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
0.15 and 1.22 | |
0.19 and 6.60 | |
1.22 and 0.15 | |
6.60 and 0.19 |
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}
\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
Inorganic silts of high plasticity with liquid limit more than 50% | |
Inorganic silts of low plasticity with liquid limit less than 50% | |
Inorganic clays of high plasticity with liquid limit less than 50% | |
Inorganic clays of low plasticity with liquid limit more than 50% |
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 _____
50 | |
75 | |
90 | |
100 |
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
283.6 m | |
284.5 m | |
285 m | |
285.6 m |
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
CL | |
CH | |
CL-ML | |
CI |
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.
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
Halloysite | |
Illite | |
Kaolinite | |
Montmorillonite |
Question 7 |
As per Indian Standard Soil Classification System (IS: 1498-1970), an expression for A-line is
I_{p}=0.73(w_{L}-20) | |
I_{p}=0.70(w_{L}-20) | |
I_{p}=0.73(w_{L}-10) | |
I_{p}=0.70(w_{L}-10) |
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
CH | |
CI | |
CL | |
CL-ML |
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
FM_{Q}=\sqrt{FM_{P} \times FM_{R}} | |
FM_{Q}=0.5(FM_{P}+FM_{R}) | |
FM_{P}\gt FM_{Q}\gt FM_{R} | |
FM_{P}\lt FM_{Q}\lt FM_{R} |
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)
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
CL | |
CI | |
CH | |
CL-ML |
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.
\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.
Question number 2 and 3 are of geometric engineering.
Yes. Change the Question.
Good 👍job.
Thank You Rajesh Kumar Singh,
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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