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Question 1
ప్రశ్న 1
Which statement correctly represents the three-phase soil system and the usual weight assumption in fundamental soil-phase calculations?
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Explanation:
A natural soil mass is commonly represented by solid, water and air phases. Its total volume equals the sum of the three individual phase volumes. The void volume consists of the water volume plus the air volume. The weight of air is normally negligible in elementary geotechnical calculations. The weights of soil solids and water are included when calculating bulk unit weight.
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Question 2
ప్రశ్న 2
Which combination correctly defines water content w, void ratio e and porosity n for a soil sample?
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Explanation:
Water content is the ratio of water weight to dry soil-solids weight. Void ratio compares the volume of voids with the volume of solids. Porosity compares the volume of voids with the total sample volume. These properties are dimensionless, although water content and porosity are frequently expressed as percentages. The different denominators must be maintained when performing soil-phase calculations.
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Question 3
ప్రశ్న 3
A moist soil sample has a mass of 315 g and a mass of 280 g after oven drying. What is its gravimetric water content?
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Explanation:
Oven drying removes the water from the soil sample under the assumed test conditions. The mass of water is 315 - 280 = 35 g. The dry soil-solids mass is 280 g. Water content is therefore 35/280 times 100 = 12.5%. Dividing by the original moist mass instead would not give the conventional gravimetric water content.
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Question 4
ప్రశ్న 4
A soil has a void ratio of 0.75. What is its porosity, expressed as a percentage?
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Explanation:
Void ratio and porosity are related by n = e/(1 + e). Substituting e = 0.75 gives n = 0.75/1.75. The result is approximately 0.42857. Expressed as a percentage, the porosity is 42.86%. Porosity is lower than void ratio when the void ratio is expressed as a decimal because the two ratios have different denominators.
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Question 5
ప్రశ్న 5
A soil sample has a porosity of 40%. What is its void ratio?
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Explanation:
The given porosity must first be written as the decimal fraction 0.40. Void ratio is calculated using e = n/(1 - n). Substituting gives e = 0.40/0.60. The result is approximately 0.667. The answer is dimensionless and represents the ratio of void volume to solid volume.
వివరణ:
Question 6
ప్రశ్న 6
A soil has specific gravity of solids G_s = 2.70, void ratio e = 0.80 and water content w = 20%. Using Se = wG_s, what is its degree of saturation?
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Explanation:
The fundamental phase relationship is Se = wG_s when water content and saturation are decimal fractions. The water content of 20% is therefore written as 0.20. Rearranging gives S = 0.20 times 2.70 divided by 0.80. This equals 0.675, or 67.5%. The result indicates that approximately 67.5% of the void volume is occupied by water.
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Question 7
ప్రశ్న 7
A partially saturated soil has G_s = 2.65, e = 0.70 and S = 80%. What is its water content approximately?
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Explanation:
The relationship Se = wG_s can be rearranged as w = Se/G_s. The degree of saturation is 0.80 and the void ratio is 0.70. Substitution gives w = 0.80 times 0.70 divided by 2.65. The result is approximately 0.21132, or 21.13%. Water content differs from degree of saturation because the former is based on weights while the latter is based on void volumes.
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Question 8
ప్రశ్న 8
A soil specimen has total volume 1.00 m³, solid volume 0.60 m³, water volume 0.30 m³ and air volume 0.10 m³. Which combination gives its degree of saturation, air content and percentage air voids, respectively?
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Explanation:
The void volume is the sum of water and air volumes, giving 0.40 m³. Degree of saturation is 0.30/0.40 = 75%. Air content is the air volume divided by void volume, giving 0.10/0.40 = 25%. Percentage air voids is the air volume divided by total volume, giving 0.10/1.00 times 100 = 10%. This problem demonstrates why air content and percentage air voids are not interchangeable quantities.
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Question 9
ప్రశ్న 9
A soil has porosity 35% and air content 20%, where air content is defined as air volume divided by void volume. What is its percentage air voids, defined as air volume divided by total soil volume?
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Explanation:
Porosity indicates that voids occupy 35% of the total volume. Air content indicates that air occupies 20% of those voids. Therefore the fraction of total volume occupied by air is 0.35 times 0.20. This equals 0.07, or 7%. The air content remains 20%, but the percentage air voids is only 7% because its denominator is the total soil volume.
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Question 10
ప్రశ్న 10
The bulk unit weight of a moist soil is 19.8 kN/m³ and its gravimetric water content is 10%. What is its dry unit weight?
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Explanation:
Bulk and dry unit weights are related by gamma = gamma_d(1 + w). The water content must be expressed as the decimal 0.10. Rearranging gives gamma_d = 19.8/1.10. The resulting dry unit weight is 18.00 kN/m³. The calculation assumes that the same total sample volume is used to define both unit weights.
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Question 11
ప్రశ్న 11
A dry soil has specific gravity of solids G_s = 2.70 and void ratio e = 0.50. Taking the unit weight of water as 10 kN/m³, determine its dry unit weight.
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Explanation:
Dry unit weight is given by gamma_d = G_s gamma_w/(1 + e). The numerator is 2.70 times 10 = 27 kN/m³. The denominator is 1 + 0.50 = 1.50. Dividing gives gamma_d = 18.0 kN/m³. Dry unit weight depends on the mass of solids and total soil volume, even though the pore water is absent.
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Question 12
ప్రశ్న 12
A soil has G_s = 2.65, e = 0.60 and S = 50%. If gamma_w = 10 kN/m³, calculate its bulk unit weight.
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Explanation:
The bulk unit weight formula is gamma = (G_s + Se)gamma_w/(1 + e). The product Se is 0.50 times 0.60 = 0.30. The numerator factor is therefore 2.65 + 0.30 = 2.95. Substitution gives gamma = 2.95 times 10 divided by 1.60 = 18.4375 kN/m³. The appropriate rounded result is 18.44 kN/m³.
వివరణ:
Question 13
ప్రశ్న 13
A completely saturated soil has G_s = 2.70 and e = 0.80. If gamma_w = 10 kN/m³, what is its saturated unit weight?
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Explanation:
At complete saturation, water occupies the entire void volume. The saturated unit weight is gamma_sat = (G_s + e)gamma_w/(1 + e). Substituting gives gamma_sat = (2.70 + 0.80) times 10 divided by 1.80. This equals approximately 19.4444 kN/m³. The correct rounded answer is 19.44 kN/m³.
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Question 14
ప్రశ్న 14
A saturated soil has G_s = 2.70 and e = 0.80. Taking gamma_w = 10 kN/m³, determine its submerged or buoyant unit weight.
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Explanation:
Submerged unit weight represents the saturated unit weight after accounting for the buoyant effect of surrounding water. It is calculated using gamma' = (G_s - 1)gamma_w/(1 + e). Here gamma' = (2.70 - 1.00) times 10 divided by 1.80. The result is approximately 9.4444 kN/m³. This is also obtained by subtracting 10 kN/m³ from the saturated unit weight of approximately 19.4444 kN/m³.
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Question 15
ప్రశ్న 15
The bulk density of a soil is 1.96 Mg/m³ and its gravimetric water content is 12%. What is its dry density?
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Explanation:
Dry density is obtained from bulk density using rho_d = rho/(1 + w). The water content is 0.12 as a decimal fraction. Dividing 1.96 by 1.12 gives exactly 1.75 Mg/m³. Density and unit weight are different physical quantities, although both describe mass or weight relative to volume. Multiplication of density by gravitational acceleration would be needed to express the corresponding unit weight.
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Question 16
ప్రశ్న 16
The dry unit weight of a soil is 16.875 kN/m³. If G_s = 2.70 and gamma_w = 10 kN/m³, what is its void ratio?
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Explanation:
The relation gamma_d = G_s gamma_w/(1 + e) is rearranged to e = G_s gamma_w/gamma_d - 1. The numerator G_s gamma_w equals 27 kN/m³. Dividing 27 by 16.875 gives 1.60. Subtracting one gives e = 0.60. The calculated void ratio is dimensionless and is consistent with the specified dry unit weight.
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Question 17
ప్రశ్న 17
For a soil having G_s = 2.65 and water content w = 20%, calculate the theoretical zero-air-void dry unit weight. Take gamma_w = 10 kN/m³ and assume complete saturation.
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Explanation:
The zero-air-void condition corresponds to complete saturation with no air occupying the voids. The theoretical dry unit weight is gamma_d,ZAV = G_s gamma_w/(1 + wG_s). Substitution gives 2.65 times 10 divided by (1 + 0.20 times 2.65). This equals 26.5/1.53, or approximately 17.32 kN/m³. Under the stated assumptions, this is the upper theoretical dry unit weight at the specified water content and specific gravity.
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Question 18
ప్రశ్న 18
A laboratory report gives dry unit weight 19.0 kN/m³ at water content 25% for a soil with G_s = 2.65. With gamma_w = 10 kN/m³, the zero-air-void dry unit weight at that water content is approximately 15.94 kN/m³. What is the correct engineering interpretation?
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Explanation:
At a fixed water content and specific gravity, the zero-air-void dry unit weight represents the theoretical complete-saturation boundary. The stated zero-air-void value is approximately 15.94 kN/m³. A reported dry unit weight of 19.0 kN/m³ lies above that boundary. Such a state would require an impossible phase arrangement under the specified properties. The appropriate action is to review the water content, specific gravity, density measurements and calculations.
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Question 19
ప్రశ్న 19
A soil specimen has total volume 180 cm³ and dry mass 270 g. Its moist mass is 324 g, the specific gravity of solids is 2.70 and the density of water is 1 g/cm³. What is the degree of saturation?
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Explanation:
The volume of solids is dry mass divided by the density of solids, giving 270/2.70 = 100 cm³. The void volume is therefore 180 - 100 = 80 cm³. The water mass is 324 - 270 = 54 g, which occupies 54 cm³ at the given water density. Degree of saturation is water volume divided by void volume, or 54/80. The result is 0.675, corresponding to 67.5% saturation.
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Question 20
ప్రశ్న 20
A fully saturated soil has water content 25% and G_s = 2.60. If gamma_w = 10 kN/m³, what is its submerged unit weight approximately?
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Explanation:
For saturated soil, Se = wG_s becomes e = wG_s because S equals one. Thus e = 0.25 times 2.60 = 0.65. The submerged unit weight is gamma' = (G_s - 1)gamma_w/(1 + e). Substitution gives 1.60 times 10 divided by 1.65 = approximately 9.70 kN/m³. The corresponding saturated unit weight would be approximately 19.70 kN/m³, which is greater by the unit weight of water.
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Question 21
ప్రశ్న 21
A soil specimen contains 0.40 m³ of solids, 0.20 m³ of water and 0.10 m³ of air. Its soil-solids specific gravity is 2.50 and gamma_w = 10 kN/m³. Neglecting air weight, which pair gives the bulk unit weight and degree of saturation?
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Explanation:
The total specimen volume is 0.40 + 0.20 + 0.10 = 0.70 m³. The weight of solids is 2.50 times 10 times 0.40 = 10 kN, while the water weight is 2 kN. The bulk unit weight is therefore (10 + 2)/0.70 = approximately 17.14 kN/m³. The void volume is 0.20 + 0.10 = 0.30 m³. Degree of saturation is 0.20/0.30 times 100 = approximately 66.7%.
వివరణ:
Question 22
ప్రశ్న 22
A soil has dry unit weight 17 kN/m³, water content 16% and specific gravity of solids 2.65. Taking gamma_w = 10 kN/m³, determine its degree of saturation approximately.
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Explanation:
First calculate void ratio using e = G_s gamma_w/gamma_d - 1. Substitution gives e = 26.5/17 - 1 = approximately 0.55882. The phase relationship gives S = wG_s/e. Therefore S = 0.16 times 2.65 divided by 0.55882 = approximately 0.75874. Expressed as a percentage and rounded to one decimal place, the degree of saturation is 75.9%.
వివరణ:
Question 23
ప్రశ్న 23
Two soil specimens have the same specific gravity of solids and are evaluated using the same unit weight of water. Specimen P has a smaller void ratio than specimen Q. Which conclusion necessarily follows?
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Explanation:
Dry unit weight is inversely related to 1 + e when specific gravity and water unit weight remain constant. A smaller void ratio therefore gives a greater dry unit weight. Degree of saturation depends on the amount of water in the available void space. The void ratio alone does not establish how much of that space contains water. Consequently the dry-unit-weight comparison is certain, but the saturation comparison is not.
వివరణ:
Question 24
ప్రశ్న 24
Water is added to an initially unsaturated soil specimen without loss of solids, change in total volume or drainage. Water progressively replaces air in the existing voids. Which statement correctly describes the changes while the soil remains partially saturated?
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Explanation:
The quantity of soil solids and the total specimen volume remain constant. Consequently the volume of voids, void ratio and porosity do not change. Adding water increases water mass and the fraction of void volume occupied by water. The bulk unit weight increases because additional water weight is present in the same total volume. Dry unit weight remains unchanged because solid weight and total volume are constant.
వివరణ:
Question 25
ప్రశ్న 25
A soil specimen has total volume 1000 cm³, oven-dry mass 1620 g, specific gravity of solids G_s = 2.70 and water content 15%. Taking water density as 1 g/cm³, determine the percentage air voids and air content, respectively.
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Explanation:
The volume of solids is 1620/2.70 = 600 cm³, leaving a void volume of 400 cm³. The water mass is 15% of 1620 g, giving 243 g or 243 cm³ of water. The air volume is therefore 400 - 243 = 157 cm³. Percentage air voids is 157/1000 times 100 = 15.70%. Air content is 157/400 times 100 = 39.25%, showing the effect of using void volume rather than total volume as the denominator.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Total volume is the sum of solid, water and air volumes, while the weight of air is normally neglected. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. w = W_w/W_s, e = V_v/V_s and n = V_v/V. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 12.5%. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 42.86%. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 0.667. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 67.5%. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. 21.13%. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 75%, 25%, 10%. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 7%. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 18.00 kN/m³. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 18.0 kN/m³. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 18.44 kN/m³. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 19.44 kN/m³. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. 9.44 kN/m³. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. 1.75 Mg/m³. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 0.60. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. 17.32 kN/m³. సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. The reported phase data are physically inconsistent under the stated assumptions and should be checked for measurement or calculation errors. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: D. 67.5%. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. 9.70 kN/m³. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 17.14 kN/m³ and 66.7%. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 75.9%. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. Specimen P has a higher dry unit weight, but its degree of saturation cannot be inferred without additional water-phase information. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. Water content, degree of saturation and bulk unit weight increase, while void ratio, porosity and dry unit weight remain unchanged. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 15.70% and 39.25%. సమాధానం: B.