Test: Soil Mechanics - 1 - Civil Engineering (CE) MCQ

# Test: Soil Mechanics - 1 - Civil Engineering (CE) MCQ

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## 25 Questions MCQ Test Civil Engineering SSC JE (Technical) - Test: Soil Mechanics - 1

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Test: Soil Mechanics - 1 - Question 1

### Which of the following types of soil is transported by gravitational forces?

Detailed Solution for Test: Soil Mechanics - 1 - Question 1

Talus is a type of soil that is transported by gravitational forces. It is formed due to the accumulation of rock fragments and debris at the base of cliffs, mountains, or other steep slopes. Here are some details about talus:

- Formation: Talus is created by the weathering and erosion of rock surfaces, which causes rock fragments to break off and accumulate at the base of the slope.
- Gravitational forces: The primary force responsible for the transportation of talus is gravity. As rocks and debris break off from the slope, they are pulled down by gravity, forming a pile of loose material at the base.
- Characteristics: Talus deposits typically have a conical or fan-shaped appearance, with the slope angle being influenced by the size and shape of the rock fragments. The particles in a talus deposit can range from small pebbles to large boulders.
- Importance: Talus deposits can provide important information about the geologic history and processes in a given area. They also play a role in slope stability and can be a significant factor in the development of landslides.

Other types of soil mentioned in the question are transported by different forces:
- Loess: This is a type of soil composed predominantly of silt-sized particles, and it is typically transported by wind. Loess deposits are often found in areas adjacent to deserts or other sources of windblown sediment.
- Drift: This term refers to glacial deposits, which are transported by the movement of glaciers. Drift can include a variety of materials, such as clay, silt, sand, and gravel.
- Dune sand: As the name suggests, dune sand is transported by wind and forms sand dunes in desert environments or along coastlines. The grain size and sorting of dune sand are influenced by the strength and consistency of the wind in a given area.

Test: Soil Mechanics - 1 - Question 2

### Water content of soil can

Detailed Solution for Test: Soil Mechanics - 1 - Question 2

In soil Mechanics Water Content (w) = Weight of water/weight of Soil.

take 0.9 Kg of dry Soil and add water about 1 liter (in volume) which is about 1Kg in weight.

Then the water content of mixture = 1/0.9 = 1.11 = 111%

which is greater than 100%,

Mud or slurry tends to have water content more than 100%.

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Test: Soil Mechanics - 1 - Question 3

### Which of the following types of soil is transported by gravitational forces?

Detailed Solution for Test: Soil Mechanics - 1 - Question 3

The correct answer is talus. Let me explain in detail.

Talus is a type of soil that is transported and deposited by gravitational forces. This type of soil is typically found at the base of cliffs or steep slopes and is formed as a result of rockfall and other slope failures.

Here are some key points about talus:

• Formation: Talus is formed through the accumulation of rock fragments and debris that have fallen from a higher elevation due to weathering, erosion, or slope instability.
• Characteristics: Talus soil is generally coarse, angular, and poorly sorted, with a wide range of particle sizes. The soil is often unstable and susceptible to further movement due to its loose nature and steep slopes.
• Location: Talus is commonly found at the base of mountain cliffs or steep slopes where rockfall and other slope failures are frequent.
• Role of gravity: Gravitational forces play a crucial role in the transportation and deposition of talus soil. As rocks and debris break off from a slope, gravity pulls them downhill, causing them to accumulate and form talus deposits.

In contrast, other types of soil mentioned have different transportation mechanisms:

• Loess: Loess is a wind-blown (aeolian) deposit consisting predominantly of silt-sized particles.
• Drift: Drift soil is a general term for material deposited by glaciers or ice sheets, so its transportation mechanism is glacial movement.
• Dune sand: Dune sand is transported and deposited by wind, forming sandy mounds or ridges in desert regions or along coastlines.

In summary, among the given options, talus is the type of soil that is transported by gravitational forces, as it accumulates at the base of slopes due to the downhill movement of rock fragments and debris.

Test: Soil Mechanics - 1 - Question 4

A fully saturated soil is said to be

Detailed Solution for Test: Soil Mechanics - 1 - Question 4

A fully saturated soil is considered a two-phase system with soil and water. To understand this in detail, let's break down the components of soil and their interactions:

Soil can be categorized into three primary phases:
1. Solid phase (soil particles)
2. Liquid phase (water)
3. Gaseous phase (air)

These phases can combine to form different soil systems:
- One-phase system: Only solid soil particles are present.
- Two-phase system with soil and air: Solid soil particles and air-filled voids are present.
- Two-phase system with soil and water: Solid soil particles and water-filled voids are present.
- Three-phase system: Solid soil particles, water-filled voids, and air-filled voids are present simultaneously.

In the case of a fully saturated soil, all the voids between the soil particles are completely filled with water, leaving no space for air. This creates a two-phase system with soil and water.

The two-phase system with soil and water is essential in understanding various soil properties and behaviors, such as:
- Shear strength: The resistance of soil to deformation and failure due to the presence of water.
- Permeability: The ease with which water can flow through the soil.
- Compressibility: The change in soil volume due to the expulsion of water from voids under applied pressure.

In summary, a fully saturated soil is a two-phase system with soil and water because all the voids between the soil particles are filled with water, and no air is present in the system. This configuration plays a crucial role in determining the soil's engineering properties and behavior.

Test: Soil Mechanics - 1 - Question 5

Total number of stress components at a point within a soil mass loaded at its boundary is

Detailed Solution for Test: Soil Mechanics - 1 - Question 5

Explanation : Considering a cube will give us 6 stresses till where it’s easy to get. And rest counting as 3 shear stresses along each plane in each direction will give total 9 stresses.

Test: Soil Mechanics - 1 - Question 6

The submerged density of soil in terms of unit weight of watergw, specific gravity G and Voids
ratio e is given by the expression

Detailed Solution for Test: Soil Mechanics - 1 - Question 6

Explanation:
The submerged density (also known as submerged unit weight or buoyant unit weight) of soil is a crucial parameter in geotechnical engineering, particularly for the analysis of structures in water or structures with water present in the soil pores. It is used to calculate the effective stress in soils, as well as to evaluate the stability of slopes and retaining structures. The submerged density of soil is given by the following expression:

Submerged Density (γ_sub) = (G - 1)* g/ (1 + e)

In this expression:
- G is the specific gravity of soil solids
- gis the unit weight of water (typically 9.81 kN/m³)
- e is the voids ratio of the soil

Test: Soil Mechanics - 1 - Question 7

A soil has a bulk density of 22 kN/m3 and water content 10%. The dry density of soil is

Detailed Solution for Test: Soil Mechanics - 1 - Question 7

To calculate the dry density of the soil, we can use the following formula:

Dry Density = Bulk Density / (1 + Water Content)

Given:
- Bulk Density = 22 kN/m3
- Water Content = 10% = 0.10 (as a decimal)

Now, we can plug these values into the formula:
Dry Density = 22 / (1 + 0.10)
Dry Density = 22 / 1.10
Dry Density = 20.0 kN/m3

Hence, the correct answer is: 20.0 kN/m3

Test: Soil Mechanics - 1 - Question 8

If the voids of a soil mass are full of air only, the soil is termed as

Detailed Solution for Test: Soil Mechanics - 1 - Question 8

The correct answer is (c) dry soil.

To understand this classification better, let's break it down into the different types of soil based on their moisture content and voids:

1. Dry Soil: In dry soil, the voids are filled with air only, and there is no water present in the soil mass. This means that the soil particles are not surrounded by water, and the soil exhibits low moisture content. Dry soil can be found in arid regions or areas where the water has evaporated.

2. Partially Saturated Soil: Partially saturated soil means that the voids in the soil mass are filled with both air and water. This occurs when the soil has absorbed some water but not enough to fill all the voids completely. In this type of soil, the moisture content is higher than in dry soil but lower than in saturated soil.

3. Saturated Soil: In saturated soil, the voids are completely filled with water, and there is no air present. This can happen when the soil is in contact with a water source, such as a lake, river, or underground water table. Saturated soil has the highest possible moisture content and can be prone to various problems, such as reduced bearing capacity, increased settlement, and slope instability.

4. Air Entrained Soil: This term is typically not used to describe soil types based on moisture content. Air entrainment refers to the process of trapping air bubbles within a material, such as concrete, to improve its workability and resistance to freeze-thaw cycles. This concept is not directly applicable to soil classification.

5. Dehydrated Soil: This term is also not commonly used to describe soil types based on moisture content. Dehydration refers to the process of losing water, which can occur in any type of soil. However, it does not specifically denote a particular soil classification based on voids and moisture content.

In conclusion, when the voids of a soil mass are full of air only, the soil is termed as dry soil because it has low moisture content and the voids are not filled with water.

Test: Soil Mechanics - 1 - Question 9

Valid range for n, the percentage voids, is

Detailed Solution for Test: Soil Mechanics - 1 - Question 9

The correct answer is: 0 < n < 100

Explanation:

The percentage voids, denoted by 'n', represent the empty spaces or voids in a material as a percentage of the total volume. It is used to describe the porosity of a material, such as soil, rock, or concrete. The percentage voids can range from 0% (completely solid with no voids) to 100% (completely empty with no solid material).

To better understand the valid range for n, let's break it down:
- 0 < n: This means that the percentage voids should be greater than 0%. It indicates that there should be some voids present in the material. If n = 0%, the material would be completely solid with no voids.
- n < 100: This means that the percentage voids should be less than 100%. It indicates that there should be some solid material present. If n = 100%, the material would be completely empty with no solid material.

Therefore, the valid range for n, the percentage voids, is 0 < n < 100. This range ensures that the material has both solid and void components, which is necessary for most practical applications.

Test: Soil Mechanics - 1 - Question 10

Select the correct statement.

Detailed Solution for Test: Soil Mechanics - 1 - Question 10

The correct statement is: Unit weight of soil decreases due to sub-mergence in water.

To explain in detail:

• When soil is submerged in water, it experiences a loss of weight due to the buoyant force exerted by the water. This is because the water molecules exert an upward force on the soil particles, which opposes the gravitational force acting on them.
• This buoyant force results in a decrease in the effective weight of the soil particles, and consequently, the unit weight of the soil decreases.
• The concept of submerged (or buoyant) unit weight is often used in geotechnical engineering to account for this decrease in unit weight when soil is submerged in water. Submerged unit weight is calculated as the difference between the total unit weight of the soil and the unit weight of water.

In summary:
- Soil submerged in water experiences a buoyant force that opposes the gravitational force acting on the soil particles.
- This buoyant force decreases the effective weight of the soil particles, resulting in a decrease in the unit weight of the soil.
- The submerged unit weight is used in geotechnical engineering to account for this decrease in unit weight when soil is submerged in water.

Test: Soil Mechanics - 1 - Question 11

Voids ratio of a soil mass can

Detailed Solution for Test: Soil Mechanics - 1 - Question 11

Explanation : The void ratio is the ratio of the volume of voids (open spaces, i.e. air and water) in a soil to volume of solids. The void ratio is thus a ratio which can be greater than 1. It can also be expressed as a fraction. Void ratio and porosity differ only in the denominator.

Test: Soil Mechanics - 1 - Question 12

If the volume of voids is equal to the volume of solids in a soil mass, then the values of porosity and voids ratio respectively are

Detailed Solution for Test: Soil Mechanics - 1 - Question 12

Test: Soil Mechanics - 1 - Question 13

When the degree of saturation is zero, the soil mass under consideration represents

Detailed Solution for Test: Soil Mechanics - 1 - Question 13

The correct answer is (b) two phase system with soil and air.

Soil is generally composed of three main phases: solid particles (soil), water, and air. The degree of saturation (S) is a parameter that represents the proportion of the soil's pore space occupied by water. It is expressed as a percentage and is given by the following formula:
S = (Volume of water / Volume of voids) * 100

When the degree of saturation is zero (S = 0), it means that there is no water in the soil's pore spaces, and only air is present. In such a case, the soil mass can be considered as a two-phase system consisting of:
1. Solid particles (soil)
2. Air

This two-phase system implies that the soil is completely dry, and all the voids between the solid particles are filled with air. It is important to note that the degree of saturation can vary between 0 and 100%, depending on the soil's moisture content and environmental conditions. A soil with a degree of saturation of 100% would be a two-phase system with soil and water, while a soil with a degree of saturation between 0% and 100% would be a three-phase system, consisting of solid particles (soil), water, and air.

Test: Soil Mechanics - 1 - Question 14

In a liquid limit test, the moisture content at 10 blows was 70% and that at 100 blows was 20%. The liquid limit of the soil, is

Detailed Solution for Test: Soil Mechanics - 1 - Question 14

The liquid limit of the soil is 65 %

• Liquid limit is define as water content at which the soil possess an arbitrarily fixed small amount of shear strength.
• The result of test are represented in form of curve known as flow curve from which water content corresponding to 25 no. of blows is interpolated and is called as liquid limit.
• Graph is shown in the attached figure.

Let N be the number of blows and w be the water content at N blows.

Liquid Limit= W * (N/25)^0.121.
Where,
W = Water content at N blows.
N = No.of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
Required liquid limit = 88-23 = 65%.

Test: Soil Mechanics - 1 - Question 15

If the degree of saturated of a partially saturated soil is 60% then air content of the soil is

Detailed Solution for Test: Soil Mechanics - 1 - Question 15

Test: Soil Mechanics - 1 - Question 16

If the water content of a fully, saturated soil mass is 100%, then the voids ratio of the sample is

Detailed Solution for Test: Soil Mechanics - 1 - Question 16

Test: Soil Mechanics - 1 - Question 17

The ratio of the volume of voids to the volume of soil solids in a given soil mass, is known

Detailed Solution for Test: Soil Mechanics - 1 - Question 17

The void ratio (e) is defined as the ratio of the volume of voids (Vv) to the volume of soil solids (Vs) in a given soil mass. It can be mathematically expressed as:

e = Vv / Vs

The void ratio is a dimensionless value used to describe the relative proportion of void space to solid particles in a soil mass. It is an important parameter in geotechnical engineering, as it provides information about the soil's compressibility, permeability, and shear strength.

Test: Soil Mechanics - 1 - Question 18

Relative density of a compacted dense sand is approximately equal to

Detailed Solution for Test: Soil Mechanics - 1 - Question 18

Correct answer is option C) 0.95
Explanation:
Relative Density of Compacted Dense Sand:
- The relative density of compacted dense sand typically ranges between 0.7 to 1.0, which is equivalent to 70% to 100% relative density.
- For a compacted dense sand, the relative density is approximately equal to 0.95 (or 95% relative density). This is because dense sand has a high degree of compaction, resulting in a higher value for its relative density.
- A relative density of 0.95 indicates that the soil is well compacted and has good strength and stiffness properties. This makes it suitable for use in engineering applications, such as the construction of foundations, embankments, and retaining structures.
To summarize, the correct answer is 0.95 (option c) because a compacted dense sand typically has a high degree of compaction, resulting in a higher relative density value.

Test: Soil Mechanics - 1 - Question 19

If the sand in - situ is in its densest state, then the relative density of sand is

Detailed Solution for Test: Soil Mechanics - 1 - Question 19

Correct answer is option B) 1

Explanation:
The relative density, also known as the density index, of a granular soil like sand is a measure of its compactness compared to its loosest and densest possible states. The relative density is defined as:

Relative Density (Dr) = (emax - e) / (emax - emin)
where:
- emax is the maximum void ratio, corresponding to the loosest state of the sand
- emin is the minimum void ratio, corresponding to the densest state of the sand
- e is the in-situ void ratio, which is the ratio of the volume of voids to the volume of solids in the soil

When the sand is in its densest state (emin), the in-situ void ratio (e) equals the minimum void ratio (emin). Therefore, the equation for relative density becomes:
Dr= (emax - emin) / (emax - emin)
Since the numerator and denominator are equal, the relative density is:
Dr= 1

So, when the sand is in its densest state, the relative density of the sand is 1.

Test: Soil Mechanics - 1 - Question 20

Which of the following methods is most accurate for the determination of the water content of soil?

Detailed Solution for Test: Soil Mechanics - 1 - Question 20

The most accurate method for determining the water content of soil is the oven drying method.

Here's a detailed explanation of this method and its advantages over other methods:
Oven Drying Method:

The oven drying method is a standard laboratory procedure used for determining the water content of a soil sample. This method involves drying a soil sample in an oven at a constant temperature (usually 105°C to 110°C) for a specified period (usually 24 hours) and comparing the mass of the soil before and after drying. The water content is then calculated as the ratio of the mass of water lost to the mass of the dry soil.

Procedure:
1. Weigh a clean, dry, and empty container.
2. Place a representative soil sample in the container and weigh the container with the sample.
3. Place the container with the sample in the oven and heat it at a constant temperature (105°C to 110°C) for 24 hours.
4. After 24 hours, remove the container from the oven and allow it to cool in a desiccator to avoid absorption of moisture from the air.
5. Weigh the container with the dried soil sample.
6. Calculate the water content using the formula:

Water content (%) = [(Wet mass - Dry mass) / Dry mass] x 100

Advantages of the Oven Drying Method:
- It provides highly accurate and consistent results.
- It is a widely accepted and standardized method, which allows for easy comparison of results across different labs and studies.
- The method is applicable to a wide range of soil types and conditions.

Comparison with Other Methods:
- Sand Bath Method: This method involves heating a soil sample in a container surrounded by sand. Although it is a relatively simple method, it may not provide accurate results due to uneven heating and the potential for contamination from sand particles.
- Alcohol Method: This method uses alcohol to replace the water in a soil sample, and the water content is determined by the difference in mass before and after alcohol evaporation. However, this method can be less accurate than the oven drying method because alcohol may not completely replace all the water in the sample.
- Calcium Carbide Method: This method involves the reaction between calcium carbide and water in a sealed container to produce acetylene gas. The pressure of the gas is proportional to the water content of the sample. Although this method is quick and can be used in the field, it can be less accurate than the oven drying method due to potential errors in gas pressure measurements and the sensitivity of the reaction to sample preparation and handling.

In conclusion, the oven drying method is the most accurate method for determining the water content of soil due to its high precision, consistency, and applicability to various soil types.

Test: Soil Mechanics - 1 - Question 21

For proper field control which of the following methods is best suited for quick determination of water content of a soil mass?

Detailed Solution for Test: Soil Mechanics - 1 - Question 21

The calcium carbide method is best suited for quick determination of water content of a soil mass in the field. Here's a detailed explanation:

Calcium Carbide Method:
- The calcium carbide method, also known as the speedy moisture test, is a widely used field test for determining the moisture content of soil quickly and accurately.
- It is based on the chemical reaction between calcium carbide and water, which produces acetylene gas. The amount of gas produced is directly proportional to the water content in the soil sample.

- Quick results: The calcium carbide method provides results in a matter of minutes, whereas other methods like oven drying and sand bath may take several hours or even days.
- Field-friendly: It can be easily performed in the field without any need for electricity or lab equipment.
- Accurate: The method provides accurate results when performed correctly, and the calibration chart accounts for different soil types and their specific characteristics.

Limitations:
- The accuracy of the test depends on the proper calibration of the testing device and the skill of the operator.
- The method may not be suitable for soils with very low or very high moisture content.
- The presence of certain chemicals or minerals in the soil may interfere with the accuracy of the test.

In conclusion, the calcium carbide method is best suited for the quick determination of water content in a soil mass for proper field control. However, it's essential to follow the correct procedure and maintain the testing equipment to ensure accurate results.

Test: Soil Mechanics - 1 - Question 22

A pycnometer is used to determine

Detailed Solution for Test: Soil Mechanics - 1 - Question 22

A pycnometer is a scientific instrument used to determine the specific gravity of a liquid or a solid. It is most commonly used for determining the water content and specific gravity of soil particles. The correct answer is (c) water content and specific gravity.

Here's a detailed explanation of how a pycnometer works and its importance in soil analysis:

Working Principle of a Pycnometer:
- A pycnometer is a glass or metal container with a known volume, equipped with a stopper or lid that has a small hole to allow excess liquid to escape when the container is filled.
- The pycnometer is first weighed empty (W1) and then filled with a liquid (usually water or a suspension of soil particles in water) and weighed again (W2).
- The liquid is then replaced with a solid sample (soil particles), and the pycnometer is weighed once more (W3). The pycnometer can also be filled with a soil-water suspension to determine the water content of the soil.
- The specific gravity of the soil particles can be determined using the known volume of the pycnometer and the weights obtained during the experiment.

Importance in Soil Analysis:
- Determining the water content and specific gravity of soil particles is essential for various soil tests and engineering applications.
- Water content is a crucial parameter in understanding the behavior of soil, as it affects the soil's strength, compressibility, and permeability.
- Specific gravity is an important property that describes the density of soil particles relative to the density of water. It helps in characterizing the soil and understanding its engineering behavior.
- By determining the water content and specific gravity of soil particles, engineers can make informed decisions about the suitability of the soil for construction projects, the design of foundations, and the assessment of slope stability, among other applications.

Test: Soil Mechanics - 1 - Question 23

Stoke’s law is valid only if the size of particle is

Detailed Solution for Test: Soil Mechanics - 1 - Question 23

• It is assumed that particles show discrete settling (i.e. grains of different sizes fall through a liquid at different velocities).
• Particles assumed spherical in shape.
• Medium is assumed infinite.
• Particles size range 0.2 mm to 0.0002 mm.

Limitations in the use of Stoke's Law in Sedimentation analysis:

1. Stoke's law is applicable for spherical particles only. The fine clay particles are not spherical in shape. While applying Stoke's law, the concept of equivalent diameter is used. The equivalent diameter of a soil particle is defined as the diameter of an imaginary sphere that has the same specific gravity as the soil particle and settles with the same terminal velocity as that of the soil particle.

2. It is assumed that every particle settles independently without interference from other particles as well as from the sides of the jar. To minimize error due to this assumption, it is recommended that not more than 50 gms of soil particles be taken in 1000 ml of soil suspension.

3. The soil particles in the soil suspension may have different values of specific gravity. But in the computations, an average value of G is used.

4. The lower limit of particle size for the validity of Stoke's Law is 0.0002 mm. However, the upper limit for the same is 0.2 mm. For particles of size less than 0.0002 mm, Brownian movement affects their settlement and in the case of particles larger than 0.2 mm, turbulence affects the settlement.

5. Stokes's law is valid upto a maximum Reynolds number of 1.

Test: Soil Mechanics - 1 - Question 24

In hydrometer analysis for a soil mass

Detailed Solution for Test: Soil Mechanics - 1 - Question 24

The correct answer is c: meniscus correction is additive and dispersing agent correction is subtractive.

Hydrometer analysis is a laboratory method used to determine the particle size distribution of fine-grained soils, such as clay and silt. This method involves the use of a hydrometer, which measures the density of a soil-water suspension at specific time intervals. To accurately determine the particle size distribution, corrections must be made for meniscus and dispersing agent effects.

- A meniscus is the curved surface of a liquid in a container, such as a graduated cylinder or burette.
- The meniscus correction is applied to account for the effect of the meniscus on the hydrometer readings.
- When the hydrometer is placed in the soil suspension, the liquid level is higher around the hydrometer stem due to the meniscus effect.
- This makes the hydrometer reading slightly lower than the actual value.
- Therefore, the meniscus correction is added to the hydrometer reading to obtain the correct value.

Dispersing Agent Correction (Subtractive):
- A dispersing agent is a chemical substance added to the soil suspension to separate and stabilize the soil particles, ensuring that they remain uniformly distributed in the suspension.
- The dispersing agent increases the density of the liquid, which in turn affects the hydrometer reading.
- The dispersing agent correction is applied to account for the effect of the dispersing agent on the hydrometer readings.
- Since the dispersing agent increases the density of the liquid, the hydrometer reading will be higher than the actual value.
- Therefore, the dispersing agent correction is subtracted from the hydrometer reading to obtain the correct value.
In summary, the meniscus correction is additive, while the dispersing agent correction is subtractive in hydrometer analysis for soil mass, ensuring accurate determination of the particle size distribution of fine-grained soils.

Test: Soil Mechanics - 1 - Question 25

The hydrometer method of sedimentation analysis differs from the pipette analysis mainly in

Detailed Solution for Test: Soil Mechanics - 1 - Question 25

The sedimentation analysis is a process used to determine the particle size distribution of fine-grained soils, such as silts and clays. Two common methods used for sedimentation analysis are the hydrometer method and the pipette method. Both methods involve the preparation of a soil suspension, allowing particles to settle over time, and taking observations to calculate the distribution of particle sizes. However, the main difference between these two methods lies in the method of taking observations.

Hydrometer Method:
- In the hydrometer method, a hydrometer is used to measure the density of the soil suspension at various time intervals.
- The hydrometer measures the relative density of the soil suspension by comparing it to the density of water.
- The hydrometer sinks into the suspension to a depth that depends on the density of the suspension; the denser the suspension, the less the hydrometer sinks.
- The hydrometer has a calibrated scale that allows for the direct reading of the relative density or the percentage of finer particles in the suspension.
- The observations are taken at specific time intervals, and the results are used to calculate the particle size distribution.

Pipette Method:
- In the pipette method, a pipette is used to extract a sample of the soil suspension at various depths and time intervals.
- The pipette is first calibrated to extract a known volume of the suspension.
- At each time interval, the pipette is lowered to a specific depth and the suspension is drawn into the pipette.
- The extracted sample is then transferred to a container, and the sediment in the sample is allowed to dry and weighed.
- The weight of the sediment in each sample is used to calculate the percentage of finer particles in the suspension and determine the particle size distribution.

In summary, the main difference between the hydrometer method and the pipette method of sedimentation analysis is in the method of taking observations. While the hydrometer method uses a hydrometer to measure the density of the soil suspension, the pipette method involves extracting and weighing samples of the soil suspension using a pipette. Both methods have their advantages and disadvantages, and the choice between them depends on factors such as the soil type, equipment availability, and accuracy requirements.

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