Q: Topographic interpretation — closed contour lines whose elevations decrease towards the centre indicate which landform?

Introduction / Context: Reading contour maps is foundational for site selection, alignment design, drainage planning, and hazard assessment. The pattern and sequencing of contour values convey landform types such as hills, valleys, depressions, and saddles. This question tests recognition of a classic pattern: decreasing contour values toward a closed centre. Given Data / Assumptions: Contours are closed and labelled with elevation values. There are no form lines or supplementary features overriding interpretation. Standard cartographic conventions are used. Concept / Approach: Closed contours where elevation values decrease towards the centre represent a topographic depression (basin, sink). The opposite pattern—values increasing towards the centre—marks a hill. A saddle shows two opposing highs connected by a low pass, producing hourglass-shaped contours. A river bed is indicated by V-shaped contours pointing upstream, not by concentric decreases unless the river occupies a closed basin. Step-by-Step Solution: 1) Trace the innermost contour and read its value.2) Compare with surrounding contours; observe decreasing values inward.3) Conclude the centre is the lowest point locally → a depression.4) Optionally look for hachures (tick marks), used in some maps to emphasize depressions. Verification / Alternative check: Standard mapping references adopt this interpretation; many maps add short ticks on depression contours pointing downslope to avoid ambiguity. Why Other Options Are Wrong: Hill — requires increasing values inward. Saddle — needs two ridges separated by a pass, not nested decreasing loops. River bed — generally open V shapes pointing upstream, not closed loops. Common Pitfalls: Misreading the sign of elevation change; overlooking closed contour hachures; confusing depression with crater versus sinkhole—both are depressions by contour logic. Final Answer: a depression

Q: Hydrographic surveying — hydrographic surveys primarily deal with the mapping and charting of which domain?

Introduction / Context: Hydrographic surveys provide depth, seabed characteristics, shoreline features, and hazards necessary for navigation, coastal engineering, dredging, and resource management. Unlike terrestrial topographic surveys, they are conducted over water and require specialized methods and instruments to determine underwater topography (bathymetry). Given Data / Assumptions: Focus is on oceans, seas, lakes, reservoirs, estuaries, and navigable rivers. Survey outcomes include bathymetric charts, nautical charts, and coastal maps. Measurements include depths (soundings), tides, currents, and seabed sediment types. Concept / Approach: Hydrography is the science of surveying and charting large water bodies. It uses echo sounders, sonar, GPS/RTK positioning, tide gauges, and current meters to produce accurate bathymetric datasets. Terrestrial tools (levels, total stations) are used only for coastal tie-ins. Products support safe navigation and engineering works such as breakwaters, ports, and dredging projects. Step-by-Step Solution: 1) Identify the environment: over-water domains (oceans, lakes, rivers).2) Identify measurements: depth and seabed properties, along with tidal and current data.3) Recognize outputs: charts and maps for navigation and engineering.4) Conclude that hydrographic surveys map large water bodies. Verification / Alternative check: International standards (e.g., IHO S-44) define hydrographic survey requirements for nautical charting—explicitly focused on water bodies. Why Other Options Are Wrong: Heavenly bodies — astronomy, not surveying. Mountainous region — terrestrial topography. Canal system — a subset of water infrastructure; hydrography broadly covers larger natural water bodies. Movement of clouds — meteorology. Common Pitfalls: Confusing hydrography with hydrology (scientific study of water in the environment); hydrography is about mapping and charting. Final Answer: large water bodies

Q: In quadrantal bearing system, back bearing of a line may be obtained from its forward bearing, by

changing the cardinal points, i.e. substituting N for S and E for W and vice-versa

Q: In magnetism and surveying, the vertical angle made by the longitudinal axis of a freely suspended magnetic needle with the horizontal at its pivot is called what? Select the correct geomagnetic term.

Introduction / Context: Geomagnetic terminology is foundational for surveying and navigation. A magnetic needle aligns with the local magnetic field, which is generally not horizontal. The vertical angle from the horizontal is a key parameter measured with a dip circle or equivalent instrument. Given Data / Assumptions: The needle is free to swing in a vertical plane aligned with magnetic meridian. We are distinguishing among several related angular terms. Standard surveying definitions apply. Concept / Approach: “Dip” (magnetic inclination) is the vertical angle between the needle and the horizontal. “Declination” is the horizontal angle between true north and magnetic north. “Azimuth” is a horizontal direction angle from a reference meridian. “Bearing” is the acute horizontal angle from a reference line in a quadrant system. “Deviation” often refers to compass error due to local attraction. Thus, the only vertical-plane term in the list is dip. Step-by-Step Solution: Identify plane: vertical angle from horizontal → inclination.Match term: inclination = dip.Eliminate horizontal-plane terms (declination, azimuth, bearing) and local error (deviation). Verification / Alternative check: Survey texts define dip as the angle of the earth’s magnetic field with respect to the horizontal; instruments like the dip circle are used to measure it. Why Other Options Are Wrong: Declination/azimuth/bearing are horizontal angles; deviation is not the standard name for the vertical angle. Common Pitfalls: Confusing declination (horizontal) with dip (vertical); mixing “inclination” and “dip”—they are synonyms here. Final Answer: Dip

Q: Spherical excess in geodetic surveying – angular sum difference on the Earth’s surface The sum of the interior angles of a polygon laid on the Earth’s surface exceeds that of the corresponding plane figure by about 1 second for approximately how muc…

Introduction / Context: On a sphere, the sum of the interior angles of a polygon is greater than on a plane. This excess—called spherical excess—becomes relevant in geodetic surveying, triangulation, and large-area mapping. The question asks for the approximate area on Earth that corresponds to a spherical excess of 1 second of arc in angle sum. Given Data / Assumptions: Spherical excess E (in radians) ≈ Area / R^2 for small regions. R is Earth’s mean radius, about 6370 km. We want E expressed as 1 second of arc (1'), and the corresponding area in sq. km. Concept / Approach: Convert 1 second to radians, then solve Area = E * R^2. One second = 1/206265 of a radian. Using R ≈ 6370 km gives R^2 ≈ 4.06 * 10^7 km^2. Multiplying yields Area ≈ (1/206265) * 4.06 * 10^7 ≈ 197 sq. km, commonly rounded to about 200 sq. km in surveying handbooks. Thus, a spherical polygon of roughly 200 sq. km has an angle-sum excess of about 1' compared to the plane case. Step-by-Step Solution: 1) E = 1' = 1/206265 rad.2) R ≈ 6370 km → R^2 ≈ 4.06 * 10^7 km^2.3) Area = E * R^2 ≈ (1/206265) * 4.06 * 10^7 ≈ 197 km^2.4) Rounded engineering value → ≈ 200 sq. km. Verification / Alternative check: Many surveying texts quote 1' ≈ 197–200 sq. km for Earth. Using more precise R marginally changes the value but not the chosen option. Why Other Options Are Wrong: 100 or 150 sq. km: Underestimates the area corresponding to 1' of spherical excess. None of these: Incorrect because ≈ 200 sq. km is a standard approximation. Common Pitfalls: Confusing seconds with degrees; using diameter instead of radius; forgetting to express E in radians before multiplying by R^2. Final Answer: 200 sq. km

Q: Optical square – fixed angle between the two mirrors In an optical square used for setting out right angles, the acute angle between the two mirrors is equal to which value?

Introduction / Context: An optical square is a simple reflective instrument widely used in chain surveying to set out perpendiculars (right angles). It consists of two mirrors oriented at a specific acute angle so that due to reflection geometry, the user can align one sight along the chain line and set a perpendicular to it. This question asks for the fixed angle between the mirrors that yields a 90° observation. Given Data / Assumptions: The device uses the law of reflection: angle of incidence equals angle of reflection. Two-mirror reflection rotates a ray by twice the angle between the mirrors. We desire a right-angle deflection between the two observed directions. Concept / Approach: If the angle between mirrors is α, a ray incident on the first mirror and then reflected by the second is deflected by 2α. To set out a right angle (90°), we require 2α = 90°, hence α = 45°. Optical squares are therefore manufactured with mirrors inclined at 45°, producing a perceived perpendicular when the reflected and direct images coincide correctly. Step-by-Step Solution: 1) Desired deflection between chain line and offset = 90°.2) Two reflections → net deflection = 2α.3) Set 2α = 90° → α = 45°.4) Therefore mirror angle = 45°. Verification / Alternative check: Instrument descriptions of optical squares and prism squares align with this 45° mirror setting to achieve perpendicular offsets without a theodolite. Why Other Options Are Wrong: 20°, 30°, 60°: Would produce 40°, 60°, or 120° deflections, not a right angle. 90°: Would lead to 180° deflection, not usable for perpendicular offsets. Common Pitfalls: Confusing the acute mirror angle with the desired right angle; mixing prism square (which uses refraction) with the mirror-based optical square. Final Answer: 45°

Question 1

Survey accessories — identify the correct uses: box sextant, cross-staff, gradiometer, and line ranger in routine field practice.

Accepted Answer

Introduction / Context: Efficient field surveying relies on simple yet specialized instruments that speed up routine tasks. Knowing which accessory to use for a specific operation prevents mistakes and improves productivity. This question checks recognition of the standard functions of four common tools: box sextant, cross-staff, gradiometer, and line ranger. Given Data / Assumptions: Field tasks include setting right angles, measuring small angles, establishing gradients, and lining in intermediate points. Only the primary/common use of each device is considered. Instrumentation is assumed to be in standard condition. Concept / Approach: Box sextant: a compact optical instrument for measuring relatively small horizontal (and sometimes vertical) angles by sextant principle. Cross-staff: a simple device with perpendicular slots/vanes used to set out right angles to a chain line (for offsets). Gradiometer: used to set and check required gradients/slopes rapidly on site (e.g., drains). Line ranger: an optical device that helps position intermediate points directly on a straight survey line by visually aligning with end stations. All statements listed are correct uses, so the comprehensive answer is ‘‘All the above.’’ Step-by-Step Solution: 1) Match each instrument with its core task in field operations.2) Confirm that no statement contradicts standard practice.3) Recognize that the set collectively describes correct applications.4) Choose ‘‘All the above’’ accordingly. Verification / Alternative check: Training manuals and practical surveying guides specify these exact functions for the respective tools, confirming correctness. Why Other Options Are Wrong: If a single statement were chosen, it would omit other equally true statements; the question explicitly groups multiple correct uses. Common Pitfalls: Confusing line rangers with ranging rods; using a cross-staff where higher angular precision (e.g., theodolite) is required. Final Answer: All the above

Question 2

Traverse bearings — ABCD is a parallelogram with ∠BAD = 60°. If the bearing of AB is 30°, what is the whole-circle bearing of side CD?

Accepted Answer

Introduction / Context: Parallelogram geometry implies that opposite sides are parallel and equal. In a traverse, bearings of parallel lines are either identical or differ by 180°, depending on direction of travel along the side. This problem asks for the bearing of CD given the bearing of AB in a regular parallelogram ABCD with a known interior angle at A. Given Data / Assumptions: Parallelogram ABCD in order A→B→C→D. Bearing of AB = 30° (whole-circle). Opposite sides AB ∥ CD and AD ∥ BC. Concept / Approach: Because AB and CD are parallel, their direction lines are the same. However, when traversing from C to D, CD is oriented opposite to AB from A to B. Therefore, the bearing of CD is the back bearing of AB, i.e., AB + 180° (mod 360°). The given interior angle ∠BAD = 60° confirms a plausible acute angle between AB and AD but does not alter the parallelism conclusion for AB and CD. Step-by-Step Solution: 1) Note AB ∥ CD ⇒ direction lines align.2) Determine traversal direction: along CD from C to D is opposite to A to B.3) Compute CD bearing = 30° + 180° = 210°.4) Report whole-circle bearing 210°. Verification / Alternative check: Sketch a simple rectangle (special case of a parallelogram) to visualize that the top side bearing is 180° opposite to the bottom side when traversed in the polygonal order. Why Other Options Are Wrong: 90°/270° — correspond to east–west bearings unrelated to the given 30° base. 120° — would be a direction parallel to AD/BC rather than AB/CD. Common Pitfalls: Confusing forward and back bearings; assuming opposite sides always have identical numeric bearings without accounting for direction of travel. Final Answer: 210°

Question 3

Stadia staff handling — how does the observed staff intercept change with staff tilt relative to the line of sight in tacheometry?

Accepted Answer

Introduction / Context: In tacheometric observations, the horizontal distance is computed from the staff intercept (difference between upper and lower stadia-hair readings). The value of this intercept depends not only on instrument constants but also on how the staff is oriented with respect to the line of sight. This question addresses the effect of staff tilt on the measured intercept. Given Data / Assumptions: Fixed-hair stadia method is used. The line of sight makes an angle with the staff according to the staff's tilt. No staff-head prism or automatic normalizing device is assumed. Concept / Approach: The stadia intercept is proportional to the projection of the actual staff length between the stadia hairs onto a line perpendicular to the line of sight. When the staff is held truly normal (perpendicular) to the line of sight, this projection is maximum, yielding the largest intercept reading. If the staff is tilted away from the normal, the effective projected length reduces by the cosine of the deviation angle, thereby decreasing the intercept and leading to underestimated distances if uncorrected. Step-by-Step Solution: 1) Consider a small staff segment subtended by the stadia hairs.2) The observed intercept equals true length × cos(θ), where θ is deviation from normal.3) At θ = 0° (staff normal), cos θ = 1 → maximum intercept.4) As |θ| increases, cos θ < 1 → intercept decreases. Verification / Alternative check: Field practice requires the staffman to rock the staff through the normal position and freeze at the point of maximum reading; this ensures correct intercept capture. Why Other Options Are Wrong: Minimum at normal — opposite of the projection principle. Decreases if tilted away — true but incomplete; the key maximum condition is at true normal. Increases if tilted towards normal — only until normal is reached; beyond that it decreases again. Common Pitfalls: Accepting a reading without rocking the staff; ignoring that small tilt can introduce systematic underestimation of distance. Final Answer: is maximum if the staff is held truly normal to the line of sight.

Question 4

Topographic interpretation — closed contour lines whose elevations decrease towards the centre indicate which landform?

Accepted Answer

Introduction / Context: Reading contour maps is foundational for site selection, alignment design, drainage planning, and hazard assessment. The pattern and sequencing of contour values convey landform types such as hills, valleys, depressions, and saddles. This question tests recognition of a classic pattern: decreasing contour values toward a closed centre. Given Data / Assumptions: Contours are closed and labelled with elevation values. There are no form lines or supplementary features overriding interpretation. Standard cartographic conventions are used. Concept / Approach: Closed contours where elevation values decrease towards the centre represent a topographic depression (basin, sink). The opposite pattern—values increasing towards the centre—marks a hill. A saddle shows two opposing highs connected by a low pass, producing hourglass-shaped contours. A river bed is indicated by V-shaped contours pointing upstream, not by concentric decreases unless the river occupies a closed basin. Step-by-Step Solution: 1) Trace the innermost contour and read its value.2) Compare with surrounding contours; observe decreasing values inward.3) Conclude the centre is the lowest point locally → a depression.4) Optionally look for hachures (tick marks), used in some maps to emphasize depressions. Verification / Alternative check: Standard mapping references adopt this interpretation; many maps add short ticks on depression contours pointing downslope to avoid ambiguity. Why Other Options Are Wrong: Hill — requires increasing values inward. Saddle — needs two ridges separated by a pass, not nested decreasing loops. River bed — generally open V shapes pointing upstream, not closed loops. Common Pitfalls: Misreading the sign of elevation change; overlooking closed contour hachures; confusing depression with crater versus sinkhole—both are depressions by contour logic. Final Answer: a depression

Question 5

Hydrographic surveying — hydrographic surveys primarily deal with the mapping and charting of which domain?

Accepted Answer

Introduction / Context: Hydrographic surveys provide depth, seabed characteristics, shoreline features, and hazards necessary for navigation, coastal engineering, dredging, and resource management. Unlike terrestrial topographic surveys, they are conducted over water and require specialized methods and instruments to determine underwater topography (bathymetry). Given Data / Assumptions: Focus is on oceans, seas, lakes, reservoirs, estuaries, and navigable rivers. Survey outcomes include bathymetric charts, nautical charts, and coastal maps. Measurements include depths (soundings), tides, currents, and seabed sediment types. Concept / Approach: Hydrography is the science of surveying and charting large water bodies. It uses echo sounders, sonar, GPS/RTK positioning, tide gauges, and current meters to produce accurate bathymetric datasets. Terrestrial tools (levels, total stations) are used only for coastal tie-ins. Products support safe navigation and engineering works such as breakwaters, ports, and dredging projects. Step-by-Step Solution: 1) Identify the environment: over-water domains (oceans, lakes, rivers).2) Identify measurements: depth and seabed properties, along with tidal and current data.3) Recognize outputs: charts and maps for navigation and engineering.4) Conclude that hydrographic surveys map large water bodies. Verification / Alternative check: International standards (e.g., IHO S-44) define hydrographic survey requirements for nautical charting—explicitly focused on water bodies. Why Other Options Are Wrong: Heavenly bodies — astronomy, not surveying. Mountainous region — terrestrial topography. Canal system — a subset of water infrastructure; hydrography broadly covers larger natural water bodies. Movement of clouds — meteorology. Common Pitfalls: Confusing hydrography with hydrology (scientific study of water in the environment); hydrography is about mapping and charting. Final Answer: large water bodies

Question 6

Levelling with inverted and upright sights — when one staff reading is taken below the plane of collimation and another above it, the difference of level equals the sum of the two readings; what fixed staff error is introduced in the result?

Accepted Answer

Introduction / Context: In precise levelling, staff readings must reference the same physical zero. If the staff's zero graduation does not coincide exactly with the foot (shoe) of the staff—i.e., if there is a fixed offset a—certain combinations of readings can accumulate this error into the computed difference of level (D.L.). This question concerns the case of one reading below and one above the line of collimation, where D.L. is obtained by adding the two readings. Given Data / Assumptions: Let the fixed offset between the staff foot and the zero graduation be a. One sight is an inverted (or ceiling) reading; the other is a normal upright reading. The computed difference of level equals the sum of the two scale readings. Concept / Approach: If each individual staff reading includes the constant zero offset a (because the true zero is a units away from the physical foot), then when adding two readings, the error a is included twice. Consequently, the D.L. is biased by +2a. Recognizing this helps surveyors avoid mixed-sight techniques with miscalibrated staffs or to apply the proper correction when unavoidable. Step-by-Step Solution: 1) Let reading R₁ = true value + a (below collimation).2) Let reading R₂ = true value + a (above collimation).3) D.L. = R₁ + R₂ = (true sum) + 2a.4) Hence the error in the D.L. equals 2a (twice the zero offset). Verification / Alternative check: Staff calibration checks (comparing to a standard) quantify a; the correction of −2a removes the bias in the add-case scenario. Why Other Options Are Wrong: Single offset a — underestimates the accumulated bias when two readings are added. Thrice the distance — no basis for triple accumulation in this configuration. None — incorrect because a fixed bias clearly exists. Common Pitfalls: Ignoring fixed staff errors; mixing inverted and upright sights without awareness of how the zero offset propagates into the D.L. Final Answer: twice the distance between the zero of graduation and the foot of the staff

Question 7

Chain surveying scope — in traditional chain (tape) surveying, field work is primarily limited to which type(s) of measurement?

Accepted Answer

Introduction / Context: Chain (or tape) surveying is the simplest form of surveying used for relatively small, level areas with uncomplicated details. Its economy and speed come from restricting field operations to direct distance measurements along straight lines and short offsets. This question asks what types of measurements chain surveying is limited to in the field. Given Data / Assumptions: No theodolite or total station is used; angular work is not fundamental. Perpendiculars are set out with simple devices (cross-staff/optical square) but are treated as linear offsets from a chain line. Objective is to plot detail by linear ties and offsets to a framework of chain lines. Concept / Approach: Chain surveying relies on linear measurements only: chaining along baselines, tie lines, and check lines; taking offsets to locate features. Although devices like optical squares help set out right angles, the method does not involve measuring angles for computation in the way compass traverses or theodolite surveys do. Accuracy control is achieved by good layout (well-conditioned triangles/rectangles), check measurements, and adherence to limiting offset lengths for the selected plotting scale. Step-by-Step Solution: 1) Establish a network of chain lines (baseline, ties, checks).2) Measure distances with chain/tape; record offsets to features.3) Avoid angular computations; perpendiculars are constructed, not measured as angles.4) Plot directly to scale using linear data. Verification / Alternative check: Surveying texts define chain surveying explicitly as a method involving only linear measurements, differentiating it from compass/theodolite methods which are angular. Why Other Options Are Wrong: Angular only — contradicts the essence of chain surveying. Both linear and angular — would describe traverses or plane-table with measured angles. All the above — logically impossible. Common Pitfalls: Over-reliance on long offsets beyond the limiting length for the chosen scale; poor chain-line layout which reduces overall accuracy. Final Answer: linear measurements only

Question 8

Diopter is the power of a lens having a focal length of

Accepted Answer

100 cm

Question 9

In quadrantal bearing system, back bearing of a line may be obtained from its forward bearing, by

Accepted Answer

changing the cardinal points, i.e. substituting N for S and E for W and vice-versa

Question 10

If the radius of a simple curve is R, the length of the chord for calculating offsets by the method of chords produced, should not exceed.

Accepted Answer

R/20

Question 11

In magnetism and surveying, the vertical angle made by the longitudinal axis of a freely suspended magnetic needle with the horizontal at its pivot is called what? Select the correct geomagnetic term.

Accepted Answer

Introduction / Context: Geomagnetic terminology is foundational for surveying and navigation. A magnetic needle aligns with the local magnetic field, which is generally not horizontal. The vertical angle from the horizontal is a key parameter measured with a dip circle or equivalent instrument. Given Data / Assumptions: The needle is free to swing in a vertical plane aligned with magnetic meridian. We are distinguishing among several related angular terms. Standard surveying definitions apply. Concept / Approach: “Dip” (magnetic inclination) is the vertical angle between the needle and the horizontal. “Declination” is the horizontal angle between true north and magnetic north. “Azimuth” is a horizontal direction angle from a reference meridian. “Bearing” is the acute horizontal angle from a reference line in a quadrant system. “Deviation” often refers to compass error due to local attraction. Thus, the only vertical-plane term in the list is dip. Step-by-Step Solution: Identify plane: vertical angle from horizontal → inclination.Match term: inclination = dip.Eliminate horizontal-plane terms (declination, azimuth, bearing) and local error (deviation). Verification / Alternative check: Survey texts define dip as the angle of the earth’s magnetic field with respect to the horizontal; instruments like the dip circle are used to measure it. Why Other Options Are Wrong: Declination/azimuth/bearing are horizontal angles; deviation is not the standard name for the vertical angle. Common Pitfalls: Confusing declination (horizontal) with dip (vertical); mixing “inclination” and “dip”—they are synonyms here. Final Answer: Dip

Question 12

Highway and railway engineering – choosing the ideal transition curve In route geometric design, the ideal transition curve that provides a linear rate of change of curvature (from a tangent to a circular curve) is which of the following?

Accepted Answer

Introduction / Context: In highway and railway geometric design, a transition curve is interposed between a straight (tangent) and a circular curve so that the curvature increases gradually. The goal is to provide comfort, safety, and structural relief by avoiding sudden lateral acceleration (centrifugal force) and abrupt changes in superelevation and steering. This question asks you to identify the ideal transition curve used because of its linear curvature property. Given Data / Assumptions: A transition must change curvature smoothly from 0 at the tangent to 1/R at the circular arc. Vehicle dynamics demand a controlled rate of change of centrifugal acceleration (often termed jerk). Practical construction also needs a curve that is easy to set out with standard methods. Concept / Approach: The clothoid (also called an Euler spiral or Cornu spiral) has the defining property that curvature is directly proportional to its arc length from the beginning of the curve. This provides a constant rate of change of radial acceleration with respect to time for nearly uniform speed, which is highly desirable for passenger comfort and track/roadway performance. Because of this linear relationship, superelevation runoff and transition length calculations integrate neatly with the clothoid form. Step-by-Step Solution: 1) Requirement: curvature k should vary from 0 to 1/R smoothly.2) Ideal criterion: k ∝ s (arc length), ensuring constant jerk at constant speed.3) Curve satisfying k = a * s: the clothoid spiral.4) Therefore, among listed options, the ideal transition curve is the clothoid spiral. Verification / Alternative check: Design manuals for highways and railways standardize the clothoid for transition curves because it simplifies setting out, transitions superelevation smoothly, and minimizes discomfort due to jerk. While other spirals exist, none meets the ideal linear-curvature condition as directly and conveniently as the clothoid in routine practice. Why Other Options Are Wrong: Cubic parabola / Cubic spiral: Usable approximations in some contexts but do not perfectly satisfy k ∝ s over the entire length. True spiral: Generic term; without specifying the curvature law, it lacks the ideal k–s property that defines the clothoid. Common Pitfalls: Confusing any spiral with the ideal transition; assuming a cubic curve is automatically optimal; neglecting jerk criteria in comfort analysis. Final Answer: Clothoid spiral

Question 13

Levelling operations – coordinating instrument and staff movement During standard differential levelling in fieldwork, which of the following statements are correct?

Accepted Answer

Introduction / Context: Differential levelling relies on precise backsight and foresight observations taken on stable points so that height differences are accurate. Coordination between the instrument operator and the staff holder is therefore critical. This question tests your understanding of best practice during instrument shifts and staff movements to avoid blunders in reduced levels. Given Data / Assumptions: A change point is a firm point used to transfer the line of collimation between setups. Backsight (BS) and foresight (FS) must refer to the same physical point during an instrument shift. Readings are invalid if either the staff or the instrument is moved at the wrong moment. Concept / Approach: The essence of levelling is that the BS and FS defining a change must be observed on a staff kept immovable until both shots are completed. Similarly, while a staff is walked forward, the instrument must remain stationary so that the reading corresponds to a fixed line of sight. Any breach introduces errors in the height of collimation or in the computed reduced level at subsequent stations. Step-by-Step Solution: 1) During an instrument shift, hold the staff fixed on the change point until both FS (old setup) and BS (new setup) are taken.2) While carrying the staff to the next point, do not move the instrument; otherwise the line of sight would change between observations.3) Therefore, both statements (a) and (b) reflect correct levelling practice. Verification / Alternative check: Standard levelling procedures emphasize stability at the change point and prohibit simultaneous movements of instrument and staff. Field check notes and arithmetic (summing BS and FS) corroborate correctness when these rules are followed. Why Other Options Are Wrong: Neither (a) nor (b): Conflicts with fundamental practice and would cause inconsistencies. Common Pitfalls: Moving the staff prematurely; failing to call out “staff fixed” before shifting; forgetting to re-center and re-level the instrument before taking the new BS. Final Answer: Both (a) and (b)

Question 14

Plane table and drafting tools – function of a pantograph In mapping and drafting practice, a pantograph is primarily used for which operation on maps and plans?

Accepted Answer

Introduction / Context: Classical surveying and drafting employ mechanical aids to prepare drawings at different scales. Among them, the pantograph enables the accurate reproduction of a plan at an enlarged or reduced scale without redrawing each feature by hand. This question asks you to identify the chief function of a pantograph in the workflow of mapping and drafting. Given Data / Assumptions: A pantograph is a linked-arm mechanism with adjustable pivot points. It transfers motion proportionally from a tracing point to a drawing point. Scale ratio is controlled by the geometry of the linkages and fixed points. Concept / Approach: By fixing the tracer, pivot, and pencil points at specific holes in its arms, a pantograph creates a mechanical similarity transformation. Tracing over an original plan produces a copy whose lengths are multiplied by a constant factor (greater than 1 for enlargement, less than 1 for reduction). This is particularly handy for producing plan sets at different scales for presentation or site use. Step-by-Step Solution: 1) Choose the required scale, e.g., reduction to 1:2.2) Set the pantograph arm points to achieve that ratio.3) Trace the original lines with the tracer; the pencil reproduces them at the new scale.4) Hence, the primary use is enlarging or reducing plans. Verification / Alternative check: Drafting manuals list the pantograph under copying and scaling devices, distinct from planimeters (for area measurement) or chains/tapes (for distance measurement). Why Other Options Are Wrong: Measuring distances/areas: Distances are measured with scales or tapes; areas with planimeters or counting methods, not a pantograph. Setting out right angles: That is done with optical squares, theodolites, or cross-staffs. Common Pitfalls: Confusing pantographs with planimeters; expecting direct area outputs rather than scaled copies; ignoring setup geometry that controls the scale factor. Final Answer: Enlarging or reducing existing plans to a different scale

Question 15

Spherical excess in geodetic surveying – angular sum difference on the Earth’s surface The sum of the interior angles of a polygon laid on the Earth’s surface exceeds that of the corresponding plane figure by about 1 second for approximately how muc…

Accepted Answer

Introduction / Context: On a sphere, the sum of the interior angles of a polygon is greater than on a plane. This excess—called spherical excess—becomes relevant in geodetic surveying, triangulation, and large-area mapping. The question asks for the approximate area on Earth that corresponds to a spherical excess of 1 second of arc in angle sum. Given Data / Assumptions: Spherical excess E (in radians) ≈ Area / R^2 for small regions. R is Earth’s mean radius, about 6370 km. We want E expressed as 1 second of arc (1'), and the corresponding area in sq. km. Concept / Approach: Convert 1 second to radians, then solve Area = E * R^2. One second = 1/206265 of a radian. Using R ≈ 6370 km gives R^2 ≈ 4.06 * 10^7 km^2. Multiplying yields Area ≈ (1/206265) * 4.06 * 10^7 ≈ 197 sq. km, commonly rounded to about 200 sq. km in surveying handbooks. Thus, a spherical polygon of roughly 200 sq. km has an angle-sum excess of about 1' compared to the plane case. Step-by-Step Solution: 1) E = 1' = 1/206265 rad.2) R ≈ 6370 km → R^2 ≈ 4.06 * 10^7 km^2.3) Area = E * R^2 ≈ (1/206265) * 4.06 * 10^7 ≈ 197 km^2.4) Rounded engineering value → ≈ 200 sq. km. Verification / Alternative check: Many surveying texts quote 1' ≈ 197–200 sq. km for Earth. Using more precise R marginally changes the value but not the chosen option. Why Other Options Are Wrong: 100 or 150 sq. km: Underestimates the area corresponding to 1' of spherical excess. None of these: Incorrect because ≈ 200 sq. km is a standard approximation. Common Pitfalls: Confusing seconds with degrees; using diameter instead of radius; forgetting to express E in radians before multiplying by R^2. Final Answer: 200 sq. km

Question 16

Theodolite theory – effect of centering error on observed angles Centering error (imperfect placement of the instrument over the station) produces what kind of error in the measured horizontal angles?

Accepted Answer

Introduction / Context: Accurate centering of a theodolite over a ground station is essential to minimize angular errors. If the instrument is off the true station point, rays sighted to targets subtend slightly different angles than intended. This question explores how such centering error propagates into the measured horizontal angles. Given Data / Assumptions: The instrument is displaced laterally by a small distance e from the true station. Sights are taken to targets at various bearings and distances. Small-angle approximations are valid for surveying-level precision analysis. Concept / Approach: The angular error due to centering depends on the geometry: the greater the sight length, the smaller the angular deviation caused by a fixed lateral shift at the instrument. Furthermore, the sign and magnitude depend on the pointing direction relative to the displacement vector. Hence, errors vary with azimuth and are inversely proportional to target distance, so distant targets reduce centering effects. Step-by-Step Solution: 1) Model station shift by vector e.2) For a sight of length L, the apparent direction shifts by roughly e/L in radians projected onto the sight direction.3) Therefore, angle errors differ for each pointing and scale approximately with 1/L.4) Conclusion: error varies with direction and inversely with sight length. Verification / Alternative check: Adjustment theory derives the same relationship, and field practice reduces centering sensitivity by taking long sights and balancing directions. Why Other Options Are Wrong: Equal in all angles / direction-independent: Contradicts geometry; effect clearly depends on pointing and distance. None of these: Incorrect because option (c) states the known relationship. Common Pitfalls: Ignoring that longer sights mitigate centering error; assuming centering only affects orientation but not angle magnitudes between any two rays. Final Answer: Error that varies with the direction of pointing and inversely with the length of sight

Question 17

Levelling instruments – bubble-tube sensitivity per division For a spirit-level bubble tube: if the length of one graduation (division) on the tube is x and R is the radius of curvature of the internal surface, what is the angular value φ (in second…

Accepted Answer

Introduction / Context: The sensitivity of a level is the angle through which the line of collimation tilts when the bubble moves by one division on the bubble tube. Designers and field engineers use this relation to evaluate instrument precision and to compute staff intercept versus bubble displacement during adjustments. This question asks for the angular value per division in seconds as a function of the division length and the tube’s radius of curvature. Given Data / Assumptions: Length of one division on the bubble tube arc = x (linear measure along the arc). Radius of curvature of the internal surface = R. Small-angle approximation holds: arc length s ≈ R * θ for θ in radians. We want φ in arc-seconds for one-division movement. Concept / Approach: When the bubble moves by one division, the arc length swept equals x. For small angles, θ (in radians) = s / R = x / R. To convert to seconds, multiply radians by 206265 (since 1 radian = 206265 arc-seconds). Therefore, φ = 206265 * (x / R). This directly links physical tube geometry to angular sensitivity per division. Step-by-Step Solution: 1) For one-division shift, s = x.2) θ (radians) = s / R = x / R.3) Convert to seconds: φ = θ * 206265 = (206265 * x) / R.4) Hence the required expression is φ = (206265 * x) / R. Verification / Alternative check: Instrument adjustment notes use the same formula to relate bubble movement, target displacement on the staff, and the length of sight, confirming the dependence on x/R and the conversion factor 206265 for seconds. Why Other Options Are Wrong: Option B: Gives φ in degrees, not seconds. Option C: Inverts the correct proportionality; sensitivity increases with larger x and decreases with larger R. Option D: Has wrong dimensional dependence (product x*R) and lacks the seconds-per-radian factor placement. Common Pitfalls: Confusing division length x with total tube length; forgetting to convert from radians to seconds; assuming linear displacement along the chord instead of the arc (for small angles, arc and chord are nearly equal, which justifies the approximation). Final Answer: φ = (206265 * x) / R

Question 18

Levelling concepts – meaning of a central bubble in a level tube When the bubble of a level’s bubble tube is centered (under permanent adjustments), which statement correctly describes the resulting geometry?

Accepted Answer

Introduction / Context: Levelling depends on the instrument’s internal relationships (permanent adjustments) that link the bubble axis, the vertical axis, and the line of collimation. Understanding what “bubble centered” guarantees is essential for correct interpretations of readings. This question asks which line is truly horizontal when the bubble is centered in a properly adjusted instrument. Given Data / Assumptions: The instrument has been correctly adjusted: bubble axis is perpendicular to the vertical axis, and the line of collimation is parallel to the bubble axis. Bubble is observed to be central in its run. Standard levelling definitions apply. Concept / Approach: With permanent adjustments, centering the bubble aligns the bubble axis horizontally. Since the line of collimation is made parallel to the bubble axis during adjustment, the line of collimation is therefore horizontal whenever the bubble is centered. The terms “axis of the telescope” or “geometrical axis” can be ambiguous, whereas “line of collimation” is the precise line through the cross-hairs and optical center defining the sight line used for readings. Step-by-Step Solution: 1) Bubble centered → bubble axis horizontal.2) By permanent adjustment, line of collimation ∥ bubble axis.3) Therefore, line of collimation is horizontal.4) This ensures level readings reflect a true horizontal line of sight. Verification / Alternative check: Instrument manuals specify that in a properly adjusted level, the line of collimation becomes horizontal when the bubble is centered; two-peg tests and collimation checks verify this relationship empirically. Why Other Options Are Wrong: Axis of the telescope / geometrical axis: These are not the operative definitions in levelling; optical and mechanical axes can differ and are not guaranteed to be aligned to the bubble axis. “Line of sight” (colloquial): Often used loosely; the precise term in levelling is “line of collimation.” Common Pitfalls: Using imprecise terminology; assuming bubble centering guarantees every conceivable axis is horizontal; neglecting the need for periodic permanent adjustment checks. Final Answer: Line of collimation is horizontal

Question 19

Optical square – fixed angle between the two mirrors In an optical square used for setting out right angles, the acute angle between the two mirrors is equal to which value?

Accepted Answer

Introduction / Context: An optical square is a simple reflective instrument widely used in chain surveying to set out perpendiculars (right angles). It consists of two mirrors oriented at a specific acute angle so that due to reflection geometry, the user can align one sight along the chain line and set a perpendicular to it. This question asks for the fixed angle between the mirrors that yields a 90° observation. Given Data / Assumptions: The device uses the law of reflection: angle of incidence equals angle of reflection. Two-mirror reflection rotates a ray by twice the angle between the mirrors. We desire a right-angle deflection between the two observed directions. Concept / Approach: If the angle between mirrors is α, a ray incident on the first mirror and then reflected by the second is deflected by 2α. To set out a right angle (90°), we require 2α = 90°, hence α = 45°. Optical squares are therefore manufactured with mirrors inclined at 45°, producing a perceived perpendicular when the reflected and direct images coincide correctly. Step-by-Step Solution: 1) Desired deflection between chain line and offset = 90°.2) Two reflections → net deflection = 2α.3) Set 2α = 90° → α = 45°.4) Therefore mirror angle = 45°. Verification / Alternative check: Instrument descriptions of optical squares and prism squares align with this 45° mirror setting to achieve perpendicular offsets without a theodolite. Why Other Options Are Wrong: 20°, 30°, 60°: Would produce 40°, 60°, or 120° deflections, not a right angle. 90°: Would lead to 180° deflection, not usable for perpendicular offsets. Common Pitfalls: Confusing the acute mirror angle with the desired right angle; mixing prism square (which uses refraction) with the mirror-based optical square. Final Answer: 45°

Question 20

Optical-reading theodolites – behavior of vertical circle graduations In optical-reading instruments, which of the following statements about the vertical circle readings is generally correct?

Accepted Answer

Introduction / Context: Optical-reading theodolites and total stations have vertical circles whose graduations and zero settings are defined by instrument design. Understanding how the reading changes with telescope elevation for each face is important for correct booking and for applying face-left/face-right procedures that reduce systematic errors. This question reviews standard conventions found in many optical-reading instruments. Given Data / Assumptions: Instrument uses a continuous vertical circle (0°–359°) rather than quadrantal indexing. Face-left and face-right refer to standard positioning of the telescope and vertical circle relative to the observer. We assume the common convention used by many instruments and textbooks. Concept / Approach: Many optical-reading instruments adopt a continuous vertical circle to avoid ambiguity in elevation/zenith angles. With a particular zero reference and circle orientation, elevating the telescope increases the readout in face-left, while the same elevation produces a decreasing readout in face-right (because the circle is viewed from the opposite side and the index orientation reverses). These trends underpin booking rules that ensure consistency across faces and enable averaging to eliminate certain instrumental errors. Step-by-Step Solution: 1) Recognize continuous circle minimizes index jumps.2) In face-left, elevating the telescope advances the reading upward.3) In face-right, elevating the telescope causes the reading to reduce symmetrically.4) Hence, all listed statements are consistent with typical instrument behavior. Verification / Alternative check: Instrument manuals and surveying texts present tables showing opposite-sense changes in vertical readings between faces, confirming the described increase/decrease rules and continuous-circle practice. Why Other Options Are Wrong: Each individual statement is correct under the stated convention, so the combined option is the most accurate. Common Pitfalls: Assuming every brand uses identical zero references; forgetting to note face position during booking; mixing elevation angle with zenith angle definitions. Final Answer: All of the above

Surveying Questions