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Question 1
ప్రశ్న 1
Which statement correctly expresses Darcy's law for steady, saturated, one-dimensional flow through a soil specimen under conditions where the law is valid?
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Explanation:
Darcy's law is expressed as Q = kiA for the stated flow conditions. Here Q is volumetric discharge, k is hydraulic conductivity or coefficient of permeability, and i is hydraulic gradient. The area A is the total cross-sectional area normal to the direction of flow. The relationship assumes conditions under which discharge is proportional to hydraulic gradient. It is commonly applied to steady laminar seepage through saturated soils.
వివరణ:
Question 2
ప్రశ్న 2
The difference in total hydraulic head between two observation points in a saturated soil specimen is 1.8 m. Their separation measured along the direction of flow is 0.9 m. What is the magnitude of the hydraulic gradient?
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Explanation:
Hydraulic gradient is the total head loss per unit length along the flow path. The supplied total head difference is 1.8 m. The flow-path length is 0.9 m. Therefore i = 1.8/0.9 = 2.0. Hydraulic gradient is dimensionless because it is the ratio of two lengths.
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Question 3
ప్రశ్న 3
A saturated soil has coefficient of permeability 2.0 × 10^-5 m/s and carries steady flow under hydraulic gradient 0.50. If the gross area normal to flow is 3.0 m², what is the volumetric discharge?
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Explanation:
Darcy's equation gives Q = kiA. The coefficient of permeability is 2.0 × 10^-5 m/s and the gradient is 0.50. The gross area is 3.0 m². Substitution gives Q = 2.0 × 10^-5 × 0.50 × 3.0 = 3.0 × 10^-5 m³/s. The resulting unit is cubic metres per second because hydraulic gradient is dimensionless.
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Question 4
ప్రశ్న 4
Water flows steadily through a saturated soil having porosity 0.40. The total discharge is 8.0 × 10^-5 m³/s through a gross area of 2.0 m². Assuming the entire pore space participates in the average flow, determine the average seepage velocity.
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Explanation:
The discharge velocity is obtained by dividing the discharge by gross area. Thus v = 8.0 × 10^-5/2.0 = 4.0 × 10^-5 m/s. Average seepage velocity is obtained by dividing discharge velocity by porosity. Therefore v_s = 4.0 × 10^-5/0.40 = 1.0 × 10^-4 m/s. This is an average pore-space velocity and does not represent the velocity along every individual tortuous flow path.
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Question 5
ప్రశ్న 5
A saturated soil has k = 2.0 × 10^-4 m/s, hydraulic gradient 0.75 and porosity 0.30. Assuming Darcy flow and the standard average-pore-velocity approximation, what are its discharge velocity and average seepage velocity, respectively?
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Explanation:
Darcy discharge velocity is v = ki. Substitution gives v = 2.0 × 10^-4 × 0.75 = 1.5 × 10^-4 m/s. Average seepage velocity is v_s = v/n, where n is porosity. Dividing by 0.30 gives v_s = 5.0 × 10^-4 m/s. The average seepage velocity exceeds the discharge velocity because water flows through the pore space rather than the entire gross section.
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Question 6
ప్రశ్న 6
Which pairing of dimensions and physical meanings is correct for Darcy's law in SI units?
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Explanation:
Coefficient of permeability, also called hydraulic conductivity, has dimensions of length per unit time. Hydraulic gradient is the ratio of hydraulic head difference to flow-path length and is therefore dimensionless. The cross-sectional area has dimensions of length squared. Their product in Darcy's law has dimensions of volume per unit time. Intrinsic permeability is a separate porous-medium property having dimensions of length squared.
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Question 7
ప్రశ్న 7
Which observation would most directly indicate that Darcy's linear law should not be applied without further verification to flow through a coarse granular soil?
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Explanation:
Darcy's linear law assumes a proportional relationship between discharge velocity and hydraulic gradient. This behaviour is generally associated with laminar flow under suitable soil conditions. At sufficiently high velocities in coarse materials, inertial effects can become important. A nonlinear discharge-gradient relationship indicates that a constant Darcy coefficient may no longer describe the flow adequately. The engineer should then investigate the flow regime and use a model appropriate to the observed behaviour.
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Question 8
ప్రశ్న 8
In a constant-head permeability test, 300 cm³ of water is collected in 60 s. The soil specimen is 15 cm long, has gross cross-sectional area 50 cm², and experiences a constant head loss of 30 cm. What is the coefficient of permeability?
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Explanation:
For a constant-head test, k = QL/(Aht). The collected volume is 300 cm³ and the specimen length is 15 cm. The denominator is 50 × 30 × 60 = 90000 in the corresponding centimetre-second units. The numerator is 300 × 15 = 4500. Therefore k = 4500/90000 = 0.050 cm/s.
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Question 9
ప్రశ్న 9
A constant-head permeameter has specimen length 20 cm and cross-sectional area 100 cm². Under a maintained head difference of 10 cm, 12 cm³ of water is collected over 240 s. What is the coefficient of permeability expressed in m/s?
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Explanation:
The constant-head formula is k = QL/(Aht). Substitution in centimetre-second units gives k = 12 × 20/(100 × 10 × 240). The resulting value is 0.001 cm/s. Since 1 cm/s equals 0.01 m/s, the value becomes 1.0 × 10^-5 m/s. The conversion from centimetres to metres is necessary because the options are expressed in SI units.
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Question 10
ప్రశ్న 10
Which selection of Indian Standard laboratory permeability methods is technically appropriate for the stated materials?
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Explanation:
IS 2720 (Part 17) addresses laboratory determination of permeability using constant-head and falling-head arrangements. IS 2720 (Part 36) specifically addresses constant-head testing of suitable granular soils. Constant-head arrangements are commonly convenient when measurable discharge can be maintained through relatively permeable specimens. Falling-head arrangements are commonly useful for less permeable soils because changes in standpipe head can be measured over time. The specimen characteristics and detailed applicability of the relevant test standard must still be checked before choosing the method.
వివరణ:
Question 11
ప్రశ్న 11
A saturated soil has coefficient of permeability 0.004 cm/s, hydraulic gradient 0.50 and porosity 0.35. Assuming the average seepage velocity equals Darcy velocity divided by porosity, what is the seepage velocity?
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Explanation:
Darcy velocity is calculated by multiplying the permeability coefficient by hydraulic gradient. Thus v = 0.004 × 0.50 = 0.002 cm/s. Average seepage velocity is v_s = v/n. Substituting porosity 0.35 gives 0.002/0.35 = 0.005714 cm/s. Rounding to three significant figures gives approximately 0.00571 cm/s.
వివరణ:
Question 12
ప్రశ్న 12
A falling-head test uses a standpipe of area 2 cm² and a soil specimen of area 100 cm² and length 20 cm. The head falls from 100 cm to 50 cm in 1000 s. Using k = (aL/At) ln(h1/h2), what is the coefficient of permeability?
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Explanation:
The falling-head relationship is k = (aL/At) ln(h1/h2). The geometric and time factor equals 2 × 20/(100 × 1000) = 0.0004 cm/s. The head ratio is 100/50 = 2. The natural logarithm of two is approximately 0.693147. Therefore k = 0.0004 × 0.693147 = approximately 2.77 × 10^-4 cm/s.
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Question 13
ప్రశ్న 13
During a falling-head permeability test, the standpipe area is 1 cm², specimen area is 100 cm² and specimen length is 20 cm. If k = 1.0 × 10^-4 cm/s, approximately how long will the head take to fall from 80 cm to 40 cm?
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Explanation:
The falling-head formula can be rearranged as t = (aL/Ak) ln(h1/h2). Substituting the given dimensions gives aL/(Ak) = 1 × 20/(100 × 0.0001) = 2000 s. The head ratio is 80/40 = 2. Multiplying 2000 by ln(2), or approximately 0.693147, gives 1386.3 s. The closest listed answer is 1386 s.
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Question 14
ప్రశ్న 14
For the same saturated soil specimen, standpipe and testing conditions, a falling-head test records time t for head to decrease from 80 cm to 40 cm. What time is theoretically required for the head to decrease from 80 cm to 20 cm if Darcy flow remains valid?
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Explanation:
Falling-head test duration is proportional to the natural logarithm of the initial-to-final head ratio. For the first measurement, the head ratio is 80/40 = 2. For the second measurement, the ratio is 80/20 = 4. Since ln(4) = 2 ln(2), the second duration is twice the first. This conclusion assumes that the specimen permeability and standpipe dimensions remain unchanged during both measurements.
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Question 15
ప్రశ్న 15
In a falling-head test, the standpipe cross-sectional area is doubled without changing the soil specimen, initial and final heads, fluid properties or permeability. How does the time required for the specified head reduction change?
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Explanation:
The falling-head duration is t = (aL/Ak) ln(h1/h2). The standpipe area a appears directly in the numerator. All the other terms are held constant by the question. Therefore doubling the standpipe area doubles the duration for the same head reduction. A larger standpipe contains more water volume for a given fall in water level.
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Question 16
ప్రశ్న 16
A saturated soil has measured hydraulic conductivity 1.50 × 10^-4 m/s at temperature T, when the dynamic viscosity of water is 0.80 mPa·s. At a reference temperature the viscosity is 1.00 mPa·s. Assuming unchanged fluid density and intrinsic permeability, what is the hydraulic conductivity at the reference temperature?
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Explanation:
For the same porous medium and approximately unchanged water density, hydraulic conductivity varies inversely with dynamic viscosity. The reference conductivity is k_ref = k_T times mu_T/mu_ref. The viscosity ratio is 0.80/1.00 = 0.80. Therefore k_ref = 1.50 × 10^-4 × 0.80 = 1.20 × 10^-4 m/s. The calculation keeps intrinsic soil permeability fixed and changes only the specified fluid viscosity.
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Question 17
ప్రశ్న 17
For a saturated soil whose pore structure and fluid density remain approximately unchanged, which statement correctly describes the usual effect of increasing water temperature on measured hydraulic conductivity?
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Explanation:
Water viscosity generally decreases as temperature increases over ordinary geotechnical testing ranges. Hydraulic conductivity depends on both porous-medium characteristics and fluid properties. Lower viscosity generally permits a larger flow rate for the same hydraulic gradient. Consequently the measured hydraulic conductivity tends to increase as water temperature rises. Intrinsic permeability can remain unchanged because it reflects the geometry of the connected pore space rather than the fluid viscosity.
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Question 18
ప్రశ్న 18
A stratified saturated soil deposit consists of two horizontal layers. Layer 1 is 2 m thick with horizontal permeability 1.0 × 10^-4 m/s, while Layer 2 is 3 m thick with horizontal permeability 4.0 × 10^-4 m/s. For flow parallel to bedding, what is the equivalent horizontal permeability?
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Explanation:
For flow parallel to horizontal layers, the equivalent permeability is the thickness-weighted arithmetic mean. The formula is k_H = (k1H1 + k2H2)/(H1 + H2). Substitution gives (1.0 × 10^-4 × 2 + 4.0 × 10^-4 × 3)/5. The numerator is 14.0 × 10^-4 in the corresponding thickness-weighted units. Dividing by five gives k_H = 2.80 × 10^-4 m/s.
వివరణ:
Question 19
ప్రశ్న 19
Two horizontal saturated soil layers are arranged in series for vertical flow. Their thicknesses are 2 m and 3 m, and their vertical permeabilities are 1.0 × 10^-4 m/s and 4.0 × 10^-4 m/s, respectively. What is the equivalent vertical permeability?
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Explanation:
For vertical flow through horizontal layers, the same discharge per unit area passes through each layer. The equivalent vertical permeability is k_V = (H1 + H2)/(H1/k1 + H2/k2). The denominator is 2/(1.0 × 10^-4) + 3/(4.0 × 10^-4) = 27500 s. The total thickness is 5 m. Therefore k_V = 5/27500 = approximately 1.82 × 10^-4 m/s.
వివరణ:
Question 20
ప్రశ్న 20
A saturated deposit has two horizontal layers of equal thickness with isotropic permeabilities k1 = 1.0 × 10^-5 m/s and k2 = 9.0 × 10^-5 m/s. What are the equivalent permeabilities parallel and perpendicular to bedding, respectively?
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Explanation:
For equal-thickness layers, horizontal equivalent permeability is their arithmetic mean. This gives k_H = (1.0 × 10^-5 + 9.0 × 10^-5)/2 = 5.0 × 10^-5 m/s. Vertical equivalent permeability is their harmonic mean. Thus k_V = 2/(1/k1 + 1/k2) = 1.8 × 10^-5 m/s. The horizontal value exceeds the vertical value because the different layers conduct water in parallel in one direction and in series in the other.
వివరణ:
Question 21
ప్రశ్న 21
For steady one-dimensional saturated flow through a horizontally stratified soil deposit, which statement correctly compares hydraulic conditions for flow parallel and perpendicular to the bedding planes?
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Explanation:
For horizontal flow through continuous horizontal layers, the hydraulic gradient along the common flow length is the same in each layer. Individual layer discharges differ because their permeabilities and thicknesses differ. For vertical flow, the layers act in series. Continuity requires the same discharge per unit cross-sectional area through each layer when there is no storage or lateral leakage. These conditions lead to arithmetic averaging for parallel flow and harmonic averaging for perpendicular flow.
వివరణ:
Question 22
ప్రశ్న 22
Water flows horizontally through two saturated soil layers of thicknesses 1 m and 2 m and common width 2 m. Their permeabilities are 1.0 × 10^-4 m/s and 3.0 × 10^-4 m/s, respectively. If the head loss is 1 m over a horizontal flow length of 10 m, what is the combined discharge?
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Explanation:
The hydraulic gradient common to both layers is 1/10 = 0.10. Layer 1 has flow area 1 × 2 = 2 m², while Layer 2 has area 2 × 2 = 4 m². Darcy's law gives Q1 = 1.0 × 10^-4 × 0.10 × 2 = 2.0 × 10^-5 m³/s. The second layer carries Q2 = 3.0 × 10^-4 × 0.10 × 4 = 1.2 × 10^-4 m³/s. The combined discharge is Q1 + Q2 = 1.40 × 10^-4 m³/s.
వివరణ:
Question 23
ప్రశ్న 23
Water flows vertically through two saturated horizontal soil layers, each 1 m thick. The upper layer has permeability 1.0 × 10^-5 m/s and the lower layer has permeability 4.0 × 10^-5 m/s. If total head loss across both layers is 0.50 m and steady flow occurs without storage, what is the head loss in the upper layer?
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Explanation:
Vertical flow through the two layers occurs in series, so the Darcy discharge velocity is equal in both. Head loss in each layer is proportional to its thickness divided by permeability. The corresponding resistances are 1/(1.0 × 10^-5) and 1/(4.0 × 10^-5), giving a ratio of 4 to 1. Therefore the upper layer receives four-fifths of the total 0.50 m head loss. Its head loss is 0.50 × 4/5 = 0.40 m.
వివరణ:
Question 24
ప్రశ్న 24
Two laboratory specimens have identical grain-size distributions and total porosities, but one contains appreciable trapped air while the other is fully saturated. Which conclusion is most defensible when assessing water permeability?
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Explanation:
The amount and connectivity of pore space available to flowing water affect hydraulic conductivity. Trapped air can occupy parts of the pore network and obstruct water movement. Thus two specimens with the same total porosity need not exhibit the same measured conductivity to water. Laboratory permeability methods require attention to specimen saturation and elimination of air-related testing errors. Soil structure, void ratio, particle arrangement and pore-fluid properties are also important factors affecting permeability.
వివరణ:
Question 25
ప్రశ్న 25
A saturated soil's pore geometry and intrinsic permeability remain unchanged when the permeating liquid is replaced by another liquid of equal density but half the original dynamic viscosity. Assuming Darcy flow and no chemical interaction with the soil, how does the hydraulic conductivity change?
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Explanation:
Hydraulic conductivity is related to intrinsic permeability by k = K_int rho g/mu. Here K_int is intrinsic permeability, rho is fluid density, g is gravitational acceleration and mu is dynamic viscosity. The question keeps intrinsic permeability, density and gravitational acceleration unchanged. Reducing viscosity to half its original value therefore doubles hydraulic conductivity. This demonstrates that hydraulic conductivity depends on both the soil pore structure and the properties of the permeating fluid.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Discharge equals the coefficient of permeability multiplied by hydraulic gradient and the gross cross-sectional area normal to flow. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 2.0. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 3.0 × 10^-5 m³/s. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 1.0 × 10^-4 m/s. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 1.5 × 10^-4 m/s and 5.0 × 10^-4 m/s. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. Coefficient of permeability has dimensions of velocity, hydraulic gradient is dimensionless, and volumetric discharge has dimensions L³/T. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. Increasing hydraulic gradient produces a distinctly nonlinear discharge-gradient relationship because inertial effects have become significant. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 0.050 cm/s. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 1.0 × 10^-5 m/s. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. IS 2720 (Part 17) covers general constant-head and falling-head laboratory methods, while IS 2720 (Part 36) addresses constant-head testing of suitable granular soils. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. 0.00571 cm/s. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: C. 2.77 × 10^-4 cm/s. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 1386 s. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. 2t. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. It becomes twice the original time. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 1.20 × 10^-4 m/s. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. Hydraulic conductivity generally increases as water viscosity decreases, even though the soil's intrinsic permeability is unchanged. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. 2.80 × 10^-4 m/s. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 1.82 × 10^-4 m/s. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. 5.0 × 10^-5 m/s and 1.8 × 10^-5 m/s. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. Parallel flow has the same hydraulic gradient in each layer, whereas perpendicular flow has the same discharge per unit area through successive layers. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 1.40 × 10^-4 m³/s. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 0.40 m. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. Trapped air can obstruct connected water-flow paths and reduce measured water conductivity, so saturation condition must be controlled during testing. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. It becomes twice the original hydraulic conductivity because conductivity is inversely proportional to dynamic viscosity. సమాధానం: A.