Question 1

Contour Gradient from a Point on a Uniformly Inclined Surface From any chosen point on a surface with a given (uniform) inclination, how many distinct contour gradients (lines of constant slope) can be drawn through that point?

Accepted Answer

Introduction: A contour gradient is a ground line along which the slope (rise over horizontal run) remains constant. In route location and drainage design, engineers often need to trace such lines from a given point on a hillside to maintain a prescribed gradient for roads, canals, or pipelines. Given Data / Assumptions: The surface has a uniform inclination (a plane surface). A prescribed gradient value is fixed (e.g., 1 in n). A specific starting point is chosen on the surface. Concept / Approach: On a plane surface, all lines making the same angle with the horizontal have the same slope magnitude. Through a given point of a plane, infinitely many straight lines can be drawn in different directions; among these, an indefinite set can maintain the same slope magnitude, although their directions differ. Hence, for a specified gradient magnitude on a uniformly inclined plane, there is not just one unique path—there are infinitely many possible contour gradients through the point. Step-by-Step Solution: Model the surface as z = ax + by + c, a plane with constant gradient magnitude.For a chosen slope S, lines through the point can be oriented to maintain vertical rise/run = S.Because line direction is a free parameter, multiple solutions exist.Therefore, the number of contour gradients through the point is indefinite. Verification / Alternative check: Graphically, equal-slope lines on a plane correspond to a family of straight lines; field setting with a clinometer can trace many such lines emanating from the same point. Why Other Options Are Wrong: (a) and (b) artificially restrict possibilities; (d) cannot hold since options are mutually exclusive; (e) is false because contour gradients are meaningful on uniform slopes. Common Pitfalls: Confusing “contour line” (constant elevation) with “contour gradient” (constant slope); assuming unique alignment when only slope magnitude is specified. Final Answer: An indefinite number of contour gradients are possible

Question 2

Chaining Errors – Situations Causing ~1 in 1000 Relative Error with a 20 m Chain Which of the following situations typically introduces an error on the order of 1 part in 1000 when using a 20 m chain for distance measurement?

Accepted Answer

Introduction: Understanding the magnitude of common field errors helps surveyors judge when corrections are necessary. With a 20 m chain, small mishandlings can create significant relative errors near 1 in 1000, impacting layout and quantity computations if not controlled. Given Data / Assumptions: Nominal chain length = 20 m. Small deviations in length, alignment, or slope are considered. Relative error is approximated as error / measured length. Concept / Approach: Each listed situation yields an error roughly 20 mm over 20 m, which is 1/1000. Length error is direct. Alignment and slope errors can be estimated using small-angle approximations: the excess due to using a sloping or misaligned line rather than true horizontal/straight alignment is about d^2/(2L), which with the given numbers is near 0.02 m (20 mm) over 20 m. Step-by-Step Solution: Length error: 20 mm over 20 m ⇒ 0.02/20 = 1/1000.Alignment (one end 0.9 m off): error ≈ d^2/(2L) = 0.9^2/(40) ≈ 0.020 m ⇒ ~1/1000.Slope (0.9 m vertical over 20 m): horizontal shortfall ≈ h^2/(2L) = 0.9^2/40 ≈ 0.020 m ⇒ ~1/1000.Middle offset 0.45 m: similar magnitude by geometry of bowing/offsets. Verification / Alternative check: Exact computations using Pythagoras or alignment triangles confirm each is about 20 mm over 20 m, matching the 1 in 1000 scale. Why Other Options Are Wrong: Each single case indeed produces the specified magnitude; hence selecting any one alone would be incomplete. The correct holistic choice is “All of the above.” Common Pitfalls: Ignoring small alignment errors as “negligible”; forgetting to reduce slope distances to horizontal; overlooking periodic standardization of chains. Final Answer: All of the above

Question 3

Traverse Coordinates – Sign Convention for Total Latitude of a Point In traverse computations, the total latitude (algebraic north–south displacement) of a point is taken as positive under which positional condition relative to the reference paralle…

Accepted Answer

Introduction: Traverse closure and coordinate computations use latitudes (north–south components) and departures (east–west components). A consistent sign convention is vital for balancing and plotting. This question checks the standard algebraic sign of latitude for a point's position. Given Data / Assumptions: Latitude is the projection along the meridian (N–S axis). Departure is the projection along the east–west axis. Reference parallel/meridian define the coordinate origin or baseline. Concept / Approach: By convention in surveying, northings are positive and southings are negative. Hence, any net displacement north of the reference parallel yields a positive total latitude. Eastings are positive departures; westings are negative, but these affect departure, not latitude. Step-by-Step Solution: Resolve each traverse leg into latitude L * cos θ (with sign by quadrant).Sum latitudes algebraically to get total latitude of a point.Interpret sign: positive sum implies the point is north of the reference parallel.Therefore select option (a). Verification / Alternative check: Check a simple case: a single northward leg yields positive latitude; a southward leg yields negative—consistent with convention across textbooks and practice. Why Other Options Are Wrong: (b) reverses the sign; (c) and (d) refer to departures, not latitudes; (e) is unrelated to the sign definition. Common Pitfalls: Confusing latitude with departure; mixing local grid conventions with geodetic signs—always confirm the project standard. Final Answer: When the point lies north of the reference parallel (northing)

Question 4

Offsets by Eye – Maximum Practical Length for Judging Perpendicularity Without Instruments In basic chain surveying, up to what offset length can a right angle from the chain line generally be judged by eye with acceptable accuracy before a cross-st…

Accepted Answer

Introduction: Setting out perpendicular offsets from a baseline is routine in chain surveying. While instruments like cross-staffs and optical squares ensure right angles, short offsets are sometimes judged by eye to save time. Knowing the practical limit avoids unacceptable plan errors. Given Data / Assumptions: Flat, unobstructed ground. Competent observer with normal visual acuity. Acceptable plotting accuracy for small-scale plans. Concept / Approach: Angular error in visual judgment increases the lateral error at the feature by approximately offset_length * sin(error_angle). For larger offsets, even small angular misjudgments create significant displacement. Field practice commonly limits by-eye perpendiculars to about 5 m; beyond this, a cross-staff or optical square is recommended to control angular error. Step-by-Step Solution: Adopt a reasonable visual angular uncertainty (e.g., 1°–2°).Compute lateral error at 5 m: ≈ 5 * sin(1°) ≈ 0.087 m, typically tolerable for rough detail on small scales.At 10 m: error roughly doubles and may exceed plotting tolerances.Hence, limit by-eye offsets to ~5 m; use instruments beyond that. Verification / Alternative check: Surveying manuals cite ~5 m as the practical upper limit for judging right angles by eye, aligning with common site experience. Why Other Options Are Wrong: 10–20 m offsets magnify angular errors; 2 m is overly conservative and unnecessarily slows work. Common Pitfalls: Relying on visual judgment for long offsets; not rechecking with an instrument in built-up or high-accuracy work. Final Answer: 5 m

Question 5

Lemniscate Curve Geometry – Characteristic Angle Ratio at the Polar Ray End For a lemniscate transition curve, the ratio of (i) the angle between the tangent at the end of the polar ray and the straight, to (ii) the angle between the polar ray and t…

Accepted Answer

Introduction: Lemniscate transition curves (e.g., Bernoulli lemniscate used in some route designs) possess useful geometric properties governing how the tangent direction evolves from the initial straight. Recognizing these relationships aids in quick checks of layout angles without re-deriving full equations. Given Data / Assumptions: A lemniscate is used as the transition between a tangent and a circular arc. The polar ray is drawn from the point of commencement along a fixed direction (the straight). Angles considered: (i) between tangent at the polar-ray end and the straight; (ii) between the polar ray and the straight. Concept / Approach: Classical results for the lemniscate show that, at the end of the polar ray, the tangent deflection from the straight equals twice the angle that the polar ray itself makes with the straight. This 2:1 ratio is a handy rule of thumb used in approximate setting-out where lemniscates are adopted instead of clothoids. Step-by-Step Solution: Identify the two relevant angles with respect to the initial tangent (straight).Recall the lemniscate property: tangent deflection at the ray end = 2 × (polar-ray angle).Form the ratio: (tangent–straight angle) / (polar-ray–straight angle) = 2.Hence select option (a) = 2. Verification / Alternative check: Standard geometric derivations of the lemniscate in polar coordinates confirm this proportionality at the specified point, which is routinely quoted in transportation engineering references. Why Other Options Are Wrong: Other integers or fractional values do not match the inherent geometry of the lemniscate; choosing them would contradict established properties of the curve. Common Pitfalls: Confusing lemniscate properties with those of a clothoid (Cornu spiral), which follow different angle–arc length relations. Final Answer: 2

Question 6

Levelling Arithmetic – Expressions for Total Change in Level Along a Line In differential levelling, the total change in level between the first and last stations can be expressed equivalently by which of the following?

Accepted Answer

Introduction: Levelling computations can be carried out by either the Rise and Fall method or the Height of Instrument (HI) method. Both lead to equivalent expressions for the overall change in elevation from start to end of a levelling run. Recognizing the equivalence is useful for checks and error detection. Given Data / Assumptions: Back sight (BS) and fore sight (FS) readings are taken properly. Intermediate sights (IS) are handled according to the chosen method. Reduced levels (RLs) are computed consistently from a known benchmark. Concept / Approach: In the HI method, net change in level = ΣBS − ΣFS. In the Rise and Fall method, net change in level = ΣRises − ΣFalls. By definition of RLs, net change also equals RL_last − RL_first. Therefore, all three statements are equivalent ways of expressing the same total elevation difference across the line. Step-by-Step Solution: From HI: HI = RL + BS; RL_next = HI − FS ⇒ sum across all setups yields ΣBS − ΣFS.From Rise/Fall: RL_next − RL = Rise − Fall ⇒ Σ(RL_next − RL) = ΣRises − ΣFalls.Direct RL difference: RL_last − RL_first equals the same net change.Hence, options (a), (b), and (c) are equivalent, making (d) correct. Verification / Alternative check: Well-kept field books include arithmetical checks ensuring ΣBS − ΣFS = ΣRises − ΣFalls = RL_last − RL_first. Why Other Options Are Wrong: “None of the above” is false since all listed relations are standard and equivalent. Common Pitfalls: Mixing signs on rises/falls; misplacing an IS as BS or FS; inconsistent RL reference causing mismatched totals. Final Answer: All of the above

Question 7

Instrument Reversal – Relationship Between Apparent Error and Actual Error When an instrument reading is taken in one position and then with the instrument reversed (e.g., telescope transited or vernier read on the opposite), the apparent error obse…

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Introduction: Reversal (face left/right or direct/reverse) is a standard technique to detect and eliminate certain instrumental errors such as index error or collimation error. Understanding how the observed discrepancy relates to the true error enables correct adjustment or averaging procedures. Given Data / Assumptions: The instrument has a small constant error (e.g., line of sight not exactly perpendicular to trunnion). Readings are taken in two opposite positions that reverse the sign of the error. The mean of the two readings is used to eliminate the error. Concept / Approach: When the instrument is reversed, the error contribution changes sign. The difference between the two readings (apparent error on reversal) equals 2 * (actual error). Hence, the true error is half the observed discrepancy, and averaging the two readings cancels the error entirely in the mean value. Step-by-Step Solution: Let true value = T, actual error = e.Reading 1 (position A) = T + e; Reading 2 (reversed position B) = T − e.Apparent error on reversal = (T + e) − (T − e) = 2e.Therefore, actual error = (apparent error)/2 and averaging gives T. Verification / Alternative check: The same logic underlies double centering and face-left/face-right procedures in theodolite observations to remove collimation and index errors. Why Other Options Are Wrong: Equal, thrice, or half the actual error do not match the algebra of sign reversal; “none” is inapplicable because a precise relation exists. Common Pitfalls: Confusing the observed difference with the true error; forgetting to take the mean to eliminate the error in reported values. Final Answer: Twice the actual error

Question 8

Small-angle geometry in surveying: What is the ratio of the linear displacement at the end of a line (arc chord) subtended by an angle of 1 second, to the length of that line? Choose the standard constant used in precise angular-to-linear conversion…

Accepted Answer

Introduction / Context: In precise surveying and astronomy, small angles are frequently converted to linear displacements along an arc or chord. Knowing the conversion factor for one arc-second is essential when translating instrument sensitivities (seconds of arc) into ground linear errors over long sight lengths. Given Data / Assumptions: Angle considered = 1 arc second. Small-angle approximation applies: arc length s ≈ R * θ with θ in radians. Comparison is a pure ratio: linear displacement : length of line. Concept / Approach: One radian equals 180/π degrees, or 206 265 arc-seconds. Therefore, 1 arc-second equals (π / 648 000) radians ≈ 4.8481 * 10^-6 rad. The corresponding linear displacement at the end of a line is the line length multiplied by this angular value. Hence the ratio displacement:length is approximately 1:206 265, commonly rounded as 1:206 300 for practical surveying constants. Step-by-Step Solution: Convert 1″ to radians: θ = (π / 648000) rad ≈ 4.8481e-6.For a line of length L, displacement ≈ L * θ.Thus, displacement : L ≈ 1 : (1/θ) ≈ 1 : 206265 ≈ 1 : 206300 (rounded constant). Verification / Alternative check: Using calculators or tables of astronomical constants yields the same figure. Many survey texts adopt 1:206 265; rounding to 1:206 300 keeps field arithmetic simple without affecting design-level accuracy. Why Other Options Are Wrong: 1:3440 pertains to 1 minute (since 60′ ≈ 1° and 1 rad ≈ 57.3°). 1:57 refers to 1 radian in degrees, not seconds. 1:100 and 1:108 000 are unrelated magnitudes. Common Pitfalls: Forgetting to use radians; mixing up seconds with minutes; using chord vs arc distinctions when angles are not sufficiently small. Final Answer: 1:206 300

Question 9

Setting out a simple circular curve: For a curve of radius R = 600 m, what is the maximum practical chord length commonly adopted for calculating offsets (to control accuracy and field convenience)?

Accepted Answer

Introduction / Context: Offsets from chords or from the long chord are used extensively to set out simple circular curves. The choice of peg interval (chord length) influences both accuracy (through small-angle approximations) and site productivity. For moderate to large radii, standard practice fixes an upper bound on the chord length for reliable offset computations. Given Data / Assumptions: Simple circular curve with R = 600 m. Offsets computed by common field formulas such as mid-ordinate y ≈ c^2 / (8R) and ordinates to chord. Need a practical maximum chord length used in routine highway/rail setting out. Concept / Approach: For typical highway curves with R in hundreds of meters, peg intervals of 20–30 m are standard. The small-angle approximations used in offset formulae remain sufficiently accurate at c ≤ 30 m for R around 600 m, while longer chords would reduce point density and may degrade fit at transitions or in rugged terrain. Hence many manuals specify 30 m as the maximum peg interval for offsets on such curves, with 10–20 m used where finer control is desired. Step-by-Step Solution: Select a chord length that keeps c/R small (c ≪ R).For R = 600 m and c = 30 m, c/R = 0.05 → small-angle relations hold well.Therefore adopt 30 m as the practical maximum for calculating offsets. Verification / Alternative check: Using y = c^2/(8R): with c = 30 m, y ≈ 900/(4800) ≈ 0.1875 m, an easily measurable offset; errors from chord-arc substitution are negligible at this scale. Why Other Options Are Wrong: Shorter lengths (10–25 m) are acceptable but not the requested maximum commonly adopted; choosing them increases point density without necessity for R = 600 m. Common Pitfalls: Using excessive peg spacing on sharp curves; forgetting that terrain or accuracy needs may still warrant shorter chords. Final Answer: 30 m

Question 10

Chain surveying terminology: What does “ranging” mean in the context of chaining a straight line between two fixed end points?

Accepted Answer

Introduction / Context: “Ranging” is a basic operation in chain surveying used to align intermediate points on a straight line between two end stations. Correct understanding of this term is necessary to coordinate the roles of the leader and follower and to ensure straight-line measurement across undulating ground. Given Data / Assumptions: Two end points of the line are known and intervisible or made intervisible. Objective is to place intermediate marks so that the chain can be laid straight. Standard ranging methods: direct, reciprocal, and by eye or with ranging rods. Concept / Approach: Ranging is the act of fixing points that lie on the straight line joining two given points. It involves aligning a ranging rod (or temporary mark) so that it visually falls on the line of sight connecting the end stations. Direct ranging is used when both stations are intervisible; reciprocal ranging is employed when an intervening obstruction prevents direct alignment from one end. Step-by-Step Solution: Set up rods at both ends of the intended line.Guide the intermediate rod until it visually aligns with both end rods.Confirm alignment by parallax checking from both ends or by reciprocal adjustments if needed. Verification / Alternative check: Field checks include sighting back and forth and ensuring that the intermediate marks remain on line when the observer changes position slightly. Why Other Options Are Wrong: Option A refers to an off-line point; C confuses ranging with mere distance computation; D describes offsetting, a different operation; E is not applicable. Common Pitfalls: Leaving intermediate points slightly off-line over rolling ground; not using reciprocal ranging when direct line of sight is blocked. Final Answer: Establishing an intermediate point on the line

Question 11

Levelling field procedure: Identify the correct statement about the sequence of sights on change points and at the start/close of a levelling line.

Accepted Answer

Introduction / Context: Levelling book work follows a strict sequence to avoid blunders. Understanding which readings are back sights (B.S.) and fore sights (F.S.) at the start, at change points (turning points), and at the close of a run ensures that checks like ΣB.S. − ΣF.S. match the elevation differences. Given Data / Assumptions: A levelling line starts on a benchmark (known R.L.). At change points, the instrument is moved and the same point is sighted from both setups. The run ends at a final point or benchmark. Concept / Approach: By convention, the first staff reading taken after setting up is a back sight to establish the height of instrument (H.I.). The last reading before moving the instrument from its current setup is a fore sight to a change point. After moving, the first reading is again a back sight on that change point. A levelling line, therefore, commences with a back sight and must close with a fore sight to the final point. Step-by-Step Solution: Start at BM: take B.S. → compute H.I.Before shifting instrument: take F.S. on C.P. → close that setup.At new setup: first read B.S. on same C.P. → continue sequence.At final point: take F.S. to close the run. Verification / Alternative check: The arithmetic check ΣB.S. − ΣF.S. = R.L.(last) − R.L.(first) is satisfied only when the line starts with B.S. and ends with F.S., aligning with this convention. Why Other Options Are Wrong: Option A can be ambiguously interpreted; the safe universally correct statement about the line is (d). Options B and C contradict standard procedure. Common Pitfalls: Mislabeling a C.P. reading; forgetting to close with an F.S.; mixing up the sequence which breaks the standard book check. Final Answer: The line commences with a back sight and closes with a fore sight

Question 12

Standardization pull for a 30 m metallic tape: What pull (tension) is typically applied so that the tape reads its standardized length during measurement?

Accepted Answer

Introduction / Context: Tapes are standardized to read true length at a specified temperature and tension (pull). Applying the correct pull minimizes systematic error from elastic stretch or slack. Metallic tapes (linen or synthetic fabric reinforced with fine wires) require relatively small standardized pulls compared with steel tapes. Given Data / Assumptions: Tape length = 30 m (metallic tape, not a steel tape). Field temperature is assumed close to the standard (or temperature correction handled separately). Pull is applied steadily without jerks. Concept / Approach: Survey practice distinguishes metallic tapes from steel tapes. Metallic tapes are lighter, more flexible, and standardized at lower tensions, commonly around 2 kg, to limit permanent set and to reflect their elastic characteristics. Using a much higher pull (typical of steel tapes) risks overstretching and increases random error if the pull cannot be held steady. Therefore, for a 30 m metallic tape, a pull of about 2 kg is typically adopted as the standardization pull unless a specific calibration certificate specifies otherwise. Step-by-Step Solution: Identify tape type: metallic (fabric/metal composite), not steel.Recall standardization: lower pull values (≈ 2 kg) are specified.Adopt 2 kg as the correct pull for standard length. Verification / Alternative check: Calibration data from manufacturers and many surveying manuals list ~2 kg for metallic tapes and higher values (often 5–8 kg) for steel tapes, affirming this choice. Why Other Options Are Wrong: 1 kg may be too low to remove slack; 3–5 kg reflects steel tape standards and risks overstretching a metallic tape. Common Pitfalls: Applying steel-tape pulls to metallic tapes; ignoring the calibration sheet; failing to use a spring balance, leading to inconsistent tension. Final Answer: 2 kg

Question 13

Included angle at B from bearings: If the whole-circle bearing of AB is 190° and that of CB is 260° 30', what is the included angle ABC at station B (the smaller interior angle between BA and BC)?

Accepted Answer

Introduction / Context: Converting whole-circle bearings (WCB) to included angles at a traverse station is a routine computation. Careful handling of back bearings is needed when bearings are given for directions that do not originate at the station of interest. Given Data / Assumptions: Bearing of AB (from A to B) = 190°. Bearing of CB (from C to B) = 260° 30'. Required: included angle at B between BA and BC (smaller angle). Concept / Approach: At station B we need bearings of BA and BC. These are the back bearings of AB and CB respectively. Back bearing = fore bearing ± 180° (adjust to 0–360°). Then take the smaller difference between the two resulting bearings to get the interior angle at B. Step-by-Step Solution: Back bearing of AB → BA = 190° − 180° = 10°.Back bearing of CB → BC = 260°30' − 180° = 80°30'.Included angle at B = |80°30' − 10°| = 70°30'. Verification / Alternative check: A sketch with rays at 10° and 80°30' from a common origin (B) visibly confirms an acute included angle of 70°30'. Why Other Options Are Wrong: 99°30' and 80°30' arise from misinterpreting fore/back bearings; 120°30' is unrelated; “None” is incorrect because the computation is definite. Common Pitfalls: Forgetting to convert to back bearings at the correct station; taking the reflex angle instead of the smaller interior angle. Final Answer: 70° 30'

Question 14

Classification of surveys by purpose and content: Surveys carried out to depict relief and planimetric features such as mountains, rivers, lakes, wooded areas, and cultural details are called:

Accepted Answer

Introduction / Context: Surveying projects are categorized by objective—boundary definition, engineering design, mapping relief and culture, or reconnaissance. Recognizing the correct category helps in selecting appropriate instruments, scales, and contouring methods. Given Data / Assumptions: The mapping must show both natural features (relief, hydrography, vegetation) and cultural features (roads, settlements). Contours or other relief-representation methods are required. Output is a general map suitable for planning and analysis. Concept / Approach: Topographical surveys aim to represent the three-dimensional form of the land on a two-dimensional map. They include contours or spot heights for relief and depict planimetric details like rivers, lakes, forests, roads, and buildings. This differs from cadastral surveys (property boundaries), city surveys (urban detailed layouts at large scales), and simple guide maps (schematic, not precise). “Plane surveys” refers to an assumption about earth curvature, not a purpose-based class. Step-by-Step Solution: Identify requirement to show relief → implies topographic mapping with contours.Confirm inclusion of natural and cultural details → characteristic of topographical surveys.Select the category “Topographical surveys”. Verification / Alternative check: National mapping agencies label sheets with “Topographical” where contours and cultural details are systematically compiled. Why Other Options Are Wrong: Cadastral: boundary/legal emphasis; City: urban planning at very large scale; Guide maps: not survey-grade; Plane surveys: a geometric assumption used in many types, not a purpose. Common Pitfalls: Confusing “plane survey” with “planimetric map”; assuming cadastral always includes contours (it generally does not). Final Answer: Topographical surveys

Question 15

Optics of instruments: The ratio of the distances at which a stated small length (detail) can be just distinguished by a telescope and by the unaided human eye, respectively, is called the:

Accepted Answer

Introduction / Context: When specifying surveying telescopes (levels, theodolites, total stations), several optical performance terms arise: magnification, resolving power, brightness, field of view. Understanding these distinctions ensures correct selection for sight length and target size. Given Data / Assumptions: A “stated small length” is the test object (e.g., a line pair or small graduated mark). Two distances are compared: at which the telescope resolves it and at which the unaided eye resolves it. All other conditions (illumination, contrast) are comparable. Concept / Approach: Magnification (angular magnifying power) can be defined operationally as the ratio of distances at which a given fine detail is just distinguishable through the telescope and by the naked eye. If the eye can just distinguish a mark at distance d_e, and with the telescope it can be distinguished at d_t, then magnification M ≈ d_t / d_e (equivalently, the ratio of the angular sizes after the instrument). Resolving power, in contrast, concerns the smallest angular separation between two points the instrument can distinguish, and brightness deals with exit pupil/illumination—not with this distance ratio definition. Step-by-Step Solution: Define a small standard detail.Find d_e where the eye resolves it unaided; find d_t where the telescope resolves it.Compute M = d_t / d_e → the requested ratio. Verification / Alternative check: Instrument datasheets list magnification (e.g., 20x, 30x), which aligns with practical “distance ratio” tests for just resolvable details. Why Other Options Are Wrong: Resolving power is a threshold angular separation, not a distance ratio against the eye. Brightness/field factor describe other optical qualities. Common Pitfalls: Interchanging magnification with resolution; assuming higher magnification always improves resolution irrespective of optical quality and seeing conditions. Final Answer: Magnification of the telescope

Question 16

Highway/railway transition geometry – lemniscate condition for being transitional throughout A lemniscate curve inserted between two tangents will act as a transition curve for its entire length if the polar deflection angle at the apex has what val…

Accepted Answer

Introduction / Context: The lemniscate is sometimes used as a transition curve in highway and railway alignment when the overall deflection between tangents is comparatively large. For a transition curve, the curvature should start from zero at the tangent and vary gradually to match the circular curve portion. This question asks for the specific polar-deflection requirement at the apex that ensures the lemniscate remains transitional throughout, without any portion behaving like a simple circular arc. Given Data / Assumptions: A pair of straight tangents meet with intersection (deflection) angle Δ. A single symmetric lemniscate is used between the tangents. Pola r deflection angle at the apex is the maximum polar angle of the curve (measured from the initial tangent to the radius vector at the apex). We require curvature to increase and then decrease in the correct proportion so that throughout the used length it behaves as a transition (no constant curvature segment). Concept / Approach: Geometric relations of the lemniscate yield a fixed ratio between the deviation (deflection) angle of the alignment and the maximum polar deflection at the apex for which the curvature law remains transitional everywhere. Classical surveying results show that for a lemniscate used between tangents, transitional behavior throughout the curve requires the apex polar deflection to be one-sixth of the overall intersection angle Δ. Step-by-Step Solution: Let Δ be the tangent intersection angle.Let α_apex be the apex polar deflection.For a lemniscate to remain transitional throughout, use the relation α_apex = Δ/6.Therefore the correct required value is Δ/6. Verification / Alternative check: Surveying references commonly tabulate the lemniscate condition as apex polar deflection = Δ/6 for fully transitional behavior; larger apex angles cause part of the curve to act effectively non-transitional over a segment, which is undesirable for comfort and safety at speed. Why Other Options Are Wrong: Δ/2, Δ/3, Δ/4: these overstate the apex polar angle, leading to nonuniform transition behavior and possible entry into a pseudo-constant-curvature region. Common Pitfalls: Confusing the apex polar deflection with the spiral angle; mixing up the lemniscate with the clothoid (cubic parabola) where different relationships apply; assuming any lemniscate is automatically transitional throughout without checking the apex condition. Final Answer: Δ/6

Question 17

Plane table surveying – naming the operation that makes all drawn lines parallel to ground directions The process of revolving the plane table about its vertical axis until every line on the drawing sheet becomes parallel to the corresponding line o…

Accepted Answer

Introduction / Context: Plane tabling is a rapid field-plotting method. Several preparatory operations are performed at each station: centering, levelling, and orientation. Many learners confuse these steps. This question focuses on the specific action of revolving the board so that all plotted lines are parallel to the corresponding ground directions, which is essential for direct graphical plotting. Given Data / Assumptions: The plane table is already leveled in the sense of being made horizontal. The instrument is placed at a known station on the ground whose plotted position on the sheet is also known. The board can be rotated about its vertical axis until alignment is achieved. Concept / Approach: Levelling makes the board horizontal. Centering places the plotted station exactly over the ground station (usually by plumbing fork or optical plummet). Orientation is the action of rotating the board to make the directions of all plotted lines parallel to their actual directions on the ground, which can be done by backsighting a known point or by matching the magnetic meridian if magnetic orientation is used. Step-by-Step Solution: Level the board with the spirit level.Center the board so the plotted point is vertically above the station.Rotate the board until the line on the sheet to a known point becomes parallel to the line of sight to that point.This rotation process is the orientation of the plane table. Verification / Alternative check: If correctly oriented, rays plotted to multiple visible points should coincide with their ground directions; misorientation produces systematic parallel shifts in plotted detail. Why Other Options Are Wrong: Levelling: addresses horizontality, not directional parallelism. Centering: locates the station position, not the directions. Setting: a generic term sometimes used loosely, but not the precise technical term for this rotation step. Common Pitfalls: Attempting to begin detail plotting without verifying orientation; confusing magnetic orientation with backsight orientation; ignoring local attraction when using the magnetic method. Final Answer: Orientation

Question 18

Traverse balancing – quadrant of the closing error from sign of latitude and departure sums In a closed traverse, the sum of south latitudes exceeds the sum of north latitudes and the sum of east departures exceeds the sum of west departures. In whi…

Accepted Answer

Introduction / Context: In coordinate (latitude–departure) computations for traverses, the algebraic sums of latitudes (northings/southings) and departures (eastings/westings) diagnose misclosure. Their signs directly reveal the quadrant in which the closing error vector lies. This question tests your ability to infer that quadrant correctly from the given inequalities. Given Data / Assumptions: ΣS > ΣN, so the net latitude is southing. ΣE > ΣW, so the net departure is easting. The resultant closing line is drawn from the computed end point back to the start point. Concept / Approach: Latitude is the north–south component; departure is the east–west component. If the algebraic sum indicates net southing and net easting, the closing error vector must point toward the south–east direction. Therefore, the closing line lies in the south–east quadrant (SE). After identifying the quadrant, balancing (e.g., Bowditch or transit rule) distributes corrections proportionally to remove the misclosure. Step-by-Step Solution: Check latitude sums: ΣS > ΣN → net southing → negative latitude.Check departure sums: ΣE > ΣW → net easting → positive departure.Combine components: (E, S) → south–east quadrant.Hence, the closing line is in the SE quadrant. Verification / Alternative check: Plot a simple vector with components (east positive, south negative). The directional quadrant immediately shows as SE, confirming the verbal reasoning. Why Other Options Are Wrong: NE requires net northing and easting; NW requires northing and westing; SW requires southing and westing. None match the given sign pattern. Common Pitfalls: Mixing signs for latitude and departure; drawing the closing line in the opposite sense; forgetting that ΣLat and ΣDep are computed algebraically with sign conventions. Final Answer: South–east quadrant

Question 19

Optics in surveying instruments – aberration properties of a Huygens eyepiece The Huygens eyepiece used in many optical instruments is best described as:

Accepted Answer

Introduction / Context: Eyepieces influence image quality and reading ease in surveying telescopes and microscopes. Two common negative eyepieces are Huygens and Ramsden. Their aberration characteristics determine which is preferable for specific instruments and observing tasks. This question asks what property the Huygens eyepiece is noted for in classical instrument design. Given Data / Assumptions: Huygens eyepiece consists of two plano-convex lenses with plane faces inward. It is a negative eyepiece; the real image forms before the eyepiece. We are comparing its control of chromatic and spherical aberrations qualitatively. Concept / Approach: The Huygens arrangement provides partial achromatism: the separation and focal powers of the two elements are chosen so longitudinal chromatic aberration is minimized for the eye lens and field lens combination. However, the Huygens eyepiece is not aplanatic; spherical aberration and coma are not fully corrected to the same extent. Hence standard texts describe it as achromatic (to a first order) but not aplanatic, making option describing it as achromatic the best statement among the choices. Step-by-Step Solution: Identify eyepiece design: two plano-convex lenses, plane faces toward the eye and the objective side respectively.Assess chromatic behavior: element spacing and focal powers reduce net color error → achromatic behavior.Assess aplanatic behavior: spherical/comatic errors are not eliminated → not aplanatic.Therefore, classify Huygens as achromatic (not aplanatic). Verification / Alternative check: Instrument manuals routinely recommend Huygens where color correction is valued with simple construction, while advanced wide-field, aplanatic eyepieces require more complex designs. Why Other Options Are Wrong: Aplanatic: overstates correction; Huygens does not remove spherical/comatic errors fully. Both aplanatic and achromatic: too strong; not supported by design limitations. Neither: too weak; it does achieve useful achromatization. Common Pitfalls: Confusing Huygens with Ramsden; assuming any two-lens eyepiece is both achromatic and aplanatic; overlooking the negative eyepiece property that keeps cross-wires sharply defined. Final Answer: Achromatic (chromatic aberration minimized by design)

Question 20

Levelling applications – purpose of profile levelling along a line Profile levelling is typically carried out in order to determine which of the following for design and earthwork planning?

Accepted Answer

Introduction / Context: Levelling serves many survey tasks: general mapping, contouring, cross sections, and detailed longitudinal profiles. A design team often needs the ground elevations along a proposed centerline for a road, canal, pipeline, or transmission line. This procedure is known as profile (or longitudinal) levelling. The question checks your understanding of its specific deliverable. Given Data / Assumptions: A line (straight or laid out by chainage) is chosen on the ground. Staff readings are taken at regular intervals and at critical points (change of gradient, culverts, intersections). Reduced levels are computed for all chainages. Concept / Approach: Profile levelling outputs a longitudinal section—elevation versus chainage—used to design gradients, earthwork, and drainage. It is different from contouring, which gives a plan of equal-elevation lines over an area, and from property surveys, which define legal boundaries. Reservoir capacity estimation relies on area–elevation curves derived from contouring or bathymetry, not merely a single profile. Step-by-Step Solution: Establish a benchmark and start chainage.Take BS, IS, FS readings at stations along the line.Compute reduced levels (e.g., height of instrument or rise-and-fall methods).Plot RL vs chainage to obtain the longitudinal profile for design. Verification / Alternative check: Design drawings for linear projects always include a longitudinal profile and usually cross sections, confirming the purpose of profile levelling as elevations along a line. Why Other Options Are Wrong: Contours of an area: requires systematic area levelling or photogrammetry. Reservoir capacity: derived from contour areas, not a single profile. Boundaries: a cadastral task relying more on traversing than on detailed profiles. Common Pitfalls: Using too coarse an interval, missing breaks in slope, and neglecting cross profiles where side earthworks matter. Final Answer: Elevations along a straight (or selected) line

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