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Question 1
ప్రశ్న 1
For n successive earthwork cross sections having areas A1, A2, ..., An at equal longitudinal spacing d, which expression correctly gives the total volume by the composite trapezoidal method?
Explanation:
The composite trapezoidal method estimates volume using the areas of successive cross sections. For every interval, the average of its two end-sectional areas is multiplied by the longitudinal spacing. When the spacing is uniform, the individual interval expressions can be combined into one formula. The first and last sectional areas appear once in the combined bracket, while each intermediate area appears twice. Therefore, the required expression is V = (d/2)[A1 + An + 2(A2 + A3 + ... + A(n-1))]. This formula is equivalent to applying the average-end-area method separately over all consecutive intervals.
వివరణ:
Question 2
ప్రశ్న 2
Two successive road-cutting cross sections have areas of 14 m2 and 26 m2. The longitudinal distance between the sections is 30 m. What is the estimated cutting volume using the trapezoidal method?
Explanation:
The trapezoidal method uses the arithmetic mean of the areas at the two ends of an interval. The first sectional area is 14 m2 and the second sectional area is 26 m2. Their average is (14 + 26)/2 = 20 m2. The distance between the two sections is 30 m. Therefore, the estimated cutting volume is 20 x 30 = 600 m3. The same result is obtained directly from V = L(A1 + A2)/2.
వివరణ:
Question 3
ప్రశ్న 3
Four consecutive earthwork cross sections have areas of 8 m2, 14 m2, 20 m2 and 12 m2. The longitudinal spacing between each pair of successive sections is 25 m. What is the total volume calculated using the trapezoidal rule?
Explanation:
Four cross sections form three longitudinal intervals. For equal spacing, the composite trapezoidal expression is V = d[A1/2 + A2 + A3 + A4/2]. Substituting the given areas gives V = 25[8/2 + 14 + 20 + 12/2]. The weighted sectional-area sum is 4 + 14 + 20 + 6 = 44 m2. Therefore, the total volume is 25 x 44 = 1,100 m3. The intermediate sections receive full weighting because each belongs to two adjoining intervals.
వివరణ:
Question 4
ప్రశ్న 4
Five consecutive earthwork cross sections are spaced at uniform 20 m intervals. Their areas are 10 m2, 16 m2, 12 m2, 18 m2 and 24 m2 respectively. What is the estimated earthwork volume by the composite trapezoidal formula?
Explanation:
Five consecutive sections define four equal longitudinal intervals. The composite trapezoidal formula gives V = d[A1/2 + A2 + A3 + A4 + A5/2]. The weighted area sum is 10/2 + 16 + 12 + 18 + 24/2. This is equal to 5 + 16 + 12 + 18 + 12 = 63 m2. Multiplying by the common spacing of 20 m gives V = 63 x 20 = 1,260 m3. The first and last sections have half-weighting in the combined expression, while all three internal sections have full weighting.
వివరణ:
Question 5
ప్రశ్న 5
Earthwork cross sections at chainages 0+000, 0+025, 0+065 and 0+090 have areas of 12 m2, 18 m2, 10 m2 and 20 m2 respectively. Using the trapezoidal method with the ACTUAL interval lengths, what is the total volume?
Explanation:
The longitudinal intervals are 25 m, 40 m and 25 m. The first interval volume is 25(12 + 18)/2 = 375 m3. The second interval volume is 40(18 + 10)/2 = 560 m3. The third interval volume is 25(10 + 20)/2 = 375 m3. Adding the three quantities gives 375 + 560 + 375 = 1,310 m3. The simplified equal-interval trapezoidal formula cannot be applied using a single spacing because the station intervals are unequal.
వివరణ:
Question 6
ప్రశ్న 6
Six consecutive cross sections of a canal excavation have areas of 6 m2, 10 m2, 14 m2, 18 m2, 22 m2 and 26 m2. Their longitudinal spacing is uniformly 15 m. What is the total excavation volume by the trapezoidal rule?
Explanation:
Six cross sections form five intervals of 15 m each. The composite trapezoidal formula gives half-weighting to the first and last sectional areas. The weighted area sum is 6/2 + 10 + 14 + 18 + 22 + 26/2. Simplifying gives 3 + 10 + 14 + 18 + 22 + 13 = 80 m2. Multiplication by the common interval length gives V = 80 x 15 = 1,200 m3. The computation uses all six cross sections rather than only the two extreme sections.
వివరణ:
Question 7
ప్రశ్న 7
Three equally spaced earthwork cross sections have areas of 8 m2, A m2 and 20 m2. The spacing is 25 m, and the total volume calculated by the trapezoidal method is 650 m3. What is the unknown intermediate sectional area A?
Explanation:
The composite trapezoidal formula for three sections is V = d(A1/2 + A2 + A3/2). Substituting the known information gives 650 = 25(8/2 + A + 20/2). Simplification produces 650 = 25(14 + A). Dividing by 25 gives 26 = 14 + A. Therefore, the unknown intermediate area is A = 12 m2. Substitution verifies the result because 25(4 + 12 + 10) = 650 m3.
వివరణ:
Question 8
ప్రశ్న 8
Four consecutive earthwork sections have areas of 10 m2, 14 m2, 20 m2 and 16 m2. The sections are equally spaced, and their total trapezoidal volume is 1,175 m3. What is the common longitudinal interval?
Explanation:
The common interval can be determined by rearranging the composite trapezoidal formula. The weighted sectional-area sum is 10/2 + 14 + 20 + 16/2. This gives 5 + 14 + 20 + 8 = 47 m2. Since V = d x 47, the spacing is d = 1,175/47. Therefore, the common interval is 25 m. The calculation can be verified by multiplying 47 x 25 to obtain the stated volume of 1,175 m3.
వివరణ:
Question 9
ప్రశ్న 9
Two survey stations on a pipeline excavation alignment are situated at chainages 0+080 and 0+135. Their respective excavation cross-sectional areas are 9 m2 and 15 m2. What is the trapezoidal earthwork volume between these stations?
Explanation:
The first chainage represents 80 m and the second represents 135 m along the alignment. Their separation is 135 - 80 = 55 m. The average of the two sectional areas is (9 + 15)/2 = 12 m2. Multiplying this mean area by the actual interval gives 12 x 55 = 660 m3. Therefore, the trapezoidal earthwork volume is 660 m3. Using the absolute chainage value instead of the difference between chainages would give an incorrect estimate.
వివరణ:
Question 10
ప్రశ్న 10
A road embankment has a formation width of 8 m and symmetrical side slopes of 1.5 horizontal to 1 vertical over level transverse ground. The filling heights at two consecutive sections are 1 m and 2 m, with a longitudinal spacing of 30 m. What volume is obtained by the trapezoidal method?
Explanation:
For a symmetrical embankment on level transverse ground, the cross-sectional area is A = Bd + md^2. At the first section, A1 = 8 x 1 + 1.5 x 1^2 = 9.5 m2. At the second section, A2 = 8 x 2 + 1.5 x 2^2 = 22 m2. The mean of these two complete sectional areas is (9.5 + 22)/2 = 15.75 m2. The trapezoidal volume is therefore 15.75 x 30 = 472.5 m3. The method requires averaging the end-sectional areas rather than first averaging the filling heights.
వివరణ:
Question 11
ప్రశ్న 11
A cutting has a formation width of 6 m, a left side slope of 1 horizontal to 1 vertical and a right side slope of 2 horizontal to 1 vertical. The cutting depths at two consecutive sections are 1.5 m and 2.5 m over level transverse ground. If the interval is 20 m, what is the earthwork volume by the trapezoidal method?
Explanation:
For unequal side slopes on level transverse ground, the sectional area is A = Bd + [(m1 + m2)/2]d^2. The combined side-slope coefficient is (1 + 2)/2 = 1.5. At the first section, A1 = 6 x 1.5 + 1.5 x 1.5^2 = 12.375 m2. At the second section, A2 = 6 x 2.5 + 1.5 x 2.5^2 = 24.375 m2. Their arithmetic mean is (12.375 + 24.375)/2 = 18.375 m2. Multiplying by the 20 m spacing gives a trapezoidal volume of 367.5 m3.
వివరణ:
Question 12
ప్రశ్న 12
A straight excavation has vertical lateral boundaries. Its width is 5.0 m and depth is 1.2 m at the first section, while its width is 7.0 m and depth is 1.8 m at the second section. The longitudinal spacing is 40 m. What volume is estimated by the trapezoidal method?
Explanation:
The stated cross sections are rectangular because their lateral boundaries are vertical. The first sectional area is 5.0 x 1.2 = 6.0 m2. The second sectional area is 7.0 x 1.8 = 12.6 m2. Their arithmetic mean is (6.0 + 12.6)/2 = 9.3 m2. Multiplying by the interval length gives 9.3 x 40 = 372 m3. This is the required trapezoidal estimate, irrespective of whether the actual intervening geometry produces exactly linear variation in sectional area.
వివరణ:
Question 13
ప్రశ్న 13
Two consecutive excavation sections have trapezoidal geometry. The first section has a bottom width of 6 m, a top width of 10 m and a depth of 2 m. The second section has a bottom width of 6 m, a top width of 14 m and a depth of 3 m. What is the trapezoidal-method volume if the sections are 30 m apart?
Explanation:
Each excavation cross section is a trapezoid, so its area is the average of its parallel widths multiplied by its depth. The first sectional area is (6 + 10) x 2/2 = 16 m2. The second sectional area is (6 + 14) x 3/2 = 30 m2. The arithmetic mean of the two sectional areas is (16 + 30)/2 = 23 m2. Applying the longitudinal spacing gives V = 23 x 30 = 690 m3. The calculation distinguishes the trapezoidal geometry of an individual cross section from the trapezoidal method used to combine successive sectional areas.
వివరణ:
Question 14
ప్రశ్న 14
Over a 50 m earthwork interval, the SIGNED cross-sectional area varies linearly from +15 m2 of cutting at the first section to -10 m2 of filling at the last section. Using the trapezoidal method separately before and after the zero-area transition, what are the cutting and filling volumes?
Explanation:
The signed sectional area changes from +15 m2 to -10 m2 over a distance of 50 m. Under the stated linear-area assumption, the zero-area point lies 50 x 15/(15 + 10) = 30 m from the first station. The cutting reach is therefore 30 m long, while the filling reach is 20 m long. Applying the trapezoidal method to the cutting portion gives (15 + 0) x 30/2 = 225 m3. Applying it separately to the filling portion gives (0 + 10) x 20/2 = 100 m3. These quantities must remain separate because cutting and filling are different earthwork operations.
వివరణ:
Question 15
ప్రశ్న 15
Two consecutive 40 m-spaced cross sections each contain both cutting and filling. The first section has 6 m2 of cutting and 2 m2 of filling, while the second has 14 m2 of cutting and 8 m2 of filling. Assuming linear variation of the respective sectional areas, what are the trapezoidal cutting and filling volumes?
Explanation:
The cutting and filling portions must be computed independently. The average cutting area is (6 + 14)/2 = 10 m2. The cutting volume is therefore 10 x 40 = 400 m3. The average filling area is (2 + 8)/2 = 5 m2. The filling volume is consequently 5 x 40 = 200 m3. The two results should not be algebraically combined into one net volume because they represent distinct measurable operations.
వివరణ:
Question 16
ప్రశ్న 16
At two consecutive excavation cross sections 50 m apart, the respective soil-cutting areas are 12 m2 and 18 m2, while the rock-cutting areas are 3 m2 and 5 m2. What are the SOIL and ROCK excavation quantities using the trapezoidal method separately for each material class?
Explanation:
Different excavation material classes should be measured separately when the estimate specifies different classifications. The mean soil-cutting area is (12 + 18)/2 = 15 m2. The soil excavation volume is therefore 15 x 50 = 750 m3. The mean rock-cutting area is (3 + 5)/2 = 4 m2. The rock excavation volume is 4 x 50 = 200 m3. Although their combined physical volume is 950 m3, the quantities should remain classified separately for estimating and applicable rate selection.
వివరణ:
Question 17
ప్రశ్న 17
A cutting cross section has zero area at the beginning of a 30 m reach and an area of 24 m2 at the end. If the trapezoidal method is specifically prescribed, what is its ESTIMATED volume?
Explanation:
The trapezoidal method uses the arithmetic mean of two consecutive end-sectional areas. The first area is zero and the second area is 24 m2. Their average is (0 + 24)/2 = 12 m2. Multiplying by the specified interval length gives 12 x 30 = 360 m3. This is the estimated volume under the prescribed trapezoidal formula. It should not automatically be treated as the exact geometric volume of every possible shape that tapers to a zero-area section.
వివరణ:
Question 18
ప్రశ్న 18
For a 30 m earthwork interval, the end-sectional areas are 10 m2 and 30 m2, and the actual midpoint sectional area is 18 m2. By how much does the TRAPEZOIDAL volume exceed the PRISMOIDAL volume?
Explanation:
The trapezoidal volume is 30(10 + 30)/2 = 600 m3. The prismoidal formula is V = L(A1 + 4Am + A2)/6. Substituting the supplied areas gives V = 30(10 + 4 x 18 + 30)/6. The expression inside the brackets is 112, giving a prismoidal volume of 560 m3. Therefore, the trapezoidal result exceeds the prismoidal result by 600 - 560 = 40 m3. This comparison illustrates how the actual midpoint area can affect the estimated volume.
వివరణ:
Question 19
ప్రశ్న 19
Consider the following statements about the trapezoidal method of earthwork computation. Statement I: The method gives the exact geometric volume when cross-sectional area varies linearly with longitudinal distance. Statement II: If the actual midpoint area equals the arithmetic mean of the two end areas, the trapezoidal and prismoidal formulas give identical volumes. Which option is correct?
Explanation:
The trapezoidal method is exact when cross-sectional area varies linearly with longitudinal distance. Under that condition, the integral of sectional area equals the interval length multiplied by the mean of the end areas. Statement I is therefore correct. When the actual midpoint sectional area equals the arithmetic mean of the end areas, substitution into the prismoidal formula produces the same result as the trapezoidal formula. Statement II is consequently also correct. The agreement of the methods depends on the stated sectional-area relationship and is not guaranteed for every irregular earthwork profile.
వివరణ:
Question 20
ప్రశ్న 20
A symmetrical embankment has a formation width of 10 m and side slopes of 2 horizontal to 1 vertical on level transverse ground. The end filling depths are 0.5 m and 1.5 m over a 20 m interval. By how much does the TRAPEZOIDAL volume exceed the MID-SECTIONAL volume when the midpoint depth is taken as the mean of the end depths?
Explanation:
The end-sectional area formula is A = Bd + md^2, where B = 10 m and m = 2. At a depth of 0.5 m, the first area is 10 x 0.5 + 2 x 0.5^2 = 5.5 m2. At a depth of 1.5 m, the second area is 10 x 1.5 + 2 x 1.5^2 = 19.5 m2. The trapezoidal volume is therefore (5.5 + 19.5) x 20/2 = 250 m3. The midpoint depth is 1.0 m, giving a midpoint area of 10 x 1 + 2 x 1^2 = 12 m2 and a midpoint volume of 240 m3. The trapezoidal estimate consequently exceeds the midpoint estimate by 10 m3.
వివరణ:
Question 21
ప్రశ్న 21
Five consecutive earthwork cross sections have areas of 6 m2, 10 m2, 14 m2, 10 m2 and 6 m2. All four longitudinal intervals are 20 m. Which total volume is obtained using the composite trapezoidal rule?
Explanation:
The five sections form four equal longitudinal intervals. The first and last areas receive half-weighting in the combined trapezoidal formula. The weighted sectional-area sum is 6/2 + 10 + 14 + 10 + 6/2. This is equal to 3 + 10 + 14 + 10 + 3 = 40 m2. Multiplying by the common spacing of 20 m gives a total volume of 800 m3. The intermediate cross-sectional areas should not be omitted because they describe changes along the alignment.
వివరణ:
Question 22
ప్రశ్న 22
Four consecutive earthwork sections have areas of 8 m2, 12 m2, 20 m2 and 28 m2 at equal 10 m intervals. An estimator incorrectly takes the simple mean of ALL FOUR areas and multiplies it by the total 30 m length. How does this result compare with the correct composite trapezoidal volume?
Explanation:
The correct composite trapezoidal volume is 10[8/2 + 12 + 20 + 28/2]. The weighted area sum is 4 + 12 + 20 + 14 = 50 m2, giving 500 m3. The incorrect simple mean of all four areas is (8 + 12 + 20 + 28)/4 = 17 m2. Multiplying that mean by the full 30 m length produces 510 m3. The incorrect procedure therefore overestimates the volume by 510 - 500 = 10 m3. This demonstrates why the composite trapezoidal method requires appropriate endpoint and intermediate-section weighting.
వివరణ:
Question 23
ప్రశ్న 23
Four successive earthwork cross sections have areas of 5 m2, 9 m2, 15 m2 and 21 m2. Their respective longitudinal intervals are 10 m, 20 m and 30 m. What is the total volume by the trapezoidal method?
Explanation:
The longitudinal intervals are unequal, so each pair of adjoining sections must be evaluated separately. The first interval volume is (5 + 9) x 10/2 = 70 m3. The second interval volume is (9 + 15) x 20/2 = 240 m3. The third interval volume is (15 + 21) x 30/2 = 540 m3. The total is therefore 70 + 240 + 540 = 850 m3. Applying one common spacing to all four sectional areas would not represent the given survey data correctly.
వివరణ:
Question 24
ప్రశ్న 24
A vertical-sided excavation strip has a constant width of 6 m and a horizontal formation level of RL 100.00 m. At five consecutive stations spaced 20 m apart, the existing ground levels are RL 101.00 m, RL 101.50 m, RL 102.00 m, RL 101.00 m and RL 100.50 m. What is the total cutting volume by the composite trapezoidal method?
Explanation:
All ground levels are above the specified formation level, so the earthwork is cutting throughout the reach. The cutting depths at the five stations are 1.00 m, 1.50 m, 2.00 m, 1.00 m and 0.50 m. With a constant excavation width of 6 m and vertical sides, the sectional areas are 6 m2, 9 m2, 12 m2, 6 m2 and 3 m2. The composite trapezoidal formula gives V = 20[6/2 + 9 + 12 + 6 + 3/2]. The weighted sum is 3 + 9 + 12 + 6 + 1.5 = 31.5 m2. The total cutting volume is therefore 31.5 x 20 = 630 m3.
వివరణ:
Question 25
ప్రశ్న 25
A road earthwork alignment has six successive cross sections. Their data, in order, are: cutting area 12 m2 at chainage 0+000; cutting area 18 m2 at 0+020; cutting area 8 m2 at 0+050; zero earthwork area at 0+070; filling area 10 m2 at 0+100; and filling area 20 m2 at 0+130. Assuming linear variation of sectional area within each interval and using the trapezoidal method separately for cutting and filling, what are the TOTAL cutting and filling volumes?
Explanation:
The cutting portion extends from chainage 0+000 to the zero-area station at 0+070. Its three interval volumes are (12 + 18) x 20/2 = 300 m3, (18 + 8) x 30/2 = 390 m3 and (8 + 0) x 20/2 = 80 m3. Therefore, the total cutting volume is 300 + 390 + 80 = 770 m3. The filling portion extends from chainage 0+070 to 0+130. Its two interval volumes are (0 + 10) x 30/2 = 150 m3 and (10 + 20) x 30/2 = 450 m3. The total filling volume is consequently 150 + 450 = 600 m3. Cutting and filling must be recorded separately because their construction operations and applicable rates may differ.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. V = (d/2)[A1 + An + 2(A2 + A3 + ... + A(n-1))]. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. 600 m3. సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: D. 1,100 m3. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: A. 1,260 m3. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 1,310 m3. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 1,200 m3. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: D. 12 m2. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: A. 25 m. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 660 m3. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 472.5 m3. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: D. 367.5 m3. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: A. 372 m3. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. 690 m3. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 225 m3 cutting and 100 m3 filling. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: D. 400 m3 cutting and 200 m3 filling. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: A. 750 m3 soil and 200 m3 rock. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. 360 m3. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 40 m3. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: D. Both Statement I and Statement II. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: A. 10 m3. సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: B. 800 m3. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. The incorrect method overestimates the volume by 10 m3. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: D. 850 m3. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: A. 630 m3. సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: B. 770 m3 cutting and 600 m3 filling. సమాధానం: B.
