Question 1

Plane vs geodetic assumptions: which statements are valid? Decide which statements about plumb lines, curvature, and survey extent are correct for practical surveying.

Accepted Answer

Introduction / Context: Survey computations rest on geometric assumptions about the Earth. Over small extents, the Earth can be approximated as a plane; over larger extents, it must be treated as a curved surface. Plumb lines (gravity directions) also vary slightly in direction across the globe. This question checks your ability to choose appropriate assumptions by survey scale. Given Data / Assumptions: Plane surveying approximations apply over small areas. Geodetic principles are required over large areas. Plumb lines point toward the Earth’s center of mass direction locally; they are not parallel across distance. Concept / Approach: For small projects (e.g., site plans), neglecting curvature and treating level surfaces as planes yields negligible error. For long lines and large networks, curvature and convergence of meridians matter. The non-parallelism of plumb lines reflects Earth’s curvature and gravity anomalies; while the effect is tiny locally, it is fundamental conceptually. Step-by-Step Solution: Recognize scale dependence: small → plane, large → geodetic.Acknowledge plumb line directions vary with position.Combine to select the answer including all three truths. Verification / Alternative check: Textbooks define plane surveying limits (often within a few kilometres) and recommend geodetic treatment for state/national networks. Why Other Options Are Wrong: Any subset misses at least one correct statement; hence “All of the above” is the best choice. Common Pitfalls: Applying plane formulas across long distances; assuming plumb lines are exactly parallel everywhere. Final Answer: All of the above

Question 2

Slope correction for a 100-link chain on a small angle α (radians) For a measured chain length of 100 links laid along a uniform slope of α radians, what is the approximate subtractive slope correction per chain length to obtain the horizontal dista…

Accepted Answer

Introduction / Context: When a line is measured along a slope, the recorded length exceeds the required horizontal distance. A slope correction must therefore be subtracted. For small slopes, a convenient quadratic approximation in the slope angle provides fast and accurate corrections in the field for a 100-link chain (20.1168 m for a Gunter’s chain or 20 m for a metric chain, depending on convention). Given Data / Assumptions: Measured along-slope chain length per segment = 100 links. Uniform slope angle = α radians, with α small. Horizontal distance L_h = L * cos α; Correction C_s = L_h − L (negative). Concept / Approach: Use the identity cos α ≈ 1 − α^2/2 for small α. Then L_h = L * (1 − α^2/2), so the subtractive slope correction magnitude is |C_s| = L * (1 − cos α) ≈ L * (α^2/2). For L = 100 links, the magnitude becomes 100 * α^2 / 2 = 50 * α^2 links. The correction is subtractive from the measured sloping length to recover the horizontal length. Step-by-Step Solution: Start with L_h = L * cos α.Use cos α ≈ 1 − α^2/2 → L_h ≈ L * (1 − α^2/2).Compute C_s = L_h − L ≈ − L * (α^2/2).For L = 100 links: C_s ≈ − 50 * α^2 links (subtractive). Verification / Alternative check: Exact correction is L * (cos α − 1). For small α (e.g., ≤ 5°), the quadratic approximation differs negligibly from the exact value for most chaining work. Why Other Options Are Wrong: Linear (α) or cubic (α^3) dependence is incorrect; 100/α is dimensionally wrong; zero ignores real slope effects. Common Pitfalls: Using degrees in place of radians in the series; forgetting that the correction is subtractive; mixing links and metres without consistent conversion. Final Answer: 50 * α^2 links (i.e., 100 * α^2 / 2)

Question 3

Behavior of contours at identical elevation: what is possible? For a normal ground surface represented on a topographic map, two contour lines corresponding to the same elevation may:

Accepted Answer

Introduction / Context: Contours are level curves. Their geometry follows from the requirement that each elevation corresponds to a unique locus on a single-valued ground surface (no overhangs). Understanding when same-elevation contours can merge or cross avoids misinterpretation of terrain, especially around ridges, valleys, and flats. Given Data / Assumptions: Ground surface is single-valued with no overhanging cliffs represented. Contours are drawn at a fixed interval. Standard cartographic conventions apply. Concept / Approach: Contours of identical elevation cannot cross, because crossing would imply two different elevations at the same plan location. However, two segments of the same-elevation contour can unite (join) because they are simply different parts of the same closed level curve coming together. On complex terrain, a contour may split and rejoin, but it always remains a single closed line overall (possibly outside the sheet). Step-by-Step Solution: Recall: contour = locus of equal RL; unique RL at any plan point.Crossing would violate uniqueness → disallowed.Joining represents continuity of the same contour → allowed. Verification / Alternative check: Examples of ridges or basins show same-elevation contours looping and sometimes rejoining, but never crossing. Why Other Options Are Wrong: Crossing (A, D, E) contradicts level-curve properties on a single-valued surface. C categorically forbids uniting, which is not correct; the same contour can join itself. Common Pitfalls: Confusing special cases like overhangs (not shown on standard maps) with general ground; misreading close spacing as an intersection. Final Answer: Never cross, but they may unite (join) to form a single contour

Question 4

Achieving high accuracy in geodetic surveys: which treatment is essential? Select the statement that best supports the higher accuracy sought in geodetic (large-area) surveys.

Accepted Answer

Introduction / Context: Geodetic surveys cover long lines and large areas where the assumption of a flat Earth becomes invalid. To attain high positional accuracy, the computations must reflect the true geometry of the Earth, including curvature and, where required, the ellipsoidal model used for a datum. Given Data / Assumptions: Survey spans are large enough that plane approximations introduce non-negligible error. Angles are observed between lines on a curved surface (geodesics). Standard geodetic reductions, including curvature and possibly refraction, are applied. Concept / Approach: High accuracy requires modelling the Earth as an ellipsoid (or sphere as a first approximation). Distances must be reduced to the ellipsoid; angles must consider the convergence of meridians and curvature. Plane surveying formulas can be used only after appropriate map projections and geodetic reductions are performed. Step-by-Step Solution: Identify scale: large extent → geodetic treatment necessary.Incorporate curvature corrections and, if needed, projection conversions.Choose the option that explicitly includes curvature: (b). Verification / Alternative check: National control networks universally use ellipsoidal models and geodetic reductions—evidence that curvature must be accounted for to attain sub-arcsecond accuracies. Why Other Options Are Wrong: Ignoring curvature or using only plane formulas leads to bias over long extents; (c) is incomplete without corrections. Common Pitfalls: Applying plane computations directly to geodetic observations; neglecting projection scale factors when moving to map coordinates. Final Answer: Account for the Earth’s curvature in angles and distances

Question 5

Best practice with a plane table: how should rays be drawn from stations? When plotting on a plane table from an instrument station, which practice yields the most reliable intersections and clean plotting?

Accepted Answer

Introduction / Context: Plane tabling integrates observation and plotting. Ray drawing from stations must be disciplined to keep the sheet legible and to maintain accuracy in intersections, resections, and radiations. Excessively long rays clutter the sheet and complicate later work. Given Data / Assumptions: Proper orientation, levelling, and centering of the plane table are performed. Lines (rays) are drawn to represent directions to detail points. Good drafting hygiene is required for clarity and precision. Concept / Approach: The recommended practice is to draw only short rays just long enough to mark the required point. This limits overlapping, maintains readability, and focuses attention on the usable segment of each ray near the plotted point. Continuous full-length rays from multiple stations create dense linework and can cause confusion or misidentification during intersection and checking. Step-by-Step Solution: Sight the detail with the alidade and mark a short ray in the correct direction.Plot the point where the short rays intersect (intersection method) or at a scaled distance (radiation method).Repeat only as necessary, avoiding unnecessary extensions across the sheet. Verification / Alternative check: Standard plane table manuals explicitly advise short rays to reduce clutter and avoid masking of subsequent work. Why Other Options Are Wrong: Long continuous rays (A) clutter the drawing; avoiding second rays (C) defeats intersection accuracy; indiscriminate distant sights (D) can degrade precision due to atmospheric and plotting limits; constant redrawing (E) is inefficient and risks wear on the sheet. Common Pitfalls: Forgetting to re-orient between stations; failing to label rays lightly to keep the sheet tidy. Final Answer: Draw only short rays sufficient to contain the required points

Question 6

Traverse bearing computation: apply interior angle at station B Given bearing of AB = 152° 30′. The interior angle ABC measured clockwise at B is 124° 28′ (from BA to BC). Compute the whole-circle bearing (WCB) of BC.

Accepted Answer

Introduction / Context: Traverse computations often require converting a known line bearing and an interior angle at the next station into the bearing of the following line. Careful handling of fore and back bearings and the clockwise/anticlockwise sense of the measured interior angle is critical for correct results. Given Data / Assumptions: Bearing of AB (WCB): 152° 30′. Angle ABC at station B is measured clockwise from BA to BC and equals 124° 28′. Whole-circle bearings are measured clockwise from north (0° to 360°). Concept / Approach: At station B, the reference direction along BA has the back bearing of AB. Compute BA’s bearing at B by adding 180° to AB (modulo 360°). Then, turn the given interior angle clockwise from BA to reach BC. Normalize to 0°–360° as needed. Step-by-Step Solution: Back bearing of BA at B: BB(BA) = 152° 30′ + 180° = 332° 30′.Apply interior angle clockwise to get BC: 332° 30′ + 124° 28′ = 456° 58′.Normalize to 0°–360°: 456° 58′ − 360° = 96° 58′.Thus, WCB(BC) = 96° 58′. Verification / Alternative check: A quick sketch with north lines at A and B confirms the quadrant and magnitude. The result lies in the first quadrant, consistent with adding a 124° 28′ clockwise turn from a bearing near north-west (332° 30′) past north into the east quadrants. Why Other Options Are Wrong: 27° 52′, 148° 08′, and 186° 58′ correspond to incorrect addition/subtraction or wrong use of fore vs back bearing; 316° 58′ is a common error from subtracting instead of adding the interior angle. Common Pitfalls: Forgetting to convert AB to BA at station B; mixing clockwise/anticlockwise sense; failing to wrap angles to 0°–360° correctly. Final Answer: 96° 58′

Question 7

Thin-lens (geometrical optics) applicability of the lens formula: Under what conditions is the standard lens equation 1/f = 1/u + 1/v considered valid for engineering surveying optics and simple optical instruments?

Accepted Answer

Introduction / Context: The thin-lens equation, written as 1/f = 1/u + 1/v, is used in geometrical optics to relate object distance u, image distance v, and focal length f. Surveying instruments (levels, theodolites, and telescopes) rely on this relation when forming images on cross-hair diaphragms. Knowing when this formula applies avoids errors in instrument design, adjustment, and basic optics questions in civil engineering exams. Given Data / Assumptions: Lenses are idealized as thin, so thickness is negligible in first-order calculations. Distances are measured along the principal axis (paraxial approximation). Required accuracy is at the geometrical-optics level; wave effects and thick-lens corrections are not demanded. Concept / Approach: The thin-lens model assumes the principal planes coincide (negligible thickness) and rays make small angles with the axis (paraxial rays). Under these conditions, the simple reciprocal relation predicts image location and magnification accurately enough for most educational and field-calibration purposes. Off-axis points, high-aperture systems, or thick compound lenses need more advanced models (principal planes separated, nodal points, or matrix optics). Step-by-Step Solution: Recognize the thin-lens assumptions: negligible lens thickness and paraxial rays.State the applicable geometry: object and image distances measured along the principal axis only.Adopt the thin-lens formula: 1/f = 1/u + 1/v for computing conjugate positions.Confirm accuracy suffices for geometrical optics problems in simple instruments. Verification / Alternative check: For a simple telescope objective forming a sharp image on the diaphragm, predicted image distances closely match actual values for small apertures and modest fields, validating thin-lens use in most surveying optics questions. Why Other Options Are Wrong: Each individual statement (thin assumption, axial distances, geometrical accuracy) is correct but incomplete alone; all together define the applicability domain. Common Pitfalls: Applying the formula to thick lenses without principal plane corrections; using skew-ray distances; expecting diffraction-limited predictions from a purely geometrical model. Final Answer: All of the above

Question 8

Highway/rail curves – long chord of a simple circular curve: For a simple circular curve of radius R and central (deflection) angle Δ, what is the expression for the length of the long chord joining the curve’s end points?

Accepted Answer

Introduction / Context: In highway and railway alignment, basic elements of circular curves—tangent length, long chord, mid-ordinate—are frequently needed. The long chord connects the points of tangency and depends on the curve radius R and the central angle Δ subtended at the curve center. Given Data / Assumptions: Simple circular curve of radius R. Central (deflection) angle Δ in degrees or radians (consistent units). We seek the long chord length between curve end points. Concept / Approach: The end points and the circle center form an isosceles triangle with vertex angle Δ at the center and equal sides of length R. The half-chord is opposite the half-angle Δ/2. Using elementary trigonometry, the half-chord equals R * sin(Δ/2), so the full chord is twice that value. Step-by-Step Solution: Construct triangle: sides R, R, and chord C; vertex angle at center = Δ.Use sine in the right triangle formed by bisecting the isosceles triangle: (C/2) / R = sin(Δ/2).Compute C/2 = R * sin(Δ/2).Therefore C = 2R * sin(Δ/2). Verification / Alternative check: Check limits: as Δ → 0, sin(Δ/2) ≈ Δ/2 (in radians), so C ≈ R * Δ, which matches the arc approximation for small angles. Why Other Options Are Wrong: R * cos(Δ/2), 2R * cos(Δ/2), R * sin(Δ/2) represent other curve elements (e.g., mid-ordinates or half-chords) but not the full long chord. Common Pitfalls: Using degrees in calculators set to radians; confusing chord with arc length; forgetting the Δ/2 factor. Final Answer: 2R * sin(Δ/2)

Question 9

Transit orientation and azimuth readings: If a transit is oriented by setting Vernier A to the back azimuth of the previous line, taking a back sight, and then rotating to the forward station, what will Vernier A read?

Accepted Answer

Introduction / Context: Precise traverse work relies on careful orientation of the transit (theodolite). One standard method is to set the plate reading to the known back azimuth of the previous line and sight the back point. Understanding what the vernier reads when subsequently sighting the forward station avoids confusion in azimuth-based traverses. Given Data / Assumptions: Vernier A is set initially to the back azimuth of the previous line. A backsight is taken on the previous station to align the instrument. The telescope is then rotated to sight the forward station. Concept / Approach: If the transit circle is set to equal the back azimuth while sighting the back point, the instrument is oriented to the reference meridian. Rotating the telescope to the forward point changes the vernier reading to the azimuth of the forward line directly because azimuths are measured clockwise from the reference meridian and the orientation has been transferred to the plate reading. Step-by-Step Solution: Set Vernier A to back azimuth of the previous line.Backsight the previous station (plate now oriented to the meridian).Rotate to the forward station.Read Vernier A: it now equals the azimuth of the forward line. Verification / Alternative check: Repeat in reverse: if set to forward azimuth and fore-sight the forward point, a backsight to the previous point yields the back azimuth—confirming symmetry. Why Other Options Are Wrong: Bearing may differ from azimuth depending on system (quadrantal vs whole-circle). The back bearing of the forward line would be read if sighting opposite direction. 360° − azimuth results from reading on the opposite face or different zero reference. Common Pitfalls: Mixing whole-circle bearings with azimuths; failing to clamp correctly before rotating; ignoring face-left/face-right conventions. Final Answer: Azimuth of the forward line

Question 10

Topographic interpretation rule: Contours of different elevations can cross each other only under which special terrain condition?

Accepted Answer

Introduction / Context: Contours represent loci of equal elevation and ordinarily never cross or merge. Recognizing the few exceptional cases prevents serious drafting and interpretation mistakes in topographic surveying and civil site planning. Given Data / Assumptions: Contours are drawn at uniform vertical intervals. Each contour corresponds to a unique elevation value. We consider accurate, properly drafted maps. Concept / Approach: Contours cross only when the surface above a given plan location is present at two or more elevations vertically above the same plan point. This happens with an overhanging cliff (undercut face), where higher rock strata project over lower ground. On a vertical cliff, contours become coincident (touch) but do not cross; on saddles and inclined planes they remain separate and ordered. Step-by-Step Solution: Recall the rule: one plan point cannot have two elevations—so contours cannot cross.Identify exception: overhang yields multiple elevations above the same plan location.Drafting outcome: contours of different elevations appear to cross in plan for overhangs.Other features (vertical cliff, saddle, plane) do not produce crossing; they show coincidence or characteristic spacing only. Verification / Alternative check: Field sketches of caves or overhangs confirm that the plan projection overlaps different elevation edges, matching the crossing appearance on contour maps. Why Other Options Are Wrong: Vertical cliff: contours coincide (merge) because horizontal spacing is zero, but they do not cross. Saddle and inclined plane: contours remain distinct and ordered; no crossing occurs. Common Pitfalls: Confusing “coinciding contours” on a vertical cliff with “crossing contours”; misreading dense contour spacing as overlap; ignoring elevation labels. Final Answer: An overhanging cliff (undercut face)

Question 11

Basic concept in surveying: The operation of determining the difference in elevation (RL difference) between two points on the Earth’s surface is called what?

Accepted Answer

Introduction / Context: Surveying distinguishes among several levelling procedures, but all serve the core purpose of finding elevation differences. Understanding the umbrella term and its specific variants helps select the right method for site work, profiles, and transfer of benchmarks. Given Data / Assumptions: Two points need an elevation difference. Standard levelling instruments (automatic/dumpy level) and staff are used. Terminology per conventional surveying texts. Concept / Approach: Levelling is the general operation of determining elevations or differences in level. Within it, simple levelling uses a single set-up visible to both points; differential levelling uses a series of set-ups via change points to connect points far apart; longitudinal (profile) levelling determines elevations along a route for design sections. The question asks for the general name of the operation, hence “Levelling.” Step-by-Step Solution: Identify need: find RL difference between two points.Choose the umbrella term covering all applicable methods: levelling.Note that specific contexts may require simple, differential, or profile levelling.Report the general term as the answer. Verification / Alternative check: Standard definitions: “Levelling is the art of determining elevations of points or their differences of level.” Differential, simple, and longitudinal are subtypes. Why Other Options Are Wrong: Simple, differential, and longitudinal levelling are specific methods. The stem is generic; therefore, the broad term “Levelling” is most accurate. Common Pitfalls: Assuming the question always intends “differential levelling”; overlooking visibility constraints that distinguish methods. Final Answer: Levelling (general process)

Question 12

Linear measurement scales: For a direct vernier scale, how do the graduations relate to the main scale and what direction do they increase?

Accepted Answer

Introduction / Context: Vernier scales enhance precision by subdividing the main scale. Surveying instruments use either direct or retrograde verniers. Knowing their relative graduation direction prevents misreading lengths or angles during precise work. Given Data / Assumptions: We compare the orientation of vernier graduations to the main scale. Definitions follow standard metrology conventions. Concept / Approach: In a direct vernier, the graduations increase in the same direction as those of the main scale, and the vernier’s least division is slightly smaller than the main scale’s least division so that a fixed number of vernier divisions equals a different number of main divisions. A retrograde vernier increases in the opposite direction. The direction of increase governs how readings are added or subtracted when aligning vernier and main scale marks. Step-by-Step Solution: Identify vernier type: direct vernier.State the rule: graduations increase in the same direction as the main scale.Recall dimensional relation: vernier least division is slightly smaller than the main scale’s least division (not larger).Hence, option stating same-direction increase is correct. Verification / Alternative check: Examine any engineer’s vernier caliper or transit circle with a direct vernier; the numbering increases in the same sense along both scales. Why Other Options Are Wrong: Opposite direction defines a retrograde vernier, not a direct one. Larger vernier least division would not produce vernier coincidence behavior for fine interpolation. Common Pitfalls: Confusing “direct” with “retrograde”; misinterpreting which side of the index line to read when scales are mirrored. Final Answer: Vernier graduations increase in the same direction as those of the main scale

Question 13

Levelling instrument care – permanent adjustments of a dumpy/automatic level: How many permanent adjustments are classically recognized for a level instrument used in surveying?

Accepted Answer

Introduction / Context: Permanent adjustments ensure that a level reads true horizontals when the bubble is centered. Recognizing how many such adjustments exist helps technicians maintain instruments and diagnose errors quickly in the field. Given Data / Assumptions: Instrument is a dumpy or automatic level used with a levelling staff. “Permanent adjustments” are those maintained at all times, not temporary set-up steps. Concept / Approach: Classically, a level has two permanent adjustments: (1) The axis of the bubble tube (or the line of the compensator in an automatic level) must be perpendicular to the vertical axis. (2) The line of collimation (line of sight) must be parallel to the axis of the bubble tube. When both are satisfied, centering the bubble makes the line of sight horizontal. Other checks (like focussing to eliminate parallax) are routine but not permanent “adjustments.” Step-by-Step Solution: State adjustment 1: bubble axis ⟂ vertical axis.State adjustment 2: line of collimation ∥ bubble axis.Conclude: exactly two permanent adjustments satisfy horizontal sight when bubble is centered.Apply in field by two-peg test and bubble-axis checks. Verification / Alternative check: Two-peg test detects collimation error; symmetrical readings before and after leveling confirm correct permanent adjustments. Why Other Options Are Wrong: 3, 4, or 6 adjustments overstate the number for a level; such counts might apply to other instruments (e.g., theodolite). Common Pitfalls: Confusing temporary levelling (foot-screw operations) with permanent adjustments; assuming automatic compensator removes the need for collimation checks. Final Answer: 2 in number

Question 14

Metric degree of curve using a 30 m chord: For a simple circular curve with radius R = 300 m and the specified standard chord length c = 30 m, what is the degree of the curve (D)?

Accepted Answer

Introduction / Context: In metric practice, the degree of curve D is often defined as the central angle subtended by a standard chord length (commonly 30 m). This provides a convenient field measure linked directly to chords used during curve setting out by Rankine’s method. Given Data / Assumptions: Radius R = 300 m. Standard chord c = 30 m. D is the central angle subtended by chord c (in degrees). Concept / Approach: Geometry gives the chord–angle relation: c = 2R * sin(D/2). Solving for D yields D = 2 * arcsin(c / (2R)). Substituting the given values produces the numeric degree of curve appropriate for field deflection-angle calculations. Step-by-Step Solution: Write formula: c = 2R * sin(D/2).Solve: D = 2 * arcsin(c / (2R)).Plug numbers: c / (2R) = 30 / (2 * 300) = 30 / 600 = 0.05.Compute: D = 2 * arcsin(0.05) ≈ 2 * 2.864789° ≈ 5.7296° ≈ 5.73°. Verification / Alternative check: Approximate small-angle check: arcsin x ≈ x (radians). Using radians: 0.05 rad ≈ 2.8648°, doubled gives ≈ 5.7296°, matching the exact computation. Why Other Options Are Wrong: 5.37°, 3.75°, 3.57° are inconsistent with the chord–radius relation for c = 30 m and R = 300 m. Common Pitfalls: Using degrees in a calculator set to radians; confusing “degree of curve by arc length” with “degree by chord.” Final Answer: 5.73°

Question 15

Cartographic rule for wide rivers (double-line water features): How should contours be drawn where a river is represented by double lines on the map?

Accepted Answer

Introduction / Context: On topographic maps, water bodies affect how contours are depicted. Streams drawn as a single line are narrow enough that contours can often continue, but wide rivers drawn with double banks require a different convention to avoid implying known elevations beneath water where the bed profile is uncertain or variable. Given Data / Assumptions: The river is wide enough to be shown by two parallel lines (double-line symbol). Bank elevations are known from surveyed ground; the bed elevation is typically not contoured. Standard mapping practice is followed. Concept / Approach: Contours represent ground surfaces of known elevation. In wide water bodies, the ground surface is submerged and its elevation is not mapped as part of the land topography; hence contours terminate at the banks. This avoids misrepresenting the channel bed and preserves clarity for hydrologic interpretation and design tasks such as levee placement and floodplain analysis. Step-by-Step Solution: Identify the mapping symbol: double-line river indicates appreciable width.Follow convention: stop land-surface contours at each bank line.If needed, annotate water surface elevation separately (e.g., spot heights, stages).Do not draw across the water unless the bed is surveyed and separately depicted (not standard for contours). Verification / Alternative check: Inspect standard topographic sheets: contours neatly terminate at banks for wide rivers, while bridges or fords may show related spot levels but not continuous contouring under water. Why Other Options Are Wrong: Drawing across the water implies known ground elevations under water—incorrect for standard contour maps. Parabolic arcs at the channel center are not a contouring convention. “Edge of water only” is ambiguous; the correct reference is the mapped banks. Common Pitfalls: Letting contours cross into the channel; forgetting to close contours correctly at embankments or levees. Final Answer: Contours are stopped at the banks of the river

Surveying Questions