Question 1

Sewer setting-out with sight rails and boning rods: A sewer runs from manhole A to manhole B, 250 m apart, on a gradient of 1 in 125 falling from A to B. If the reduced level (RL) of the invert at A is 205.75 m and the boning rod height is 3.00 m, w…

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Introduction / Context: For sewer construction, sight rails and boning rods establish a straight line parallel to the designed invert. The sight-rail level must be set to account for gradient and the boning-rod height so that when the rod top aligns with the sight line, the invert is at the correct RL. Given Data / Assumptions: Horizontal distance A→B = 250 m. Gradient = 1 in 125, falling from A to B. RL(invert at A) = 205.75 m. Boning-rod height = 3.00 m. Concept / Approach: A fall of 1 in 125 means a drop of 1 m per 125 m. Over 250 m, the invert fall is 250 / 125 = 2.00 m. Therefore RL(invert at B) = 205.75 − 2.00 = 203.75 m. The sight rail at B is set above the invert by the boning-rod height so that the rod top aligns with the sight string: RL(sight rail at B) = RL(invert at B) + 3.00 m. Step-by-Step Solution: Compute fall: 250 / 125 = 2.00 m.Find RL(invert at B): 205.75 − 2.00 = 203.75 m.Add boning-rod height: 203.75 + 3.00 = 206.75 m.State the RL of the sight rail at B: 206.75 m. Verification / Alternative check: If instead the gradient rose toward B, the invert would be higher and the sight rail RL would increase accordingly; here it decreases then adds the rod height. Why Other Options Are Wrong: 208.75 m / 211.75 m: Overestimate; do not reflect the 2 m fall correctly. 202.75 m: Below even the invert at B; ignores the added boning-rod height. Common Pitfalls: Confusing invert fall direction; forgetting to add the boning-rod height to get the sight-rail level. Final Answer: 206.75 m

Question 2

Intersection from two transit stations: For locating a distant object visible from two transit (theodolite) stations, which method is usually preferred?

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Introduction / Context: When a remote point cannot be occupied or measured directly, its position can be fixed by observations from two known stations. Several combinations of observed quantities are possible, but one is generally most efficient and precise for distant targets. Given Data / Assumptions: The distant point is intervisible from both transit stations. Station coordinates or relative position are known. Angular measurements are typically more reliable than long-distance linear measures in difficult terrain. Concept / Approach: Angular intersection uses the measured directions (bearings or angles) to the unknown point from both stations, forming two rays that intersect at the point's position. It avoids the challenges of long EDM/tape distances and often yields higher accuracy with good geometry (wide intersection angles). Step-by-Step Solution: Set up at station A and measure the direction to the target.Set up at station B and measure the direction to the same target.Plot/compute the intersection of the two rays to obtain coordinates.Check geometry: prefer intersection angles between about 30° and 150° for stability. Verification / Alternative check: Redundancy by adding a third station or repeating angle sets improves confidence and provides blunder detection. Why Other Options Are Wrong: Angles + distances: Feasible but unnecessary for most cases and costlier. Distances only: Requires distance to the remote point from both stations, often impractical. Mixed angle/distance: Works but loses the simplicity and robustness of pure angular intersection. Common Pitfalls: Poor intersection geometry (very acute/obtuse angles) magnifies position error; choose well-spaced stations. Final Answer: Measure only angles from the two stations (angular intersection)

Question 3

Scope of chain surveying: Chain (tape) surveying is best suited for which type of area and site condition?

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Introduction / Context: Chain surveying is the simplest form of plane surveying. It relies primarily on linear measurements with chains or tapes and uses well-conditioned triangles as the framework. Its applicability depends strongly on ground conditions and the complexity of features to be recorded. Given Data / Assumptions: Minimal need for angular measurements. Open ground where chaining and ranging lines are feasible. Plan is of relatively small extent to maintain accuracy without advanced instruments. Concept / Approach: Because chain surveying lacks precise angular control and depends on unobstructed straight lines, it is ideal for small, open, and relatively simple sites (e.g., fields, plots). Complex or crowded details, and large extents, introduce alignment difficulties, cumulative errors, and plotting complications best handled with theodolite/EDM or total station methods. Step-by-Step Solution: Assess site: open vs obstructed, simple vs complex detail.If small and open with simple detail, adopt chain survey with check lines.If large or crowded, prefer higher-order methods (traverse, tachymetry, total station).Ensure well-conditioned triangles and independent checks for quality control. Verification / Alternative check: Textbook guidance consistently restricts chain surveying to small, simple, open areas; modern practice often replaces it with total stations for efficiency in complex sites. Why Other Options Are Wrong: Crowded details: Offsets and chaining become cumbersome and error-prone. Large areas: Accumulated errors and logistics make chain-only methods unsuitable. Difficult details: Require precise angular control and instruments. Common Pitfalls: Attempting chain surveys in obstructed urban environments; results are slow and inaccurate. Final Answer: Small areas in open ground with simple details

Question 4

Levelling Methods – Limitation of the Line of Collimation (Height of Instrument) Method In differential levelling, the line of collimation method (also called the Height of Instrument method) is quick but offers fewer internal checks. Which of the f…

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Introduction / Context: The line of collimation (Height of Instrument) method is a popular approach for reducing level-book observations. It is efficient because it computes a height of instrument for each setup and then derives reduced levels directly. However, compared to the Rise and Fall method, it offers fewer built-in diagnostic checks on individual observations. This question asks which item is not automatically checked by the method. Given Data / Assumptions: Standard levelling terminology: back sight (BS), intermediate sight (IS), fore sight (FS), reduced level (RL). Computation pathway: HI = RL_bench + BS; then RL_point = HI − staff_reading (IS or FS). An arithmetic check is usually: ΣBS − ΣFS = RL_last − RL_first. Concept / Approach: In the HI method, the principal arithmetic check is the equality ΣBS − ΣFS = net change in RL. This validates the overall run but does not test each IS individually. In contrast, the Rise and Fall method generates rise/fall between successive points, providing a line-by-line internal check that can reveal a single bad IS reading. Therefore, the HI method lacks an inherent check on intermediate sights. Step-by-Step Solution: Identify what the HI method computes directly: HI and then RLs.Note that IS values affect RLs but are not independently cross-checked.Recognize that the method still supports an overall arithmetic check using ΣBS − ΣFS.Conclude that intermediate sights do not receive an independent internal verification. Verification / Alternative check: Compare with the Rise and Fall method, where each consecutive difference produces a rise or fall, creating a redundant check on every staff observation, including IS. This confirms the relative weakness of the HI method for detecting a single erroneous IS. Why Other Options Are Wrong: Fore sights and back sights are part of the ΣBS − ΣFS arithmetic check; reduced levels are computed consistently from HI. The arithmetic check itself is available in the HI method and is not the missing check. Common Pitfalls: Assuming that a successful ΣBS − ΣFS check validates all IS readings; it does not. A faulty IS can remain hidden if BS and FS accidentally balance the total. Final Answer: Intermediate sights

Question 5

Chain & Tape Work – Best Tool for Measuring Short Offsets from a Baseline In basic chain surveying, very short perpendicular offsets to locate details (walls, hedges, building corners) are typically measured using which tool under ordinary site accu…

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Introduction / Context: Short offsets are common in detail surveying. Choosing the right measuring tool affects speed, flexibility, and accuracy. This question checks your practical knowledge of which tape or chain is commonly used for short, frequent offsets in routine topographic work. Given Data / Assumptions: Offsets are short (a few metres). Required accuracy is that of ordinary chain surveying. Environment may involve frequent contact with ground and obstacles. Concept / Approach: Metallic tapes (linen or synthetic fabric reinforced with metallic wires) are light, flexible, and easy to handle repeatedly at short lengths. They are fast for small offsets and sufficiently accurate for ordinary detail work. Steel tapes are more accurate but heavier and typically reserved for longer measurements where precision justifies handling effort. Invar tapes are for high-precision baselines and are unnecessary for short offsets. Ordinary chains are cumbersome for very small perpendiculars and do not drape well around obstacles. Step-by-Step Solution: Identify the task: frequent short offsets from a chain line.Select an instrument that is lightweight and quickly deployable.Balance accuracy with practicality: metallic tape suits common practice best.Therefore, choose the metallic tape for short offsets. Verification / Alternative check: Field handbooks recommend metallic tapes (or light fibreglass tapes) for rapid offsetting, reserving steel/invar tapes for higher precision tasks. Why Other Options Are Wrong: Ordinary chains are unwieldy for short offsets; invar is overkill; steel tapes are accurate but slower for repeated short measures; fibreglass is acceptable but not the standard answer in classical chain surveying texts. Common Pitfalls: Using a heavy steel tape when speed is the priority; attempting short offsets with a bulky chain leading to alignment errors. Final Answer: A metallic tape

Question 6

Circular Curves – Approximate Formula for Perpendicular (Radial) Offsets from the Tangent For setting out a simple circular curve, what is the commonly used approximate expression for the perpendicular (radial) offset y from the tangent at a point l…

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Introduction / Context: In highway and railway curve setting, quick field layout often relies on approximate formulas for offsets. Perpendicular (sometimes called radial) offsets from the tangent are especially handy near the tangent point, where the curve deviates gradually from the tangent. This question asks you to recall the standard approximation used by surveyors. Given Data / Assumptions: Simple circular curve of radius R. Point on the tangent located at distance x from the point of tangency T. Offset y is measured perpendicular to the tangent to reach the curve. Concept / Approach: From circle geometry, the exact offset from the tangent at distance x is y_exact = R − √(R^2 − x^2). For x ≪ R (near T), a binomial expansion gives √(R^2 − x^2) ≈ R − x^2/(2R). Therefore, y ≈ x^2/(2R). This approximation is robust for small x/R and is widely used for peg setting at short intervals from T in the field. Step-by-Step Solution: Start with the exact relation: y = R − √(R^2 − x^2).Apply expansion: √(R^2 − x^2) ≈ R − x^2/(2R) for small x/R.Subtract from R: y ≈ x^2/(2R).Use this to compute quick offsets for setting curve pegs near T. Verification / Alternative check: Numerical test: for R = 300 m and x = 10 m, exact y ≈ 0.1667 m; approximation gives x^2/(2R) = 100/600 ≈ 0.1667 m, confirming high accuracy for small x. Why Other Options Are Wrong: y = x^2/R and y = x^2/(3R) over/under-estimate the true offset; the cubic term y = x^3/(6R^2) is not the leading-order term; y = 2 R x is dimensionally incorrect and grossly wrong. Common Pitfalls: Using the approximation too far from T (large x/R), where errors grow; mixing up perpendicular offsets with radial offsets from the long chord method. Final Answer: y = x^2 / (2R)

Question 7

Chaining on Slopes – Approximate Slope Correction per 100 Links for a Gradient of 1 in n If a 100-link chain (20 m) is laid along a uniform slope with a rise of 1 unit in n horizontal units (tan θ = 1/n), what is the approximate correction per chain…

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Introduction / Context: When a chain is laid along a slope, the measured sloping distance exceeds the horizontal projection. For mapping and setting out, distances must be reduced to the horizontal. This question asks for the commonly used approximate correction per 100 links for a slope of 1 in n, a staple result in elementary surveying. Given Data / Assumptions: Chain length along slope L = 100 links. Gradient 1 in n ⇒ tan θ = 1/n with small θ (n is reasonably large). We need the correction to subtract from the sloping length to get horizontal length. Concept / Approach: The exact horizontal length is H = L * cos θ, so the correction per chain is C_exact = L − H = L (1 − cos θ). For small θ, use cos θ ≈ 1 − θ^2/2 and tan θ ≈ θ (in radians). With tan θ = 1/n, take θ ≈ 1/n. Then 1 − cos θ ≈ θ^2/2 ≈ 1/(2 n^2). Therefore, C_approx ≈ L * (1/(2 n^2)). With L = 100 links, C_approx ≈ 100/(2 n^2) = 50/n^2 links. This is the standard approximate slope correction per 100 links to be subtracted from the measured sloping length. Step-by-Step Solution: Write exact correction: C_exact = 100 (1 − cos θ) links.Use small-angle approximation: 1 − cos θ ≈ θ^2/2.Relate θ to n: θ ≈ 1/n (radians) for tan θ = 1/n.Compute C_approx = 100 * (1/(2 n^2)) = 50/n^2 links (subtract from slope distance). Verification / Alternative check: If n = 20, C_approx = 50/400 = 0.125 links (≈ 0.025 m), consistent with practical expectations for gentle slopes over 20 m. Why Other Options Are Wrong: 100/n^2 doubles the correction; 100 (1 − cos θ) is exact form but the question asks for the approximate value in terms of n; 100 (sec θ − 1) relates to forward shift in stepping, not horizontal reduction; 25/n has wrong dependence on n. Common Pitfalls: Confusing the forward shift formula with reduction to horizontal; forgetting that the correction is to be subtracted; mixing degrees with radians in approximations. Final Answer: 50 / n^2 links (to be subtracted)

Question 8

Theodolite Eccentricity – Identifying and Dealing with Non-coincident Axes Which statements about circle/vernier eccentricity in a theodolite are correct when the axes of rotation of the graduated circle and the verniers are not coincident?

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Introduction / Context: Eccentricity in a theodolite refers to geometric misalignment where the center of the vernier system does not coincide with the center of the graduated circle. Understanding its identification and mitigation is essential for accurate angle measurement and is a common exam topic in surveying instrumentation. Given Data / Assumptions: A horizontally graduated circle read by two opposite verniers. Possible constant offset between vernier readings due to eccentricity. Instrument otherwise adjusted and functioning normally. Concept / Approach: When eccentricity is present, individual vernier readings are biased depending on where the circle is stopped. Using two opposite verniers and averaging their readings cancels the symmetric eccentricity effect, yielding the true angle. If the difference between the two verniers is a constant, a single vernier may be used by applying the appropriate constant correction. Recognizing that non-coincident axes define eccentricity solidifies option (a). Step-by-Step Solution: Identify eccentricity by noting a constant difference between opposite verniers across positions.Use the mean of the two readings to eliminate the bias instantly.Alternatively, adopt one vernier with a consistent correction if the difference is stable.Confirm that non-coincidence of axes is the root cause. Verification / Alternative check: Instrument manuals recommend reading both verniers and averaging as standard practice, explicitly to nullify circle/vernier eccentricity. Why Other Options Are Wrong: Since (a), (b), and (c) are all valid, options denying these statements are incorrect; therefore “All of the above” is the correct collective choice. Common Pitfalls: Forgetting to average opposite verniers; assuming random error rather than recognizing a repeatable systematic difference. Final Answer: All of the above

Question 9

Chain Surveying – Prolonging a Chain Line Across an Obstruction When a chain line is obstructed by a pond, building, or other obstacle, which of the following techniques may be used in practice to prolong the line across the obstruction?

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Introduction / Context: Obstructions commonly interrupt straight chain lines. Competent surveyors must know several field methods to regain the alignment beyond the obstacle while maintaining accuracy. This question assesses recognition of acceptable techniques used in routine practice. Given Data / Assumptions: There is no direct path for chaining. Nearby ground permits auxiliary setups or offsets. A straight line must be continued past the obstruction. Concept / Approach: Options include: (1) Angular measurement around the obstacle using a theodolite/compass with short traverse legs, then resecting the continuation; (2) Offset methods using perpendiculars/parallel lines established by chain or tape to port the line; (3) Triangulation by establishing auxiliary points and solving triangles to compute the continuation. Each method is standard and chosen based on site constraints and required accuracy. Step-by-Step Solution: Evaluate obstruction size and surroundings to choose a method.If angles can be observed, traverse around and compute the forward point.If space allows, construct perpendicular offsets and parallels to leapfrog the obstacle.Alternatively, fix two accessible points and compute the forward alignment by triangle solution. Verification / Alternative check: Field manuals include detailed sketches for each technique, confirming they are all accepted solutions for prolonging a chain line. Why Other Options Are Wrong: “None of the above” is incorrect because each listed method is a legitimate approach used by surveyors. Common Pitfalls: Failing to close and check a small traverse around the obstacle; poor right-angle formation when using chain-only methods; neglecting to tie in control points for verification. Final Answer: All of the above

Question 10

Topographic Terms – Where Is a Bergschrund Found? A “bergschrund” (also spelled bergschrund) is a distinct topographical feature associated with which kind of terrain?

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Introduction / Context: Surveyors and engineers often interpret landforms from maps and field reconnaissance. Recognizing glacial features is important in mountainous route location, hydrology, and hazard assessment. The term “bergschrund” is specific to glacier morphology and signals particular ground conditions. Given Data / Assumptions: Feature name: bergschrund. Context: alpine or polar glaciation. Map evidence can include contours, shading, or symbols near glacier heads. Concept / Approach: A bergschrund is a large, often semicircular crevasse that forms where a moving glacier or cirque glacier pulls away from the headwall, separating active ice from stagnant or firn ice above. It commonly marks the upper boundary of ice movement and can be deep and hazardous. Therefore, it is unequivocally a feature of glaciated regions, not plains, ordinary hills, water bodies, or dune fields. Step-by-Step Solution: Identify the landform category as glacial based on the term.Recall location: near the head of a glacier or cirque.Associate with headwall retreat and ice flow dynamics.Select “Glaciated region”. Verification / Alternative check: Glaciology references and mountaineering guides consistently define a bergschrund as a crevasse at the upper limit of glacier flow. Why Other Options Are Wrong: Plains, non-glaciated hills, water bodies, and deserts do not create the characteristic crevasse formed by differential ice movement against a headwall. Common Pitfalls: Confusing bergschrund with crevasses anywhere on a glacier; a bergschrund specifically occurs near the head where the glacier pulls away from the bounding rock. Final Answer: Glaciated region

Question 11

Short Sights & Centering – Techniques to Reduce Centering Error for Very Short Traverse Legs When working with very short traverse legs where centering error can dominate, which practices are commonly adopted to reduce the centering and pointing err…

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Introduction / Context: On very short legs, the angular miscentering can create disproportionately large errors in computed bearings and coordinates. Practical field techniques aim to reduce the centering and pointing components so that instrument precision is actually realized. This question reviews several accepted practices. Given Data / Assumptions: Traverse legs are short enough that a few millimetres of centering error matter. Standard theodolite with optical plummet is available. Simple targets (e.g., chain pins) may be used for very short sights. Concept / Approach: Mitigation strategies include: ensuring precise instrument centering with an optical plummet; using fixed targets on tripods so the instrument can be re-set exactly over the same point; employing chain pins as close-range sighting objects to reduce pointing uncertainty for very short observations. Collectively, these practices minimize lateral target definition errors and instrument centering offsets. Step-by-Step Solution: Set the theodolite precisely over the station using the optical plummet.Use rigid targets (tripod-mounted) to avoid wobble and establish repeatable positions.For very short lines, chain pins can serve as clear, small targets for quick, repeatable pointing.Combine these methods as needed to keep centering error below the required tolerance. Verification / Alternative check: Survey practice guides endorse optical centering and stable targets to control short-sight errors; numerous field crews adopt chain pins for convenience on very short legs. Why Other Options Are Wrong: All listed practices are valid in the stated context; therefore “All of the above” is correct and “None” is incorrect. Common Pitfalls: Ignoring centering on short legs; using handheld targets that move; failing to re-check centering after tightening screws. Final Answer: All of the above

Question 12

Angular Observations – Why Keep the Line of Sight High Above Ground? In theodolite work, the line of sight is kept as high above the ground surface as practicable to reduce errors in observed angles primarily due to which effects?

Accepted Answer

Introduction / Context: Atmospheric effects near the ground can degrade angular observations. Temperature gradients and air turbulence are most severe within the lowest metre above the surface, causing apparent target movement (shimmering) and bending of the light path (refraction). Raising the line of sight mitigates these errors, improving precision in angle reading and pointing. Given Data / Assumptions: Observations made over sun-heated ground or roads are prone to turbulence. Targets may be at low height unless elevated on tripods or poles. Theodolite and staff/target heights can be adjusted in the field. Concept / Approach: Shimmering arises from turbulent cells with varying refractive indices, making the image dance and blurring the cross-hair coincidence. Horizontal refraction results from lateral gradients in refractive index close to the ground, displacing the apparent horizontal direction. Elevating the line of sight moves the observation path into more stable air with weaker gradients, reducing both effects simultaneously. Vertical refraction is a significant concern in vertical angle/levelling over long sights, but the question emphasizes minimization by higher line of sight for angle observations, especially horizontal components. Step-by-Step Solution: Raise instrument height and target heights where possible.Ensure sight lines clear heated surfaces, exhausts, and pavements.Observe during cooler periods to further reduce turbulence.Repeat pointings and average to lessen residual random effects. Verification / Alternative check: Field experience and optical surveying texts consistently advise higher sight lines to minimize scintillation and near-ground refraction on warm days. Why Other Options Are Wrong: Focusing on only one effect ignores the compound benefit; instrument collimation error is an instrument adjustment issue not solved by raising the line of sight. Common Pitfalls: Observing at low target heights over hot surfaces; attributing all image motion to instrument faults rather than atmospheric causes. Final Answer: Both shimmering and horizontal refraction

Question 13

Measuring Along Upgrade – Forward Shift Needed with a 20 m Chain on Slope θ While measuring distance uphill with a 20 m chain, the forward (upper) end of the chain must be shifted ahead by what amount so that each stepped position represents a true …

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Introduction / Context: On sloping ground, stepping with a chain is often used to measure horizontal distances. If the chain is laid along the slope, a geometric adjustment is required so that the horizontal projection equals the nominal chain length. This adjustment is implemented by advancing the forward end by a specific shift each step. Given Data / Assumptions: Nominal chain length L = 20 m. Slope angle = θ (upgrade). Goal: ensure each position corresponds to a 20 m horizontal distance. Concept / Approach: If the chain lies along the slope, its horizontal projection equals L cos θ. To make the horizontal projection equal exactly 20 m, the actual sloping length between pegs must be 20 sec θ. Therefore, the forward end must be advanced beyond 20 m by the amount Δ = 20 sec θ − 20 = 20 (sec θ − 1). This shift ensures each step measures 20 m horizontally despite the slope. Step-by-Step Solution: Required sloping length per step S = horizontal / cos θ = 20 / cos θ = 20 sec θ.Nominal chain length along the ground = 20 m.Additional forward shift Δ = S − 20 = 20 (sec θ − 1).Thus, move the forward end by Δ each step to achieve true 20 m horizontal intervals. Verification / Alternative check: For small θ, sec θ ≈ 1 + θ^2/2 ⇒ Δ ≈ 10 θ^2 (metres), a small positive value, matching practical expectations for gentle slopes. Why Other Options Are Wrong: 20 (cos θ − 1) is negative; 20 (sin θ − 1) and 20 (cosec θ − 1) are unrelated to horizontal projection; 20 tan θ is not the necessary shift derived from the projection requirement. Common Pitfalls: Forgetting that we must increase the ground step length to compensate for the slope; confusing the forward shift formula with slope reduction formulas used for converting slope distance to horizontal. Final Answer: 20 (sec θ − 1)

Question 14

Bearings: Convert a whole-circle bearing (W.C.B.) of 180° to the reduced (quadrantal) bearing format. State the correct reduced bearing for this exact due-south direction.

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Introduction / Context: Bearing systems in surveying are commonly expressed in two ways: whole-circle bearings (0° to 360° measured clockwise from true north) and reduced or quadrantal bearings (measured within 0° to 90° referenced to the nearest cardinal direction, N or S, and directed toward E or W). Converting between these is routine when reducing field notes or combining data from instruments and maps. Given Data / Assumptions: Whole-circle bearing (W.C.B.) of the line is exactly 180°. Reduced (quadrantal) bearing is requested. Cardinal directions: N (0°/360°), E (90°), S (180°), W (270°). Concept / Approach: W.C.B. 180° points exactly to the south. In quadrantal form, directions exactly along a cardinal are written simply by the cardinal letter itself. Some texts also write S 0° E or S 0° W for the limiting case, both meaning the same due-south line. However, the most compact and standard reduced form for a pure cardinal line is just the cardinal letter, here S. Step-by-Step Solution: Identify the direction of 180° → due south.Express in reduced bearing: the line coincides with the south cardinal.Adopt the concise reduced form: S. Verification / Alternative check: Converting back: reduced S corresponds uniquely to W.C.B. 180°. This matches map and instrument conventions for cardinal bearings. Why Other Options Are Wrong: S 0° E and S 0° W are verbose forms sometimes seen but the canonical reduced form for a cardinal line is S. N is incorrect; “None” is unnecessary because S is valid. Common Pitfalls: Writing an arbitrary small angle with the cardinal when the value is exactly 180°; confusing W.C.B. with reduced bearing notation. Final Answer: S

Question 15

Chain surveying practice: In indirect (reciprocal) ranging between two end stations that are not intervisible, how many ranging rods are required as a minimum to carry out the process effectively?

Accepted Answer

Introduction / Context: Ranging ensures that intermediate points lie on a straight line between two end stations. When the end stations are not intervisible due to intervening ground, indirect (reciprocal) ranging is used to keep the line straight. Knowing the minimum equipment—especially the number of ranging rods—helps organize the field party efficiently. Given Data / Assumptions: Two fixed end points A and B define the line. Direct line of sight between A and B is blocked. Standard reciprocal ranging procedure is followed. Concept / Approach: Reciprocal ranging involves placing an intermediate rod so that it appears in line from each end alternately. Typically, one rod is held at A, another at B, and a third is moved back and forth in between until both end observers agree that it is aligned. This iterative process usually requires at least three rods to maintain references at both ends while adjusting the intermediate point. Step-by-Step Solution: Place rods at A and B to act as fixed references.Move a third rod near the midpoint; observer at A signals alignment relative to B.Observer at B then refines alignment relative to A; repeat until centered on the straight line. Verification / Alternative check: Textbook procedures and field manuals consistently specify three rods as the practical minimum for reciprocal ranging; more rods can be used for longer lines to maintain intermediate reference points. Why Other Options Are Wrong: One or two rods cannot maintain simultaneous end references; four or five rods may help on long lines but exceed the minimum required. Common Pitfalls: Attempting reciprocal ranging with only two rods, leading to repeated re-sighting and loss of end references; poor signaling causing the intermediate rod to wander off line. Final Answer: 3

Question 16

Plane table orientation with conical hill stations: A and B are two inaccessible conical hill stations. Choose a point C in line with AB and orient the table at C by bringing line ab parallel to AB. Draw a back ray toward a ground point P (roughly s…

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Introduction / Context: Plane table orientation can be done by compass, by resection, or by back-ray. When using distant conical peaks and a single alignment (AB), care must be taken about the inherent 180° ambiguity: a straight line has no sense unless an additional reference fixes the direction. This question probes understanding of the multiplicity of valid orientations in such a setup. Given Data / Assumptions: A and B are inaccessible, conical (distinct) hill stations. Point C lies on AB and is used to set ab ∥ AB. From C a back ray is drawn toward a ground point P; at P the table is oriented by back ray to CP. Map scale and plotting are otherwise error-free. Concept / Approach: Bringing ab parallel to AB at C establishes orientation with respect to the line AB but does not by itself fix the sense along AB; rotating the board by 180° maintains ab ∥ AB. When the board is then oriented at P by the back-ray method with only that single line constraint, two orientations separated by 180° can satisfy the condition equally well. Both orientations are geometrically correct with respect to AB, i.e., manifold (non-unique), though additional resections or known points would remove the ambiguity. Step-by-Step Solution: At C: enforce ab ∥ AB → two possible board orientations (0° and 180°).Draw back ray toward P; at P use back ray to align to CP.Both 0° and 180° orientations can satisfy the back-ray condition → multiplicity persists. Verification / Alternative check: Including a second non-collinear resection line (e.g., to another identifiable point) would yield a unique orientation, confirming that the single-line constraint allows two solutions. Why Other Options Are Wrong: “Unique and correct” ignores the 180° ambiguity; “incorrect” and “not reliable” misstate the geometry; “unique but approximate” is not applicable here. Common Pitfalls: Forgetting the sense ambiguity of parallel alignment; failing to add a second ray or a compass check to fix unique orientation. Final Answer: Manifold and correct (two possible orientations 180° apart)

Question 17

Circular-curve geometry: For a simple circular curve of radius R with deflection (intersection) angle Δ, what is the ratio of the long chord length to the tangent length? Give your answer as a function of Δ.

Accepted Answer

Introduction / Context: Highway and railway curve computations repeatedly use relations among the long chord, tangent length, and central angle. Recognizing compact ratios simplifies checking field calculations and designing peg intervals for setting out curves with deflection methods. Given Data / Assumptions: Simple circular curve of radius R. Total central (deflection) angle is Δ. Long chord T1–T2 spans the curve; tangents from the point of intersection are of equal length. Concept / Approach: Standard formulas for a simple circular curve are: long chord = 2 * R * sin(Δ/2) and tangent length = R * tan(Δ/2). The desired ratio is therefore (2 R sin(Δ/2)) / (R tan(Δ/2)) = 2 * sin(Δ/2) / tan(Δ/2) = 2 * cos(Δ/2). This tidy identity is useful for quick checks without substituting R explicitly. Step-by-Step Solution: Write long chord Lc = 2 * R * sin(Δ/2).Write tangent length T = R * tan(Δ/2).Form ratio Lc/T = (2R sin(Δ/2)) / (R tan(Δ/2)) = 2 cos(Δ/2). Verification / Alternative check: Check limiting cases: Δ → 0 ⇒ cos(Δ/2) → 1, so Lc/T → 2, consistent with near-straight geometry where two tangents almost coincide. For Δ = 180°, cos(90°) = 0, ratio vanishes because the tangent length tends to infinity while long chord is finite. Why Other Options Are Wrong: sin(Δ/2), cos(Δ/2), tan(Δ/2), and 2 sin(Δ/2) do not match the derived ratio; only 2 cos(Δ/2) is correct. Common Pitfalls: Mixing the long chord with a sub-chord; confusing tangent length with external distance or length of curve. Final Answer: 2 cos(Δ/2)

Question 18

Earth’s magnetism: During secular variation of the magnetic meridian, how do the oscillation characteristics (range and period) behave at different places on Earth?

Accepted Answer

Introduction / Context: Magnetic declination at a given location changes slowly over years and decades, a phenomenon known as secular variation. Surveyors must understand these long-term trends to interpret historical bearings, update maps, and convert between magnetic and true directions reliably. Given Data / Assumptions: Secular variation arises from changes in Earth’s core-generated geomagnetic field. Both magnitude (range) and timing (period) of oscillations may differ regionally. Short-term disturbances (diurnal, storms) are not the subject here. Concept / Approach: Secular variation is not globally uniform. Observatories record that the amplitude (range) of declination change and the time between successive extrema (period) vary with latitude, longitude, and epoch. Consequently, correction tables and isogonic charts are published with regional specificity, and survey data must consider the appropriate local curve of variation rather than a single global constant. Step-by-Step Solution: Recognize secular variation as a slow drift modulated by core dynamics.Identify that both range and period differ geographically and temporally.Conclude that neither is constant worldwide: both vary. Verification / Alternative check: Magnetic observatory records (e.g., IGRF/WMM models) show location-dependent drift rates and varying quasi-periodic behaviors, supporting the conclusion. Why Other Options Are Wrong: Any statement claiming constancy of range or period contradicts empirical records; asserting only one varies is incomplete. Common Pitfalls: Confusing secular variation with diurnal variation; assuming a single fixed annual correction applies everywhere. Final Answer: Range and period of oscillation both vary

Question 19

Errors and mistakes in surveying: Mistakes that can produce a very serious effect on final results most commonly arise due to which of the following?

Accepted Answer

Introduction / Context: Measurement quality depends not only on instrument precision and method but also on human factors. Random errors can be mitigated statistically, but mistakes (blunders) can severely corrupt results if not detected and corrected through good practice and redundancy. Given Data / Assumptions: Mistakes are preventable human errors, not random noise. They include misreadings, transpositions, misidentification of points, and incorrect instrument setup. We aim to identify their usual origins. Concept / Approach: Mistakes typically stem from human factors such as inattention (distraction, fatigue), inexperience (insufficient training to recognize pitfalls), and carelessness (haste, poor checking). Robust field procedures, check measurements, and standardized note-keeping are critical defenses. Therefore, all listed causes contribute materially to serious mistakes in surveying. Step-by-Step Solution: List common blunders: wrong staff reading, incorrect station identification, mis-set verniers, wrong sign conventions.Map causes to blunders: inattention, inexperience, and carelessness can each produce these errors.Conclude that all are contributors → choose “All of these”. Verification / Alternative check: Field QA/QC manuals emphasize training, rest, double-entry checks, and independent checks specifically to combat these human-factor sources. Why Other Options Are Wrong: Picking any single cause ignores the multifactorial nature of mistakes; “None” is plainly incorrect. Common Pitfalls: Confusing mistakes with random errors; assuming instrument accuracy alone prevents blunders. Final Answer: All of these

Question 20

Instrument testing: The method of reversal is used for what purpose in surveying instrument adjustments and checks?

Accepted Answer

Introduction / Context: Accurate surveying depends on specific geometric relationships between instrument axes (line of sight, horizontal axis, vertical axis). The method of reversal is a classical test technique to reveal misalignments and verify perpendicularity or parallelism by observing symmetry under reversed configurations. Given Data / Assumptions: Reversal means interchanging positions (e.g., face left/face right, turning telescope through 180°, swapping ends). Correct geometry yields invariant readings under reversal. Any difference between the pair of readings indicates an error relationship. Concept / Approach: By taking observations in two opposite configurations, systematic geometric errors change sign while true values remain the same. Averaging or comparing the two reveals whether a component is truly parallel/perpendicular to another and makes any erroneous relationship evident. The method is widely used for collimation checks, trunnion axis tests, and bubble line alignment, among others. Hence both statements (a) and (b) accurately describe its purpose and effect. Step-by-Step Solution: Observe a target in face left and face right; note readings.Compare and compute half-difference → error estimate; half-sum → true value.Use result to diagnose parallel/perpendicular relations or apply adjustments. Verification / Alternative check: Instrument adjustment procedures in manuals employ reversal for line-of-sight error, index error, and bubble axis alignment, confirming its generality. Why Other Options Are Wrong: Restricting to a single error type (e) is too narrow; choosing only (a) or only (b) omits part of the method’s function; “neither” contradicts standard practice. Common Pitfalls: Failing to center or level consistently between reversals; not recording both configurations accurately, which obscures diagnostics. Final Answer: Both (a) and (b)

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