Question 1

Highway / railway curves — the “Point of Curve (PC)” of a simple circular curve refers to which location on the alignment? Choose the most accurate definition.

Accepted Answer

Introduction / Context: On a simple circular curve, key points define geometry and setting-out: PI (point of intersection), PC (point of curve), and PT (point of tangency). Understanding these labels is crucial for field layout and for interpreting design drawings in road and rail engineering. Given Data / Assumptions: Simple circular curve between two tangents. Standard U.S./international notation: PC = start of curve, PT = end of curve, PI = intersection of tangents. Mid-point of the arc is not a standard control point name. Concept / Approach: The Point of Curve (PC) is where the back tangent ends and the circular arc begins. At the far end, the curve terminates at the Point of Tangency (PT), where the forward tangent begins. The PI is where the two tangents meet if extended. Correct identification avoids stakeout mistakes and ensures correct application of tangent lengths and offsets. Step-by-Step Solution: Identify tangents: back tangent approaching the curve; forward tangent leaving it.Define PC: the first point on the arc from the back tangent.Define PT: the last point on the arc before the forward tangent.Define PI: the intersection of the two tangents (not on the arc). Verification / Alternative check: Stakeout formulae use tangent length from PI to PC/PT; field notes and textbooks confirm these definitions universally. Why Other Options Are Wrong: Point of tangency: That is PT, not PC. Point of intersection: Different point (PI). Mid-point of curve: No standard control name in basic curve geometry. Common Pitfalls: Mixing PC and PT; remember PC = beginning, PT = end of the arc. Final Answer: Point of commencement (beginning of curve)

Question 2

Theodolite adjustments — a theodolite is said to be in perfect adjustment when which conditions hold? Select the most comprehensive option.

Accepted Answer

Introduction / Context: Theodolites measure horizontal and vertical angles with high precision. Proper geometry among the instrument axes is essential; otherwise, observed angles will contain systematic errors. The “perfect adjustment” conditions summarize the required orthogonality and plane constraints among axes and sighting lines. Given Data / Assumptions: Definitions: vertical (rotation) axis, trunnion (horizontal) axis, and line of collimation. Instrument is correctly leveled so the vertical axis is truly vertical. Cross-hairs are properly centered and focused. Concept / Approach: For perfect adjustment: the vertical axis must be perpendicular to the trunnion axis so that vertical motion is truly in a vertical plane; the line of collimation must be perpendicular to the trunnion axis so that vertical circle motions do not change collimation; and as the telescope is raised or lowered, the line of sight should remain in a single vertical plane—ensuring correct vertical angle measurement and elimination of lateral sweeping errors. Step-by-Step Solution: Confirm vertical axis ⟂ trunnion axis via standard checks (spirit levels/auto levels).Adjust the line of collimation to be ⟂ trunnion axis using collimation tests (e.g., two-peg or reciprocal checks for tilt scopes).Verify vertical plane sweep by sighting a high and low point on the same vertical line; adjust if necessary.Re-check after adjustments as changes can couple between conditions. Verification / Alternative check: Reversing face observations (face left/right) and obtaining consistent angles indicate near-perfect adjustment. Why Other Options Are Wrong: Any single condition alone is insufficient; all must hold simultaneously to avoid systematic angular errors. Common Pitfalls: Assuming temporary leveling alone ensures perfect geometry; internal axis relationships also require adjustment. Final Answer: All of the above

Question 3

Area by ordinates — Simpson’s 1/3 rule for a curvilinear boundary divided into an even number of equal-width strips states that the area equals: Identify the correct coefficient pattern for extreme, odd, and even offsets.

Accepted Answer

Introduction / Context: When a boundary is irregular, its area can be estimated from offset ordinates at equal intervals using numerical integration rules. Simpson’s 1/3 rule gives a higher-accuracy estimate than the trapezoidal rule by weighting odd and even ordinates differently. This is a standard technique in surveying computations of areas from field measurements. Given Data / Assumptions: Offsets (ordinates) are taken at equal spacing (strip width = d). The number of strips is even (number of ordinates is odd). Offsets are ordered from the first extreme to the last extreme. Concept / Approach: Simpson’s 1/3 rule formula for area A uses weighting 1 for the two extreme offsets, 4 for odd-numbered interior offsets, and 2 for even-numbered interior offsets. The multiplier outside is d/3. This produces improved accuracy for smooth curves by approximating the boundary with parabolic segments over pairs of strips. Step-by-Step Solution: Let d be strip width; O0 and On be extreme offsets.Let the interior offsets be O1, O2, O3, ..., On−1.Compute A = (d/3) * [O0 + On + 4*(O1 + O3 + O5 + ...) + 2*(O2 + O4 + O6 + ...)].Ensure the number of strips is even to apply the rule directly. Verification / Alternative check: Cross-verify by splitting into parabolic pairs and confirming the 1:4:2 pattern across each pair, then summing for the whole boundary. Why Other Options Are Wrong: Options (a), (b), and (d) use incorrect coefficients and/or multipliers; they do not match the 1:4:2 with d/3 rule. (e) is wrong because (c) is exactly the Simpson’s 1/3 rule. Common Pitfalls: Applying Simpson’s rule with an odd number of strips or confusing which offsets are “odd” vs “even” (count interior positions from the first extreme). Final Answer: One-third the strip width * [sum of extreme offsets + 4*(sum of remaining odd) + 2*(sum of even)]

Question 4

Reciprocal levelling — which source of error is NOT completely eliminated by reciprocal levelling across an obstacle? Choose the correct option.

Accepted Answer

Introduction / Context: Reciprocal levelling is performed across rivers, valleys, or other obstacles where equal backsight and foresight distances from one setup are not possible. By observing from both banks (stations) and combining readings, several systematic errors cancel out. However, one atmospheric effect is variable and not always fully removed. Given Data / Assumptions: Two setups on opposite sides of an obstacle. Readings taken reciprocally on staves at each end. Aim is to eliminate systematic errors depending on sight length and instrument imperfections. Concept / Approach: By exchanging the instrument position, the effects of Earth’s curvature and collimation error (line of sight not truly horizontal) tend to cancel because they act with opposite signs over equal long sights. Refraction, however, depends on instantaneous atmospheric gradients along the line of sight and can differ between observations, so it is not completely eliminated in practice, only reduced on average. Step-by-Step Solution: Set up at A, read on staff at A and B; then set up at B, read on staff at A and B.Compute the level difference using the reciprocal formula; curvature and collimation errors cancel.Note that refraction may vary with time and path, leaving residual error.Adopt procedures to minimize refraction (shorter sights near ground avoided, cool stable air preferred). Verification / Alternative check: Field practice often repeats reciprocal readings and averages results to further reduce refraction effects. Why Other Options Are Wrong: Earth’s curvature: cancels with reciprocal observations. Non-adjusted collimation: cancels as readings are swapped. Bubble tube non-adjustment: largely manifests as a collimation tilt and is mitigated similarly. Common Pitfalls: Assuming a single reciprocal set guarantees perfect cancellation; strong temperature gradients can still bias results. Final Answer: Refraction

Question 5

Verniers — if the smallest division of a vernier scale is longer than the smallest division of the main (primary) scale, what type of vernier is it? Select the correct name.

Accepted Answer

Introduction / Context: Vernier scales enable readings finer than the main scale divisions by comparing slightly different pitches. The naming depends on whether the vernier divisions are shorter or longer than those of the main scale and on the direction in which the vernier graduations are laid out. Given Data / Assumptions: Main scale division length is taken as the reference. Vernier’s smallest division is longer than the main scale’s smallest division. Least count is the absolute difference between one main division and one vernier division. Concept / Approach: When each vernier division is longer than one main division, the vernier must be laid out in the opposite (backward) direction to maintain readability. This configuration is termed a retrograde vernier. In contrast, a direct vernier has vernier divisions shorter than the main scale division and is laid out in the same direction. Step-by-Step Solution: Let one main division = M; one vernier division = V.Given V > M → least count = V − M (positive difference).To read conveniently, vernier graduations are set opposite to the main scale direction → retrograde layout.Hence, the vernier type is retrograde. Verification / Alternative check: Instrument manuals distinguish direct and retrograde verniers exactly by this longer/shorter criterion and orientation. Why Other Options Are Wrong: Direct vernier: V < M and same direction. Double vernier: Refers to two verniers for eliminating errors, not division length. Simple vernier: Generic term; not specific to V > M. Common Pitfalls: Confusing least count sign; remember least count is magnitude, not algebraic sign, and layout dictates the name. Final Answer: Retrograde vernier

Question 6

Tacheometry — why is vertical staff holding generally preferred over normal (perpendicular) staffing during tacheometric observations? Choose the most appropriate reason.

Accepted Answer

Introduction / Context: Tacheometry (stadia surveying) computes distances and elevations from staff intercepts and vertical angles observed through a tacheometer. The way the staff is held affects the intercept and thus the derived quantities. Practitioners often prefer a particular holding because it reduces sensitivity to small errors in staff orientation. Given Data / Assumptions: Stadia readings are taken on a graduated staff. Observer measures upper and lower stadia hair coincidences and the vertical angle. Choice between vertical staff holding and “normal” staffing (perpendicular to the line of sight). Concept / Approach: With the staff held truly vertical, small deviations by the staffman cause smaller proportional errors in the stadia intercept than when trying to hold the staff normal to the oblique line of sight, which is harder to judge. Thus, vertical staff holding minimizes the effect of careless holding and simplifies routine instructions to the staffman, improving reliability of results. Step-by-Step Solution: Adopt vertical staff holding as the default instruction.Read upper and lower stadia hairs quickly; record vertical angle at the instrument.Reduce observations using standard tacheometric formulae with vertical staff assumption.Benefit: errors from slight staff tilt have reduced impact on the computed distance/elevation. Verification / Alternative check: Field practice and textbooks recommend vertical staffing for routine tacheometry to control human-factor errors in staff orientation. Why Other Options Are Wrong: Ease of reduction: Possible advantage, but the principal benefit is error minimization. Facility of holding: Vertical is simple, but the core reason is reduced sensitivity to tilt. None: Incorrect because a strong reason exists. Common Pitfalls: Instructing staff to be “normal to line of sight,” which is difficult to judge and leads to larger, inconsistent errors. Final Answer: Minimum effect of careless holding on the computed result

Question 7

Mass haul (mass diagram) — identify the correct statements about slope sign, volume interpretation, and haul from a mass diagram. Choose the most comprehensive option.

Accepted Answer

Introduction / Context: Mass diagrams (mass haul curves) are cumulative-volume plots used in highway and railway earthwork planning. They help identify where cut balances fill, the quantity to be moved, the direction of haul, and average haul distances. Correct interpretation of slopes, extrema, and loop areas is crucial for economical earthmoving strategies. Given Data / Assumptions: Mass diagram ordinate represents cumulative earthwork relative to a datum. Sign convention: increasing ordinate with cut; decreasing with fill (embankment). Balancing line connects points of equal cumulative volume for haul analysis. Concept / Approach: A downward slope with chainage typically indicates net embankment (fill), while an upward slope indicates net cutting. Vertical differences between characteristic points yield total volumes in each section, and the area of a loop (between curve and balancing line) represents the haul, i.e., volume * average haul distance in that direction. Step-by-Step Solution: Identify maxima and minima: rises to maxima with cutting; falls to minima with filling.Volume of embankment between successive maxima equals the drop (vertical difference) between the ordinates.Volume of cutting between a minimum and the next maximum equals the rise in ordinate between them.Compute haul from loop area with respect to the balancing line to plan movement and equipment selection. Verification / Alternative check: Check that total cut equals total fill when the balancing line is correctly placed; loop areas on either side quantify hauls in opposite directions. Why Other Options Are Wrong: Each individual statement is correct; the only fully correct choice is the combined option that includes all of them. Common Pitfalls: Misinterpreting sign conventions; always confirm how the mass curve was constructed before drawing conclusions. Final Answer: All of the above

Question 8

Geomagnetism in surveying — what is the term for an imaginary line that passes through all points on the Earth having equal magnetic declination? Choose the correct term.

Accepted Answer

Introduction / Context: Magnetic declination is the horizontal angle between magnetic north and true north. Surveyors must account for declination to relate compass bearings to true meridians. Several families of geomagnetic lines describe equal values of different magnetic parameters across the Earth’s surface. Given Data / Assumptions: Declination varies with location and time. Terminology relates to maps of Earth’s magnetic field. We seek the term associated specifically with equal declination. Concept / Approach: An isogon (or isogonic line) connects points of equal magnetic declination. An agonic line is a special isogon where declination is zero (magnetic north equals true north). An isoclinic line connects points of equal magnetic dip (inclination), a different parameter from declination. Step-by-Step Solution: Identify parameter: declination (horizontal angular difference).Map to term: isogon / isogonic line.Note special case: agonic = zero declination only.Exclude inclination-related terms (isoclinic) as they denote dip, not declination. Verification / Alternative check: Geomagnetic charts publish isogonic lines with values in degrees east or west; agonic line is labeled 0°. Why Other Options Are Wrong: Agonic: Only where declination is zero, not a general equal-declination line. Isoclinic: Equal dip (inclination), not declination. None: Incorrect because “isogon” is the established term. Common Pitfalls: Confusing declination with inclination; they are orthogonal components of the geomagnetic field. Final Answer: Isogon (isogonic line)

Question 9

Level vs horizontal — which statements correctly describe level surfaces/lines and horizontal surfaces/lines in geodetic terms? Choose the most comprehensive option.

Accepted Answer

Introduction / Context: “Level” and “horizontal” are often used casually as synonyms, but in geodesy and precise surveying they have distinct meanings tied to the Earth’s gravity field. Understanding the difference is essential when transferring elevations and when aligning instruments over significant distances. Given Data / Assumptions: Direction of gravity is defined by the plumb line (local vertical). Level surfaces approximate equipotential surfaces of the Earth’s gravity field. Horizontal planes/lines are defined locally at a point. Concept / Approach: A level surface is everywhere perpendicular to gravity (plumb line) and thus curves with the Earth. A level line lies on this surface. A horizontal surface is a plane perpendicular to gravity at a single point; extended far from that point, it departs from the level surface due to curvature. A horizontal line is a straight line in that plane and is tangential to the level surface at the point of tangency. Step-by-Step Solution: Define gravity direction at a point via the plumb line.Describe level surface: everywhere normal to gravity → curved surface.Define horizontal surface: plane normal to gravity only at that point.Relate lines: a horizontal line is tangent to the level surface at that point; a level line lies on the level surface. Verification / Alternative check: Precise levelling over long distances uses level lines (curved) rather than straight horizontal lines; curvature corrections exemplify this distinction. Why Other Options Are Wrong: Each of A–D is correct; only the combined option fully captures the relationship. Common Pitfalls: Assuming a horizontal plane is globally level; the “level” concept is inherently tied to the curved equipotential surface. Final Answer: All of the above

Question 10

Earthwork Quantity Computation – Valid Methods Used in Highway and Irrigation Projects In engineering surveying and quantity estimation, the volume of earthwork between two cross-sections (or between multiple sections) may be calculated using which …

Accepted Answer

Introduction: Accurate earthwork quantity estimation underpins cost, planning, and payment in road, railway, and canal works. Surveyors compute volumes between successive cross-sections using standard mathematical methods. This question tests recognition of the commonly accepted formulas and when they are used. Given Data / Assumptions: Two or more cross-sections at known chainages are available. Areas are measured from field notes or CAD. Ground varies more or less regularly between sections. Concept / Approach: The 'end areas' and 'trapezoidal' approaches assume linear variation, while the 'prismoidal' formula fits a parabolic variation of the surface between sections and is more accurate when curvature exists. The 'mean areas' framing is a practical variant when intermediate shapes are averaged. Choice depends on section spacing and smoothness of terrain, but all are legitimate in practice and in specifications. Step-by-Step Solution: For two sections with areas A1 and A2 spaced by distance L, the trapezoidal (end areas) volume is V = L * (A1 + A2) / 2.When the intermediate area Ai at mid-distance is available or estimated, the prismoidal volume is V = L * (A1 + 4 * Ai + A2) / 6.Mean areas method uses V = L * Amean, where Amean is the average of representative areas if justified by geometry.For multiple sections, apply the chosen rule segment by segment and sum volumes. Verification / Alternative check: Using both trapezoidal and prismoidal methods on a curved surface typically shows the prismoidal result slightly smaller (more accurate) than trapezoidal when areas curve concavely between sections, aligning with numerical integration theory. Why Other Options Are Wrong: Each named method is valid under suitable assumptions; therefore, selecting a single one would be incomplete. 'All of the above' correctly captures accepted practice. Common Pitfalls: Applying the trapezoidal rule over long intervals where ground curvature is significant; ignoring the necessity of closer section spacing on irregular terrain. Final Answer: All of the above

Question 11

Optical Survey Instruments – Features of an Internal Focusing Telescope Which of the following statements correctly describe an internal focusing telescope used on theodolites and levels?

Accepted Answer

Introduction: Modern surveying telescopes often employ internal focusing for robustness and sealed optics. Understanding which element moves and where it is placed helps in diagnosing parallax, collimation, and focusing issues in the field. Given Data / Assumptions: An internal focusing design uses an additional focusing lens rather than moving the entire objective. The cross-hair diaphragm is fixed relative to the objective mount. Survey tasks require quick, smooth focusing without shifting external lengths. Concept / Approach: In internal focusing telescopes, the objective and diaphragm maintain a fixed separation to preserve image plane stability. Focus is adjusted by moving a negative (divergent) lens inside the tube, generally around the mid-section. This motion forms a virtual image at the cross-hair plane for targets at varying distances, while keeping the instrument compact and better sealed against dust and moisture. Step-by-Step Solution: Confirm that the objective does not translate during focusing (fixed to the tube front).Identify the internal sliding element as a divergent lens whose displacement changes the conjugate focus.Note typical placement near the tube’s middle for optical balance and mechanical convenience.Hence, all three statements are consistent with an internal focusing layout. Verification / Alternative check: Manufacturer schematics and optics texts diagram internal focusing with a negative focusing lens and a fixed objective–diaphragm spacing. Why Other Options Are Wrong: 'Only the objective lens moves' describes external focusing; it does not apply here. Common Pitfalls: Confusing removal of parallax (eye-piece and objective focusing) with the mechanical means by which focus is achieved inside the tube. Final Answer: All of the above

Question 12

Plane Table Resection – When Is the Three-Point “Fix” Most Reliable? While fixing a plane-table station by the three-point problem with three known points plotted on the sheet, the quality of the fix is best under which geometric condition?

Accepted Answer

Introduction: The three-point problem determines the position and orientation of a plane table by sighting three well-defined points whose plan locations are already plotted. The reliability of the 'fix' depends on the geometry formed by the points relative to the unknown station. Given Data / Assumptions: Three well-spaced, non-collinear control points are visible. The station to be fixed lies within the triangle of these points (preferred). The 'middle' point refers to the point that lies between the other two when viewed from the plane-table station. Concept / Approach: A well-conditioned triangle with the station near its interior reduces angular intersection sensitivity. Classical plane-table guidance states that the fix improves when the middle point is the nearest, yielding better angular spread and smaller intersection errors on the table. Collinearity or extreme skinny triangles degrade accuracy severely. Step-by-Step Solution: Ensure points form a sizable triangle surrounding the station.Identify the middle point (between the other two in bearing order).Prefer configurations where this middle point is at the shortest range to enhance angular subtenses.This reduces the error ellipse and gives a strong, reliable fix. Verification / Alternative check: Textbook rules for good resection condition lists include: station inside the control triangle, good spread of rays, and middle point nearest to the station to strengthen the solution. Why Other Options Are Wrong: Making the middle point farthest tends to flatten angular spread; choosing a side point as nearest skews geometry; collinear points give no unique solution. Common Pitfalls: Placing the plane table outside the control triangle; using poorly separated control points; ignoring identification of the middle point in the field. Final Answer: The middle (intermediate) control point is nearest to the plane-table station

Question 13

Telescope Eyepieces – Construction of the Ramsden Eyepiece Used in Survey Instruments Which component arrangement correctly describes a Ramsden eyepiece commonly used on surveying telescopes?

Accepted Answer

Introduction: The eyepiece enables comfortable viewing and carries the cross-hair diaphragm at or near the focal plane. The Ramsden eyepiece is favored in surveying for its relatively flat field and manageable aberrations at modest cost. Given Data / Assumptions: Eyepiece must permit focusing of the observer’s eye on the cross-hairs. It should minimize distortion and chromatic blur for instrument use. Construction details matter for image quality and parallax control. Concept / Approach: A Ramsden eyepiece consists of two identical plano-convex lenses, with their convex surfaces facing each other and separated by a short distance. The cross-hair diaphragm is placed near the common focal plane, providing a clear image with reduced spherical aberration compared to a simple single lens. This configuration is standard in many theodolites and levels. Step-by-Step Solution: Identify the lens types: plano-convex × 2.Orient them: convex faces toward each other to position the focal plane near the diaphragm.Recognize benefits: acceptable field flatness and controlled aberrations for survey work.Therefore, option (d) matches the Ramsden design. Verification / Alternative check: Instrument optics references diagram Ramsden eyepiece as P-C + P-C with convex faces inward, distinguishing it from Huygens (two plano-convex with plane faces inward). Why Other Options Are Wrong: Generic 'two convex/concave' descriptions lack the specific orientation; mixed convex–concave is not Ramsden; a single doublet does not define this eyepiece type. Common Pitfalls: Confusing Ramsden with Huygens orientation; forgetting the diaphragm position relative to the focal plane. Final Answer: Two plano-convex lenses placed close together with convex faces facing each other

Question 14

Parallax in Survey Telescopes – Correct Procedure to Remove It Completely To eliminate parallax so that the image of the object and the cross-hairs lie in the same focal plane, which focusing steps must be correctly performed?

Accepted Answer

Introduction: Parallax causes apparent relative motion between the target image and the cross-hairs when the observer’s eye shifts. It leads to pointing errors in angle and staff readings. The remedy requires a specific focusing sequence to make both images coincide in one plane. Given Data / Assumptions: Cross-hair diaphragm position is fixed inside the eyepiece assembly. Eyepiece focuses the observer’s eye on the cross-hairs (forming a virtual image at infinity). Objective forms the target image at the diaphragm plane when correctly focused. Concept / Approach: First, remove ocular accommodation effects by focusing the eyepiece until the cross-hairs appear razor sharp against a bright background. This sets the eye at the correct optical condition. Next, use the objective focusing to bring the external object’s image exactly onto the cross-hair plane. When both coincide, shifting the eye causes no relative motion, indicating zero parallax. Step-by-Step Solution: Point the telescope to the sky or a plain surface; adjust eyepiece for sharp cross-hairs.Aim at the object and focus the objective slowly until the object image is sharp and steady against the cross-hairs.Verify by gently moving the eye; absence of relative motion confirms parallax removal.Repeat if necessary for different observers. Verification / Alternative check: Standard theodolite and level manuals prescribe the same two-step focusing order for parallax removal. Why Other Options Are Wrong: Eyepiece only or objective only will not ensure both images share a plane; rotating the telescope does nothing to parallax. Common Pitfalls: Focusing the objective first; relying on eye accommodation which reintroduces parallax for other users. Final Answer: Focus the eyepiece first for sharp cross-hairs, then focus the objective on the target

Question 15

Chaining on Slopes – When Can Slope Correction Be Neglected in Practice? Under typical site accuracy requirements, the slope-to-horizontal correction in chaining can be ignored in which of the following practical cases?

Accepted Answer

Introduction: Distance measured on a slope exceeds the corresponding horizontal projection. Reducing slope distances to horizontal improves accuracy, but for small slopes the correction may be negligible relative to instrument and reading limits. This question asks when it is acceptable to ignore the correction in routine work. Given Data / Assumptions: Small-slope scenarios: less than about 3° or gradient near 1 in 19. Required accuracy is typical of ordinary chaining, not geodetic surveys. Line lengths are modest so that cumulative effect remains within tolerance. Concept / Approach: For small slope angle θ, horizontal distance H ≈ s * cos θ ≈ s * (1 − θ^2/2). The fractional error is about θ^2/2 in radians. At θ ≈ 3° ≈ 0.0524 rad, θ^2/2 ≈ 0.00137, or about 1.37 mm per metre (1.37 m per kilometre). For short runs in ordinary work, this can fall within permissible error, justifying neglect of the correction. Step-by-Step Solution: Express reduction: H = s * cos θ.Approximate for small θ: cos θ ≈ 1 − θ^2/2.Estimate error at 3° to confirm it is minor for routine site lengths.Conclude that slopes under ~3° (≈ 1 in 19) allow neglecting correction in ordinary chaining. Verification / Alternative check: A quick check for a 100 m line at 3°: reduction ≈ 0.137 m. If this exceeds tolerance for your task, apply correction; otherwise, many field handbooks accept ignoring it below this threshold for rough work. Why Other Options Are Wrong: 'Neither' is incorrect because both listed small-slope cases are typical thresholds; 'exactly zero' is unnecessarily restrictive. Common Pitfalls: Ignoring correction on steeper slopes; forgetting that cumulative error over many segments can become significant even if each is small. Final Answer: Both (a) and (b)

Question 16

Transit Vernier Reading – Smallest Readable Angle from Plate and Vernier Data The main (horizontal) plate of a transit is divided into 1080 equal divisions. If 60 vernier divisions coincide exactly with 59 main plate divisions, what is the smallest …

Accepted Answer

Introduction: Least count quantifies the fineness of angular reading on a vernier theodolite. It depends on the main scale division (MSD) and the vernier design. This problem requires applying the standard relation when the vernier spans one less main division than its own number of divisions. Given Data / Assumptions: Main plate: 360° divided into 1080 divisions ⇒ 1 MSD = 360° / 1080 = 1/3° = 20′ (minutes). Vernier: 60 vernier divisions coincide with 59 MSD ⇒ 1 VSD = 59/60 MSD. Least count LC = 1 MSD − 1 VSD for a direct vernier. Concept / Approach: Compute LC from the difference between MSD and VSD values. Convert minutes to seconds for the final answer consistent with instrument precision statements. Step-by-Step Solution: 1 MSD = 20′.1 VSD = (59/60) * 20′ = 19′ 40″.LC = 1 MSD − 1 VSD = 20′ − 19′ 40″ = 20″.Therefore, the smallest readable angle is 20 seconds. Verification / Alternative check: Using LC = MSD / number of vernier divisions = 20′ / 60 = 1/3′ = 20″ gives the same result. Why Other Options Are Wrong: 5″, 10″, 15″ reflect finer verniers than stated; 30″ is coarser than the computed capability. Common Pitfalls: Dividing 360° by 60 directly; forgetting that the vernier spans 59 MSD in this configuration. Final Answer: 20″ (seconds)

Question 17

Bearings – Convert a Whole Circle Bearing to a Reduced (Quadrantal) Bearing A line has a whole circle bearing (WCB) of 290°. What is its reduced bearing (RB) expressed in quadrant notation?

Accepted Answer

Introduction: Bearings can be given as whole circle bearings (0°–360° from true or magnetic north, clockwise) or reduced bearings (quadrantal form referenced to the nearest north or south toward east or west). Converting between forms is routine in traverse computation. Given Data / Assumptions: WCB = 290°. RB format: N/S θ E/W depending on the quadrant. Angles measured from the north or south reference toward east or west. Concept / Approach: Locate the quadrant: 270°–360° corresponds to the northwest (NW) quadrant. The reduced bearing is the acute angle from the north toward the west for this quadrant: RB = N (360° − 290°) W = N 70° W. Step-by-Step Solution: Determine quadrant: 290° lies between 270° and 360° ⇒ NW quadrant.Compute acute angle to nearest pole: 360° − 290° = 70°.State RB with correct letters: N 70° W.Check reasonableness: angle large from north toward west, consistent with NW direction. Verification / Alternative check: Alternative: from west reference, RB could be W 20° N, which is equivalent; however, standard RB convention prefers N/S first, so N 70° W is selected. Why Other Options Are Wrong: N 20° E is NE quadrant; N 20° W is too close to north; S 70° E and S 20° W are southern quadrants. Common Pitfalls: Subtracting from 180° instead of 360°; mixing RB letter order (N/S should be first). Final Answer: N 70° W

Question 18

Simple Survey Tools – Primary Use of a Cross-Staff in Chain Surveying In chain and offset surveying, the basic cross-staff is primarily used for which operation on the ground?

Accepted Answer

Introduction: A cross-staff is a simple optical square used in basic surveying to project perpendiculars from a baseline. It is light, inexpensive, and quick to use for setting offsets to locate details such as fences, buildings, and hedges. Given Data / Assumptions: Survey context: chain and tape measurements of a baseline. Need to locate features by perpendicular offsets to the baseline. Tool: cross-staff or optical square with right-angle sighting slots or prisms. Concept / Approach: By aligning one slit with the chain line and turning the instrument until the other slit sights the object, the cross-staff ensures a right angle between the baseline and the offset line. Offsets are then measured with tape along the perpendicular to plot features accurately on the plan. Step-by-Step Solution: Station the cross-staff on the chain line at the desired chainage.Align one set of sights with the chain line.Rotate until the perpendicular sights align with the ground point or feature.Measure the offset distance along the perpendicular with a tape. Verification / Alternative check: Where higher precision is required, an optical square provides the same 90° function with better optics but the principle remains identical. Why Other Options Are Wrong: Cross-staff does not measure gradients (clinometers are used), cannot take levels like a dumpy level, and does not measure long distances by itself. Common Pitfalls: Using the cross-staff off the chain line; neglecting to check instrument orientation causing skewed offsets. Final Answer: Setting out right angles to establish offsets from a chain line

Question 19

Vernier Scales – Definition of Least Count for a Direct Vernier What is the correct expression for the least count of a vernier scale in relation to the main (primary) scale divisions?

Accepted Answer

Introduction: The least count (LC) expresses the smallest difference in measurement that a vernier instrument can resolve. Understanding LC is essential for reading angles and lengths accurately and for estimating observational precision in field notes. Given Data / Assumptions: A direct vernier is used in which n vernier divisions coincide with (n − 1) main scale divisions. Let 1 MSD be the value of one main scale division. Let 1 VSD be the value of one vernier scale division. Concept / Approach: For a direct vernier, n VSD = (n − 1) MSD ⇒ 1 VSD = ((n − 1)/n) MSD. The least count equals the difference between 1 MSD and 1 VSD, which simplifies to LC = MSD / n. Therefore, one practical way to compute LC is to divide the value of one main division by the number of vernier divisions, matching the given option. Step-by-Step Solution: Start from n VSD = (n − 1) MSD.Compute 1 VSD = ((n − 1)/n) MSD.Find LC = 1 MSD − 1 VSD = MSD − ((n − 1)/n) MSD = MSD / n.Hence, LC equals the value of one main division divided by the number of vernier divisions. Verification / Alternative check: Worked examples for typical theodolites: if 60 vernier divisions span 59 main divisions and 1 MSD = 20′, LC = 20′ / 60 = 20″, consistent with field instruments. Why Other Options Are Wrong: Sum or product have no meaning in this context; option (c) inverts the relation; 'none' is invalid since a standard expression exists. Common Pitfalls: Confusing LC with the smallest graduation seen on the main scale; overlooking that LC depends on both scales’ geometry. Final Answer: Value of one division of the main (primary) scale divided by the total number of divisions on the vernier

Question 20

Surveying and error theory: how do random (accidental) errors accumulate as the number of independent observations/operations increases? Choose the most accepted proportionality used for precision estimates in civil engineering measurements.

Accepted Answer

Introduction / Context: Random (accidental) errors arise from small, uncontrollable variations in observation conditions—sight clarity, hand steadiness, reading interpolation, and atmospheric flicker. In surveying and geospatial engineering, understanding how such errors accumulate is essential for planning redundancy, estimating precision, and choosing appropriate adjustment methods. Given Data / Assumptions: Errors are random with zero mean and finite variance. Individual observations/operations are independent and similarly distributed. No dominant systematic (bias) or gross errors are present. Concept / Approach: For independent random errors, variances add. If each operation has standard deviation sigma, then over n operations, the resultant standard deviation is sigma_total = sqrt(n) * sigma. Hence the overall random error grows with the square root of the number of operations, not linearly with n. This square-root law underpins the familiar improvement in precision by averaging multiple observations (precision improves only as 1/sqrt(n)). Step-by-Step Solution: Model each operation i with error e_i having E[e_i]=0 and Var[e_i]=sigma^2.Resultant error E_R over n operations has variance Var[E_R] = n * sigma^2 (independence).Therefore, standard deviation SD(E_R) = sqrt(n) * sigma → proportional to sqrt(n).Practical takeaway: doubling observations reduces standard error by about 1/sqrt(2) ≈ 0.707. Verification / Alternative check: Least-squares theory uses the same variance-addition principle; traverse closures and leveling misclosures follow root-n behavior when random direction/reading errors dominate. Why Other Options Are Wrong: Linear with n: implies systematic drift, not random accumulation. Reciprocal with n or cube root: do not reflect variance addition for independent errors. “None” ignores established statistics. Common Pitfalls: Confusing random with systematic errors; assuming averaging eliminates error completely; ignoring correlation which can violate sqrt(n) scaling. Final Answer: square root of the number of operations involved

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