# 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”

1. Question number 2 and 3 are of geometric engineering.

• Yes. Change the Question.

2. Good 👍job.

• Thank You Rajesh Kumar Singh,