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Question 1
ప్రశ్న 1
A surveyor must transfer the reduced level of an established benchmark to a distant pumping-station site through several instrument setups and stable change points. Which levelling procedure is most appropriate?
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Explanation:
Differential levelling determines the difference in elevation between separated points using one or more instrument setups. A known benchmark supplies the initial vertical reference. At each setup, a backsight establishes the height of collimation. A foresight transfers the computed elevation to a stable change point or final station. Repeating this procedure extends vertical control to the pumping-station site. The arithmetic relation sum BS minus sum FS equals final RL minus initial RL provides an internal check.
వివరణ:
Question 2
ప్రశ్న 2
A differential levelling run starts from BM of RL 100.225 m. Setup 1: BS on BM = 1.345 m and FS on CP1 = 2.115 m. Setup 2: BS on CP1 = 0.965 m and FS on CP2 = 1.485 m. Setup 3: BS on CP2 = 1.755 m and FS on endpoint E = 1.025 m. What is RL(E)?
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Explanation:
The first setup gives RL(CP1) = 100.225 + 1.345 - 2.115 = 99.455 m. The second setup gives RL(CP2) = 99.455 + 0.965 - 1.485 = 98.935 m. The third setup gives RL(E) = 98.935 + 1.755 - 1.025 = 99.665 m. Independently, the backsights total 4.065 m and the foresights total 4.625 m. Their difference is -0.560 m, which equals 99.665 - 100.225 m. The endpoint is therefore 0.560 m below the starting benchmark, and the level-book arithmetic balances.
వివరణ:
Question 3
ప్రశ్న 3
For planning the gravity sewer along a proposed road, a surveyor records ground RLs at successive chainages along the sewer centreline. Which product is primarily obtained from these observations?
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Explanation:
Longitudinal or profile levelling establishes elevations at successive points along a selected alignment. The horizontal location is commonly identified by chainage measured from an adopted starting station. The recorded RLs can be plotted against chainage to construct a ground profile. Such a profile is useful for designing sewer gradients, pipelines, canals and roads. It helps determine the relationship between existing ground levels and proposed formation or invert levels. Cross-section levelling supplies additional transverse information rather than replacing the longitudinal profile.
వివరణ:
Question 4
ప్రశ్న 4
During profile levelling, the starting benchmark has RL 101.500 m and the backsight is 1.220 m. Without shifting the instrument, intermediate sights are taken at chainages 0 m, 20 m and 40 m as 1.550 m, 1.890 m and 2.145 m respectively. A foresight of 1.780 m is then taken on a change point at chainage 60 m. What is the ground RL at chainage 40 m?
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Explanation:
The height of collimation is 101.500 + 1.220 = 102.720 m. The intermediate sight at chainage 40 m is 2.145 m. Its reduced level is therefore 102.720 - 2.145 = 100.575 m. The earlier intermediate readings are not added or subtracted again when using the HI method. The foresight on the change point gives its separate RL of 100.940 m. Chainage identifies the horizontal station and does not change the elevation-reduction formula.
వివరణ:
Question 5
ప్రశ్న 5
The existing ground RL at chainage 0 m of a proposed pipeline is 102.800 m, while at chainage 150 m it is 102.050 m. Assuming a uniform gradient between these points, what is the average ground slope in the direction of increasing chainage?
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Explanation:
The elevation difference is 102.050 - 102.800 = -0.750 m. The horizontal chainage difference is 150 m. The magnitude of the gradient is 0.750 divided by 150, giving 0.005. This equals 0.5 percent or 1 in 200. The negative elevation difference indicates a fall in the direction of increasing chainage. This is the average existing ground gradient and need not equal the proposed pipeline invert gradient.
వివరణ:
Question 6
ప్రశ్న 6
A sewer alignment crosses a side-sloping road corridor, and the designer needs the ground levels at several offsets to the left and right of the centreline at a specified chainage. Which field procedure best provides these data?
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Explanation:
Cross-section levelling determines elevations across an alignment at selected chainages. Staff points are normally located at known offsets from a reference centreline. The offsets should be recorded with their side or signed direction and the associated RLs. The resulting transverse profiles reveal road camber, side slopes, depressions and changes in ground shape. These data support earthwork estimation and the design of road, drain and pipeline corridors. A single centreline elevation cannot reliably describe an irregular transverse ground surface.
వివరణ:
Question 7
ప్రశ్న 7
In a cross-section survey, the centreline point has RL 100.250 m and its staff reading is 1.625 m. From the SAME undisturbed level setup, the readings at 8 m left and 6 m right of the centreline are 2.175 m and 1.840 m respectively. What are the corresponding RLs of the two offset points?
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Explanation:
The common height of collimation is 100.250 + 1.625 = 101.875 m. The left-offset RL is 101.875 - 2.175 = 99.700 m. The right-offset RL is 101.875 - 1.840 = 100.035 m. The left point is therefore 0.550 m below the centreline. The right point is 0.215 m below the centreline. Lateral offsets locate the staff stations horizontally but do not enter the HI subtraction directly.
వివరణ:
Question 8
ప్రశ్న 8
Which statement correctly distinguishes a longitudinal profile from the cross-sections collected for the same pipeline alignment?
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Explanation:
A longitudinal profile represents elevation changes along the direction of the project alignment. Chainage provides the main horizontal coordinate for this representation. Cross-sections represent ground levels across the alignment at selected stations. They ordinarily use lateral offsets from the centreline to define the observation positions. Both surveys should use compatible vertical control and the same specified datum. Together they support grade design, drainage planning and calculation of earthwork quantities.
వివరణ:
Question 9
ప్రశ్న 9
A wide reservoir prevents the instrument from being placed between two benchmarks A and B on opposite banks. A surveyor can establish the level close to each bank but must take a long sight across the water. Which method is most suitable for determining their difference in elevation?
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Explanation:
Reciprocal levelling is particularly useful across rivers, reservoirs and deep valleys. The level is first placed near one endpoint and readings are taken on both endpoints. The instrument is then moved near the other endpoint and the observations are repeated. Combining the two apparent height differences eliminates the common systematic long-sight contribution under the usual assumptions. The method compensates for collimation error and the effects of curvature and refraction when these remain sufficiently stable. It does not eliminate arbitrary observation mistakes or unpredictable atmospheric variation.
వివరణ:
Question 10
ప్రశ్న 10
In reciprocal levelling, the instrument is first placed near A, giving readings a1 on A and b1 on B. It is then placed near B, giving readings a2 on A and b2 on B. Assuming negligible near-sight errors and stable systematic long-sight effects, which formula gives the corrected elevation difference RL(B) - RL(A)?
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Explanation:
The apparent difference RL(B) minus RL(A) from the first setup is a1 - b1. The corresponding apparent difference from the second setup is a2 - b2. A systematic error affecting the long sight enters the two apparent differences with opposite signs. Taking their arithmetic mean therefore cancels that common contribution. The corrected difference is [(a1 - b1) + (a2 - b2)] divided by two. A positive result indicates that B is higher, while a negative result indicates that B is lower.
వివరణ:
Question 11
ప్రశ్న 11
Two benchmarks A and B are on opposite banks of a river. With the level near A, staff readings are A = 1.320 m and B = 2.150 m. With the level near B, staff readings are A = 0.785 m and B = 1.455 m. What is the corrected elevation difference between A and B by reciprocal levelling?
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Explanation:
The apparent difference RL(B) minus RL(A) from the setup near A is 1.320 - 2.150 = -0.830 m. The apparent difference from the setup near B is 0.785 - 1.455 = -0.670 m. Their mean is (-0.830 - 0.670) / 2 = -0.750 m. The negative sign indicates that B lies below A. The two apparent differences are not individually adopted as the corrected result because their long-sight errors may differ in sign. Averaging the reciprocal observations gives a fall of 0.750 m from A to B.
వివరణ:
Question 12
ప్రశ్న 12
Reciprocal levelling gives the following readings: instrument near A, staff A = 1.855 m and staff B = 1.265 m; instrument near B, staff A = 2.015 m and staff B = 1.355 m. If RL(A) = 84.375 m, what is the corrected RL of B?
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Explanation:
The first apparent elevation difference is 1.855 - 1.265 = +0.590 m. The second apparent difference is 2.015 - 1.355 = +0.660 m. Their average is (0.590 + 0.660) / 2 = +0.625 m. Therefore B is 0.625 m higher than A. Adding this difference to RL(A) gives 84.375 + 0.625 = 85.000 m. The reciprocal average is used rather than either uncorrected individual difference.
వివరణ:
Question 13
ప్రశ్న 13
In reciprocal levelling across a river, the apparent value of RL(B) - RL(A) is -1.410 m when the level is near A and -1.280 m when it is near B. Assuming equal-magnitude, opposite-sign systematic long-sight contributions, what is the MAGNITUDE of that contribution to either apparent difference?
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Explanation:
The two apparent elevation differences are -1.410 m and -1.280 m. Their average is -1.345 m, representing the corrected value under the stated assumptions. The departure of the first apparent difference from this mean is -1.410 - (-1.345) = -0.065 m. The second difference departs by +0.065 m. Thus the systematic contribution to either apparent difference has a magnitude of 0.065 m. The full difference between the two apparent results is 0.130 m, or twice that contribution.
వివరణ:
Question 14
ప్రశ్న 14
Which statement describes an important limitation of reciprocal levelling across a long river span?
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Explanation:
The reciprocal method uses observations from near both endpoints to counteract systematic long-sight effects. Collimation inclination and curvature-related contributions can be cancelled through the reciprocal average under the usual assumptions. Atmospheric refraction is less predictable because it depends on conditions along the optical path. If refraction changes materially between the reciprocal observations, its effects need not cancel exactly. Random staff-reading errors and movement of the reference points also remain possible. Prompt reciprocal observations and appropriate field checks help reduce these risks.
వివరణ:
Question 15
ప్రశ్న 15
In long-sight geometric levelling, which statement correctly describes the effect of EARTH CURVATURE on an observed staff reading when compared with the corresponding reading free from curvature?
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Explanation:
The Earth's level surface is curved, while an ideal local horizontal sight line is tangent to it at the instrument. Over a long distance, the straight horizontal line lies above the corresponding curved level surface. The distant staff therefore gives an excessively large reading due to curvature alone. The correction applied to that observed staff reading is negative. Its magnitude grows approximately with the square of sight distance. Atmospheric refraction acts in the opposite direction and is treated separately.
వివరణ:
Question 16
ప్రశ్న 16
For ordinary surveying calculations using mean Earth radius, which expression gives the curvature correction Cc to be ADDED ALGEBRAICALLY to an observed staff reading, if d is the horizontal sight distance in kilometres and Cc is in metres?
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Explanation:
The approximate geometrical curvature departure is d squared divided by twice the Earth radius. With the mean Earth radius taken as approximately 6370 km, its magnitude becomes 0.0785d^2 metres when d is in kilometres. Curvature causes an excessive distant staff reading in geometric levelling. Therefore the correction applied to the observed reading is subtractive. The signed expression is Cc = -0.0785d^2 m. This is the curvature-only correction and does not include atmospheric refraction.
వివరణ:
Question 17
ప్రశ్న 17
A levelling staff is observed at a horizontal distance of 1.600 km from the instrument. Neglect atmospheric refraction and take the curvature correction to a staff reading as Cc = -0.0785d^2 m, where d is in kilometres. What is the correction?
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Explanation:
The given sight distance is 1.600 km, so d squared is 2.5600. The curvature correction is -0.0785 multiplied by 2.5600. This gives Cc = -0.200960 m. The negative sign indicates that the correction is subtracted from the observed staff reading. It is not a correction to be added directly to an observed foresight. Atmospheric refraction has been expressly excluded from this calculation.
వివరణ:
Question 18
ప్రశ్న 18
Under the conventional average-atmosphere approximation used in levelling, which statement correctly describes the correction for ATMOSPHERIC REFRACTION applied to an observed staff reading?
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Explanation:
Atmospheric refraction causes the optical ray to curve approximately downward under the conventional model. This effect partially counteracts the apparent increase in staff reading caused by Earth curvature. The standard approximate refraction correction applied to the staff reading is positive. Its magnitude is conventionally taken as about one-seventh of the curvature effect. Accordingly Cr = +0.0112d^2 m when d is measured in kilometres. Actual refraction varies with atmospheric conditions, so the coefficient is an approximation rather than an exact universal constant.
వివరణ:
Question 19
ప్రశ్న 19
A sight length is 2.500 km. Using the conventional refraction correction Cr = +0.0112d^2 m, where d is in kilometres, what correction should be applied to the observed staff reading due to refraction ALONE?
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Explanation:
The horizontal sight length is 2.500 km. Squaring this distance gives 6.2500 km squared as the numerical input to the stated formula. The correction is 0.0112 multiplied by 6.2500 = 0.0700 m. The sign is positive because the conventional refraction correction is additive to the staff reading. This result considers atmospheric refraction alone and excludes Earth curvature. Combining both effects would produce a negative net correction under the standard approximation.
వివరణ:
Question 20
ప్రశ్న 20
For a long horizontal sight under the conventional average atmospheric conditions, which expression gives the COMBINED correction due to Earth curvature and atmospheric refraction that must be added algebraically to an observed staff reading? Take d in kilometres and correction in metres.
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Explanation:
The standard correction for curvature is -0.0785d^2 m. The standard correction for refraction is +0.0112d^2 m. The combined correction is the algebraic sum of these two quantities. Consequently C = -0.0785d^2 + 0.0112d^2 = -0.0673d^2 m. The negative sign means that the combined correction is subtracted from the observed staff reading. This expression assumes the conventional average refraction model and uses kilometres for sight distance.
వివరణ:
Question 21
ప్రశ్న 21
A distant staff is observed at a horizontal distance of 1.500 km. Using C = -0.0673d^2 m for the combined curvature-and-refraction correction to an observed staff reading, what is its value?
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Explanation:
The sight distance is 1.500 km, so d squared equals 2.2500. The combined correction is -0.0673 multiplied by 2.2500. This gives -0.151425 m. The negative value is the signed correction applied to the observed staff reading. The separate curvature correction is -0.176625 m, while the separate refraction correction is +0.025200 m. Their algebraic sum independently confirms the combined result of -0.151425 m.
వివరణ:
Question 22
ప్రశ్న 22
An observed distant staff reading is 2.6800 m, and the sight length is 2.000 km. If the combined curvature-and-refraction correction to the reading is -0.0673d^2 m, what is the corrected staff reading?
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Explanation:
The combined correction is -0.0673 multiplied by 2.000 squared. This gives -0.2692 m. The corrected staff reading is the observed reading plus the signed correction. Therefore the corrected value is 2.6800 - 0.2692 = 2.4108 m. Adding 0.2692 m would incorrectly reverse the conventional combined correction. The calculation assumes average refraction and that no other staff-reading corrections are required.
వివరణ:
Question 23
ప్రశ్న 23
A level is set up close to benchmark A of RL 150.000 m, and the near backsight is 1.450 m. A distant foresight on point B is 2.880 m at a horizontal distance of 1.200 km. Neglect the curvature-and-refraction effect on the near backsight. Using the combined correction -0.0673d^2 m on the distant foresight, what is the corrected RL of B to three decimal places?
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Explanation:
The height of collimation from the near backsight is 150.000 + 1.450 = 151.450 m. The combined correction on the distant foresight is -0.0673 multiplied by 1.200 squared. This gives -0.096912 m, so the corrected foresight is 2.880 - 0.096912 = 2.783088 m. The corrected RL of B is 151.450 - 2.783088 = 148.666912 m. Rounded to three decimal places, the result is 148.667 m. Equivalently, the uncorrected RL of 148.570 m increases by 0.096912 m after correcting the distant foresight.
వివరణ:
Question 24
ప్రశ్న 24
From ONE setup, the observed backsight on A is 1.660 m at a sight distance of 0.500 km, while the observed foresight on B is 2.140 m at 1.500 km. Apply the combined correction C = -0.0673d^2 m separately to BOTH staff readings. What is the corrected elevation difference RL(B) - RL(A)?
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Explanation:
The combined correction on the backsight is -0.0673 multiplied by 0.500 squared, giving -0.016825 m. The corrected backsight is therefore 1.643175 m. The combined correction on the foresight is -0.0673 multiplied by 1.500 squared, giving -0.151425 m. The corrected foresight becomes 2.140 - 0.151425 = 1.988575 m. Hence RL(B) - RL(A) = 1.643175 - 1.988575 = -0.345400 m. The unequal sight lengths prevent the curvature-and-refraction effects from cancelling, so both observations require their respective corrections.
వివరణ:
Question 25
ప్రశ్న 25
In a TWO-PEG test, pegs A and B are 80 m apart. From a midpoint setup, readings on A and B are 1.225 m and 1.875 m. The level is then placed very close to A, giving readings A = 1.145 m and B = 1.811 m. Assuming negligible near-sight distance and neglecting curvature and refraction, what should the corrected reading on B be from the second setup if the collimation error is removed?
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Explanation:
The midpoint setup has equal sights, so the assumed linear collimation error cancels in the difference between readings. The true elevation difference RL(B) - RL(A) is 1.225 - 1.875 = -0.650 m. From the second setup close to A, the corrected reading on B must satisfy 1.145 - corrected B reading = -0.650 m. Therefore the corrected B reading is 1.145 + 0.650 = 1.795 m. The observed distant reading of 1.811 m exceeds this value by 0.016 m, showing a positive distance-dependent reading error under the stated assumptions. The two-peg test identifies this collimation discrepancy, while balanced backsight and foresight distances help cancel its effect during ordinary levelling.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Differential levelling with successive backsights and foresights. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. 99.665 m. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: B. A longitudinal ground profile showing elevation variation with chainage. సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: D. 100.575 m. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: A. Fall of 1 in 200, equivalent to 0.5%. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. Carry out cross-section levelling at known lateral offsets from the centreline. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. Left = 99.700 m; right = 100.035 m. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. A longitudinal profile follows the alignment by chainage, while cross-sections show transverse elevation variation at selected chainages. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. Reciprocal levelling using observations from near both A and B. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. [(a1 - b1) + (a2 - b2)] / 2. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. B is 0.750 m lower than A. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. 85.000 m. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. 0.065 m. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. It relies on sufficiently stable systematic effects between reciprocal observations and cannot guarantee cancellation of variable atmospheric refraction. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. Curvature makes the observed staff reading larger and requires a negative correction. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: D. Cc = -0.0785d^2. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: C. -0.200960 m. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. It is positive and approximately one-seventh of the curvature correction in magnitude. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. +0.0700 m. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. -0.0673d^2. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. -0.151425 m. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: A. 2.4108 m. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 148.667 m. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: B. -0.3454 m. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1.795 m. సమాధానం: A.