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Question 1
ప్రశ్న 1
In preparing a contour map by an indirect surveying method, what is meant by INTERPOLATION OF CONTOURS?
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Explanation:
Contour interpolation determines where selected elevations occur between surveyed points of known position and reduced level. The procedure is commonly used after indirect contouring observations have been plotted. The principal traditional interpolation methods are estimation, arithmetic calculation and graphical interpolation. Arithmetic interpolation ordinarily assumes a uniform ground gradient between the two guide points. The resulting contour positions are plotted and joined to represent the terrain. Interpolation cannot reliably identify unobserved abrupt changes in ground elevation.
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Question 2
ప్రశ్న 2
Two surveyed ground points A and B are separated by a horizontal distance of 48 m. Their reduced levels are 107.200 m and 113.200 m respectively. Assuming a uniform gradient along AB, what is the interpolated RL of a point 18 m from A toward B?
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Explanation:
The elevation difference between B and A is 113.200 - 107.200 = 6.000 m. The fraction of the distance from A to the required point is 18/48 = 0.375. Under linear interpolation, the corresponding rise is 6.000 multiplied by 0.375 = 2.250 m. Adding this rise to RL(A) gives 107.200 + 2.250 = 109.450 m. The result lies between the two observed reduced levels, as expected. This calculation assumes the elevation changes linearly along AB.
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Question 3
ప్రశ్న 3
Two surveyed points A and B are 60 m apart horizontally, with RL(A) = 122.600 m and RL(B) = 134.600 m. Under LINEAR interpolation, at what distance from A will the 128.600 m contour intersect the line AB?
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Explanation:
The total elevation difference between A and B is 134.600 - 122.600 = 12.000 m. The required contour lies 128.600 - 122.600 = 6.000 m above A. The proportional distance from A is therefore 6/12 of the total horizontal length. Multiplying 60 m by 6/12 gives 30 m. Thus the 128.600 m contour intersects AB at its midpoint. The calculation is valid under the explicitly stated uniform-gradient assumption.
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Question 4
ప్రశ్న 4
Along a straight ground line, the 95 m and 105 m contours intersect at points separated by a horizontal distance of 80 m. Assuming uniform slope between these intersections, what is the RL of a point 28 m from the 95 m contour toward the 105 m contour?
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Explanation:
The elevation difference between the contours is 105 - 95 = 10 m. The specified point is 28 m along the 80 m interval. Its proportional elevation gain is (28/80) multiplied by 10 = 3.5 m. Therefore the interpolated RL is 95 + 3.5 = 98.500 m. The point is closer to the lower contour and consequently has an elevation below the midpoint value of 100 m. The solution assumes a constant ground gradient between the two contour intersections.
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Question 5
ప్రశ్న 5
Which assumption and limitation should be recognised when applying ordinary ARITHMETICAL interpolation between two spot levels?
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Explanation:
Arithmetic interpolation distributes the known elevation difference proportionately along a line between two surveyed points. This procedure assumes a uniform gradient over the interpolated segment. Natural ground may contain ridges, depressions or abrupt changes in slope between observations. Such variations cannot necessarily be detected from only the two endpoint elevations. Additional spot levels improve the representation of irregular terrain. Therefore interpolation is a mathematical estimate rather than proof of the actual elevation at every intervening position.
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Question 6
ప్రశ్న 6
Which statement correctly compares the ESTIMATION, ARITHMETICAL and GRAPHICAL methods of contour interpolation?
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Explanation:
Interpolation by estimation locates contour positions approximately by visual judgement. The arithmetic method calculates proportional distances using known elevations and horizontal separations. Graphical interpolation uses a suitable graphical construction, such as a prepared tracing sheet with elevation divisions. Both arithmetic and conventional graphical methods generally represent the ground between guide points by an assumed uniform gradient. Arithmetic calculations are reproducible and useful where numerical precision is important. None of these methods can guarantee the true shape of irregular ground without adequate field observations.
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Question 7
ప్రశ్న 7
Surveyed points A and B are 75 m apart horizontally and have RLs of 91 m and 106 m respectively. Assuming linear variation in elevation, what are the distances from A to the intersections of the 95 m, 100 m and 105 m contours, respectively?
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Explanation:
The total elevation difference is 106 - 91 = 15 m across a horizontal distance of 75 m. Therefore each metre of elevation change corresponds to 75/15 = 5 m of horizontal distance. The 95 m contour is 4 m above A and occurs at 4 multiplied by 5 = 20 m. The 100 m contour is 9 m above A and occurs at 45 m. The 105 m contour is 14 m above A and occurs at 70 m. These distances follow directly from linear interpolation and are all within the surveyed segment.
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Question 8
ప్రశ్న 8
On a contour plan drawn to a scale of 1:2000, two successive contours having a vertical interval of 2.5 m are separated by 3 cm on the map, measured along the direction of maximum slope. Assuming uniform ground slope, what is the gradient?
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Explanation:
At a map scale of 1:2000, one centimetre represents 20 m on the ground. A map distance of 3 cm therefore represents 60 m horizontally. The vertical difference between the successive contours is 2.5 m. The slope magnitude is 2.5/60 = 1/24. Expressed as a percentage, this is approximately 4.17 percent. The direction of rise or fall must be determined separately from the contour elevation labels.
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Question 9
ప్రశ్న 9
A proposed road alignment must follow a constant rising gradient of 1 in 125. If the contour interval on the survey map is 5 m, what horizontal distance along the proposed alignment corresponds to the rise between successive contour levels?
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Explanation:
A rising gradient of 1 in 125 means an elevation increase of 1 m for every 125 m of horizontal distance along the alignment. The required change between successive contour levels is 5 m. Therefore the horizontal distance is 5 multiplied by 125 = 625 m. This is the horizontal distance travelled along the constant-gradient alignment between those elevations. The straight-line separation of the two contour intersections may be shorter if the alignment bends. The calculation is used when developing trial alignments with prescribed gradients.
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Question 10
ప్రశ్న 10
A proposed access road must rise by 12 m between two terminal points. The maximum permitted rising gradient is 1 in 300, expressed as vertical rise divided by horizontal alignment length. What is the MINIMUM horizontal length required to satisfy this gradient restriction?
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Explanation:
A gradient of 1 in 300 permits a rise of 1 m for every 300 m of horizontal alignment length. The total required rise is 12 m. At the limiting gradient, the minimum horizontal length is 12 multiplied by 300 = 3600 m. A shorter route would have a steeper average rising gradient and violate the stated limit. A longer route may satisfy the restriction if its local gradients are also controlled. The terrain and feasible alignment geometry determine whether such a route can actually be provided.
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Question 11
ప్రశ్న 11
A contour map has a scale of 1:10000 and a contour interval of 2 m. For plotting a trial road alignment at a uniform gradient of 1 in 100, what MAP DISTANCE represents the required horizontal alignment length between successive contour elevations?
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Explanation:
The specified gradient is 1 m vertical in 100 m horizontal. A contour interval of 2 m therefore corresponds to 2 multiplied by 100 = 200 m horizontally. At a map scale of 1:10000, 1 cm represents 100 m on the ground. The map length corresponding to 200 m is consequently 200/100 = 2 cm. This length can be used as a divider setting while tracing a trial constant-gradient alignment between successive contours. Actual alignment suitability must also account for ground conditions and engineering constraints.
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Question 12
ప్రశ్న 12
An engineer is tracing a preliminary road alignment of prescribed constant gradient on a contour plan. Which procedure is technically appropriate?
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Explanation:
A route with a prescribed nonzero gradient must change elevation at a controlled rate along its length. The required horizontal alignment distance between successive contour levels can be calculated from the contour interval and gradient. This distance is used to identify possible contour crossings when tracing a trial route. Following one contour would ordinarily give a zero ground gradient instead of the specified nonzero gradient. Practical route selection must also consider terrain, drainage crossings, stability and construction constraints. A contour-based trial alignment requires subsequent field verification and detailed design.
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Question 13
ప్రశ్న 13
The proposed formation RL of an access road at chainage 0 m is 95.000 m. The road is designed with a uniform RISING gradient of 1 in 200 in the direction of increasing chainage. What is the formation RL at chainage 400 m?
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Explanation:
The specified gradient gives a rise of 1 m for every 200 m of horizontal chainage. Over 400 m, the elevation increase is 400/200 = 2.000 m. The initial formation RL is 95.000 m. The formation RL at chainage 400 m is therefore 95.000 + 2.000 = 97.000 m. The positive change follows the explicitly stated rising direction. This is a designed formation elevation and is not necessarily equal to the existing ground RL at the same chainage.
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Question 14
ప్రశ్న 14
To determine whether two proposed surveying stations are INTERVISIBLE using a contour map, which procedure is most appropriate?
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Explanation:
Intervisibility requires an unobstructed line of sight between the relevant observation points. A contour map can be used to obtain an approximate ground profile along the line joining the stations. The elevations of the observer and target are determined by adding their respective heights above ground to their ground RLs. The connecting sight line is compared with the intervening terrain profile. If the terrain rises above the sight line, direct visibility is obstructed. Curvature, refraction and other obstructions may also require consideration for long-distance or high-precision applications.
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Question 15
ప్రశ్న 15
Two stations A and B are 400 m apart horizontally. Ground RL(A) = 105.000 m and ground RL(B) = 111.000 m. The observer's eye is 1.500 m above A, and a target is 2.000 m above B. At a point 160 m from A, an intervening ridge has RL 109.200 m. Assuming a straight line of sight, negligible curvature and refraction, and no other obstructions, what is the visibility condition?
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Explanation:
The observer's eye RL is 105.000 + 1.500 = 106.500 m. The target RL is 111.000 + 2.000 = 113.000 m. The sight line therefore rises by 6.500 m over the 400 m horizontal separation. At 160 m from A, the sight-line RL is 106.500 + (160/400) multiplied by 6.500 = 109.100 m. The ridge RL is 109.200 m, which is 0.100 m above the sight line. Therefore the ridge obstructs the direct line of sight under the stated assumptions.
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Question 16
ప్రశ్న 16
Stations P and Q are 500 m apart horizontally. Ground RL(P) = 100.000 m and ground RL(Q) = 102.000 m. The observer's eye at P is 1.600 m above ground, and the target at Q is 1.400 m above ground. The highest intervening ground on the straight path is at the midpoint and has RL 102.250 m. All other intervening terrain lies below the sight line. Neglecting curvature and refraction, what is the vertical clearance at the midpoint?
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Explanation:
The observer's eye RL is 100.000 + 1.600 = 101.600 m. The target RL is 102.000 + 1.400 = 103.400 m. At the midpoint of a straight sight line, its elevation is the mean of the endpoint elevations. Therefore the sight-line RL at the midpoint is (101.600 + 103.400)/2 = 102.500 m. The intervening ground has RL 102.250 m, giving a clearance of 102.500 - 102.250 = 0.250 m. The target is geometrically intervisible under the stated assumptions because the other intervening terrain also remains below the line of sight.
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Question 17
ప్రశ్న 17
A contour map shows a stream crossing successive contours marked 180 m, 170 m and 160 m, with their V-shaped bends pointing toward the 180 m contour. Which interpretation of the drainage direction is correct?
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Explanation:
A stream normally flows from higher ground elevations toward lower elevations. The given contour elevations decrease from 180 m through 170 m to 160 m. Contours crossing a typical valley form bends that point toward higher ground or upstream. Therefore the bends point approximately opposite to the downstream flow direction. The stream flows generally from the 180 m contour toward the 160 m contour. This interpretation is useful for identifying drainage directions and locating suitable outlets from contour maps.
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Question 18
ప్రశ్న 18
When delineating the TOPOGRAPHIC CATCHMENT AREA contributing surface runoff to a proposed check-dam site, which principle is correct?
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Explanation:
A topographic catchment is the land area whose surface runoff drains toward a specified outlet. Its boundary is generally traced along the surrounding watershed or drainage divides. These divides commonly follow ridge crests and other relatively high ground separating adjacent drainage systems. The boundary may cross contour lines of different elevations and is not itself necessarily a contour line. Contour maps help identify ridges, valleys and the direction of surface drainage. In complicated terrain, subsurface drainage and artificial diversions may cause actual contributing areas to differ from the purely topographic catchment.
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Question 19
ప్రశ్న 19
A drainage catchment boundary has been measured on a contour map of scale 1:10000. The enclosed PLAN AREA measured on the map is 12 cm^2. What is the corresponding horizontal ground area in hectares?
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Explanation:
At a scale of 1:10000, 1 cm on the map represents 100 m on the ground. Therefore 1 cm^2 on the map corresponds to 100 multiplied by 100 = 10000 m^2 on the ground. Since 1 hectare equals 10000 m^2, each square centimetre represents 1 hectare at this scale. The measured map area is 12 cm^2. The corresponding horizontal catchment area is consequently 12 hectares. This is planimetric area rather than the larger actual sloping surface area.
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Question 20
ప్రశ్న 20
An engineer delineates the topographic catchment of a proposed stormwater drainage outlet using a contour plan. Which additional information is essential before estimating design storm runoff reliably?
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Explanation:
Contours can help identify the contributing drainage area, terrain slopes and likely surface flow paths. Catchment area alone does not determine the quantity or peak rate of stormwater runoff. Rainfall intensity, duration and temporal distribution influence runoff generation. Infiltration, land cover, soil conditions, surface storage and drainage characteristics also affect runoff response. Appropriate hydrological parameters and a suitable design method are therefore required. Contour information supports the analysis but cannot independently establish the design runoff.
వివరణ:
Question 21
ప్రశ్న 21
For a proposed reservoir, the horizontal water-spread areas enclosed by successive elevation contours at RL 100 m and RL 105 m are 1800 m^2 and 3100 m^2 respectively. Using the TRAPEZOIDAL rule, what is the approximate storage volume between these two elevations?
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Explanation:
The vertical interval between the two contour levels is 105 - 100 = 5 m. The enclosed horizontal areas are 1800 m^2 and 3100 m^2. Under the trapezoidal rule, the volume is the average of the two end areas multiplied by the vertical interval. Therefore V = (5/2) multiplied by (1800 + 3100). This gives 2.5 multiplied by 4900 = 12250 m^3. The result is an approximate incremental storage volume between the stated elevations rather than the total capacity of the entire reservoir.
వివరణ:
Question 22
ప్రశ్న 22
The horizontal water-spread areas enclosed by three successive reservoir contours at elevations 100 m, 105 m and 110 m are 1000 m^2, 2000 m^2 and 5000 m^2 respectively. Using the COMPOSITE TRAPEZOIDAL rule with a uniform contour interval, what is the storage volume between RL 100 m and RL 110 m?
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Explanation:
The common contour interval is 5 m. For three successive enclosed areas, the composite trapezoidal formula is V = (h/2)(A1 + 2A2 + A3). Substituting the data gives V = (5/2)(1000 + 2 multiplied by 2000 + 5000). The bracketed sum is 1000 + 4000 + 5000 = 10000 m^2. Hence the estimated storage volume is 2.5 multiplied by 10000 = 25000 m^3. The intermediate area receives a factor of two because it contributes to both adjacent trapezoidal layers.
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Question 23
ప్రశ్న 23
Three successive reservoir contour areas, at equally spaced elevations, are 1000 m^2, 4000 m^2 and 6000 m^2. The vertical contour interval is 6 m. Using SIMPSON'S ONE-THIRD or the PRISMOIDAL formula, what is the estimated storage volume between the first and third contours?
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Explanation:
There are three contour areas and two equal vertical intervals of 6 m each. Simpson's one-third rule can therefore be applied over the two intervals. The formula is V = (h/3)(A1 + 4A2 + A3). Substitution gives V = (6/3)(1000 + 4 multiplied by 4000 + 6000). The bracketed value is 23000 m^2, so V = 2 multiplied by 23000 = 46000 m^3. The trapezoidal estimate would be 45000 m^3, illustrating that the two approximation methods need not produce identical volumes.
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Question 24
ప్రశ్న 24
A contour plan is being used to estimate the water-storage capacity of a proposed reservoir up to its specified FULL RESERVOIR LEVEL. Which statement correctly distinguishes the water-spread area from the storage volume?
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Explanation:
The water-spread area at a specified reservoir level is the horizontal area covered by water at that elevation. Storage capacity is a volume and has units of cubic metres. It depends on how the enclosed horizontal area varies with elevation from the lower reservoir levels to the specified upper level. A contour map provides the area-elevation information required for approximate integration. Trapezoidal and prismoidal rules are commonly used for such preliminary volume calculations. Estimating usable or live storage additionally requires relevant operating levels and allowances for factors such as sedimentation.
వివరణ:
Question 25
ప్రశ్న 25
A rectangular waterworks site measures 20 m by 20 m horizontally. Spot RLs derived from the contour survey at its four corners are 102.000 m, 103.000 m, 104.000 m and 105.000 m. The entire site is to be excavated to a horizontal formation RL of 101.000 m. Assuming BILINEAR interpolation of the existing ground surface between the four corners, what is the estimated volume of excavation?
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Explanation:
The corner excavation depths are obtained by subtracting the proposed formation RL from each existing ground RL. These depths are 1.000 m, 2.000 m, 3.000 m and 4.000 m. Under the stated bilinear interpolation assumption, the average depth over the rectangle equals the arithmetic mean of the four corner depths. The average excavation depth is therefore (1 + 2 + 3 + 4)/4 = 2.500 m. The horizontal site area is 20 multiplied by 20 = 400 m^2. The estimated excavation volume is consequently 400 multiplied by 2.500 = 1000 m^3, excluding bulking, shrinkage and other construction allowances.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Locating the positions of required contour elevations between surveyed points having known reduced levels. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 109.450 m. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 30 m. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 98.500 m. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Elevation is assumed to vary linearly along the selected segment, so unobserved breaks in slope may cause interpolation errors. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. Estimation is approximate, arithmetic interpolation calculates proportional positions, and graphical interpolation uses a prepared graphical scale or construction. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 20 m, 45 m and 70 m. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: D. 1 in 24, approximately 4.17%. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: B. 625 m. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 3600 m. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 2.0 cm. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. Use the horizontal distance required for each contour-interval rise or fall to trace a feasible route, adjusting the alignment for terrain and engineering constraints. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 97.000 m. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. Construct the intervening ground profile along the line joining the stations and compare it with the line of sight connecting the observation heights. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: C. The ridge is 0.100 m above the line of sight, so the target is obstructed. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: D. 0.250 m. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. Water flows generally from the 180 m contour toward the 160 m contour, opposite to the upstream-pointing contour bends. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. The catchment boundary generally follows drainage divides along surrounding high ground, separating runoff directed toward the selected outlet from runoff toward neighbouring basins. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. 12 hectares. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. Rainfall characteristics and appropriate runoff or loss parameters, in addition to catchment properties. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. 12250 m^3. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 25000 m^3. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: C. 46000 m^3. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: D. The water-spread area represents the horizontal area enclosed at the specified water level, while storage volume is estimated by integrating the enclosed areas over the corresponding elevation range. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: B. 1000 m^3. సమాధానం: B.