Question 1 |

The correct match between the physical states of the soils given in Group I and
the 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}

\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}

1-S, 2-P, 3-Q, 4-R | |

1-Q, 2-S, 3-P, 4-R | |

1-Q, 2-P, 3-R, 4-S | |

1-S, 2-Q, 3-P, 4-R |

Question 2 |

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?

SW, SP, ML, CH | |

SW, SP, ML, CH | |

SP, SW, CH, ML | |

ML, SP, CH, SW |

Question 2 Explanation:

Hydraulic conductivity Order.

Gravel \gt Sand \gt silt \gt lay

Gravel \gt Sand \gt silt \gt lay

Question 3 |

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 3 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 4 |

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 4 Explanation:

Question 5 |

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 5 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 6 |

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 6 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 7 |

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 7 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 8 |

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

Halloysite | |

Illite | |

Kaolinite | |

Montmorillonite |

Question 9 |

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 10 |

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 10 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.

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,

Please share this among all your friends and join our free education for all campaign.

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

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.

Thank You Jishna K S,

We have updated the answer.