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Question 1
ప్రశ్న 1
Which statement best explains capillary rise of water above the groundwater table in fine-grained soils?
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Explanation:
Capillary rise results from surface tension acting at the water-air interface together with wetting interactions involving the soil particles. These forces can support a column of water above the free groundwater level. Smaller connected pores generally permit a greater capillary rise under otherwise similar conditions. Water in the capillary zone may have negative gauge pore-water pressure. The actual rise in soil depends on pore geometry, connectivity, fluid properties and wetting characteristics.
వివరణ:
Question 2
ప్రశ్న 2
Water rises in an ideal circular capillary tube of diameter 0.20 mm. If surface tension is 0.072 N/m, contact angle is zero and unit weight of water is 10 kN/m³, what is the capillary rise?
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Explanation:
For an ideal circular tube, capillary rise is h = 4T cos(theta)/(gamma_w d). The contact angle is zero, so its cosine equals one. Converting units gives diameter 0.00020 m and water unit weight 10000 N/m³. Substitution gives h = 4 times 0.072 divided by (10000 times 0.00020). The resulting capillary rise is 0.144 m.
వివరణ:
Question 3
ప్రశ్న 3
Two ideal capillary tubes contain the same wetting liquid at the same temperature and have equal contact angles. If the diameter of Tube P is half that of Tube Q, how does the capillary rise in Tube P compare with that in Tube Q?
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Explanation:
Capillary rise in a circular tube is inversely proportional to its diameter when all other relevant quantities are fixed. The diameter of Tube P is one-half the diameter of Tube Q. Therefore its capillary rise is twice that of Tube Q. This follows directly from the inverse-diameter relationship. Natural soils have irregular pores, so the result is an idealized comparison rather than an exact prediction for every soil.
వివరణ:
Question 4
ప్రశ్న 4
A point lies 1.8 m vertically above a static water table within an idealized fully saturated capillary fringe. Taking gamma_w = 10 kN/m³, what is the gauge pore-water pressure at that point?
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Explanation:
Gauge pore-water pressure is zero at a static water table connected to atmospheric pressure. Within an idealized saturated capillary fringe, pore-water pressure becomes negative above that level. Its magnitude is the unit weight of water multiplied by vertical height above the water table. Therefore u = -10 times 1.8 = -18 kPa. The negative sign represents water tension or suction relative to atmospheric pressure.
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Question 5
ప్రశ్న 5
Under static hydrostatic conditions with atmospheric gauge pressure taken as zero, which pore-water pressure distribution is correct for a saturated soil and its idealized saturated capillary fringe?
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Explanation:
The water table is the reference level at which pore-water pressure is atmospheric under the stated conditions. Its gauge pressure is therefore zero. Below the water table, hydrostatic pore-water pressure increases positively with depth. Above the water table, water retained in a saturated capillary fringe is under negative gauge pressure. This simple linear distribution should not automatically be extended into a partially saturated zone without additional analysis.
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Question 6
ప్రశ్న 6
At a point in a saturated soil deposit, total vertical stress is 150 kPa and measured pore-water pressure is +45 kPa. According to Terzaghi's effective-stress principle, what is the effective vertical stress?
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Explanation:
Terzaghi's effective-stress relationship is sigma_prime = sigma - u. The total vertical stress is 150 kPa. The positive pore-water pressure is 45 kPa. Subtraction gives sigma_prime = 150 - 45 = 105 kPa. Positive pore pressure reduces the portion of total stress carried by the soil skeleton.
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Question 7
ప్రశ్న 7
A homogeneous saturated soil deposit has its water table at ground level. Its saturated unit weight is 20 kN/m³ and gamma_w = 10 kN/m³. What is the effective vertical stress at a depth of 3 m under hydrostatic conditions?
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Explanation:
The total vertical stress at 3 m is 20 times 3 = 60 kPa. Since the water table is at ground level, the positive pore-water pressure is 10 times 3 = 30 kPa. Effective stress equals total stress minus pore-water pressure. Therefore sigma_prime = 60 - 30 = 30 kPa. The same result is obtained using the submerged unit weight of 10 kN/m³ over 3 m.
వివరణ:
Question 8
ప్రశ్న 8
A homogeneous soil extends to a depth of 5 m, with the water table 2 m below ground. The moist unit weight above the water table is 18 kN/m³, saturated unit weight below it is 20 kN/m³ and gamma_w = 10 kN/m³. Neglect capillarity and seepage. What is the effective vertical stress at 5 m depth?
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Explanation:
Total vertical stress is obtained by summing the weights of the overlying soil layers. The upper 2 m contributes 2 times 18 = 36 kPa and the lower 3 m contributes 3 times 20 = 60 kPa. Therefore total vertical stress is 96 kPa. Pore-water pressure at 5 m is 3 times 10 = 30 kPa. Effective vertical stress is 96 - 30 = 66 kPa.
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Question 9
ప్రశ్న 9
A saturated deposit has the water table at ground level. The upper 2 m has saturated unit weight 19 kN/m³ and the next 3 m has saturated unit weight 21 kN/m³. Taking gamma_w = 10 kN/m³, what is the effective vertical stress at 5 m depth?
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Explanation:
The upper layer contributes 19 times 2 = 38 kPa to total stress. The lower layer contributes 21 times 3 = 63 kPa. Thus total vertical stress at 5 m is 101 kPa. With the water table at ground level, hydrostatic pore pressure is 10 times 5 = 50 kPa. Effective stress is 101 - 50 = 51 kPa.
వివరణ:
Question 10
ప్రశ్న 10
The water table in a soil deposit lies 1 m below ground level. Above it the soil is dry with unit weight 17 kN/m³, and below it the saturated unit weight is 20 kN/m³. If gamma_w = 10 kN/m³, determine the effective vertical stress at 4 m depth, neglecting capillarity.
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Explanation:
The dry soil above the water table contributes 17 times 1 = 17 kPa. The 3 m saturated portion contributes 20 times 3 = 60 kPa. Total stress is therefore 77 kPa. Hydrostatic pore-water pressure is 10 times 3 = 30 kPa. Effective vertical stress is 77 - 30 = 47 kPa.
వివరణ:
Question 11
ప్రశ్న 11
At a point 1 m above a static water table within a fully saturated capillary fringe, the total vertical stress is 35 kPa. If gamma_w = 10 kN/m³, what is the effective vertical stress using the elementary saturated-soil relationship?
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Explanation:
The point is 1 m above the water table within a saturated capillary zone. Hydrostatic gauge pore-water pressure at the point is -10 times 1 = -10 kPa. The effective-stress equation is sigma_prime = sigma - u. Substituting gives sigma_prime = 35 - (-10) = 45 kPa. Negative pore pressure increases the computed effective stress relative to the stated total stress.
వివరణ:
Question 12
ప్రశ్న 12
A homogeneous soil is fully saturated by capillary action from ground level down to a static water table at 2 m depth. Its saturated unit weight is 20 kN/m³ and gamma_w = 10 kN/m³. What is the effective vertical stress at 0.5 m depth using the idealized saturated-fringe model?
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Explanation:
At 0.5 m depth, the total stress is 0.5 times 20 = 10 kPa. The point is 1.5 m vertically above the water table. Its gauge pore-water pressure is -1.5 times 10 = -15 kPa. Effective stress is therefore 10 - (-15) = 25 kPa. This calculation assumes the soil is fully saturated at the point and the capillary pore pressure follows the specified hydrostatic idealization.
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Question 13
ప్రశ్న 13
A soil deposit is fully saturated from ground level downward, including a capillary-saturated zone above the water table located 2 m below ground. The saturated unit weight is 20 kN/m³ and gamma_w = 10 kN/m³. What is the effective vertical stress at 3 m depth under static conditions?
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Explanation:
The full 3 m of overlying soil is specified to be saturated. Therefore total vertical stress is 3 times 20 = 60 kPa. The point is 1 m below the water table, so pore-water pressure is +10 kPa. Effective vertical stress is 60 - 10 = 50 kPa. The saturated capillary zone above the water table contributes to total overburden stress and must not be treated as dry soil.
వివరణ:
Question 14
ప్రశ్న 14
A soil profile has a uniform vertical surface surcharge of 25 kPa. The water table is at 1 m depth, the moist unit weight above it is 18 kN/m³, the saturated unit weight below it is 20 kN/m³ and gamma_w = 10 kN/m³. Assuming the full surcharge acts at the point and hydrostatic pore pressure is unaffected by loading, what is the effective vertical stress at 4 m depth?
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Explanation:
The applied surcharge contributes 25 kPa to total stress. Soil overburden contributes 1 times 18 plus 3 times 20, which equals 78 kPa. Total vertical stress is therefore 25 + 78 = 103 kPa. Pore-water pressure at 4 m is 3 times 10 = 30 kPa. Effective vertical stress is 103 - 30 = 73 kPa under the specified pore-pressure assumption.
వివరణ:
Question 15
ప్రశ్న 15
At a fixed point 5 m below ground, the total vertical stress is maintained at 110 kPa. The static water table rises from 3 m to 1 m below ground, and gamma_w = 10 kN/m³. Assuming hydrostatic conditions before and after the change, how does effective vertical stress change?
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Explanation:
Initially the point lies 2 m below the water table, giving pore-water pressure of 20 kPa. The initial effective stress is 110 - 20 = 90 kPa. After the water table rises, the point lies 4 m below it, giving pore-water pressure of 40 kPa. The new effective stress is 110 - 40 = 70 kPa. Effective stress therefore decreases by 20 kPa because total stress is expressly held constant.
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Question 16
ప్రశ్న 16
The water table in a soil deposit falls from 1 m to 3 m below ground level. The moist unit weight above the water table is 18 kN/m³, the saturated unit weight below it is 20 kN/m³ and gamma_w = 10 kN/m³. Assuming hydrostatic conditions and the stated unit weights in both states, what is the change in effective vertical stress at 4 m depth?
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Explanation:
Initially total stress at 4 m is 1 times 18 plus 3 times 20 = 78 kPa. Initial pore-water pressure is 3 times 10 = 30 kPa, giving effective stress of 48 kPa. After the fall, total stress is 3 times 18 plus 1 times 20 = 74 kPa. New pore-water pressure is 1 times 10 = 10 kPa, giving effective stress of 64 kPa. The effective stress increases by 64 - 48 = 16 kPa, accounting for both the overburden-weight change and the pore-pressure change.
వివరణ:
Question 17
ప్రశ్న 17
An ideal capillary tube produces a water rise of 0.30 m. Surface tension is 0.075 N/m, contact angle is zero and gamma_w = 10 kN/m³. What is the corresponding tube diameter?
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Explanation:
The capillary-rise equation is h = 4T cos(theta)/(gamma_w d). Rearranging gives d = 4T cos(theta)/(gamma_w h). With zero contact angle, the numerator is 4 times 0.075 = 0.30 N/m. The denominator is 10000 times 0.30 = 3000 N/m². Thus d = 0.00010 m, which equals 0.10 mm.
వివరణ:
Question 18
ప్రశ్న 18
Which statement correctly describes the limitation of applying Terzaghi's saturated-soil effective-stress equation to soil above the capillary fringe that contains both pore water and pore air?
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Explanation:
Terzaghi's classical effective-stress principle is formulated for saturated soils. Partially saturated soils contain both air and water in their pore spaces. Their response can depend on pore-air pressure, pore-water pressure, matric suction and degree of saturation. The effect of suction cannot always be represented by subtracting the full negative pore-water pressure from total stress. Therefore the elementary saturated-fringe calculation should not be extended indiscriminately to partially saturated soil.
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Question 19
ప్రశ్న 19
A horizontal ground surface is covered by 2 m of standing water. Beneath it is a homogeneous saturated soil with saturated unit weight 20 kN/m³. Taking gamma_w = 10 kN/m³ and static conditions, what is the effective vertical stress at a point 3 m below the soil surface?
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Explanation:
The standing water contributes 2 times 10 = 20 kPa to total stress. The 3 m saturated soil column contributes 3 times 20 = 60 kPa. Therefore total stress is 80 kPa. The point lies 5 m below the free water surface, so pore pressure is 5 times 10 = 50 kPa. Effective stress is 80 - 50 = 30 kPa, showing that the ponded water adds equally to total and pore pressure.
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Question 20
ప్రశ్న 20
A saturated soil element experiences an additional uniformly transmitted vertical total stress of 15 kPa due to surface loading. After complete drainage and dissipation of the excess pore pressure, assume the hydrostatic groundwater level is unchanged. What is the resulting increment in effective vertical stress?
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Explanation:
The increment form of Terzaghi's equation is delta sigma_prime = delta sigma - delta u. The additional total vertical stress is 15 kPa. Complete dissipation of the excess pore pressure means its remaining increment is zero relative to the original hydrostatic condition. Therefore the effective-stress increment is 15 - 0 = 15 kPa. This result describes the specified drained final state rather than the immediate undrained response.
వివరణ:
Question 21
ప్రశ్న 21
A homogeneous saturated soil has specific gravity of solids G_s = 2.70 and void ratio e = 0.80. The water table is at ground level, gamma_w = 10 kN/m³ and groundwater is static. What is the effective vertical stress at 4 m depth approximately?
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Explanation:
The saturated unit weight is gamma_sat = (G_s + e)gamma_w/(1 + e). Substitution gives gamma_sat = (2.70 + 0.80) times 10 divided by 1.80 = 19.444 kN/m³. The submerged unit weight is 19.444 - 10 = 9.444 kN/m³. Since the water table is at ground level, effective stress at 4 m is 4 times 9.444 = 37.778 kPa. The result is approximately 37.8 kPa.
వివరణ:
Question 22
ప్రశ్న 22
A soil deposit has a static water table at 3 m depth and a fully saturated capillary fringe extending upward from 3 m to 2 m depth. The upper 2 m has moist unit weight 18 kN/m³, while the saturated fringe has unit weight 20 kN/m³. Taking gamma_w = 10 kN/m³, what is the effective vertical stress at 2.5 m depth?
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Explanation:
The upper 2 m contributes 2 times 18 = 36 kPa to total stress. The next 0.5 m of saturated capillary soil contributes 0.5 times 20 = 10 kPa. Thus total vertical stress at 2.5 m is 46 kPa. The point is 0.5 m above the water table, giving gauge pore pressure of -5 kPa. Effective stress is therefore 46 - (-5) = 51 kPa.
వివరణ:
Question 23
ప్రశ్న 23
At a point in a partially saturated soil, gauge pore-air pressure is 0 kPa and gauge pore-water pressure is -12 kPa. What is the matric suction, defined as pore-air pressure minus pore-water pressure?
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Explanation:
Matric suction is defined as the difference between pore-air pressure and pore-water pressure. The pore-air pressure is given as zero gauge. The pore-water pressure is -12 kPa. Therefore matric suction equals 0 - (-12) = 12 kPa. This positive suction value should not automatically be interpreted as an equal increase in classical saturated-soil effective stress because the soil is partially saturated.
వివరణ:
Question 24
ప్రశ్న 24
At a point in a saturated soil experiencing groundwater seepage, total vertical stress is 120 kPa and the measured gauge pore-water pressure is +55 kPa. A hydrostatic calculation based only on water-table depth would have predicted +30 kPa. What effective vertical stress should be used based on the actual measured pressure?
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Explanation:
Effective stress must be calculated using the pore-water pressure that actually acts at the point. The measured gauge pore-water pressure is +55 kPa. Terzaghi's equation therefore gives effective stress of 120 - 55 = 65 kPa. The hydrostatic estimate of 30 kPa would produce an incorrect result for the stated seepage condition. Groundwater flow can change pore-pressure distributions from their static hydrostatic values.
వివరణ:
Question 25
ప్రశ్న 25
The water table is at ground level in a saturated soil having gamma_sat = 20 kN/m³ and gamma_w = 10 kN/m³. At 4 m depth, upward seepage produces an additional 40 kPa of pore-water pressure above the hydrostatic value, without changing the total vertical stress. What is the effective vertical stress and the associated engineering implication?
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Explanation:
Total vertical stress at 4 m is 20 times 4 = 80 kPa. Hydrostatic pore-water pressure would be 10 times 4 = 40 kPa. The additional pressure from upward seepage increases actual pore pressure to 40 + 40 = 80 kPa. Effective stress is therefore 80 - 80 = 0 kPa. A zero-effective-stress condition indicates loss of effective interparticle confinement and is associated with quick conditions or boiling risk in susceptible soils.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Surface tension and wetting interactions at air-water-solid interfaces enable water to rise through sufficiently small connected pores against gravity. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 0.144 m. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. It is twice that in Tube Q. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. -18 kPa. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Gauge pore-water pressure is zero at the water table, positive below it and negative above it within the saturated capillary fringe. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 105 kPa. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 30 kPa. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 66 kPa. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 51 kPa. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 47 kPa. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 45 kPa. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 25 kPa. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 50 kPa. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 73 kPa. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. It decreases by 20 kPa. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. An increase of 16 kPa. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 0.10 mm. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. Partially saturated soil may require separate consideration of pore-air pressure, pore-water pressure and degree of saturation; the saturated-soil expression should not be applied without qualification. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 30 kPa. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 15 kPa. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 37.8 kPa. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 51 kPa. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 12 kPa. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. 65 kPa. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 0 kPa; the soil reaches a zero-effective-stress condition associated with instability or boiling risk. సమాధానం: B.