Classification of Soils and Clay Minerals

Question 1
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
A
0.15 and 1.22
B
0.19 and 6.60
C
1.22 and 0.15
D
6.60 and 0.19
GATE CE 2021 SET-2   Geotechnical Engineering
Question 1 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}
Question 2
As per the Unified Soil Classification System (USCS), the type of soil represented by 'MH' is
A
Inorganic silts of high plasticity with liquid limit more than 50%
B
Inorganic silts of low plasticity with liquid limit less than 50%
C
Inorganic clays of high plasticity with liquid limit less than 50%
D
Inorganic clays of low plasticity with liquid limit more than 50%
GATE CE 2021 SET-2   Geotechnical Engineering
Question 2 Explanation: 


Question 3
A staff is placed on a benchmark (BM) of reduced level (RL) 100.000 m and a theodolite is placed at a horizontal distance of 50 m from the BM to measure the vertical angles. The measured vertical angles from the horizontal at the staff readings of 0.400 m and 2.400 m are found to be the same. Taking the height of the instrument as 1.400m, the RL (in m) of the theodolite station is _____
A
50
B
75
C
90
D
100
GATE CE 2019 SET-1   Geotechnical Engineering
Question 3 Explanation: 


\begin{aligned} x&=50 \tan \alpha \\ 2-x&=50 \tan \alpha \\ \therefore \;\;x&=2-x \\ 2x&=2 \\ x&=1m \end{aligned}
R.L of theodolite station (P): 100 + 1.4 - 0.1 - 0.4 = 100 m
Question 4
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
A
283.6 m
B
284.5 m
C
285 m
D
285.6 m
GATE CE 2019 SET-1   Geotechnical Engineering
Question 4 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}
Question 5
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
A
CL
B
CH
C
CL-ML
D
CI
GATE CE 2016 SET-1   Geotechnical Engineering
Question 5 Explanation: 
Soil is plastic in the range of 26-48%
Hence plastic limit =26% and liquid limit =48%
Because liquid limit is in the range of 35% to 50%. Hence, intermediate compressible soil.
Question 6
The clay mineral primarily governing the swelling behavior of Black Cotton soil is
A
Halloysite
B
Illite
C
Kaolinite
D
Montmorillonite
GATE CE 2014 SET-2   Geotechnical Engineering
Question 7
As per Indian Standard Soil Classification System (IS: 1498-1970), an expression for A-line is
A
I_{p}=0.73(w_{L}-20)
B
I_{p}=0.70(w_{L}-20)
C
I_{p}=0.73(w_{L}-10)
D
I_{p}=0.70(w_{L}-10)
GATE CE 2014 SET-2   Geotechnical Engineering
Question 8
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
A
CH
B
CI
C
CL
D
CL-ML
GATE CE 2012   Geotechnical Engineering
Question 8 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.
Question 9
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
A
FM_{Q}=\sqrt{FM_{P} \times FM_{R}}
B
FM_{Q}=0.5(FM_{P}+FM_{R})
C
FM_{P}\gt FM_{Q}\gt FM_{R}
D
FM_{P}\lt FM_{Q}\lt FM_{R}
GATE CE 2011   Geotechnical Engineering
Question 9 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)
Question 10
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
A
CL
B
CI
C
CH
D
CL-ML
GATE CE 2010   Geotechnical Engineering
Question 10 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.
There are 10 questions to complete.

7 thoughts on “Classification of Soils and Clay Minerals”

  3. Great job .
    But mention the common data question. Some common data question are use first question answer are using second question in common data question ,that particular question was mention please. and great practice learning place.
    Thankyou

    Reply

  4. Answer given for question number 11 is wrong. The correct answer is SM not GM. Only 40% of the soil is retained on 4.75 mm sieve. 60% of soil particles are below 4.75 mm and 60% of them are greater than 0.075mm. So it will not be gravel. Since particle size of sand is in between 0.075 mm to 4.75mm, it will be sand.

    Reply

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