \mathrm{MDD}-\mathrm{MP} \end{array}

Q5GATE 2013MCQ

Following statements are made on compacted soils, wherein DS stands for the soils compacted on dry side of optimum moisture content and WS stands for the soils compacted on wet side of optimum moisture content. Identify the incorrect statement.

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Q6GATE 2012MCQ

Two series of compaction tests were performed in the laboratory on an inorganic clayey soil employing two different levels of compaction energy per unit volume of soil. With regard to the above tests, the following two statements are made. I The optimum moisture content is expected to be more for the tests with higher energy. II The maximum dry density is expected to be more for the tests with higher energy. The CORRECT option evaluating the above statements is

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Q7GATE 2010MCQ

In a compaction test, G, w, S and e represent the specific gravity, water content, degree of saturation and void ratio of the soil sample, respectively. If $\gamma _{w}$ represents the unit weight of water and $\gamma _{d}$ represents the dry unit weight of the soil, the equation of zero air voids line is

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📖 Explanation

The bulk unit weight of soil is given by $\gamma=\frac{(G+S e) \gamma_{w}}{1+e}$ For dry unit weight, degree of saturation, $S=0 \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+e}$ But we know that $G w=\text { Se } \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+\frac{G w}{S}}$ For zero air voids, degree of saturation, $S=100 \% \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+G w}$

Q8GATE 2009MCQ

Deposit with flocculated structure is formed when

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Q9GATE 2008MCQ

Compaction by vibratory roller is the best method of compaction in case of

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Q10GATE 2006MCQ

In a standard proctor test, 1.8 kg of moist soil was filling the mould (volume = 94.4cc) after compaction. A soil sample weighing 23 g was taken from the mould and ovendried for 24 hours at a temperature of 110 $^{\circ}$ C. Weight of the dry sample was found to be 20 g. Specific gravity of soil solids is G = 2.7. The theoretical maximum value of the dry unit weight of the soil at that water content is equal to

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📖 Explanation

Bulk density of the soil $=\frac{1.8 \times 1000}{944}=1.907 \mathrm{gm} / \mathrm{cc}$ Water content of soil $=\left(\frac{23-20}{20}\right) \times 100=15 \%$ For, theoretical maximum dry unit weight, degree of saturation should be 100 %.

\begin{aligned} S e &=G w \\ \Rightarrow \quad e &=\frac{2.7 \times 0.15}{1} \Rightarrow e=0.405 \end{aligned}

The theoretical maximum dry unit weight may be given as $\gamma_{d, \max }=\frac{G \gamma_{w}}{1+e}=\frac{2.7 \times 9.81}{1+0.405}=18.85 \mathrm{kN} / \mathrm{m}^{3}$

Q11GATE 2005MCQ

A clayey soil has a maximum dry density of 16 kN/ $m^{3}$ and optimum moisture content of 12%. A contractor during the construction of core of an earth dam obtained the dry density 15.2 kN/ $m^{3}$ and water content 11%. This construction is acceptable because

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Q12GATE 2003MCQ

Compaction of an embankment is carried out in 500 mm thick layers. The rammer used for compaction has a foot area of 0.05 $m^{2}$ and the energy imparted in every drop of rammer is 400 Nm. Assuming 50% more energy in each pass over the compacted area due to overlap, the number of passes required to develop compactive energy equivalent of Indian Standard light compaction for each layer would be

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📖 Explanation

As per Indian Standard light compaction test, a hammer of 2.6 kg is allowed to fall from a height of 310 mm and 3 layers are tamped 25 times in a mould of volume 1000 cc . $\therefore$ Energy imparted

\begin{array}{l} =2.6 \times 25 \times 3 \times 9.81 \\ \times \frac{310}{1000}=593.01 \mathrm{N}-\mathrm{m} \end{array}

For embankment compaction, the volume of soil covered by rammer

\begin{aligned} &=0.05 \times 10^{4} \times \frac{500}{10} \\ &=25000 \mathrm{cc} \end{aligned}

$\therefore$ Energy imparted in every drop of hammer

\begin{aligned} &=\left(1+\frac{50}{100}\right) \times 400\\ &=600 \mathrm{N}-\mathrm{m} \end{aligned}

If 'n' number of passes are required to develop equivalent energy to Indian Standard light compaction test, then $n \times \frac{600}{25000}=\frac{593.01}{1000}$ $\Rightarrow \quad n=24.71 \simeq 25\text{( approx.)}$ However, If the data of Standard Proctor Test are taken i.e. weight of hammer =2.495 kg; Volume of mould =944cc, then the value of 'n' will be close to 26.

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