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

There are 5 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

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.