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Question 1
ప్రశ్న 1
Which expression correctly represents the mid-ordinate or mid-sectional area method for estimating earthwork volume over several longitudinal intervals of potentially unequal lengths?
Explanation:
The mid-sectional area method estimates earthwork volume using a representative cross section near the midpoint of each longitudinal interval. The volume of an interval is obtained by multiplying its mid-sectional area by its length. For several intervals, the individual volumes are calculated and added. When intervals have unequal lengths, each midpoint area must be multiplied by its corresponding interval length. This procedure differs from the average-end-area method, which uses the arithmetic mean of the sectional areas at the two ends. The accuracy of the estimate depends on the suitability of the selected intervals and the assumed variation of the ground profile.
వివరణ:
Question 2
ప్రశ్న 2
Two consecutive earthwork survey stations are located at chainages 0+045 and 0+095. At which chainage should the representative midpoint cross section be taken for the mid-ordinate method?
Explanation:
Chainage 0+045 represents a distance of 45 m from the alignment origin. Chainage 0+095 represents a distance of 95 m. The length of the interval is 95 - 45 = 50 m. Its midpoint lies 25 m from the first station. Therefore the required midpoint chainage is 45 + 25 = 70 m, written as 0+070. The cross-sectional area at this location can be used with the 50 m interval length to estimate earthwork volume.
వివరణ:
Question 3
ప్రశ్న 3
A road embankment has a formation width of 8.00 m and symmetrical side slopes of 1.5 horizontal to 1 vertical. Its filling heights at the ends of a 20.00 m interval are 1.20 m and 2.00 m. Assuming level transverse ground and linear variation of filling height, what volume is obtained by the mid-sectional area method?
Explanation:
The representative midpoint filling height is (1.20 + 2.00)/2 = 1.60 m. For a symmetrical embankment on level transverse ground, the cross-sectional area is A = Bd + md^2. Here B = 8.00 m, d = 1.60 m and m = 1.5. The midpoint sectional area is 8 x 1.60 + 1.5 x 1.60^2 = 12.80 + 3.84 = 16.64 m2. Multiplying by the interval length gives 16.64 x 20.00 = 332.80 m3. This is the estimated volume by the specified mid-sectional method.
వివరణ:
Question 4
ప్రశ్న 4
A straight excavation has a constant width of 5.00 m and vertical side boundaries. The cutting depths at two consecutive stations are 0.80 m and 1.60 m, with a longitudinal spacing of 30.00 m. Assuming linear variation of depth, what is the volume by the mid-sectional area method?
Explanation:
The excavation has a rectangular cross section because its lateral boundaries are vertical. The midpoint cutting depth is (0.80 + 1.60)/2 = 1.20 m. The representative cross-sectional area is width multiplied by midpoint depth. Therefore the area is 5.00 x 1.20 = 6.00 m2. Multiplying by the 30.00 m interval length gives 6.00 x 30.00 = 180 m3. Because the width is constant and the depth varies linearly, the midpoint calculation also gives the exact geometric volume under the stated assumptions.
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Question 5
ప్రశ్న 5
A road cutting over level transverse ground has a formation width of 6.00 m, a representative midpoint depth of 2.00 m and symmetrical side slopes of 2 horizontal to 1 vertical. If the longitudinal interval is 25.00 m, what earthwork volume is obtained by the mid-sectional area method?
Explanation:
For a cutting with symmetrical side slopes on level transverse ground, the sectional area is A = Bd + md^2. The formation width is 6.00 m and the representative midpoint depth is 2.00 m. The side-slope coefficient is m = 2. The rectangular portion contributes 6 x 2 = 12 m2, while both side-slope triangles contribute 2 x 2^2 = 8 m2. The total midpoint area is 12 + 8 = 20 m2. The estimated volume is therefore 20 x 25 = 500 m3.
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Question 6
ప్రశ్న 6
The formation width of an embankment varies linearly from 7.00 m to 9.00 m over a 40.00 m interval. The filling depth remains 1.00 m throughout, and both side slopes are 1 horizontal to 1 vertical over level transverse ground. Using the midpoint width, what volume is obtained by the mid-sectional method?
Explanation:
The formation width changes linearly between the two specified sections. Its midpoint value is (7.00 + 9.00)/2 = 8.00 m. The filling height remains 1.00 m, and the side-slope coefficient is 1. The midpoint cross-sectional area is 8 x 1 + 1 x 1^2 = 9.00 m2. Multiplying by the interval length gives 9.00 x 40.00 = 360 m3. Since the depth and side slopes are constant while the width varies linearly, the midpoint area represents the exact average sectional area under the stated geometry.
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Question 7
ప్రశ్న 7
Three successive earthwork intervals have lengths of 20 m, 35 m and 15 m. Their corresponding measured MIDPOINT cross-sectional areas are 12 m2, 16 m2 and 10 m2 respectively. What is the total earthwork volume by the mid-ordinate method?
Explanation:
Each interval must be associated with its own representative midpoint area. The first interval volume is 12 x 20 = 240 m3. The second interval volume is 16 x 35 = 560 m3. The third interval volume is 10 x 15 = 150 m3. Adding the three results gives 240 + 560 + 150 = 950 m3. Using a common interval length would be incorrect because the specified intervals are unequal.
వివరణ:
Question 8
ప్రశ్న 8
A road embankment has a constant formation width of 6.00 m and symmetrical side slopes of 1 horizontal to 1 vertical. The filling depths at three consecutive stations are 1.00 m, 2.00 m and 3.00 m, with 20.00 m between adjacent stations. Assuming level transverse ground and linear variation of depth within each reach, what is the total volume by the mid-sectional method?
Explanation:
The three stations define two longitudinal intervals of 20.00 m each. The midpoint depths are (1.00 + 2.00)/2 = 1.50 m and (2.00 + 3.00)/2 = 2.50 m. Using A = Bd + md^2, the first midpoint area is 6 x 1.50 + 1.50^2 = 11.25 m2. The second midpoint area is 6 x 2.50 + 2.50^2 = 21.25 m2. Their respective volumes are 11.25 x 20 = 225 m3 and 21.25 x 20 = 425 m3. The total mid-sectional volume is therefore 225 + 425 = 650 m3.
వివరణ:
Question 9
ప్రశ్న 9
A cutting cross section has a formation width of 8.00 m and a midpoint depth of 2.00 m on level transverse ground. The left side slope is 1 horizontal to 1 vertical, while the right side slope is 2 horizontal to 1 vertical. If the interval length is 10.00 m, what volume is obtained by the mid-sectional method?
Explanation:
The two sides have different horizontal-to-vertical slope ratios. For level transverse ground, the area is A = Bd + [(m1 + m2)/2]d^2. The rectangular portion is 8 x 2 = 16 m2. The combined triangular area is [(1 + 2)/2] x 2^2 = 6 m2. Hence the midpoint sectional area is 16 + 6 = 22 m2. Multiplying by the longitudinal interval of 10.00 m gives 22 x 10 = 220 m3.
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Question 10
ప్రశ్న 10
A 5.00 m wide vertical-sided earthwork strip extends over a longitudinal length of 30.00 m. At the first station, the ground is 0.60 m ABOVE the formation, while at the last station it is 0.90 m BELOW the formation. Both profiles vary linearly. Applying the midpoint method separately to cutting and filling after locating the zero-depth point, what are their respective volumes?
Explanation:
The signed depth changes from 0.60 m cutting to 0.90 m filling over 30.00 m. Under linear variation, the zero-depth point lies 30 x 0.60/(0.60 + 0.90) = 12.00 m from the first station. The remaining filling reach is 30 - 12 = 18.00 m. The midpoint cutting depth is 0.30 m, giving cutting volume 5 x 0.30 x 12 = 18.00 m3. The midpoint filling depth is 0.45 m, giving filling volume 5 x 0.45 x 18 = 40.50 m3. Cutting and filling are separate quantities and must not be reduced to a single net volume.
వివరణ:
Question 11
ప్రశ్న 11
A surveyed road-cutting reach has a measured midpoint cross-sectional area of 24.00 m2. The corresponding longitudinal interval is 35.00 m. Assuming the mid-sectional area method is specified for estimating the volume, what is the earthwork quantity?
Explanation:
The question directly supplies the representative midpoint cross-sectional area. Therefore there is no need to calculate a midpoint depth from end depths. The mid-sectional method uses V = Am x L. Substituting the given values gives V = 24.00 x 35.00. The estimated earthwork quantity is 840 m3. The result assumes that the surveyed midpoint section is accepted as representative of the specified reach.
వివరణ:
Question 12
ప్రశ్న 12
At the first station of a road cutting, the ground level is RL 101.30 m and formation level is RL 100.50 m. At the second station, the ground level is RL 102.40 m and formation level is RL 100.80 m. The stations are 25 m apart, and both longitudinal profiles vary linearly. For a formation width of 6 m and symmetrical cutting slopes of 1 horizontal to 1 vertical over level transverse ground, what is the mid-sectional volume?
Explanation:
The first cutting depth is 101.30 - 100.50 = 0.80 m. The second cutting depth is 102.40 - 100.80 = 1.60 m. The representative midpoint cutting depth is (0.80 + 1.60)/2 = 1.20 m. With a formation width of 6 m and side-slope coefficient 1, the midpoint area is 6 x 1.20 + 1.20^2 = 8.64 m2. The earthwork volume by the mid-sectional method is 8.64 x 25 = 216 m3. Both stations require cutting because the ground level exceeds the formation level at each location.
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Question 13
ప్రశ్న 13
At the exact midpoint of a 20.00 m long road-cutting interval, the surveyed ground level is RL 103.75 m and the formation level is RL 102.25 m. The formation width is 8.00 m, and symmetrical cutting side slopes are 1 horizontal to 1 vertical on level transverse ground. What volume is obtained using this ACTUALLY SURVEYED midpoint section?
Explanation:
The representative depth is calculated directly from the levels surveyed at the midpoint. It equals 103.75 - 102.25 = 1.50 m of cutting. The cross-sectional area formula is A = Bd + md^2. Substitution gives A = 8 x 1.50 + 1 x 1.50^2 = 12.00 + 2.25 = 14.25 m2. The estimated interval volume is 14.25 x 20.00 = 285 m3. Actual midpoint survey data can be used directly instead of estimating the depth from the two end stations.
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Question 14
ప్రశ్న 14
An earthwork cutting has a bottom formation width of 10.00 m and a representative midpoint depth of 2.00 m. The transverse ground is level, and both cutting slopes are 0.5 horizontal to 1 vertical. What is the mid-sectional earthwork volume for a 50.00 m reach?
Explanation:
The formation width is 10.00 m and the representative cutting depth is 2.00 m. A side slope of 0.5 horizontal to 1 vertical gives m = 0.5. The midpoint sectional area is A = Bd + md^2. Substituting gives A = 10 x 2 + 0.5 x 2^2 = 20 + 2 = 22 m2. The volume for the 50.00 m reach is 22 x 50 = 1,100 m3. The formula includes the rectangular central portion and the two symmetrical side-slope triangles.
వివరణ:
Question 15
ప్రశ్న 15
A contractor prepares an earthwork computation using two consecutive reaches. The first reach is 15 m long with a midpoint sectional area of 18 m2, and the second reach is 25 m long with a midpoint sectional area of 22 m2. What is the combined mid-ordinate volume?
Explanation:
The first reach has a midpoint cross-sectional area of 18 m2. Its estimated volume is 18 x 15 = 270 m3. The second reach has a midpoint sectional area of 22 m2 and a length of 25 m. Its estimated volume is 22 x 25 = 550 m3. Adding the two interval quantities gives 270 + 550 = 820 m3. A direct arithmetic mean of the two areas multiplied by the total length would not properly account for their unequal interval lengths.
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Question 16
ప్రశ్న 16
A symmetrical embankment on level transverse ground has a formation width of 8.00 m, side slopes of 1 horizontal to 1 vertical and end filling depths of 1.00 m and 3.00 m over a 20.00 m reach. The midpoint method gives 400 m3, while the average-end-area method gives 420 m3. By how much does the average-end-area result exceed the midpoint result?
Explanation:
The question provides the two method-specific volumes as 400 m3 and 420 m3. Their difference is 420 - 400 = 20 m3. The midpoint method calculates the sectional area at the representative mean depth. The average-end-area method calculates the mean of the complete areas at the two ends. In a section with sloping sides, the area contains a term proportional to the square of depth. Consequently, the two numerical methods need not produce identical results when the depth changes along the interval.
వివరణ:
Question 17
ప్రశ్న 17
A road embankment has a representative midpoint filling depth of 2.00 m and symmetrical side slopes of 1.5 horizontal to 1 vertical over level transverse ground. For a 40.00 m longitudinal interval, what volume is contributed by the TWO SIDE-SLOPE TRIANGULAR PORTIONS ALONE?
Explanation:
The side-slope contribution to the sectional area is md^2 for symmetrical slopes on level ground. The side-slope coefficient is m = 1.5 and the midpoint filling depth is d = 2.00 m. Thus the combined area of the two triangular portions is 1.5 x 2^2 = 6.00 m2. Multiplying by the longitudinal interval gives 6.00 x 40.00 = 240 m3. This quantity excludes the central rectangular portion of the embankment. It should therefore not be interpreted as the complete earthwork volume of the reach.
వివరణ:
Question 18
ప్రశ్న 18
An embankment has formation widths of 6.00 m and 10.00 m and filling depths of 1.00 m and 2.00 m at the ends of a 20.00 m interval. Both width and depth vary linearly, with symmetrical side slopes of 1 horizontal to 1 vertical over level transverse ground. What volume is estimated using the midpoint values of width and depth?
Explanation:
The midpoint formation width is (6.00 + 10.00)/2 = 8.00 m. The midpoint filling depth is (1.00 + 2.00)/2 = 1.50 m. For symmetrical side slopes of 1 horizontal to 1 vertical, the cross-sectional area is A = Bd + d^2. Hence the midpoint area is 8 x 1.50 + 1.50^2 = 12.00 + 2.25 = 14.25 m2. The estimated volume is 14.25 x 20.00 = 285 m3. This is the midpoint-method estimate under the stated linear interpolation assumptions, not necessarily the exact geometric volume.
వివరణ:
Question 19
ప్రశ్న 19
Four consecutive earthwork intervals are each 25.00 m long. The representative midpoint cross-sectional areas are 12 m2, 18 m2, 15 m2 and 21 m2 respectively. What is the total volume using the mid-ordinate method?
Explanation:
The four intervals have the same longitudinal length of 25.00 m. The mid-ordinate method multiplies each midpoint sectional area by its corresponding interval length. For equal intervals, the common length can be multiplied by the sum of the midpoint areas. Their sum is 12 + 18 + 15 + 21 = 66 m2. The estimated total volume is therefore 66 x 25 = 1,650 m3. No half-weighting of the first or last midpoint area is used in this calculation.
వివరణ:
Question 20
ప్రశ్న 20
A 5.00 m wide vertical-sided excavation has cutting depths of 0.50 m and 1.50 m at two stations 15.00 m apart. Assuming linear variation of depth and a constant width, what volume is obtained using the mid-sectional area method?
Explanation:
The specified excavation has vertical sides and a constant width of 5.00 m. The representative cutting depth is (0.50 + 1.50)/2 = 1.00 m. The midpoint sectional area is 5.00 x 1.00 = 5.00 m2. Multiplying this area by the 15.00 m reach gives 5.00 x 15.00 = 75 m3. The method is exact for the stated geometry because sectional area varies linearly with depth. No exterior side-slope contribution is present.
వివరణ:
Question 21
ప్రశ్న 21
The estimated earthwork volume for an interval is 450 m3 by the mid-sectional method. If the accepted representative midpoint cross-sectional area is 15 m2, what must be the interval length?
Explanation:
The mid-sectional volume formula is V = Am x L. The volume is given as 450 m3 and the midpoint area as 15 m2. Rearranging the expression gives L = V/Am. Therefore the interval length is 450/15 = 30 m. The cubic-metre volume divided by the square-metre area produces a result in metres. The calculation is valid because the stated area is the representative midpoint area used to obtain the given volume.
వివరణ:
Question 22
ప్రశ్న 22
An embankment has a constant filling depth of 2.00 m, symmetrical side slopes of 1 horizontal to 1 vertical and level transverse ground. Its formation width varies LINEARLY from 6.00 m to 10.00 m along a 30.00 m interval. What is the midpoint-method volume, and how does it compare with the exact geometric volume under these assumptions?
Explanation:
The midpoint formation width is (6.00 + 10.00)/2 = 8.00 m. The filling depth is constant at 2.00 m, and the side-slope coefficient is 1. The representative sectional area is 8 x 2 + 1 x 2^2 = 20 m2. The midpoint-method volume is consequently 20 x 30 = 600 m3. Since the filling depth and side slopes remain constant and the formation width varies linearly, the sectional area also varies linearly along the reach. Therefore the midpoint-method result equals the exact geometric volume under the stated assumptions.
వివరణ:
Question 23
ప్రశ్న 23
A road embankment has a formation width of 8.00 m, symmetrical side slopes of 1 horizontal to 1 vertical and end filling depths of 1.00 m and 3.00 m over 30.00 m. The depth varies linearly, giving end areas of 9 m2 and 33 m2 and a midpoint area of 20 m2. How much GREATER is the prismoidal volume than the midpoint-method volume?
Explanation:
The midpoint-method volume is 20 x 30 = 600 m3. The prismoidal formula is V = L(A1 + 4Am + A2)/6. Substituting the given areas gives V = 30(9 + 4 x 20 + 33)/6. The bracketed sum is 9 + 80 + 33 = 122. The prismoidal volume is therefore 30 x 122/6 = 610 m3. Its excess over the midpoint estimate is 610 - 600 = 10 m3.
వివరణ:
Question 24
ప్రశ్న 24
A road embankment is divided into four equal longitudinal intervals of 20.00 m each. The MIDPOINT filling depths are 0.50 m, 1.00 m, 1.50 m and 2.00 m respectively. For all intervals, formation width is 6.00 m, symmetrical side slopes are 1 horizontal to 1 vertical, and transverse ground is level. What is the combined mid-sectional volume?
Explanation:
The area formula for each midpoint section is A = Bd + md^2, where B = 6 and m = 1. At depth 0.50 m, the area is 6 x 0.50 + 0.50^2 = 3.25 m2. At depths 1.00 m, 1.50 m and 2.00 m, the corresponding areas are 7.00 m2, 11.25 m2 and 16.00 m2. The sum of the four midpoint areas is 3.25 + 7.00 + 11.25 + 16.00 = 37.50 m2. Since every interval is 20.00 m long, total volume is 37.50 x 20.00 = 750 m3. The computation uses the given midpoint depths directly rather than treating them as end-station depths.
వివరణ:
Question 25
ప్రశ్న 25
A road embankment has a constant formation width of 8.00 m and symmetrical side slopes of 1 horizontal to 1 vertical over level transverse ground. Four consecutive stations have filling depths of 1.00 m, 2.00 m, 2.40 m and 1.60 m. The lengths of the three successive intervals are 20.00 m, 30.00 m and 25.00 m respectively. Assuming linear depth variation within each interval, what is the TOTAL earthwork volume by the mid-sectional area method?
Explanation:
The midpoint depths for the three intervals are (1.00 + 2.00)/2 = 1.50 m, (2.00 + 2.40)/2 = 2.20 m and (2.40 + 1.60)/2 = 2.00 m. The sectional area formula is A = Bd + md^2 with B = 8 and m = 1. The midpoint areas are 8 x 1.50 + 1.50^2 = 14.25 m2, 8 x 2.20 + 2.20^2 = 22.44 m2, and 8 x 2.00 + 2.00^2 = 20.00 m2. The first interval volume is 14.25 x 20.00 = 285.00 m3. The second interval volume is 22.44 x 30.00 = 673.20 m3, and the third is 20.00 x 25.00 = 500.00 m3. Adding these results gives 285.00 + 673.20 + 500.00 = 1,458.20 m3. The unequal interval lengths have been applied individually as required by the mid-sectional method.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Calculate the representative mid-sectional area for each interval, multiply it by that interval's length, and add the resulting volumes. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: B. 0+070. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: D. 332.80 m3. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: A. 180 m3. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: C. 500 m3. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. 360 m3. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: D. 950 m3. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: A. 650 m3. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 220 m3. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 18.00 m3 cutting and 40.50 m3 filling. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. 840 m3. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. 216 m3. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 285 m3. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 1,100 m3. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. 820 m3. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. 20 m3. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. 240 m3. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 285 m3. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. 1,650 m3. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: D. 75 m3. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. 30 m. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 600 m3, equal to the exact geometric volume. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. 10 m3. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: D. 750 m3. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: B. 1,458.20 m3. సమాధానం: B.
