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Test: Soil Mechanics - 2 - Civil Engineering (CE) MCQ


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25 Questions MCQ Test Civil Engineering SSC JE (Technical) - Test: Soil Mechanics - 2

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Test: Soil Mechanics - 2 - Question 1

Which of the following is a measure of particle size range?

Detailed Solution for Test: Soil Mechanics - 2 - Question 1
Measure of Particle Size Range:

  • Effective Size: Effective size is not a measure of particle size range. It is the diameter of the particle that passes through 10% by weight of the sample.

  • Uniformity Coefficient: The uniformity coefficient is a measure of the particle size range. It is calculated as the ratio of the size of the particles that make up 60% of the sample to the size of the particles that make up 10% of the sample. A lower uniformity coefficient indicates a narrower particle size range.

  • Coefficient of Curvature: The coefficient of curvature is not a measure of particle size range. It is a measure of the shape of the particle size distribution curve and indicates whether the distribution is symmetrical or skewed.

  • None of the Above: This option is incorrect as the uniformity coefficient is indeed a measure of particle size range.


Therefore, the correct answer is B: uniformity coefficient, as it is the measure of particle size range.
Test: Soil Mechanics - 2 - Question 2

Which of the following statements is correct?

Detailed Solution for Test: Soil Mechanics - 2 - Question 2

Explanation:

  • Uniformity coefficient represents the shape of the particle size distribution curve: This statement is incorrect. The uniformity coefficient (UC) is a measure of the degree of uniformity in the particle size distribution of a soil sample. It is calculated as the ratio of the size of particles at the 60% finer point to the size of particles at the 10% finer point on a grain size distribution curve.


  • For a well graded soil, both uniformity coefficient and coefficient of curvature are nearly unity: This statement is incorrect. For a well-graded soil, the uniformity coefficient is greater than 4 and the coefficient of curvature is between 1 and 3. These values indicate a wide range of particle sizes in the soil sample.


  • A soil is said to be well graded if it has most of the particles of about the same size: This statement is incorrect. A soil is considered well graded if it contains a wide range of particle sizes, from coarse to fine. Well-graded soils have a more even distribution of particle sizes compared to poorly graded soils.


Therefore, the correct answer is option D: none of the above.
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Test: Soil Mechanics - 2 - Question 3

Uniformity coefficient of a soil is

Detailed Solution for Test: Soil Mechanics - 2 - Question 3
Uniformity Coefficient of Soil

  • Definition: The uniformity coefficient of a soil is a measure of the particle size distribution within the soil sample.

  • Calculation: It is calculated as the ratio of the size of the particles that make up the soil sample. It is represented as D60/D10, where D60 is the diameter of the particle that is larger than 60% of the sample and D10 is the diameter of the particle that is larger than 10% of the sample.

  • Interpretation: A high uniformity coefficient indicates a wide range of particle sizes within the soil sample, while a low uniformity coefficient indicates a more uniform distribution of particle sizes.

  • Range: The uniformity coefficient can range from 1 to infinity.

  • Answer: The uniformity coefficient can be equal to or greater than 1, as there is no upper limit to this value.

Test: Soil Mechanics - 2 - Question 4

According to Atterberg, the soil is said to be of medium plasticity if the plasticity index PI is

Detailed Solution for Test: Soil Mechanics - 2 - Question 4

Test: Soil Mechanics - 2 - Question 5

If the natural water content of soil mass lies between its liquid limit and plastic limit, the soil mass is said to be in

Detailed Solution for Test: Soil Mechanics - 2 - Question 5
Explanation:

  • Liquid Limit: The liquid limit of a soil is the water content at which the soil changes from a plastic state to a liquid state. It is determined by specific laboratory tests.

  • Plastic Limit: The plastic limit of a soil is the water content at which the soil changes from a semi-solid state to a plastic state. It is also determined by specific laboratory tests.

  • Soil Mass Between Liquid Limit and Plastic Limit: When the natural water content of a soil mass lies between its liquid limit and plastic limit, the soil mass is said to be in a plastic state.

  • Characteristics: In this state, the soil can be easily molded and retains its shape when formed into a ball. It is neither too wet (liquid state) nor too dry (solid state).

  • Importance: Understanding the state of the soil mass is crucial in construction and engineering projects as it helps in determining the properties and behavior of the soil.

Test: Soil Mechanics - 2 - Question 6

When the plastic limit of a soil is greater than the liquid limit, then the plasticity index is reported as

Detailed Solution for Test: Soil Mechanics - 2 - Question 6

Concept:

Plasticity index (Ip):

It is the range of moisture content over which a soil exhibits plasticity.

Ip = WL - WP

where,

WL = water content at Liquid Limit.

Wp = water content at Plastic Limit.

Liquid Limit:

The minimum water content at which the soil is still in a liquid state and just develops shear strength against flowing.

Plastic Limit:

Minimum water content which makes the soil to be rolled into 3 mm diameter threads.

Note: If water content at the plastic limit is equal to water content at the liquid limit then the result of the plastic limit is also zero.

Ip = 0 ( if  WL = Wp )

Explanation:

Ip = 0 ( If or WL   W

The plasticity index of soil never is a negative value. 

∴ When the plastic limit of a soil is greater than the liquid limit, then the plasticity index is reported as zero.

Test: Soil Mechanics - 2 - Question 7

The ratio for a soil mass is called

Test: Soil Mechanics - 2 - Question 8

Toughness index is defined as the ratio of

Detailed Solution for Test: Soil Mechanics - 2 - Question 8
Toughness Index Calculation

  • Plasticity Index (PI): This is a measure of the plasticity of a soil. It is the numerical difference between the liquid limit and the plastic limit of the soil.

  • Flow Index (FI): Flow index is a measure of the flowability of a soil. It is determined by the flow curve of a soil sample.


Toughness Index Formula

  • Toughness Index (TI) = Plasticity Index (PI) / Flow Index (FI)


Explanation

  • Plasticity Index indicates the range of moisture content within which a soil exhibits plastic properties.

  • Flow Index measures the flowability of the soil and is determined by the flow curve.

  • The Toughness Index calculated by dividing the Plasticity Index by the Flow Index gives an indication of the toughness of the soil.

  • A higher Toughness Index value suggests a soil that is more resistant to flow and deformation.


Conclusion

  • Calculating the Toughness Index using the Plasticity Index and Flow Index provides valuable information about the behavior of soil under different conditions.

  • Understanding the Toughness Index can help in making informed decisions related to construction projects, agriculture, and geotechnical engineering.

Test: Soil Mechanics - 2 - Question 9

If the plasticity index of a soil mass is zero, the soil is

Detailed Solution for Test: Soil Mechanics - 2 - Question 9
Explanation:

  • Plasticity Index: The plasticity index of a soil is a measure of the range of water content within which the soil exhibits plastic behavior. It is calculated as the difference between the liquid limit and the plastic limit of the soil.

  • Zero Plasticity Index: When the plasticity index of a soil mass is zero, it means that the liquid limit is equal to the plastic limit, and there is no range of water content where the soil exhibits plastic behavior.

  • Classification: Soils with a plasticity index of zero are classified as non-plastic soils.

  • Soil Type: Since sands do not exhibit plastic behavior and have a zero plasticity index, a soil mass with a zero plasticity index is classified as sand.

Test: Soil Mechanics - 2 - Question 10

The admixture of coarser particles like sand or silt to clay causes

Detailed Solution for Test: Soil Mechanics - 2 - Question 10
Explanation:

  • Clay Soil Composition: Clay soil is composed of very fine particles that are tightly packed together.


  • Admixture of Coarser Particles: When coarser particles like sand or silt are added to clay soil:



    • Liquid Limit: The addition of coarser particles leads to a decrease in the liquid limit of the soil. This is because the presence of coarser particles allows for better drainage and reduces the water-holding capacity of the soil.


    • Plasticity Index: The plasticity index of the soil also decreases with the addition of coarser particles. This is because the coarser particles disrupt the tight packing of the clay particles, reducing the plasticity of the soil.



Therefore, the correct answer is C: decrease in both liquid limit and plasticity index.

Test: Soil Mechanics - 2 - Question 11

Select the correct statement.

Detailed Solution for Test: Soil Mechanics - 2 - Question 11


Explanation:

  • Strength and Stability: In geotechnical engineering, a non-uniform soil typically has a better strength and stability compared to a uniform soil. This is because the varying particle sizes in a non-uniform soil create interlocking arrangements, increasing shear strength.

  • Uniformity Coefficient: The uniformity coefficient is a measure of the particle size distribution in a soil sample. A higher uniformity coefficient indicates a wider range of particle sizes, making the soil more non-uniform. Therefore, a non-uniform soil with a higher uniformity coefficient often has greater strength and stability.

  • Poorly Graded vs. Well Graded Soil: The uniformity coefficient of a poorly graded soil is typically higher than that of a well-graded soil. This means that poorly graded soils are more non-uniform, leading to increased strength and stability.



Test: Soil Mechanics - 2 - Question 12

The water content of soil, which represents the boundary between plastic state and liquid state,is known as

Detailed Solution for Test: Soil Mechanics - 2 - Question 12
Explanation:

  • Liquid Limit: The water content of soil at the boundary between the plastic state and the liquid state is known as the liquid limit. It is the minimum water content at which the soil changes from the plastic state to a liquid state.

  • Plastic Limit: The plastic limit is the water content at which a soil changes from a plastic to a semisolid state. It is the water content below which the soil behaves as a solid and above which it behaves as a plastic material.

  • Shrinkage Limit: The shrinkage limit is the water content at which further loss of moisture will not cause any volume change in the soil. It is the minimum water content below which the soil will not shrink any further upon drying.

  • Plasticity Index: The plasticity index is the numerical difference between the liquid limit and the plastic limit of a soil. It gives an indication of the range of water content over which the soil exhibits plastic behavior.


Therefore, in this case, the correct answer is liquid limit as it represents the boundary between the plastic state and the liquid state of the soil.

Test: Soil Mechanics - 2 - Question 13

Which of the following soils has more plasticity index?

Detailed Solution for Test: Soil Mechanics - 2 - Question 13
Analysis:

  • Sand: Sand has a low plasticity index because it consists of larger particles that do not have the ability to stick together.

  • Silt: Silt has a moderate plasticity index compared to sand and clay. It contains smaller particles than sand but larger than clay.

  • Clay: Clay has a high plasticity index because it consists of very fine particles that can easily stick together and form a cohesive mass.

  • Gravel: Gravel has a low plasticity index similar to sand as it consists of larger particles that do not have cohesive properties.


Conclusion:

  • Among the given options, clay has the highest plasticity index due to its fine particle size and cohesive properties.

  • Therefore, the correct answer is C: clay.

Test: Soil Mechanics - 2 - Question 14

At liquid limit, all soils possess

Detailed Solution for Test: Soil Mechanics - 2 - Question 14
Explanation:

  • Liquid Limit: Liquid limit is defined as the moisture content at which the soil changes from a plastic to a liquid state.

  • Shear Strength: Shear strength is the ability of soil to resist shear stresses.

  • At liquid limit, all soils possess:


    • Same shear strength of small magnitude. This means that at the liquid limit, the shear strength of all soils is relatively low.

    • This is due to the fact that at the liquid limit, the soil particles are more separated and the soil is in a more fluid state.

    • Therefore, the shear strength of all soils at the liquid limit is similar and of small magnitude.


Test: Soil Mechanics - 2 - Question 15

According to IS classification, the range of silt size particles is

Detailed Solution for Test: Soil Mechanics - 2 - Question 15
Classification of Silt Size Particles according to IS

  • Range of Silt Size Particles: 0.075 mm to 0.002 mm


Explanation

  • IS Classification: The Indian Standard (IS) classification system categorizes particles based on their size ranges.

  • Silt Size Particles: Silt particles fall within the range of 0.075 mm to 0.002 mm according to IS classification.

  • Specific Range: The specific range for silt size particles is crucial in various industries such as construction, agriculture, and geology.

  • Importance: Understanding the size classification of silt particles helps in appropriate handling and utilization of materials in different applications.

Test: Soil Mechanics - 2 - Question 16

Highway Research Board (HRB) classification of soils is based on

Detailed Solution for Test: Soil Mechanics - 2 - Question 16
Highway Research Board (HRB) classification of soils

  • Based on: The classification is based on both particle size composition and plasticity characteristics of the soil.


Particle size composition

  • Particle size composition refers to the distribution of different sizes of particles in the soil.

  • Soils are classified based on the percentages of gravel, sand, silt, and clay present in the soil sample.

  • The classification helps in understanding the engineering properties and behavior of the soil.


Plasticity characteristics

  • Plasticity characteristics refer to the ability of the soil to undergo deformation without cracking.

  • Soils are classified based on their plasticity index, liquid limit, and plastic limit.

  • This classification is crucial for determining the suitability of soil for construction purposes.


Importance of classification based on both factors

  • By considering both particle size composition and plasticity characteristics, engineers can make informed decisions regarding the selection of appropriate soil types for various construction projects.

  • Understanding these characteristics helps in predicting the behavior of the soil under different conditions such as loading, moisture content, and compaction.

  • HRB classification provides a standardized system for categorizing soils, which is essential for ensuring the safety and stability of structures built on or with soil materials.

Test: Soil Mechanics - 2 - Question 17

Dispersed type of soil structure is an arrangement comprising particles having

Detailed Solution for Test: Soil Mechanics - 2 - Question 17
Dispersed Type of Soil Structure

  • Definition: Dispersed type of soil structure is an arrangement comprising particles having face to face or parallel orientation.

  • Characteristics:

    • Particles are arranged with their faces or edges oriented in a parallel manner.

    • The particles do not have a distinct structure and are randomly distributed.

    • This type of structure is common in clay soils.



  • Formation:

    • Dispersed soil structure can be formed due to the swelling and shrinking of clay minerals.

    • It can also result from compaction and heavy machinery use in agricultural practices.



  • Importance:

    • Dispersed soil structure affects water infiltration and root growth in plants.

    • It can lead to soil erosion and reduced soil fertility if not managed properly.



Test: Soil Mechanics - 2 - Question 18

Effective stress is

Detailed Solution for Test: Soil Mechanics - 2 - Question 18

Effective stress is a defined property which cannot be measured and is calculated by measuring the total stress and pore water pressure in the field.

Effective Stress= Total Stress- Pore water Pressure

It is not the true contact stress between particles because it uses area for division as gross area rather than Area of actual contact between particles

Test: Soil Mechanics - 2 - Question 19

The degree of saturation in soils can be defined as the ratio of

Detailed Solution for Test: Soil Mechanics - 2 - Question 19
Definition of Degree of Saturation in Soils:

  • Volume of water to volume of voids in soil: The degree of saturation in soils is defined as the ratio of the volume of water to the volume of voids in the soil.


Explanation:

  • Volume of water to volume of voids in soil: The degree of saturation is a measure of how much water is present in the soil compared to the total volume of voids in the soil. It indicates the percentage of the void space in the soil that is filled with water.

  • Importance: Understanding the degree of saturation in soils is important in various engineering and environmental applications, such as determining the permeability of soils, predicting groundwater movement, and assessing the stability of slopes and foundations.


Conclusion:

  • Volume of water to volume of voids in soil: Therefore, the correct definition of the degree of saturation in soils is the ratio of the volume of water to the volume of voids in the soil.

Test: Soil Mechanics - 2 - Question 20

Rise of water table above the ground surface causes

Detailed Solution for Test: Soil Mechanics - 2 - Question 20
Explanation:

  • Rise of water table above the ground surface causes:



  • A: equal increase in pore water pressure and total stress



  • When the water table rises above the ground surface, it increases the pore water pressure in the soil.

  • At the same time, the total stress acting on the soil also increases due to the weight of the additional water above the ground surface.

  • Therefore, both pore water pressure and total stress increase equally as a result of the rise in the water table.

  • This can have significant implications for the stability and strength of the soil, especially in areas prone to flooding or high groundwater levels.

Test: Soil Mechanics - 2 - Question 21

The total and effective stresses at a depth of 5 m below the top level of water in a swimming pool are respectively

Detailed Solution for Test: Soil Mechanics - 2 - Question 21



  • Total Stress at 5m depth: The total stress at a depth of 5m below the water surface in the swimming pool is equal to the weight of the water column above that depth. The total stress is given by the formula: Total Stress = Unit Weight of Water x Depth.


  • Effective Stress at 5m depth: The effective stress at a depth of 5m below the water surface is the difference between total stress and pore water pressure. Since the water is in equilibrium at this depth, the pore water pressure is equal to the unit weight of water multiplied by the depth.


  • Calculation:


    • Total Stress: Total Stress = Unit Weight of Water x Depth = 0.5 kg/cm2 x 5m = 2.5 kg/cm2

    • Effective Stress: Effective Stress = Total Stress - Pore Water Pressure = 2.5 kg/cm2 - 0.5 kg/cm2 = 2.0 kg/cm2



  • Conclusion: The total stress at a depth of 5m below the top level of water in the swimming pool is 2.5 kg/cm2 and the effective stress is 2.0 kg/cm2.


Therefore, the correct answer is option B: 0.5 kg/cm2 and zero.
Test: Soil Mechanics - 2 - Question 22

If the water table rises upto ground surface, then the

Test: Soil Mechanics - 2 - Question 23

The critical hydraulic gradient ic of a soil mass of specific gravity G and voids ratio e is given by

Test: Soil Mechanics - 2 - Question 24

The hydraulic head that would produce a quick condition in a sand stratum of thickness 1.5 m,specific gravity 2.67 and voids ratio 0.67 is equal to

Detailed Solution for Test: Soil Mechanics - 2 - Question 24

hydraulic gradient= (G-1)/(1+e)

G=2.67,e=0.67

we get, hydraulic gradient=1

Test: Soil Mechanics - 2 - Question 25

Physical properties of a permeant which influence permeability are

Detailed Solution for Test: Soil Mechanics - 2 - Question 25

Correct Answer :- c

Explanation : Physical properties of a permanent with influence permeability are Both viscosity and unit weight.

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