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Question 1
ప్రశ్న 1
Which feature most clearly distinguishes steady radial flow to a well in an unconfined aquifer from the corresponding ideal confined-aquifer case?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An unconfined aquifer has the water table as its upper saturated boundary. Pumping lowers this water table and therefore reduces the saturated thickness toward the well. This variation in flow thickness is the major reason the unconfined steady-flow equation contains squared-head terms. In a confined aquifer, the saturated thickness is ordinarily treated as fixed. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
In the Dupuit analysis of an unconfined aquifer over a horizontal impermeable base, what does h at radial distance r represent?
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Explanation:
For the standard Dupuit formulation, the impermeable base is commonly chosen as the elevation datum. The variable h then represents the height of the water table above that base. It is therefore also the local saturated thickness of the unconfined aquifer. Pumping reduces h near the well by lowering the water table. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which statement best represents the Dupuit assumptions used in steady unconfined-aquifer flow analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Dupuit simplified unconfined groundwater flow by treating the hydraulic gradient as the slope of the water table. For relatively small slopes, groundwater flow lines are approximated as horizontal and equipotential lines as vertical. This reduces a complicated free-surface flow problem to a tractable analytical form. The approximation becomes less accurate where vertical components become important, especially very near a pumping well. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
Under the Dupuit approximation, what is the groundwater-flow area at radial distance r from a fully penetrating well in an unconfined aquifer when the saturated thickness there is h?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Radial groundwater flow crosses an imaginary cylindrical surface surrounding the well. The circumference of that cylinder is 2πr. In an unconfined aquifer, its effective saturated height is the local saturated thickness h. Therefore the flow area is 2πrh. Unlike the confined case, this area changes with both radius and water-table elevation. Hence Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
Two observation wells in a steady unconfined aquifer are at radii r1 and r2, where r2 > r1, and the corresponding saturated thicknesses above an impermeable base are h1 and h2. Which natural-logarithm form of the Dupuit–Thiem equation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For steady unconfined radial flow, the saturated thickness varies with radial position. Applying Darcy's law under the Dupuit approximation and integrating introduces an h dh term. Integration therefore produces the difference h2² - h1². The resulting equation is Q = πK(h2² - h1²)/ln(r2/r1). Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
Which base-10 logarithm form is equivalent to Q = πK(h2² - h1²)/ln(r2/r1)?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The conversion relation is ln(x) = 2.3026 log10(x). Thus π/2.3026 is approximately 1.364. Substituting this into the natural-log form gives Q approximately equal to 1.36K(h2² - h1²)/log10(r2/r1). The coefficient 2.73 belongs to the usual confined-aquifer discharge form, not this unconfined squared-head form. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
An unconfined aquifer has K = 10 m/day. At steady state, saturated thicknesses at observation wells 10 m and 100 m from the pumping well are 20 m and 25 m respectively. Using Q = 1.36K(h2² - h1²)/log10(r2/r1), what is the pumping rate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The radius ratio is 100/10 = 10, so its common logarithm is 1. The squared-head difference is 25² - 20² = 625 - 400 = 225 m². Therefore Q = 1.36 × 10 × 225. This gives 3060 m³/day. Hence Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
During a steady pumping test in an unconfined aquifer, h1 = 15 m at r1 = 20 m and h2 = 20 m at r2 = 200 m. If Q = 1904 m³/day, what is the hydraulic conductivity using the coefficient 1.36?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The radius ratio is 200/20 = 10 and log10(r2/r1) = 1. The squared-head difference is 20² - 15² = 400 - 225 = 175 m². Rearranging gives K = 1904/(1.36 × 175). The denominator equals 238, so K = 8 m/day. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
An unconfined aquifer has K = 10 m/day and is pumped steadily at 5440 m³/day. At r1 = 10 m, h1 = 15 m, while r2 = 100 m. What is h2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because r2/r1 = 10, the common logarithm is 1. The Dupuit–Thiem equation gives h2² - h1² = Q/(1.36K). Thus the squared-head difference is 5440/(13.6) = 400 m². Therefore h2² = 225 + 400 = 625, giving h2 = 25 m. Hence Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
If H is the undisturbed saturated thickness at an ideal radius of influence R and hw is the saturated thickness at a pumping well of radius rw, which steady unconfined-aquifer relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At the radius of influence, the water table is idealized as essentially unaffected by pumping and has saturated thickness H. At the well, the saturated thickness is hw. Integration of the Dupuit radial-flow equation between rw and R gives Q = πK(H² - hw²)/ln(R/rw). The squared terms are essential because saturated thickness varies toward the well. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
An unconfined aquifer has K = 10 m/day, H = 30 m, hw = 20 m, R = 200 m and rw = 0.20 m. Using the base-10 form, what steady discharge is predicted?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The radius ratio is 200/0.20 = 1000, whose common logarithm is 3. The squared-head difference is 30² - 20² = 900 - 400 = 500 m². Therefore Q = 1.36 × 10 × 500/3. This gives approximately 2266.7 m³/day. Hence Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
If the undisturbed saturated thickness is H and the water-table drawdown at a point is s, which identity correctly expresses the squared-head deficit used in the unconfined Dupuit equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Drawdown is defined as s = H - h, so h = H - s. Therefore H² - h² = H² - (H - s)². Expanding gives H² - [H² - 2Hs + s²] = 2Hs - s². This relation is useful when the unconfined equation is expressed directly in terms of drawdown. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
An unconfined aquifer has H = 30 m and drawdown s = 6 m at a well. If K = 5 m/day and R/rw = 100, what discharge is obtained from Q = 1.36K(2Hs - s²)/log10(R/rw)?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The squared-head deficit is 2Hs - s² = 2 × 30 × 6 - 6². This equals 360 - 36 = 324 m². Also, log10(100) = 2. Thus Q = 1.36 × 5 × 324/2 = 1101.6 m³/day. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For an unconfined aquifer, K = 10 m/day, H = 30 m, hw = 20 m and Q = approximately 2266.7 m³/day. If rw = 0.20 m, what radius of influence is implied by the ideal steady Dupuit equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Rearranging the common-log form gives log10(R/rw) = 1.36K(H² - hw²)/Q. The numerator is 1.36 × 10 × 500 = 6800. Dividing by approximately 2266.7 gives 3. Therefore R/rw = 10³ = 1000, and R = 1000 × 0.20 = 200 m. Hence Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
Why does the steady unconfined Dupuit–Thiem equation contain squared heads whereas the confined Thiem equation contains a linear head difference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In a confined aquifer, the flow area contains an essentially constant saturated thickness b. The radial Darcy equation therefore integrates to a relation involving a linear difference in hydraulic head. In an unconfined aquifer, the local saturated thickness itself is h and changes with position. Integration therefore contains h dh, which produces squared-head terms. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
When drawdown s is small compared with initial saturated thickness H, which approximation follows from H² - (H - s)² = 2Hs - s²?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The exact expression is 2Hs - s². If s is small compared with H, the term s² is small relative to 2Hs. The squared-head deficit can then be approximated as 2Hs. This helps explain why small-drawdown unconfined behavior can resemble a confined-aquifer equation with an approximately constant saturated thickness H. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
For fixed observation-well heads and radii in the ideal steady Dupuit equation, what happens to pumping discharge if hydraulic conductivity is doubled?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Dupuit–Thiem discharge equation is directly proportional to hydraulic conductivity K. If the squared-head difference and radial geometry remain unchanged, doubling K doubles Q. A more permeable aquifer can therefore transmit a greater discharge under the same hydraulic-head configuration. This is consistent with Darcy's law. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
For fixed K, H, R and rw in an unconfined aquifer, which statement is correct if the pumping rate is doubled?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The exact steady Dupuit relation makes Q directly proportional to H² - hw² for fixed geometry and K. Therefore doubling Q doubles this squared-head deficit. Drawdown, however, is H - hw and has a nonlinear relation to the squared-head deficit. Only for relatively small drawdown may an approximately linear drawdown relationship be obtained. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
Why does the water-table gradient generally become steeper near a pumping well in an unconfined aquifer?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under steady radial flow, approximately the same pumping discharge crosses successive cylindrical surfaces. Their areas are 2πrh in the Dupuit representation. Both r and generally h decrease toward the well, reducing the available flow area. A greater Darcy flux and hydraulic gradient are therefore required near the well for the same discharge. Hence Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which statement best describes an important limitation of the Dupuit approximation very near a pumping well in an unconfined aquifer?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Dupuit approximation suppresses vertical hydraulic gradients and assumes essentially horizontal groundwater flow. Near a pumping well, streamlines bend strongly toward the well and vertical components can become important. The simplified predicted free surface can therefore deviate from the actual water table close to the well. The approximation is generally more satisfactory away from this strongly curved near-well region. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
What is an important advantage of using two observation wells in a steady unconfined-aquifer pumping test?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The two-observation-well Dupuit–Thiem equation uses h1, h2, r1, r2 and the pumping rate. An explicit outer radius of influence is therefore unnecessary for calculating K from those observations. The method also avoids relying on the pumped-well water level, which may include additional local well effects. Steady observations alone do not supply the time-dependent information needed to determine specific yield. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Why cannot specific yield normally be determined from the classical steady Dupuit–Thiem distance-drawdown equation alone?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Specific yield governs the quantity of water released as an unconfined aquifer drains during water-table decline. At true steady state, water levels at fixed locations no longer change with time. The steady Dupuit–Thiem equation therefore relates discharge to hydraulic conductivity and spatial head distribution without a storage term. Time-dependent pumping-test analysis is required when storage properties such as specific yield are sought. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
An unconfined aquifer has saturated thickness h = 15 m at radius r = 20 m. If the steady pumping discharge is 942.5 m³/day, what is the magnitude of Darcy flux across the cylindrical surface at that radius under the Dupuit approximation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The radial flow area is A = 2πrh. Substitution gives A = 2π × 20 × 15, which is approximately 1885 m². Darcy flux magnitude is Q/A = 942.5/1885. The resulting value is approximately 0.50 m/day. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which dimensional statement is correct for Q = πK(h2² - h1²)/ln(r2/r1)?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydraulic conductivity has dimensions L/T. The difference h2² - h1² has dimensions L². The ratio r2/r1 is dimensionless because both radii use the same type of length unit, so its logarithm is dimensionless. Multiplying K by the squared-head difference therefore gives L³/T, the correct dimensions for discharge. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about steady radial flow in an unconfined aquifer: I. Under the Dupuit approximation, hydraulic gradient is represented by the water-table slope. II. The local saturated thickness decreases toward a pumping well. III. The steady discharge equation contains squared saturated-thickness terms. IV. For small drawdown relative to saturated thickness, the unconfined relation can be approximated by a nearly linear drawdown expression. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I correctly gives the principal hydraulic-gradient approximation made by Dupuit. Statement II is correct because pumping lowers the free water table and therefore reduces saturated thickness near the well. Statement III follows from integration of radial Darcy flow when the flow thickness itself equals the variable head h. Statement IV follows because the s² term becomes relatively small when drawdown is small compared with initial saturated thickness. Therefore all four statements are correct and Option D is the valid answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. The saturated thickness decreases toward the pumping well as the water table is lowered సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. The saturated thickness or water-table elevation above the impermeable base సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. The hydraulic gradient is approximated by the water-table slope, and flow is taken as essentially horizontal for small water-table gradients సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. 2πrh సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. Q = πK(h2² - h1²) / ln(r2/r1) సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. Q ≈ 1.36K(h2² - h1²) / log10(r2/r1) సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 3060 m³/day సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 8 m/day సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: C. 25 m సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. Q = πK(H² - hw²) / ln(R/rw) సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 2267 m³/day సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. H² - h² = 2Hs - s² సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: C. 1101.6 m³/day సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 200 m సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. Because the saturated flow thickness in an unconfined aquifer varies with water-table elevation, whereas confined-aquifer thickness is treated as essentially constant సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. H² - h² ≈ 2Hs సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. It doubles సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. The quantity H² - hw² doubles, but the drawdown H - hw does not in general double exactly సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: A. The same radial discharge must pass through progressively smaller cylindrical flow areas as both radius and saturated thickness decrease toward the well సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: A. Vertical flow components and curvature of streamlines can become important, so the assumption of essentially horizontal flow becomes less accurate సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. Hydraulic conductivity can be estimated from the two saturated thicknesses and radial distances without requiring an assumed radius of influence సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. The equilibrium equation contains no time-dependent groundwater-storage term సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: B. 0.50 m/day సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. K has dimensions L/T, the squared-head difference has dimensions L², the logarithm is dimensionless, and Q has dimensions L³/T సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.