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Question 1
ప్రశ్న 1
Which statement correctly describes a conventional flow net for steady two-dimensional seepage through a homogeneous, isotropic, saturated soil?
Explanation:
A conventional flow net consists of flow lines and equipotential lines. Flow lines indicate the direction of groundwater movement, while equipotential lines join points having equal total hydraulic head. In a homogeneous isotropic soil, the two families of lines intersect approximately at right angles. A properly constructed graphical flow net generally forms curvilinear squares. These properties enable the flow net to be used for estimating seepage discharge, hydraulic gradients and pore-water pressures.
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Question 2
ప్రశ్న 2
For seepage beneath a hydraulic structure founded on homogeneous isotropic soil, how should a perfectly impermeable foundation boundary and a boundary maintained at constant total hydraulic head be represented in a flow net?
Explanation:
No groundwater can cross a perfectly impermeable boundary. Therefore the seepage direction must be tangent to that boundary, making it a flow-line boundary. A constant-head boundary has the same total hydraulic head at every point. It is consequently represented by an equipotential boundary. These conditions are essential when establishing the shape and boundary constraints of a flow net.
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Question 3
ప్రశ్న 3
A saturated soil has coefficient of permeability 4.0 × 10^-5 m/s and hydraulic gradient 0.25. Water flows steadily through a gross cross-sectional area of 5.0 m². Using Darcy's law, calculate the seepage discharge.
Explanation:
Darcy's law gives the volumetric discharge as Q = kiA. The permeability coefficient is 4.0 × 10^-5 m/s. The hydraulic gradient is 0.25, and the gross cross-sectional area is 5.0 m². Multiplication gives Q = 4.0 × 10^-5 × 0.25 × 5.0 = 5.0 × 10^-5 m³/s. The calculation assumes that Darcy's linear relationship applies under the stated saturated flow conditions.
వివరణ:
Question 4
ప్రశ్న 4
For an ideal flow net representing steady two-dimensional seepage through homogeneous isotropic soil, which expression gives the discharge per unit width perpendicular to the section?
Explanation:
The standard flow-net discharge relation is q = kH(Nf/Nd). Here k is hydraulic conductivity and H is the total head difference between the boundaries. Nf is the number of flow channels, while Nd is the number of potential drops. For a two-dimensional section, the result is discharge per unit width perpendicular to the section. The formula assumes a valid flow net for the stated homogeneous isotropic seepage problem.
వివరణ:
Question 5
ప్రశ్న 5
An ideal flow net beneath a weir has 4 flow channels and 12 potential drops. The total hydraulic head difference is 6 m, and the foundation soil has k = 2.0 × 10^-5 m/s. What is the seepage discharge per metre length of the weir?
Explanation:
The flow-net equation is q = kH(Nf/Nd). The given values are k = 2.0 × 10^-5 m/s, H = 6 m, Nf = 4 and Nd = 12. Their substitution gives q = 2.0 × 10^-5 × 6 × 4/12. The resulting discharge is 4.0 × 10^-5 m³/s per metre width. The flow-net counts are sufficient for this calculation, so an additional diagram is unnecessary.
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Question 6
ప్రశ్న 6
Two valid idealized flow-net cases have identical hydraulic conductivity, total head difference and number of flow channels. Case P has 10 potential drops, while Case Q has 20 potential drops. According to the standard flow-net discharge expression, how does the computed discharge in Case Q compare with Case P?
Explanation:
Flow-net discharge is directly proportional to the number of flow channels. It is inversely proportional to the number of potential drops when the other stated quantities remain fixed. Case Q has twice as many potential drops as Case P. Therefore its calculated discharge is one-half the discharge in Case P. The comparison assumes that both sets of counts represent valid flow nets for their respective stated cases.
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Question 7
ప్రశ్న 7
A flow net represents a total hydraulic head loss of 6 m distributed over 12 equal potential drops. What is the head loss between two successive equipotential lines?
Explanation:
A valid conventional flow net divides the total head loss into equal potential drops. The loss per drop is the overall head difference divided by the number of drops. Substituting the given values gives 6/12 = 0.50 m. Each successive equipotential interval therefore represents a 0.50 m decrease in total head along the direction of seepage. This result is independent of the number of flow channels.
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Question 8
ప్రశ్న 8
The upstream and downstream total hydraulic heads are 12 m and 3 m, respectively, relative to a common datum. A flow net has 9 equal potential drops. At a point reached after 4 potential drops from the upstream boundary, the elevation head is 2 m. If gamma_w = 10 kN/m³, what is the gauge pore-water pressure at that point?
Explanation:
The total head difference is 12 - 3 = 9 m. With nine potential drops, the loss per drop is 1 m. After four drops, the total hydraulic head is 12 - 4 = 8 m. Subtracting the elevation head gives pressure head 8 - 2 = 6 m. The gauge pore-water pressure is therefore 10 × 6 = 60 kPa.
వివరణ:
Question 9
ప్రశ్న 9
At a point beneath the impervious floor of a hydraulic structure, the total hydraulic head is 5 m above a specified datum and the point itself has an elevation head of 1 m. If gamma_w = 10 kN/m³, what is the local uplift pressure magnitude?
Explanation:
Total hydraulic head is the sum of elevation head and pressure head when velocity head is negligible. The pressure head is therefore 5 - 1 = 4 m. Pore-water pressure is obtained by multiplying pressure head by the unit weight of water. Thus u = 10 × 4 = 40 kPa. This positive pore-water pressure can produce an upward force on the underside of the structure.
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Question 10
ప్రశ్న 10
A flow net has an overall head difference of 8 m and 10 equal potential drops. At the downstream exit, the last potential drop occurs over an approximate flow-path distance of 0.40 m. What is the average exit hydraulic gradient over this interval?
Explanation:
The head loss per potential drop is 8/10 = 0.80 m. The final interval has a flow-path length of 0.40 m. Hydraulic gradient is the head loss divided by the corresponding distance along the flow direction. Thus the average exit gradient is 0.80/0.40 = 2.00. A high exit gradient requires careful assessment because upward seepage may reduce effective stress near the downstream surface.
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Question 11
ప్రశ్న 11
Two points on a seepage flow path have total hydraulic heads of 13.2 m and 9.6 m. Their separation along that path is 2.4 m. What is the magnitude of the average hydraulic gradient between the points?
Explanation:
Hydraulic gradient is the drop in total hydraulic head divided by flow-path length. The total head drop between the two points is 13.2 - 9.6 = 3.6 m. The corresponding length is 2.4 m. Therefore the average hydraulic gradient is 3.6/2.4 = 1.50. Hydraulic gradient is dimensionless because it is a ratio of lengths.
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Question 12
ప్రశ్న 12
A saturated soil experiences a uniform hydraulic gradient of magnitude 0.35. If the unit weight of water is 10 kN/m³, what is the seepage force per unit volume of soil?
Explanation:
Seepage force per unit volume is j = i gamma_w. The hydraulic gradient has a magnitude of 0.35. The unit weight of water is 10 kN/m³. Multiplication gives j = 0.35 × 10 = 3.50 kN/m³. The force acts in the direction of seepage and can either increase or decrease effective stress depending on that direction.
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Question 13
ప్రశ్న 13
A saturated soil element has a total volume of 4 m³ and experiences a uniform seepage gradient of magnitude 0.25. Taking gamma_w = 10 kN/m³, calculate the magnitude of the total seepage force acting on this element.
Explanation:
Seepage force on a soil volume equals seepage force per unit volume multiplied by the relevant bulk volume. The force per unit volume is i gamma_w = 0.25 × 10 = 2.5 kN/m³. The element volume is 4 m³. Therefore the total force magnitude is 2.5 × 4 = 10.0 kN. Its direction is the direction of groundwater seepage rather than necessarily vertically upward.
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Question 14
ప్రశ్న 14
For a fully saturated cohesionless soil, the specific gravity of solids is 2.65 and void ratio is 0.65. What is the theoretical critical upward hydraulic gradient?
Explanation:
The critical upward hydraulic gradient is i_c = (G_s - 1)/(1 + e). Substituting G_s = 2.65 gives the numerator 1.65. The denominator is also 1 + 0.65 = 1.65. Therefore i_c = 1.65/1.65 = 1.00. At this theoretical gradient, upward seepage force balances the submerged weight of the soil skeleton under the idealized conditions.
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Question 15
ప్రశ్న 15
At the downstream exit of a seepage path, the theoretical critical gradient for heave is 0.95 and the calculated upward exit gradient is 0.38. Using FS = i_c/i_exit, what is the factor of safety against the idealized zero-effective-stress condition?
Explanation:
The stated factor-of-safety expression is the critical hydraulic gradient divided by the actual exit gradient. The critical gradient is 0.95 and the exit gradient is 0.38. Their ratio is 0.95/0.38 = 2.50. The result indicates that the actual exit gradient is below the specified theoretical heave threshold. This calculation alone does not establish safety against every possible form of internal erosion or piping.
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Question 16
ప్రశ్న 16
Which statement correctly describes the idealized quicksand or boiling condition caused by upward seepage in saturated cohesionless soil?
Explanation:
Upward seepage exerts a drag force opposing the submerged weight of soil particles. As the upward hydraulic gradient increases, effective interparticle stress decreases. At the idealized critical gradient, the net effective stress can approach zero. The soil may then lose stability and behave as a fluidized mass. The term quicksand does not mean that the pore water has reached its boiling temperature.
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Question 17
ప్రశ్న 17
A saturated uniform sand layer has its top boundary at atmospheric pore-water pressure and experiences steady upward seepage with hydraulic gradient 0.60. The saturated unit weight is 20 kN/m³ and gamma_w = 10 kN/m³. What is the effective vertical stress at a point 3 m below the top boundary?
Explanation:
The submerged unit weight is 20 - 10 = 10 kN/m³. Upward seepage reduces the effective-stress gradient by i gamma_w. The reduction is 0.60 × 10 = 6 kN/m³. The net effective-stress gradient is therefore 10 - 6 = 4 kN/m³. Over 3 m depth, the effective vertical stress is 3 × 4 = 12 kPa.
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Question 18
ప్రశ్న 18
A homogeneous saturated soil column has its top boundary at atmospheric pore-water pressure and experiences steady downward seepage at hydraulic gradient 0.40. Its saturated unit weight is 20 kN/m³ and gamma_w = 10 kN/m³. What is the effective vertical stress at a depth of 2 m?
Explanation:
The submerged unit weight is gamma_sat - gamma_w = 20 - 10 = 10 kN/m³. Downward seepage adds to the effective-stress gradient under the stated boundary condition. The additional gradient contribution is i gamma_w = 0.40 × 10 = 4 kN/m³. Therefore the effective-stress gradient is 10 + 4 = 14 kN/m³. At 2 m depth, effective vertical stress is 2 × 14 = 28 kPa.
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Question 19
ప్రశ్న 19
A saturated cohesionless soil layer is 2.5 m thick and is subjected to steady upward seepage. Its specific gravity of solids is 2.70 and void ratio is 0.70. What total head loss across the layer would produce the theoretical critical upward hydraulic gradient?
Explanation:
Critical hydraulic gradient is calculated from i_c = (G_s - 1)/(1 + e). Here the numerator is 2.70 - 1 = 1.70. The denominator is also 1 + 0.70 = 1.70, giving i_c = 1.00. Hydraulic gradient equals head loss divided by the length of the seepage path. Therefore the critical head loss across the 2.5 m layer is 1.00 × 2.5 = 2.50 m.
వివరణ:
Question 20
ప్రశ్న 20
Which statement most accurately distinguishes backward-erosion piping from the idealized quicksand condition beneath a hydraulic structure?
Explanation:
Piping is an internal erosion mechanism involving movement and removal of soil particles by flowing water. Backward-erosion piping can progress from a vulnerable unfiltered seepage exit toward the upstream water source. The idealized quick condition occurs when upward seepage force reduces effective stress toward zero. These phenomena are related to groundwater flow but do not have identical initiation and progression requirements. Assessment of piping therefore requires consideration of soil erodibility, exit conditions, seepage paths and filter protection as well as hydraulic gradients.
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Question 21
ప్రశ్న 21
The pore-water uplift pressure beneath a horizontal impervious floor varies linearly from 70 kPa at the upstream end to 30 kPa at the downstream end. The floor is 8 m long in the seepage direction. What is the resultant upward water force per metre width of the floor?
Explanation:
A linearly varying uplift pressure produces a trapezoidal pressure diagram. The average pressure is (70 + 30)/2 = 50 kPa. The floor length is 8 m, and the calculation is made for a width of 1 m. The resultant upward force is 50 × 8 × 1 = 400 kN. Therefore the uplift resultant is 400 kN per metre width of the floor.
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Question 22
ప్రశ్న 22
For an unchanged homogeneous isotropic seepage domain with fixed hydraulic conductivity and the same valid flow-net shape, the upstream-to-downstream head difference increases from H to 2H. Assuming Darcy flow remains valid, what happens to the seepage discharge?
Explanation:
The flow-net discharge relation is q = kH(Nf/Nd). The soil hydraulic conductivity and the ratio of flow channels to potential drops are held constant. Therefore seepage discharge is directly proportional to the overall hydraulic head difference. Doubling H consequently doubles q. The relationship assumes that the same saturated seepage domain and linear flow conditions remain applicable.
వివరణ:
Question 23
ప్రశ్న 23
A soil foundation is hydraulically anisotropic, with horizontal and vertical permeabilities differing substantially. Which statement regarding conventional flow-net analysis is most appropriate?
Explanation:
Conventional orthogonal flow-net construction assumes hydraulic isotropy in the coordinates used. When horizontal and vertical permeabilities differ, the groundwater-flow equation reflects that directional dependence. A suitable coordinate transformation can convert a simple anisotropic problem into an equivalent isotropic form for graphical analysis. The flow and potential lines need not intersect orthogonally when viewed in the original physical coordinates. Ignoring significant anisotropy can lead to inaccurate seepage, pore-pressure and exit-gradient estimates.
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Question 24
ప్రశ్న 24
What is the principal engineering purpose of placing a properly designed graded filter near the downstream seepage exit of an earth structure?
Explanation:
A graded filter is designed to permit seepage water to pass while restricting movement of particles from the protected soil. Its particle-size distribution must be compatible with the adjacent base material. A suitable filter can reduce the risk of internal erosion and piping at vulnerable seepage exits. The filter must also maintain adequate drainage capacity without unacceptable clogging. Filter protection complements, rather than replaces, hydraulic-gradient, uplift and overall stability checks.
వివరణ:
Question 25
ప్రశ్న 25
A hydraulic structure founded on pervious soil shows excessive uplift beneath its floor, a high downstream exit gradient and signs of soil-particle movement at an unfiltered seepage outlet. Which engineering response is most appropriate?
Explanation:
High uplift pressures can reduce the net stabilizing force acting on a hydraulic structure. Excessive downstream exit gradients can cause heave or contribute to internal erosion in susceptible soils. Cutoffs may alter the seepage path, while drainage and pressure-relief measures can help control pore-water pressures. Properly designed graded filters help prevent soil-particle migration at seepage outlets. A suitable engineering solution requires coordinated evaluation of discharge, uplift, exit gradients, filter compatibility and overall foundation stability.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Flow lines and equipotential lines intersect orthogonally, and the fields are drawn approximately as curvilinear squares. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. The impermeable boundary is a flow line, while the constant-head boundary is an equipotential line. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 5.0 × 10^-5 m³/s. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. q = kH(Nf/Nd), where H is the overall head difference. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 4.0 × 10^-5 m³/s per metre. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. It is half that of Case P. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 0.50 m. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 60 kPa. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 40 kPa. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 2.00. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 1.50. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 3.50 kN/m³. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 10.0 kN. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 1.00. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 2.50. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. It occurs when upward seepage reduces the effective interparticle stress to approximately zero at the critical hydraulic gradient. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 12 kPa. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 28 kPa. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 2.50 m. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Piping involves progressive transport and erosion of soil particles along a seepage path, whereas the idealized quick condition concerns loss of effective stress under upward seepage. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 400 kN per metre width. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. It becomes twice the original discharge. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. A suitable transformed-section method may be used; flow lines and equipotentials need not be orthogonal in the original physical coordinates of an anisotropic soil. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. To allow controlled drainage while retaining vulnerable soil particles and limiting their migration. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. Reassess the seepage field and adopt appropriately designed measures such as cutoffs, pressure relief or drainage, and graded filter protection, followed by verification of uplift and erosion stability. సమాధానం: A.
