Soil Lab Reviewer-quiz For Finals
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3. Wrong spelling and incomplete answers are wrong.
Questions and Answers
- 1.
What is the title of Exp. 1?
Explanation
The title of Experiment 1 is "moisture content of soil" because the question is asking for the title of the experiment and the given answer directly states the topic being studied in Experiment 1.Rate this question:
- 2.
What is the title of Exp. 2?
- 3.
What is the title of Exp. 3?
Explanation
The title of Experiment 3 is "specific gravity of soil".Rate this question:
- 4.
What is the title of Exp. 4?
Explanation
The title of Exp. 4 is "in-place density test" or "in place density test".Rate this question:
- 5.
What is the title of Exp. 5?
Explanation
The correct answer is "compaction test" because the title of Experiment 5 is referring to the compaction test. This suggests that Experiment 5 is focused on conducting a test to determine the compaction characteristics of a material, which is an important aspect in construction and geotechnical engineering.Rate this question:
- 6.
What is the title of Exp. 6?
Explanation
The title of Experiment 6 is "sieve analysis."Rate this question:
- 7.
What is the title of Exp. 7?
Explanation
The correct answer is "hydrometer analysis" because it is the title of Experiment 7.Rate this question:
- 8.
What is the title of Exp. 8?
Explanation
The title of Experiment 8 is "water repellency test" because it is specifically mentioned in the question.Rate this question:
- 9.
What is the title of Exp. 9?
Explanation
The title of Experiment 9 is "volume expansion test."Rate this question:
- 10.
What is the title of Exp. 10?
Explanation
The title of Experiment 10 is "Visual Classification of Soils." This suggests that the experiment involves a method or process of classifying soils based on visual characteristics. It is likely that the experiment focuses on visually identifying and categorizing different types of soils based on their appearance, color, texture, or other observable features.Rate this question:
- 11.
It is the ratio, expressed as a percentage, of the mass of "pore" or "free" water in a given mass of soil to the mass of the dry soil solids.
Explanation
The correct answer is water content, moisture content. Water content and moisture content both refer to the ratio of the mass of water in a given mass of soil to the mass of the dry soil solids. It is expressed as a percentage and represents the amount of water present in the soil.Rate this question:
- 12.
How many moisture cans were used in Exp. 1?
Explanation
In Experiment 1, a total of 6 moisture cans were used.Rate this question:
- 13.
What is the formula of Moisture Content?
- A.
Ws/Ww
- B.
Ww/Ws
- C.
Vs/Vw
- D.
Vw/Vs
Correct Answer
B. Ww/WsExplanation
The formula for moisture content is Ww/Ws, where Ww represents the weight of water and Ws represents the weight of solid material. This formula is commonly used to calculate the amount of water present in a substance, such as soil or food, by comparing it to the weight of the dry material. By dividing the weight of water by the weight of solid material, we can determine the proportion of moisture in the substance.Rate this question:
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- 14.
Exp. 2 (Consistency Limits) is performed using what soil?
- A.
Fine-grained soil
- B.
Coarse-grained soil
Correct Answer
A. Fine-grained soilExplanation
The correct answer is fine-grained soil because the consistency limits test, also known as the Atterberg limits test, is used to determine the moisture content at different stages of soil consistency. This test is typically performed on fine-grained soils such as clay or silt, as these soils have a higher water-holding capacity and can exhibit different states of consistency (liquid, plastic, and solid) depending on their moisture content. Coarse-grained soils, on the other hand, have larger particle sizes and do not undergo significant changes in consistency with varying moisture content, making them unsuitable for this test.Rate this question:
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- 15.
Give one of the 2 important soil classification properties in a soil specimen which were also tested in Exp. 2.
Correct Answer
liquid limit
plastic limitExplanation
The correct answer is liquid limit and plastic limit. These two properties are important in soil classification as they help determine the behavior and characteristics of the soil. The liquid limit represents the moisture content at which the soil transitions from a plastic state to a liquid state, indicating its ability to flow. On the other hand, the plastic limit represents the moisture content at which the soil transitions from a plastic state to a semisolid state, indicating its ability to be molded. Both the liquid limit and plastic limit are crucial in understanding the consistency and plasticity of the soil.Rate this question:
- 16.
In Exp. 2, the specimen used was a soil passing sieve No. ___.
Correct Answer
40Explanation
In Experiment 2, the specimen used was a soil passing sieve No. 40. This means that the soil used in the experiment was able to pass through a sieve with a mesh size of No. 40. The sieve number indicates the size of the openings in the sieve, with a higher number indicating smaller openings. Therefore, the soil used in the experiment was relatively fine-grained and able to pass through a sieve with small openings.Rate this question:
- 17.
The apparatus used for liquid limit test in Exp. 2.
Correct Answer
casagrande apparatus
casagrande's cup
casagrandeExplanation
The correct answer for the apparatus used for liquid limit test is the Casagrande apparatus, Casagrande's cup, and Casagrande. These three options are all related to the same apparatus. The Casagrande apparatus is a device used to determine the liquid limit of a soil sample. It consists of a brass cup (Casagrande's cup) that is filled with the soil sample and a mechanism to measure the number of blows required to close a groove made in the soil. Casagrande refers to the inventor of this apparatus, Arthur Casagrande.Rate this question:
- 18.
In Exp. 2, the point on the cup that comes in contact with the base should rise to a height of __ mm.
Correct Answer
10Explanation
The height of the point on the cup that comes in contact with the base should rise to a height of 10 mm.Rate this question:
- 19.
In Exp. 2, practice using the cup and determine the correct rate to rotate the crank so that the cup drops approximately ___ times per second.
Correct Answer
2
twoExplanation
In Experiment 2, participants are instructed to practice using the cup and find the appropriate rate at which to rotate the crank. The goal is to make the cup drop approximately two times per second. This suggests that the correct answer to the question is 2, as "two" is the word that represents the numerical value of 2.Rate this question:
- 20.
The tool used to cut a clean straight groove down the center of the cup in Exp. 2.
Correct Answer
grooving toolExplanation
A grooving tool is a tool that is specifically designed to cut clean and straight grooves. In Experiment 2, the grooving tool was used to create a groove down the center of the cup. This tool would have a sharp cutting edge that can easily penetrate the material of the cup and create a precise groove. By using a grooving tool, the experimenter would be able to achieve a clean and straight groove in the cup, which would be important for the experiment.Rate this question:
- 21.
In Exp. 2, count the number of drops, N, it takes to make the two halves of the soil pat come into contact at the bottom of the groove along a distance of __ mm (1/2in.)
Correct Answer
13 - 22.
In Exp. 2, if the number of drops exceeds ___, then go directly to step eight and do not record the number of drops, otherwise, record the number of drops on the data sheet.
Correct Answer
50Explanation
If the number of drops in Experiment 2 exceeds 50, then the instruction is to skip to step eight without recording the number of drops. However, if the number of drops is 50 or less, then the instruction is to record the number of drops on the data sheet.Rate this question:
- 23.
In _______ test, you form the soil into an ellipsoidal mass. You roll the mass between the palm or the fingers and the glass plate.
Correct Answer
plastic limit
plastic
plasticityExplanation
The correct answer is plastic limit. In the plastic limit test, the soil is formed into an ellipsoidal mass and then rolled between the palm or fingers and a glass plate. This test is used to determine the moisture content at which the soil transitions from a plastic state to a semi-solid state. Plasticity refers to the ability of a material, in this case, soil, to undergo deformation without breaking or cracking. Plastic is a term used to describe materials that can be easily molded or shaped.Rate this question:
- 24.
In Exp. 2 Plastic Limit Test, use sufficient pressure to roll the mass into a thread of uniform diameter by using about ___ strokes per minute.
Correct Answer
90Explanation
In the Plastic Limit Test, it is important to apply sufficient pressure to roll the mass into a thread of uniform diameter. This can be achieved by using about 90 strokes per minute. The strokes help to compact and shape the soil sample, ensuring that it reaches its plastic limit. Using the correct number of strokes per minute is crucial in obtaining accurate and reliable results in this test.Rate this question:
- 25.
In Exp. 2 Plastic Limit Test, the thread shall be deformed so that its diameter reaches ___ mm (1/8in.), taking no more than two minutes.
Correct Answer
3.2Explanation
In the plastic limit test, the thread is subjected to deformation until its diameter reaches 3.2 mm (1/8 inch), and this should be done within a maximum time of two minutes. This specific diameter is chosen as it indicates the point at which the soil transitions from a plastic to a semi-solid state. The time limit ensures that the test is conducted efficiently and consistently.Rate this question:
- 26.
This formula is for?
Correct Answer
LL
liquid limitExplanation
This formula is used to calculate the liquid limit (LL) of a substance. The liquid limit is a property of soil that represents the moisture content at which the soil transitions from a plastic state to a liquid state. This formula helps determine the LL value by considering various factors such as the number of blows required for the soil to flow and the moisture content of the soil sample.Rate this question:
- 27.
This formula is for?
Correct Answer
PL
plastic limitExplanation
The correct answer for this question is PL, which stands for plastic limit. The term "plastic limit" refers to the moisture content at which a soil becomes plastic and can be molded into different shapes without breaking. This formula is used to determine the plastic limit of a soil sample, which is an important parameter in geotechnical engineering and soil mechanics. By knowing the plastic limit, engineers can assess the soil's behavior and make informed decisions regarding its use in construction projects.Rate this question:
- 28.
This formula is for?
Correct Answer
FI
flow indexExplanation
The correct answer is FI, flow index. The formula is used to calculate the flow index, which is a measure of the flowability of a material. The flow index indicates how easily a material can flow under a specific set of conditions.Rate this question:
- 29.
This formula is for?
Correct Answer
PI
plasticity indexExplanation
The formula is for calculating the plasticity index (PI). The plasticity index is a measure of the range of moisture content over which a soil exhibits plastic behavior. It is calculated by subtracting the liquid limit from the plastic limit. This index is important in geotechnical engineering as it helps determine the soil's behavior and suitability for construction purposes.Rate this question:
- 30.
It indicates how much heavier/lighter a material is than water.
Correct Answer
specific gravityExplanation
Specific gravity is a measurement that compares the density of a substance to the density of water. It indicates whether a material is heavier or lighter than water. If the specific gravity is less than 1, the material is lighter than water, while a specific gravity greater than 1 means the material is heavier. Therefore, the correct answer, specific gravity, accurately describes the concept of comparing the weight of a material to water.Rate this question:
- 31.
In Exp. 3, specific gravity of soil is determined using what apparatus?
Correct Answer
pycnometer
pycnometer jarExplanation
The correct answer for determining the specific gravity of soil in Exp. 3 is the pycnometer or pycnometer jar. The pycnometer is a small glass container with a known volume that is used to measure the density of a substance. In this case, it is used to measure the density of the soil sample. By comparing the weight of the pycnometer with and without the soil sample, the specific gravity of the soil can be calculated.Rate this question:
- 32.
In Exp. 3, 150 grams of soil was used passing sieve no. ___.
Correct Answer
12Explanation
In Experiment 3, the soil sample was passed through sieve no. 12. This means that the soil was sieved using a sieve with 12 mesh openings per inch. Sieving is a common method used to separate particles of different sizes in soil samples. By using sieve no. 12, the experimenters were able to remove any larger particles from the soil sample, leaving only the finer particles that were able to pass through the sieve. This step is important in order to ensure uniformity and consistency in the soil sample used for further analysis or experimentation.Rate this question:
- 33.
In Exp. 3, fill the jar with distilled water to the mark, which is ____ ml.
Correct Answer
500Explanation
The correct answer is 500 because the question states to fill the jar with distilled water to the mark, and the mark indicates a volume of 500 ml.Rate this question:
- 34.
This formula is for?
Correct Answer
SG
specific gravityExplanation
The given formula is for calculating specific gravity (SG). Specific gravity is a measurement used to compare the density of a substance to the density of a reference substance, usually water. It helps to determine the relative density or heaviness of a substance.Rate this question:
- 35.
One standard procedure used for determining the field unit weight of compaction is the _________ method.
Correct Answer
sand coneExplanation
The sand cone method is a commonly used procedure for determining the field unit weight of compaction. In this method, a hole is dug in the soil and filled with dry sand of known weight. The volume of the hole is then determined by measuring the amount of sand required to fill it. By comparing the weight of the sand to the volume of the hole, the unit weight of the soil can be calculated. This method is widely used in construction and geotechnical engineering to assess the compaction characteristics of soils.Rate this question:
- 36.
In Exp. 4, the soil specimen used was an oven-dried sand that pass through No. 30 sieve and retained on No. ___ sieve.
Correct Answer
40Explanation
The correct answer is 40. This indicates that the soil specimen used in Experiment 4 was an oven-dried sand that passed through a No. 30 sieve and was retained on a No. 40 sieve. This means that the sand particles were relatively fine, with a particle size between the sizes of the No. 30 and No. 40 sieves.Rate this question:
- 37.
The mold used in Exp. 4 (In-Place Density Test).
Correct Answer
standard proctor compaction moldExplanation
The correct answer is "standard proctor compaction mold." This type of mold is used in the In-Place Density Test, also known as the Proctor Test. This test is used to determine the maximum dry density and optimum moisture content of a soil sample. The standard proctor compaction mold is specifically designed to hold a specific volume of soil and allow for compaction using a hammer. This mold ensures that the soil sample is compacted under controlled conditions, allowing for accurate density measurements.Rate this question:
- 38.
In this test, the experiment aims to determine the relationship between the moisture content and the dry density of a soil for a specified compactive effort.
Correct Answer
compaction test
compactionExplanation
The correct answer is "compaction test, compaction." In this experiment, the researcher is trying to understand the connection between the moisture content and the dry density of a soil under a specific compactive effort. The term "compaction test" refers to the specific procedure used to measure the compaction characteristics of the soil, while "compaction" is the general process of increasing the density of the soil by applying pressure. Both terms are relevant to the experiment and its objective.Rate this question:
- 39.
In compaction test, ____________ is the amount of mechanical energy that is applied to the soil mass.
Correct Answer
compactive effortExplanation
In a compaction test, the compactive effort refers to the amount of mechanical energy that is applied to the soil mass. This energy is used to compact the soil and increase its density. The compactive effort can be controlled by adjusting factors such as the weight of the compaction equipment, the number of passes made over the soil, and the speed of the equipment. By applying the appropriate compactive effort, the soil can be compacted to achieve the desired level of density and strength.Rate this question:
- 40.
Several different methods are used to compact soil in the field. Give one example.
Correct Answer
tamping
kneading
vibration
static load compactionExplanation
The given answer lists several different methods used to compact soil in the field, which include tamping, kneading, vibration, and static load compaction. These methods involve applying pressure or force to the soil to reduce its volume and increase its density. Tamping involves repeatedly striking the soil surface with a tamper or compactor to compress it. Kneading involves using a roller or similar equipment to knead and press the soil. Vibration involves using vibrating equipment to shake and compact the soil. Static load compaction involves applying a static load or weight on the soil for a certain period to achieve compaction.Rate this question:
- 41.
The type of compaction test that was employed in the Exp. 5 is the ______________ Test.
Correct Answer
standard proctorExplanation
The correct answer for this question is "standard proctor." This suggests that in Experiment 5, the type of compaction test used was the standard proctor test. The standard proctor test is a common method used to determine the maximum dry density and optimum moisture content of a soil sample, which is important in construction and engineering projects. This test involves compacting the soil at various moisture levels and measuring the resulting density to determine the moisture content that yields the highest density.Rate this question:
- 42.
In the Standard Proctor Test, the soil is compacted by a ____ lb hammer falling a distance of one foot into a soil filled mold.
Correct Answer
5.5 - 43.
In the Standard Proctor Test, the mold is filled with three equal layers of soil, and each layer is subjected to ___ drops of the hammer.
Correct Answer
25Explanation
In the Standard Proctor Test, the mold is filled with three equal layers of soil, and each layer is subjected to 25 drops of the hammer. This is because 25 drops is the standard number of drops specified in the test procedure. The purpose of the drops is to compact the soil and simulate the effect of repeated loading on the soil in the field. By subjecting each layer to the same number of drops, it ensures consistency and allows for accurate comparison of different soil samples.Rate this question:
- 44.
In the Standard Proctor Test, the type of compaction mold that will be used for the testing is of the smaller type, which is __ inches in diameter and has a volume of about 1/30 ft3 (944 cm3).
Correct Answer
4Explanation
The correct answer is 4. In the Standard Proctor Test, a smaller type of compaction mold is used, which has a diameter of 4 inches and a volume of about 1/30 ft3 (944 cm3). This mold size is commonly used in the test to compact the soil sample and determine its maximum dry density and optimum moisture content.Rate this question:
- 45.
In Exp. 5, the soil specimen used was a 5kg of soil passing sieve no. ___.
Correct Answer
4Explanation
In this experiment, a soil specimen weighing 5kg was used. The soil specimen was required to pass through a specific sieve number. The correct answer is 4, indicating that the soil specimen used in the experiment passed through sieve number 4.Rate this question:
- 46.
In Exp. 5 Compaction Test, assume water content for the first test to be __ %.
Correct Answer
3Explanation
The correct answer is 3. This suggests that in the Compaction Test, the water content for the first test should be 3%.Rate this question:
- 47.
In Exp. 5 Compaction Test, remove the soil from the mold using a _____________ and take soil moisture content samples from the top and bottom of the specimen. Fill the moisture cans with soil and determine the water content.
Correct Answer
mechanical extruderExplanation
In the compaction test, the soil needs to be removed from the mold after the test is conducted. To do this, a mechanical extruder is used. The mechanical extruder helps in easily and efficiently removing the compacted soil from the mold. After removing the soil, moisture content samples need to be taken from the top and bottom of the specimen. This is done by filling moisture cans with soil and then determining the water content.Rate this question:
- 48.
In Exp. 5 Compaction Test, draw a smooth curve connecting the plotted points and on the same graph, draw a curve of complete saturation or “_____________ curve”.
Correct Answer
zero air voidsExplanation
In the compaction test, a smooth curve is drawn connecting the plotted points to represent the relationship between the dry density and moisture content of the soil. On the same graph, a curve of complete saturation or "zero air voids curve" is drawn to represent the maximum density that can be achieved when there are no air voids present in the soil. This curve helps to determine the optimum moisture content for compaction and ensures that the soil is compacted to its maximum potential.Rate this question:
- 49.
It is a practice or procedure used to assess the particle size distribution of a granular material.
Correct Answer
sieve analysisExplanation
Sieve analysis is a method used to determine the particle size distribution of a granular material. It involves passing the material through a series of sieves with different sized openings, and measuring the amount of material retained on each sieve. This analysis provides valuable information about the particle size distribution, which is important for various applications such as construction, agriculture, and pharmaceuticals.Rate this question:
- 50.
Sieve analysis is also called as __________ test.
Correct Answer
gradationExplanation
Sieve analysis is a test used to determine the particle size distribution of a granular material. It involves passing the material through a series of sieves with progressively smaller openings and measuring the amount of material retained on each sieve. This process helps to classify the material based on its particle size distribution, which is also known as gradation. Therefore, the correct answer for this question is gradation.Rate this question:
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