Question 1

Map Scale & Curvature – When Must Earth Curvature Be Considered? For plane surveying approximations, at what approximate area extent should curvature of the Earth be taken into account to avoid significant positional error on maps?

Accepted Answer

Introduction / Context:Small-area surveys can often neglect Earth curvature and treat the surface as a plane. As the extent grows, curvature introduces measurable discrepancies between plane and spherical geometry, affecting distance reduction, azimuth convergence, and level surfaces. Surveyors use a rule-of-thumb area threshold to decide when geodetic considerations become necessary.Given Data / Assumptions:Plane surveying approximations are acceptable only for limited extents.Target is a practical area threshold rather than a precise formula.Normal topographic accuracy and common map scales are assumed.Concept / Approach:Many standard texts suggest that for areas on the order of a few hundred square kilometres, curvature effects begin to compromise plane assumptions. An often-cited nominal limit is around 260 square kilometres. Beyond this, curvature corrections (and sometimes refraction considerations in levelling) and map projections should be applied for accuracy and consistency, particularly when joining with adjacent control networks.Step-by-Step Solution:Adopt the common threshold: take curvature into account if area ≳ 260 km^2.For smaller isolated surveys, plane methods generally suffice.For larger or long, narrow projects (e.g., highways), consider linear extent and accumulated angular distortion even below this area.Select the closest standard choice: 260 sq km.Verification / Alternative check:Specification documents and teaching references list similar thresholds; exact need also depends on accuracy class and whether the survey ties to geodetic control and projections.Why Other Options Are Wrong:100–200 sq km may still be safe for plane treatment in many engineering tasks; 400 sq km is too lax and risks noticeable distortion.Common Pitfalls:Relying on area alone; elongated corridors may require geodetic treatment sooner; forgetting projection scale factors when moving to map coordinates.Final Answer:260 sq km

Question 2

Error detection in controlled traverses (theodolite): pick the correct statements about locating gross errors in bearing or taping. Read each technique and decide which are valid ways to isolate the blundered station or leg.

Accepted Answer

Introduction / Context:In a controlled theodolite traverse, gross blunders (wrong bearing, transposed angle, wrong length) can seriously distort closure. Practical graphical diagnostics help pinpoint the fault before recomputing or recollecting data. This question asks which plotting-based techniques are valid for isolating the blundered station or leg.Given Data / Assumptions:A closed or controlled traverse with known start/end control.One dominant gross error (bearing or length) assumed.Angles and lengths otherwise consistent; plotting to a reasonable scale.Concept / Approach:Graphical error-hunting relies on the geometry of misclosure. Re-plotting from opposite ends, examining the direction of misclosure, and relating the closing-error vector to individual legs provide clues about where the blunder sits. These time-tested methods are found in traverse adjustment practice and field troubleshooting guides.Step-by-Step Solution:Method 1 (Plot from each end): If coordinates of a particular station agree when plotted forward and backward, the inconsistency leading to misclosure must lie at that station. This reveals a bearing blunder local to that point.Method 2 (Perpendicular through sector of closing line): Drawing a perpendicular to the sector of the closing line that intersects the traverse suggests the likely faulty station (bearing mistake), where angular sense changed incorrectly.Method 3 (Closing-error bearing vs. leg bearing): If length (taping) of one leg is grossly wrong, the misclosure vector tends to align roughly with that leg’s bearing; plotting at a convenient scale makes this evident.Verification / Alternative check:After identifying the suspect station/leg, recompute with corrected data or perform an independent check (repeat sight/measure). If adjusted, the traverse should close within permitted tolerances using Bowditch or transit rules.Why Other Options Are Wrong:Each of A, B, and C is a recognized diagnostic; therefore selecting anything other than “All the above” ignores valid techniques.Common Pitfalls:Assuming multiple simultaneous gross errors (these methods are most effective when a single dominant blunder exists); plotting at too small a scale to see directionality; confusing systematic drift with one-off blunders.Final Answer:All the above.

Question 3

Surveying telescope construction: where is the diaphragm (cross-hair reticle) mounted inside a telescope for correct imaging and focus?

Accepted Answer

Introduction / Context:The diaphragm (cross-hair reticle) must coincide with the real image formed by the objective to eliminate parallax and allow precise pointing. Knowing its position helps understand eyepiece and objective focusing roles in levels and theodolites.Given Data / Assumptions:Objective forms a real image at its focal plane inside the tube.Eyepiece magnifies this real image and the cross hairs together.Parallax-free viewing requires the reticle to be at the same plane as the objective’s image.Concept / Approach:The correct location is at the objective’s focal plane, which lies nearer the eyepiece end of the telescope tube for distant targets. The eyepiece is adjusted so the reticle is sharply visible; the objective focus brings the target’s image onto the same plane as the reticle.Step-by-Step Solution:Objective forms image at focal plane inside tube.Reticle (diaphragm) is fixed at that plane near the eyepiece end.Eyepiece focuses the eye on the reticle-image plane; objective focusing aligns the target image to that plane.Verification / Alternative check:Parallax test: move the eye slightly; no relative motion between image and cross hairs indicates coincidence of planes—possible only with reticle at the objective’s focal plane near the eyepiece.Why Other Options Are Wrong:Mounting inside the eyepiece or near the objective end would not coincide with the objective image plane for general targets; mid-tube is arbitrary.Common Pitfalls:Assuming the reticle is part of the eyepiece; confusing the internal focusing lens with the reticle position.Final Answer:In the telescope at the end nearer the eye-piece (at the objective’s focal plane)

Question 4

Why whole-circle bearing (WCB) is preferred over quadrantal bearing (QB) in practice? Choose the most compelling advantage that makes computations simpler.

Accepted Answer

Introduction / Context:Bearings can be recorded either as whole-circle bearings (0°–360° from north) or as quadrantal bearings (N/S, angle, E/W). Computational convenience often dictates the preferred system for traverse reductions and coordinate computations.Given Data / Assumptions:Need to compute latitudes and departures via sin and cos of a direction angle.Desire to minimize sign errors and quadrant confusion.Declination corrections are handled separately.Concept / Approach:With WCB, each direction is a single angle from 0° to 360°. Trigonometric functions and signs of the components are handled automatically by the angle's quadrant when using standard calculators/tables. Quadrantal bearings need explicit quadrant labeling (N/S ... E/W) and careful sign assignment for eastings and northings, inviting mistakes.Step-by-Step Solution:Express line direction as WCB θ (0°–360°).Compute latitude = L * cos θ; departure = L * sin θ directly.Avoid manual sign changes required in QB conversion.Verification / Alternative check:Traverse software and tables typically accept azimuth/WCB inputs directly, reducing operator intervention and errors.Why Other Options Are Wrong:A and B are vague or not decisive; C highlights an issue but WCB’s key practical advantage is computational simplicity with trig functions; E is false because WCB and azimuth are essentially the same reference system.Common Pitfalls:Mixing magnetic, grid, and true bearings; forgetting to apply convergence or declination when converting to/from WCB.Final Answer:Because its trigonometric values can be directly extracted from ordinary tables without quadrant interpretation

Question 5

Internal focusing telescopes: focusing is achieved by moving which internal optical element?

Accepted Answer

Introduction / Context:Many modern surveying telescopes use internal focusing to keep the external length fixed and to protect against dust/moisture. Understanding which lens is moved clarifies why such telescopes have stable collimation and reduced backlash.Given Data / Assumptions:Telescope is of the internal focusing type.Objective remains fixed relative to the tube ends.Focusing alters the position of the real image formed by the objective.Concept / Approach:Internal focusing designs commonly employ a movable negative (concave) lens between the objective and eyepiece. Translating this lens shifts the focal position without moving the objective, allowing focus over a wide range with minimal external movement and preserved sealing.Step-by-Step Solution:Objective forms an intermediate image.A movable concave lens alters effective focal distance to bring the image onto the reticle plane.Eyepiece then magnifies the reticle-image for viewing.Verification / Alternative check:Manufacturers’ optical diagrams and adjustment manuals confirm the use of a movable negative lens for internal focusing.Why Other Options Are Wrong:Moving the objective is characteristic of external focusing; the eyepiece is adjusted for reticle clarity, not target focus; naming a specific “plano-convex only” is too restrictive.Common Pitfalls:Confusing eyepiece focusing (reticle clarity) with target focusing; assuming objective movement in sealed internal-focus designs.Final Answer:An internal concave lens (negative focusing lens)

Question 6

Transition (clothoid) curve setting out: A 90 m transition is introduced between a straight and a circular curve of radius R = 500 m. What is the maximum deflection angle (from the tangent) used to locate the junction point (end of transition)?

Accepted Answer

Introduction / Context:For a clothoid (spiral) transition, the tangent deflection used in field setting-out increases with the square of the distance along the transition. The final or maximum deflection at the junction with the circular arc is a standard quantity required to peg the end point precisely.Given Data / Assumptions:Transition length L = 90 m.Circular-curve radius R = 500 m.Standard spiral relations apply (A^2 = R * L).Concept / Approach:For a clothoid transition laid from the tangent, the maximum deflection at the junction point (end of transition) from the original tangent is given by the well-known approximate field formula: θ_max (in radians) = L / (6R). This comes from integrating the linear curvature law along the length of the spiral and is sufficiently accurate for highway and railway work.Step-by-Step Solution:Compute θ_max (radians) = L / (6 * R) = 90 / (6 * 500) = 90 / 3000 = 0.03 rad.Convert to degrees: θ_max (degrees) = 0.03 * 57.2958 ≈ 1.7189°.Convert decimal degrees to D–M–S: 0.7189° * 60 ≈ 43.133′; fractional minutes 0.133′ * 60 ≈ 7.98″ ≈ 8″.Therefore θ_max ≈ 1° 43′ 08″.Verification / Alternative check:Using A^2 = R L gives A = √(500 * 90) ≈ 212.13. The spiral angle at end = L^2 / (2 A^2) = 8100 / (2 * 45000) = 8100 / 90000 = 0.09 rad? Correction: with correct clothoid relations, θ_end = L / (2R) for total change in tangent direction, while the field deflection from tangent to chord for setting-out is θ_max = L / (6R). The smaller value (1°43′08″) corresponds to the setting-out deflection used at the junction, matching highway practice tables.Why Other Options Are Wrong:The other three angles differ by 10″ increments but do not match the computed 0.03 rad conversion. They reflect near values but are not exact for L = 90 m, R = 500 m.Common Pitfalls:Mixing up the total spiral angle (change in tangent direction) with the setting-out deflection to the junction; using degrees directly in the formula without converting minutes/seconds correctly.Final Answer:1° 43' 08''

Question 7

Magnetic terms and lines: identify the correct statements about declination and related isogonic/agonic lines.

Accepted Answer

Introduction / Context:Converting between magnetic and true bearings requires understanding magnetic declination and the map of its spatial variation. Standard terms include declination, isogonic lines, and agonic lines, used on navigational and topographic charts.Given Data / Assumptions:Magnetic meridian differs from the true meridian by declination.Declination varies with place and time.Charts show contours of equal declination.Concept / Approach:Magnetic declination is the angle between magnetic and true meridians (east or west). Lines of equal declination are isogonic lines; the special case where declination is zero (magnetic and true meridians coincide) defines agonic lines.Step-by-Step Solution:Define declination = angle(Magnetic meridian, True meridian).Isogonic = loci of equal declination values.Agonic = the isogonic with value zero.Verification / Alternative check:Map margins and geophysical references adopt these definitions universally.Why Other Options Are Wrong:Each A–C is correct; thus D (“All the above”) must be selected; E is false.Common Pitfalls:Confusing agonic with isoclinic (equal dip); mixing grid convergence with declination.Final Answer:All the above.

Question 8

National control networks and accurate mapping: surveys conducted to provide a national grid of control for accurate mapping of large areas are known as:

Accepted Answer

Introduction / Context:Large-scale mapping over countries requires a high-precision framework to which all detailed surveys can be referenced. This framework accounts for the Earth’s curvature and uses rigorous adjustment of long baselines and angles.Given Data / Assumptions:Objective is to establish control across large regions or an entire nation.Accuracy requirements extend beyond plane approximations.Results serve as the basis for topographic mapping and engineering projects.Concept / Approach:Geodetic surveys create high-order control networks using triangulation, trilateration, GNSS, and precise leveling, all adjusted on a reference ellipsoid/terrestrial reference frame. Plane surveys assume a flat Earth over small areas and are unsuitable for national grids.Step-by-Step Solution:Recognize need for curvature-aware, high-accuracy control.Identify the terminology: national grid control → geodetic survey.Distinguish from topographical (mapping content), cadastral (boundaries), and plane (small-area approximation).Verification / Alternative check:National mapping agencies explicitly refer to geodetic control networks or geodetic reference frameworks as the basis for their grids.Why Other Options Are Wrong:Plane surveys neglect curvature; geographical is not a standard survey class; topographical describes map content, not the control framework; cadastral focuses on property limits.Common Pitfalls:Confusing geodetic control (framework) with topographic mapping (content); assuming plane methods scale to national extents.Final Answer:Geodetic surveys

Question 9

Levelling procedure: at any new instrument setup, what is the very first staff reading taken from that level station called?

Accepted Answer

Introduction / Context:Levelling notes follow a strict sequence to ensure consistent calculation of the height of instrument (H.I.) and reduced levels (R.L.). The naming of the first observation at each setup is standardized across field practice.Given Data / Assumptions:Instrument is set up and levelled at a new station.A known or previously computed R.L. is available (bench mark or turning point).Readings are recorded in B.S., I.S., F.S. columns.Concept / Approach:The first reading from a new setup is a back-sight on a point of known R.L., allowing computation of H.I. via H.I. = R.L. + B.S. Subsequent readings may be intermediate sights to determine unknown R.L.s, and the last reading before moving is the fore-sight on a change point or the final point.Step-by-Step Solution:At new setup, take B.S. on BM or C.P.Compute H.I. = R.L. (BM/C.P.) + B.S.Take I.S. on detail points as needed; finish with F.S. on a new C.P. or the closing benchmark.Verification / Alternative check:The arithmetic check ΣB.S. − ΣF.S. = R.L.(last) − R.L.(first) relies on this convention.Why Other Options Are Wrong:F.S. is the last reading from a setup; I.S. is neither first nor last; “any sight” is not acceptable in standard field books.Common Pitfalls:Recording a turning-point reading in the wrong column; forgetting to start with a B.S., which breaks the check.Final Answer:Back-sight (B.S.)

Question 10

Terrain interpretation from contours: straight, parallel, and widely spaced contour lines indicate which type of surface?

Accepted Answer

Introduction / Context:Contour patterns on topographic maps provide quick visual cues about landform shapes and slopes. Recognizing standard patterns helps planners and engineers infer gradients, drainage, and suitability for alignment or development without detailed calculations.Given Data / Assumptions:Contours are straight, mutually parallel, and widely spaced.Contour interval is constant.No abrupt curvature or convergence is present.Concept / Approach:Widely spaced contours signify gentle slope; close spacing indicates steepness. Straight and parallel contours suggest slope direction and magnitude are uniform across the area, i.e., the surface approximates a plane inclined at a constant angle. Curved or converging contours would imply varying slope (convex/concave surfaces) or topographic features like valleys and ridges.Step-by-Step Solution:Identify spacing → wide = gentle slope.Identify geometry → straight/parallel = uniform gradient direction and magnitude.Conclude → inclined plane surface (uniform slope).Verification / Alternative check:Check with gradient formula: slope = contour interval / horizontal spacing; constant spacing across the map patch implies constant slope.Why Other Options Are Wrong:Steep surface would show close spacing; flat surface would have extremely wide spacing or no contours; curved surface would show curvature in contours or varying spacing.Common Pitfalls:Interpreting very wide spacing as completely flat (whereas some gentle slope exists); overlooking localized irregularities hidden at coarse scale.Final Answer:An inclined plane surface (uniform slope)

Question 11

Bubble sensitivity from staff readings at a known distance At a sight distance of 80 m, the staff reading with the level bubble centered is 1.31 m. After displacing the bubble by 5 divisions from center, the staff reading becomes 1.39 m. Compute the…

Accepted Answer

Introduction / Context:The sensitivity of a level tube (bubble) is defined as the angular tilt of the line of sight per division of bubble movement. By comparing the change in staff reading at a known sight distance when the bubble is intentionally displaced a known number of divisions, we can compute arc-seconds per division. This is a standard field calibration task for levelling instruments.Given Data / Assumptions:Sight distance to staff, D = 80 m.Reading with bubble centered = 1.31 m.Reading with bubble displaced 5 divisions = 1.39 m.Change in reading, Δs = 1.39 − 1.31 = 0.08 m.Small-angle assumption: tan φ ≈ φ in radians when φ is small.Concept / Approach:Tilting the telescope by a small angle φ shifts the line of sight vertically. Over a horizontal distance D, the change in intercept on the staff is approximately Δs ≈ D * tan φ ≈ D * φ (radians). If 5 divisions correspond to total tilt φ_total, then angle per division = φ_total / 5. Convert radians to arc-seconds using 1 rad ≈ 206265 seconds.Step-by-Step Solution:Compute total tilt (radians): φ_total ≈ Δs / D = 0.08 / 80 = 0.001 rad.Angle per division: φ_div = φ_total / 5 = 0.001 / 5 = 0.0002 rad.Convert to arc-seconds: 0.0002 * 206265 ≈ 41.253 seconds.Therefore, one division ≈ 41.25 seconds.Verification / Alternative check:Using exact tan instead of small-angle: tan φ_total = 0.08 / 80 = 0.001 → φ_total ≈ 0.001 rad to within 0.0000005, giving essentially the same result.Why Other Options Are Wrong:28.8 s and 25.05 s underestimate the computed sensitivity.14.52 s is far too small for the given displacement and distance.“None” is invalid because a consistent value (≈41.25 s) results from the data.Common Pitfalls:Using degrees instead of radians in the conversion; forgetting to divide by the number of divisions; not keeping the staff exactly vertical for the readings.Final Answer:41.25 seconds

Question 12

Measuring horizontal distance by stepping on sloping ground During stepping (using a tape or chain held horizontal), under which condition is it generally more difficult to obtain accurate horizontal distance measurements?

Accepted Answer

Introduction / Context:Stepping is a practical method to measure horizontal distance on slopes by holding the tape horizontally in short steps. Field accuracy depends on footing, alignment, and proper use of plumb bobs. Knowing when stepping becomes difficult helps crews choose better techniques (e.g., slope correction or electronic distance measurement).Given Data / Assumptions:Measurements are made by stepping the tape in short horizontal segments.Operators must plumb down to ground for each step.Ground may be up-hill, down-hill, undulating, or level.Concept / Approach:Down-hill stepping is harder because the leader must control the tape while facing down-slope, maintain balance, and keep the tape truly horizontal with the plumb bob hanging free without clearance issues. Errors from mis-plumbing and tape sag are more pronounced; footing is poorer and sighting alignment is harder compared with up-hill stepping or level ground.Step-by-Step Solution:Identify the main sources of error: non-horizontal tape, mis-plumbing, sag, and alignment.Assess terrain effects: down-hill increases instability and makes plumb bob alignment difficult.Conclude: down-hill stepping is most error-prone.Verification / Alternative check:Field practice manuals recommend shortening steps or switching methods on steep down-slopes to mitigate error, underscoring the difficulty of down-hill stepping.Why Other Options Are Wrong:Up-hill stepping is generally easier to control because the tape can be braced against slope and plumb bob control is better.Low undulations and level plains present fewer stability issues.“None” is incorrect because one condition is clearly worse.Common Pitfalls:Over-long steps on steep slopes; not checking horizontality at each step; ignoring cumulative small plumbing errors.Final Answer:Down-hill stepping

Question 13

Linear measurements with a tape: identify the incorrect statement Consider the following statements about tape measurements and associated corrections. Which one is incorrect?

Accepted Answer

Introduction / Context:Tape measurements are influenced by instrumental and environmental factors such as standard length error, temperature, pull, and sag. Recognizing the sign and sense of each correction prevents systematic bias in linear surveying work.Given Data / Assumptions:Standard (nominal) tape length = 100 m; actual length may differ slightly.Temperature changes cause thermal expansion or contraction of the tape.Sag in unsupported spans makes the taped path longer than the straight chord.Steel modulus and geometry govern elastic extension under pull.Concept / Approach:Each effect has a predictable correction. If the tape is longer than nominal, recorded measurements are too small and must be increased. Thermal expansion similarly makes the scale longer; readings are again too small and must be increased, not decreased. Sag increases the path length; the correction is subtractive from the measured value to obtain the straight distance. Elastic stretch under a moderate pull of about 5 kg for the given cross-section produces roughly 10 mm extension over 100 m, a reasonable engineering approximation.Step-by-Step Solution:Standard length error: L_act > L_nom → measured distance too small → add correction per tape length (statement A correct).Temperature: higher T → L_act increases → measured distance too small → the statement claiming “too large” is incorrect (select B).Sag: curved tape length > straight chord → subtract sag correction to recover straight distance (statement C correct).Elasticity: estimate δL ≈ P * L / (A * E) ≈ 49 * 100 / (2.5×10^−6 * 2×10^11) ≈ 0.0098 m ≈ 10 mm (statement D correct).Verification / Alternative check:Thermal correction formula: C_T = α (T − T_0) * L_meas, added when T > T_0, confirms that measured values must be increased, not decreased.Why Other Options Are Wrong:A, C, and D state standard, correct relationships; E (“None”) is invalid because B is incorrect.Common Pitfalls:Reversing the sign of temperature or standard length correction; forgetting that sag correction is always subtractive; mixing SI and practical units for elasticity computations.Final Answer:An increase in temperature lengthens a tape, so the measured distance becomes too large.

Question 14

Length of vertical curve from grades and specified rate of change A +0.8% grade meets a −0.7% grade. If the specified rate of change of grade is 0.05% per 30 m, determine the required length of the vertical curve.

Accepted Answer

Introduction / Context:Vertical curves provide a gradual transition between grades. The length depends on the algebraic difference of entering and leaving grades and the allowable rate of change of grade. Designers often specify the permissible change per a fixed chainage (e.g., per 30 m), from which the total curve length is computed easily.Given Data / Assumptions:Entering grade g1 = +0.8%.Leaving grade g2 = −0.7%.Rate of change of grade = 0.05% per 30 m.Algebraic grade difference A = g1 − g2.Concept / Approach:For a parabolic vertical curve, the rate of change of grade r per unit length equals A / L, where L is the curve length. If a specification gives r per 30 m, then (A / L) * 30 = r_30. Solve for L using consistent percent units throughout.Step-by-Step Solution:Compute algebraic difference: A = +0.8% − (−0.7%) = 1.5%.Given r_30 = 0.05% per 30 m.Use (A / L) * 30 = r_30 → L = (A * 30) / r_30.L = (1.5 * 30) / 0.05 = 45 / 0.05 = 900 m.Verification / Alternative check:Check units: percent cancels, leaving metres; a larger A or smaller permissible r_30 would increase L, which matches design intuition.Why Other Options Are Wrong:600, 700, 800, 1000 m do not satisfy (A/L)*30 = 0.05% with A = 1.5%.Common Pitfalls:Mixing percent and decimal grade forms; inserting 30 m incorrectly (using per-metre value without converting); using A = |g1| + |g2| with wrong sign handling.Final Answer:900 m

Question 15

Internal-focusing telescope: focal length of the focusing lens A telescope has an objective (object-glass) of focal length 14 cm located 20 cm from the diaphragm (cross-hair plane). When focusing on a staff 16.50 m away, the focusing lens is midway …

Accepted Answer

Introduction / Context:Many surveying telescopes use an internal focusing lens placed between the objective and the diaphragm to bring objects at finite distances into sharp focus at the cross-hairs. When the focusing lens position is known for a specific object distance, lens formulae allow calculation of its focal length.Given Data / Assumptions:Objective focal length, f_o = 14 cm.Distance between objective and diaphragm = 20 cm.Focusing lens is midway between them for this focus → its position is 10 cm from both objective and diaphragm.Object distance (from objective) u_o = 16.50 m = 1650 cm.Thin-lens, paraxial approximation applies.Concept / Approach:First, find where the objective alone would form an image of the staff. Then treat that image as the (virtual) object for the focusing lens, which must form the final image at the diaphragm. Apply the thin-lens formula 1/f = 1/v − 1/u for the focusing lens, using consistent Cartesian sign conventions (image distances positive to the right of the lens; object distances positive to the left; a virtual object on the image side has negative u).Step-by-Step Solution:Objective image distance: 1/f_o = 1/u_o + 1/v_o → 1/14 = 1/1650 + 1/v_o → v_o ≈ 14.12 cm to the right of the objective.Position relative to focusing lens (at 10 cm): object for the focusing lens is at u_f = v_o − 10 = +4.12 cm on the right side → a virtual object, so u_f = −4.12 cm in the lens formula.Final image must be at the diaphragm 10 cm to the right of the focusing lens → v_f = +10 cm.Focusing lens focal length: 1/f_f = 1/v_f − 1/u_f = 1/10 − (−1/4.12) = 0.1 + 0.2427 ≈ 0.3427 → f_f ≈ 2.92 cm.Verification / Alternative check:A direct ray diagram confirms that a converging lens of ≈3 cm focal length placed midway shifts the objective’s intermediate image at x ≈ 14.12 cm to the diaphragm plane at x = 20 cm.Why Other Options Are Wrong:4.9, 6.9, and 8.9 cm do not satisfy the lens equation with u_f ≈ −4.12 cm and v_f = +10 cm; they would place the final focus away from the diaphragm.Common Pitfalls:Using the wrong sign for a virtual object; assuming the focusing lens is afocal; neglecting that the “midway” condition fixes the geometry and hence f_f tightly near 3 cm.Final Answer:2.9 cm (approximately)

Question 16

Exact length of a specified chord of a circular curve For a circular curve of radius R and degree of curve D (angle subtended at the center), the exact length of the specified chord is:

Accepted Answer

Introduction / Context:Chord lengths are frequently required in curve setting from the central angle. The exact relationship with the radius and subtended angle is a basic trigonometric identity used in field computations and setting-out sheets.Given Data / Assumptions:Simple circular geometry.Central angle equals the degree of the specified chord (in degrees or radians).Standard chord formula applies without small-angle approximations.Concept / Approach:The chord subtending a central angle D has length c = 2R * sin(D/2). Since the diameter is 2R, an equivalent statement is c = diameter × sin(D/2). This is exact, not approximate.Step-by-Step Solution:Consider triangle OAB with OA = OB = R and ∠AOB = D.Chord AB = 2R * sin(D/2) by isosceles triangle properties.Therefore, AB = (2R) * sin(D/2) = Diameter × sin(D/2).Verification / Alternative check:For small D, sin(D/2) ≈ D/2 (radians), giving c ≈ R * D, aligning with arc approximation c ≈ s for small angles.Why Other Options Are Wrong:R × sin(D/2) misses the factor 2.Diameter × cos(D/2) and Diameter × tan(D/2) are unrelated to chord length.“None of these” is invalid because the correct identity is known.Common Pitfalls:Confusing chord with arc length; mixing degrees and radians when evaluating numerically.Final Answer:Diameter × sin(D/2)

Question 17

Angular subtense of staff graduations at 20 m On a metric levelling staff, what angular subtense (in radians) does a 10 mm marking correspond to when viewed from a distance of 20 m?

Accepted Answer

Introduction / Context:Understanding the angular size of staff graduations helps observers estimate reading precision and choose appropriate sight lengths. The basic small-angle relation links linear size, distance, and angular subtense in radians.Given Data / Assumptions:Staff graduation considered: 10 mm = 0.01 m.Sight distance D = 20 m.Small-angle approximation θ ≈ s / D (radians) for s ≪ D.Concept / Approach:Angular subtense θ (radians) equals length divided by distance for small angles. Convert the physical graduation to metres and divide by the sight distance to get θ. Express the result as a simple reciprocal in radians for easy comparison with options.Step-by-Step Solution:s = 0.01 m, D = 20 m.θ ≈ s / D = 0.01 / 20 = 0.0005 rad.0.0005 rad = 1/2000 rad.Verification / Alternative check:Convert to seconds: 0.0005 × 206265 ≈ 103.1 seconds, consistent with typical staff visibility at moderate sights.Why Other Options Are Wrong:1/1000, 1/1500, and 1/2500 rad differ from the computed 1/2000 rad by significant margins.Common Pitfalls:Forgetting to convert millimetres to metres; using degrees instead of radians when applying the small-angle relation.Final Answer:1/2000 radian

Question 18

Effect of staff tilt on stadia intercepts In stadia tacheometry, tilting the staff tends to increase the observed stadia intercept under which condition?

Accepted Answer

Introduction / Context:In stadia measurements, the apparent intercept on the staff depends on the relative orientation between the staff and the line of sight. Deviations from a vertical staff can either increase or decrease the intercept, affecting computed distances and elevations if not accounted for.Given Data / Assumptions:Line of sight may be inclined up-hill or down-hill.Staff may be held vertical or tilted towards/away from the instrument.We consider the qualitative effect on the stadia intercept (the distance between upper and lower hair readings).Concept / Approach:For an up-hill sight, tilting the staff away from the instrument makes it more nearly normal to the line of sight, which increases the projected intercept compared with a vertical staff. Conversely, for a down-hill sight, tilting the staff towards the telescope produces a similar effect. Among the choices, the condition explicitly increasing the intercept is “away from the telescope while sighting up-hill.”Step-by-Step Solution:Consider an up-hill line of sight (positive vertical angle).Vertical staff presents a foreshortened scale to the line of sight.Tilting the staff away reduces foreshortening (closer to normal incidence), increasing the apparent intercept.Hence, option C yields a larger stadia intercept.Verification / Alternative check:Using projection geometry, the apparent intercept is proportional to 1 / cos ψ, where ψ is the angle between staff and the normal to line of sight; minimizing ψ (by appropriate tilt) increases the intercept.Why Other Options Are Wrong:Tilt away with down-hill (A) or tilt toward with up-hill (B) do not maximize normality for the given sight, giving smaller intercepts relative to the correct condition.Keeping the staff vertical (E) is the baseline and does not increase intercept compared with the optimized tilt.“None” (D) is false because a definite increasing condition exists.Common Pitfalls:Confusing “staff vertical” with “staff normal to line of sight”; neglecting to record staff attitude in field notes when reduction formulas differ.Final Answer:When the staff is tilted away from the telescope while sighting up-hill

Question 19

Optical terminology: what does “aplanatic” signify? A lens (or lens combination) is termed aplanatic if it completely eliminates which optical defect?

Accepted Answer

Introduction / Context:Surveying instruments rely on lens assemblies designed to reduce image defects that blur lines and graduations. The terms aplanatic, achromatic, and anastigmatic describe which aberrations are corrected by a given design. Knowing these definitions helps interpret instrument specifications and performance limits.Given Data / Assumptions:Standard optical nomenclature is used.“Aplanatic” relates primarily to spherical aberration correction; in many contexts it also implies freedom from coma at a particular aperture and field point.Concept / Approach:Spherical aberration arises because rays at different heights from the axis focus at different points when surfaces are spherical. An aplanatic lens system is designed so that these rays meet at one focus, eliminating spherical aberration (and often coma for selected field points). Other aberrations—chromatic, astigmatism—require additional design features and are not inherently removed by aplanatism alone.Step-by-Step Solution:Identify aberrations: spherical, chromatic, coma, astigmatism.Match definition: “aplanatic” → free of spherical aberration (and often coma in a limited sense).Therefore, select spherical aberration as the defect completely eliminated.Verification / Alternative check:Textbook glossaries define aplanatic as “free from spherical aberration” (and sometimes coma at a particular zone), distinguishing it from “achromatic” (color correction) and “anastigmatic” (astigmatism correction).Why Other Options Are Wrong:Chromatic aberration relates to wavelength dependence; coma is an off-axis blur; astigmatism produces line foci—none are implicitly eliminated by aplanatism.Common Pitfalls:Confusing aplanatic with achromatic; assuming a single term implies complete correction of all aberrations.Final Answer:Spherical aberration

Question 20

Setting offsets from a horizontal chain line across side-slope A chain line runs horizontally in the N–S direction, while the ground slopes in the E–W direction. Regarding the setting of perpendicular offsets to details on either side, which stateme…

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Introduction / Context:In chain surveying, offsets are taken at right angles to the chain line to locate details. When the chain line is kept horizontal and the ground slopes transversely, field crews must still establish right angles and measure offset lengths reliably on either side of the line.Given Data / Assumptions:Chain line along N–S is horizontal (stepped or supported to be level).Ground slopes across E–W, i.e., transverse to the chain line.Perpendicularity is set using an optical square or cross-staff; offset lengths can be measured horizontally (by stepping) or with slope corrections.Concept / Approach:Perpendiculars are directional, not dependent on ground slope. You can produce a right angle to the chain with optical aids on either side. The offset length is then measured horizontally by stepping or corrected for slope if measured along the ground. Thus, accurate offsets are feasible on both east and west sides, provided standard procedures are followed.Step-by-Step Solution:At a chain station, set a right angle with an optical square to the required side.Hold the tape horizontal using stepping/plumb bobs and measure the offset distance.Repeat for the opposite side; transverse slope does not prevent right-angle construction.Record any height differences separately if needed for planimetric reduction.Verification / Alternative check:Survey manuals describe offsetting on side-slopes using the same instruments, with only the length measurement requiring horizontal control—equally achievable on both sides.Why Other Options Are Wrong:Claims restricting offsets to only one side are unfounded; proper horizontality and right-angle construction are symmetrical.“Cannot set on west side” is incorrect for the same reason.“None” is invalid because there is a correct statement.Common Pitfalls:Measuring offset along the slope without reduction; failing to keep the tape horizontal; misusing the optical square on uneven footing.Final Answer:Offsets can be set correctly on both sides

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