Q2GATE 2024MCQ

Consider the statements P and Q. P : Soil particles formed by mechanical weathering, and close to their origin are generally subrounded. Q: Activity of the clay physically signifies its swell potential. Which one of the following options is CORRECT?

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📖 Explanation

Bulky particles are formed mostly by mechanical weathering of rocks and minerals. Geologists use such terms as angular, subangular, subrounded and rounded to describe the shape of bulky particles. Small sand particles located close to their origin are generally very angular sand particles carried by wind and water for a long distance can be subangular to rounded in shape. Activity number of clay is used to study the swelling behaviour. Activity number $(A_C) = \frac{\text{Plasticity index}}{\text{ \% of clay sized particles \in soil}}$ $A_C \lt 0.75 \rightarrow$ normal active $0.75 \lt A_C \lt 1.25 \rightarrow inactive$ $A_C \gt 1.25 \rightarrow active$ Active means more prone to volume change.

Q3GATE 2022MCQ

The correct match between the physical states of the soils given in Group I andthe governing conditions given in Group II is

\begin{array}{|l|l|} \hline \text{Group I}& \text{Group II} \\ \hline \text{1. normally consolidated soil}& \text{P. sensitivity > 16} \\ \hline \text{2. quick clay}& \text{Q. dilation angle = 0} \\ \hline \text{3. sand \in critical state}& \text{R. liquid limit > 50} \\ \hline \text{4. clay of high plasticity}& \text{S. over consolidation ratio = 1} \\ \hline \end{array}

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Q4GATE 2022MCQ

Four different soils are classified as CH, ML, SP, and SW, as per the Unified Soil Classification System. Which one of the following options correctly represents their arrangement in the decreasing order of hydraulic conductivity?

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📖 Explanation

Hydraulic conductivity Order. Gravel $\gt$ Sand $\gt$ silt $\gt$ lay

Q5GATE 2021MCQ

From laboratory investigations, the liquid limit, plastic limit, natural moisture content and flow index of a soil specimen are obtained as 60%, 27%, 32% and 27%, respectively. The corresponding toughness index and liquidity index of the soil specimen, respectively, are

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📖 Explanation

Flow index = 27

\begin{aligned} \text { Plasticity index, } \qquad I_{P}&=W_{L}-W_{P}=33 \\ \text { Toughness index, } \qquad I_{T}&=\frac{I_{P}}{I_{f}}=\frac{33}{27}=1.22 \\ \text { Liquidity index, } \qquad I_{L}&=\frac{W_{n}-W_{P}}{W_{L}-W_{P}}=0.151 \end{aligned}

Q6GATE 2021MCQ

As per the Unified Soil Classification System (USCS), the type of soil represented by 'MH' is

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Q7GATE 2019MCQ

A survey line was measured to be 285.5 m with a tape having a nominal length of 30 m. On checking, the true length of the tape was found to be 0.05 m too short. If the line lay on a slope of 1 in 10, the reduced length (horizontal length) of the line for plotting of survey work would be

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📖 Explanation

\begin{aligned} L&=30m, \; L'=30-0.05=29.95m\\ \therefore \;\;& \; \text{Correct Length}\\ l&=l' \times \frac{l'}{L}\\ &=285.5 \times \frac{29.95}{30}=285.024m\\ D&=l \cos \theta \\ \cos \theta &=\frac{10}{\sqrt{10^{2+}1^2}}=0.995\\ \therefore \;D \;&\; =\text{ correct Horizontal length }\\ &=l\cos \theta \\ &=285.024 \times 0.9950\\ &=283.6m \end{aligned}

Q8GATE 2016MCQ

A fine grained soil is found to be plastic in the water content range of 26-48%. As per Indian Standard Classification System, the soil is classified as

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Q9GATE 2014MCQ

As per Indian Standard Soil Classification System (IS: 1498-1970), an expression for A-line is

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Q10GATE 2014MCQ

The clay mineral primarily governing the swelling behavior of Black Cotton soil is

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Q11GATE 2012MCQ

As per the Indian Standard soil classification system, a sample of silty clay with liquid limit of 40% and plasticity index of 28% is classified as

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📖 Explanation

\begin{aligned} 1_{p}\text{ of soil } &=28 \% \\ \mathrm{I}_{\mathrm{P}} \text{ of A-line }&=0.73\left(\mathrm{w}_{\mathrm{L}}-20\right) \\ &=0.73(40-20)=14.6\\ \because \quad 1_{p} &\text{ of soil } \gt 1_{p}\text{ of A-line}\\ \end{aligned}

$\therefore$ It will be above A-line and also $35 \lt w_{2} \lt 50$ So it is CI.

Q12GATE 2011MCQ

The results for sieve analysis carried out for three types of sand, P, Q and R, are given in the figure. If the fineness modulus values of the three sands are given as FMP, FMQ and FMR, it can be stated that

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📖 Explanation

The term fineness modulus is used to indicate an index number which roughly proportional to average size of particle in entire quantity of aggregate $F_{M Q}=\sqrt{F_{M P} \times F_{M R}}$ (Geometric mean)

Q13GATE 2010MCQ

A fine grained soil has liquid limit of 60 and plastic limit of 20. As per the plasticity chart, according to IS classification, the soil is represented by the letter symbols

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📖 Explanation

Plasticity index.

\begin{aligned} I_{p} &=w_{L}-w_{p} \\ &=60-20=40 \end{aligned}

Equation of A line is given by

\begin{aligned} I_{p} &=0.73\left(w_{L}-20\right) \\ &=0.73(60-20) \\ &=292 \end{aligned}

Thus the given soil lies above the A-Line. The liquid limit of the soil is more than 50 , hence the soil will be CH.

Q14GATE 2009MCQ

The laboratory test results of a soil sample are given below : Percentage finer than 4.75 mm= 60 Percentage finer than 0.075 mm= 30 Liquid Limit= 35% Plastic Limit27% The soil classification is

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📖 Explanation

Percentage finer than 4.75mm = 60 Percentage finer then 0.075mm = 30 Liquid limit, $w_L=35 %$ Plastic limit, $w_P=27 %$ Plasticity Index, $I_P=w_L-w_P=37-27=8 %$ Fine grained soils are those having 50% or more pass through 75 $\mu$ sieve (0.075 mm). It is designated as sand (S) if 50% or more pass through 4.75 mm sieve otherwise designated as gravel (G). Since the percentage finer than 4.75 mm is 60%, it is designated as sand (S). Further, percentage finer than 7575 $\mu$ is 30 which is more than 12%, observe the liquid limit and plastic limit on the plasticity chart. Plasticity index corresponding to A-line, $I_P=0.73(w_L-20)=10.95$ The point corresponding to $w_L=35 %$ and $I_P=8 %$ lies below A- line on the Plasticity chart. It is designated as M or O. Therefore, the soil is classified as SM.

Q15GATE 2008MCQ

Group Symbols assigned to silty sand and clayey sand are respectively

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📖 Explanation

First part of the dual symbol indicates gradation whereas second part indicates the nature of fineness. $\Rightarrow$ Silty sand =SM $\Rightarrow$ Clayey sand =SC

Q16GATE 2007MCQ

Sieve analysis on a dry soil sample of mass 1000 g showed that 980 g and 270 g of soil pass through 4.75 mm and 0.075 mm sieve, respectively. The liquid limit and plastic limits of the soil fraction passing through 425 $\mu$ sieves are 40% and 18%, respectively. The soil may be classified as

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📖 Explanation

% retained on $4.75 \mathrm{mm}=\frac{20}{1000} \times 100=2 \%$ % passing $75 \mu$ sieve $=\frac{270}{1000} \times 100=27 \%$ $\therefore \quad \gt 50 \%$ of coarse fraction is between 4.75 mm and $75 \mu$ Hence, sandy soil and % fines $\gt$ 12% Soil is clayey sand (SC).

Q17GATE 2006MCQ

Laboratory sieve analysis was carried out on a soil sample using a complete set of standard IS sieves. Out of 500g of soil used in the test, 200g was retained on $IS \; 600\mu$ sieve, 250g was retained on $IS \; 500\mu$ sieve and the remaining 50g was retained on $IS \; 425\mu$ sieve. The classification of the soil is

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📖 Explanation

More than 50% of the soil pass through $600 \mu$ sieve which means that even a greater percentage of the soil passes through 4.75 mm sieve. Hence, the soil is definitely sandy soil. Also, we know that for a sand to be considered well graded $\mathrm{C}_{\mathrm{u}}$ should be greater than 6 Here, in this case $C_{u}=1.2.$ Thus, the soil is poorly graded sand.

Q18GATE 2005MCQ

A soil mass contains 40% gravel, 50% sand and 10% silt. This soil can be classified as

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Q19GATE 2002MCQ

Data from a sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 micron IS sieve. The liquid limit and plastic limit of the finer fraction was found to be 45 and 33 percents respectively. The group symbol of the given soil as per IS:1498-1970 is

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📖 Explanation

% passing $75 \mu$ sieve $\gt 50 \%$ So the soil might be, silt (M) or clay (C)

\begin{aligned} I_{p} \text { of soil } &=w_{L}-w_{p} \\ &=45-33=12 \end{aligned}

Equation of A line is given by

\begin{aligned} I_{p} &=0.73\left(w_{L}-20\right) \\ &=0.73(45-20)=18.25 \end{aligned}

$\because \mathrm{I}_{\mathrm{p}}\text{ of soil } \lt \mathrm{I}_{\mathrm{p}}\text{ of A-line}$ $\therefore$ It will lie below A-line and also $35 \lt w_{L} \lt 50$ So it is MI.

Q20GATE 2001MCQ

The following two statments are made with respect to different sand samples having the same relative density. Identify if they are TRUE or FALSE. I. Poorly graded sands will have lower friction angle than the well graded sands. II. The particle size has no influence on the friction angle of sand.

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Classification Of Soils And Clay Minerals Practice Questions

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