Question 1

Magnetic Declination 10° W – Correct Corrections for Bearings (Whole-Circle and Quadrantal) If the local magnetic declination is 10° W (magnetic north is 10° west of true north), which statements about converting magnetic bearings to true bearings a…

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Introduction / Context: Magnetic declination links magnetic bearings (MB) to true bearings (TB). Correctly applying the sign convention is crucial, especially when working with reduced (quadrantal) bearings where the reference is N or S and the angle is measured toward E or W. Here declination is 10° W, meaning magnetic north lies west of true north. Given Data / Assumptions: Declination = 10° W. Whole-circle bearing (WCB) is measured clockwise from north (0°–360°). Quadrantal (reduced) bearing (RB) is measured from N or S toward E or W by the acute angle. Concept / Approach: The standard conversion for whole-circle bearings is TB = MB + declination, taking east positive and west negative. Thus with 10° W: TB = MB − 10°. For reduced bearings, one must consider how the RB is derived from the azimuth: different quadrants flip the reference between north and south, which changes the sign of the correction in alternating quadrants. Step-by-Step Solution: Whole-circle: TB = MB − 10° → statement (a) is correct.Quadrantal NE (0–90): RB_T = RB_M − 10° → correction negative in 1st quadrant.Quadrantal SE (90–180): RB_T = RB_M + 10° → correction positive in 2nd quadrant.Quadrantal SW (180–270): RB_T = RB_M − 10° → correction negative in 3rd quadrant.Quadrantal NW (270–360): RB_T = RB_M + 10° → correction positive in 4th quadrant.Hence, options (b) and (c) as stated are reversed; only the whole-circle rule is correctly given, so (e) is the best choice. Verification / Alternative check: Test with a sample MB = 80° (NE): TB = 70° (lower), confirming negative correction in NE. For MB = 260° (SW): TB = 250°; SW reduced bearing decreases by 10°, consistent with the derived signs. Why Other Options Are Wrong: (b) and (c) assert wrong signs; (d) cannot hold because only (a) is correct; choosing (a) alone would ignore the explicit quadrantal statements, making (e) the precise selection. Common Pitfalls: Forgetting the “east positive, west negative” convention; mixing up azimuth adjustments with reduced-bearing quadrant changes. Final Answer: Only the whole-circle rule is correct; the stated quadrantal signs are reversed

Question 2

Levelling Computation – RL by Height of Instrument Method with Given BS and FS A back sight of 2.685 m is taken on a benchmark of RL 500.000 m. A fore sight of 1.345 m is then taken on a point. Compute the reduced level (RL) of the point.

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Introduction / Context: The Height of Instrument (HI) method in levelling computes RLs by first establishing the instrument height above a known benchmark and then subtracting staff readings at other points. It is fast and reliable when BS and FS are correctly identified and the level is properly adjusted. Given Data / Assumptions: Benchmark RL = 500.000 m. Back sight (BS) on BM = 2.685 m. Fore sight (FS) on point = 1.345 m. Instrument is correctly levelled; no collimation error. Concept / Approach: Using HI method: HI = RL_BM + BS. Then RL_point = HI − FS. Careful sign discipline (add BS, subtract FS) avoids frequent mistakes. Units remain in metres throughout. Step-by-Step Solution: Compute instrument height: HI = 500.000 + 2.685 = 502.685 m.Compute point RL: RL = 502.685 − 1.345 = 501.340 m.Report RL to three decimal places as in standard levelling records.Therefore, RL of the point = 501.340 m. Verification / Alternative check: Perform the standard arithmetic check after a full run: ΣBS − ΣFS should equal RL_last − RL_first. With one setup, 2.685 − 1.345 = 1.340 which matches 501.340 − 500.000 = 1.340. Why Other Options Are Wrong: 502.685 m is HI, not the point RL; 501.345 m is a rounding/typo; 504.030 m is impossible; 502.585 m incorrectly subtracts 0.100 m. Common Pitfalls: Subtracting BS instead of adding; mixing IS with FS; transposing digits when copying field-book entries. Final Answer: 501.340 m

Question 3

Optical Square – Correct Hand Position for Offsets on Right Side of Chain Line When taking offsets with an optical square on the right-hand side of the chain line, how should the instrument be held to observe a right angle correctly?

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Introduction / Context: Optical squares are used in chain surveying to set out or check right-angled offsets quickly. Because the field of view and mirror orientation are fixed relative to the casing, the way you hold the instrument depends on which side of the chain you are observing, ensuring the reflected and direct images coincide correctly at 90 degrees. Given Data / Assumptions: Offset required on the right-hand side of the chain line. Standard optical square with fixed mirror arrangement. Observer stands on or near the chain line. Concept / Approach: For offsets to the right, the optical square is typically held in the right hand, upright, so that the direct image of the ranging rod on the chain line and the reflected image of the offset point align at the etched line when a right angle is achieved. Reversing hand or inverting the instrument causes awkward viewing geometry and possible misreadings in routine practice. Step-by-Step Solution: Stand at the chain line point where the offset is to be taken.Hold the optical square upright in the right hand.Sight the chain line directly and the offset point by reflection.Move along the chain line until images coincide, fixing the foot of the perpendicular. Verification / Alternative check: Field manuals depict right-hand/upright for right-side offsets and left-hand/upright for left-side offsets, keeping the instrument orientation consistent with the side of observation. Why Other Options Are Wrong: Upside-down holding changes reference lines; left-hand for right-side offsets complicates alignment; claiming orientation does not matter ignores the optical design. Common Pitfalls: Tilting the instrument, introducing parallax; standing off the chain line; confusing right vs left arrangement. Final Answer: By the right hand, held upright

Question 4

Plane Table – Displacement on Plan Due to Centering Error Across a 40 m Ray While setting a plane table, the station was mis-centered so that the plotted station is 30 cm perpendicular to a 40 m ground ray. If the mapping scale is 1 cm = 2 m, what i…

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Introduction / Context: In plane table surveying, small centering errors at the plotted station produce proportional displacements at plotted detail points along rays. Quantifying this displacement helps decide whether to re-center or accept the error for the given scale and required accuracy. Given Data / Assumptions: Ground ray length = 40 m. Centering (perpendicular) error at station = 30 cm = 0.30 m. Map scale = 1 cm represents 2 m. Error is perpendicular to the ray, producing maximum plotted shift at the ray end. Concept / Approach: The displacement at the end of a plotted ray caused by a perpendicular centering error e is proportional to the ratio e / ground_ray_length, applied to the plotted ray length. This is similar to similar-triangle scaling of the small lateral error along the ray. Step-by-Step Solution: Convert scale: 1 cm ↔ 2 m, so 40 m ↔ 20 cm on plan.Compute proportional shift: shift_on_plan = (e / ground_length) * plotted_length.Substitute values: shift_on_plan = (0.30 / 40) * 20 cm = (0.0075) * 20 cm = 0.15 cm.Thus the plot of the ray end is displaced by 0.15 cm from its true location. Verification / Alternative check: As a ratio: 0.30 m over 40 m is 1 in 133.33; 20 cm / 133.33 ≈ 0.15 cm, corroborating the result. Why Other Options Are Wrong: 0.02 cm and 0.10 cm understate the proportional effect; 0.02 m is a ground unit, not plan; 0.50 cm overstates the error by more than 3 times. Common Pitfalls: Confusing ground and plan units; using linear instead of proportional scaling; not ensuring that the error is perpendicular to the ray, which gives the maximum displacement case. Final Answer: 0.15 cm

Question 5

Curve Geometry – Definition of Angle of Intersection (I) In highway/railway curve terminology, the angle of intersection I is defined as the angle between which pair of lines at the point of intersection (PI)?

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Introduction / Context: Simple circular curves are defined with respect to tangents that meet at the Point of Intersection (PI). The “angle of intersection” I directly controls several layout elements such as tangent length, external distance, and degree of curve. Precise definition prevents confusion during design and field layout. Given Data / Assumptions: Back and forward tangents are straight lines meeting at the PI when extended appropriately. Long chord connects points of tangency, not necessarily passing through the PI. Standard curve geometry conventions apply. Concept / Approach: The angle of intersection I is the included angle between the two tangents at the PI. In phrasing, this is often stated as the angle between the prolongation of the back tangent and the forward tangent. This definition underlies formulas such as tangent length T = R * tan(I / 2) and long chord LC = 2 * R * sin(I / 2). Step-by-Step Solution: Identify the PI as the meeting point of tangents when extended.Take the angle between the extended back tangent and the forward tangent at PI.Recognize this as the intersection angle I used in curve formulae.Hence select option (b). Verification / Alternative check: Substituting I into standard formulas reproduces correct curve elements; using long chords instead of tangents would not yield those relations, confirming the proper definition. Why Other Options Are Wrong: (a) is imprecise unless one explicitly states “at their intersection when prolonged”; (c) and (d) involve chords, not the defining tangents; (e) is not a standard definition. Common Pitfalls: Confusing angle at the PI with central angle at the circle’s center; mixing up “external distance” and “mid-ordinate” formulas. Final Answer: Prolongation of the back tangent and the forward tangent

Question 6

Alignment Error – 100 m Tape with Mid-point 0.50 m Off Line (Two 50 m Straight Sections) A 100 m measurement is made in two straight 50 m sections, but the 50 m joint point is 0.50 m off the true straight line (forming a small kink). What is the res…

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Introduction / Context: Alignment errors occur when a tape measurement deviates laterally from the intended straight line. A common case is when two consecutive tape lengths meet at a small lateral offset at the joint. Estimating the induced excess length is important for quality control in baseline or detail measurements. Given Data / Assumptions: Total measurement nominally 100 m as two straight 50 m segments. Joint (50 m point) is laterally 0.50 m off the ideal straight line. Each 50 m segment is straight; the kink occurs only at the joint. Concept / Approach: Each 50 m segment forms the hypotenuse of a right triangle whose base lies along the true straight and whose perpendicular is d/2 for the first and d/2 for the second in opposite directions when the kink is centered, or equivalently consider two chords subtending a small angle. For a small lateral deviation d at the joint, the excess over the straight is approximately d^2 / (2L) for each segment of length L, giving a total excess of d^2 / L for two segments. With L = 50 m and d = 0.50 m, total error is 0.25 / 50 = 0.005 m (5 mm). Step-by-Step Solution: Use small-offset alignment error per 50 m segment: e_segment ≈ d^2 / (2L) with d = 0.50 m.Compute per segment: 0.25 / (2 * 50) = 0.25 / 100 = 0.0025 m.Total for two segments: e_total = 2 * 0.0025 m = 0.005 m.Therefore, the measured 100 m is too long by 0.005 m. Verification / Alternative check: Pythagoras for one segment: √(50^2 + 0.5^2) − 50 ≈ 0.0025 m; doubling gives 0.005 m, confirming the approximation. Why Other Options Are Wrong: 0.500 m and 0.050 m are grossly too large; “effectively zero” underestimates measurable error; 0.025 m is fivefold too high. Common Pitfalls: Using d instead of d^2 scaling; forgetting that two segments double the single-segment error. Final Answer: 0.005 m

Question 7

Stadia Tacheometry – Multiplying Constant in Terms of Optics In a stadia tacheometer, if i is the stadia hair interval (at the diaphragm), f is the focal length of the objective, and d is the distance between the objective and the instrument’s verti…

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Introduction / Context: Stadia tacheometry relates the staff intercept s to the line-of-sight distance D via D = k * s + C, where k is the multiplying constant and C is the additive constant. Understanding the optical origins of k and C helps diagnose instrument behavior and confirms correct constants for field computations. Given Data / Assumptions: i = stadia hair separation (at the diaphragm). f = focal length of the objective. d = distance from the objective to the instrument axis (adds to additive constant). Concept / Approach: From the geometry of similar triangles in the telescope optics, the ratio of image intercept to object intercept is governed by the focal length and the stadia hair interval. This yields D proportional to s with constant of proportionality k = f / i. The additive constant C accounts for f + d depending on the internal focusing arrangement and is independent of k. Step-by-Step Solution: Start with stadia optics: object distance relates to staff intercept via imaging with focal length f.Derive proportionality → multiplying constant k = f / i.Identify additive constant C = f + d for internal focusing instruments.Therefore, D = (f / i) * s + (f + d). Verification / Alternative check: Calibration sheets for tacheometers list k ≈ 100 for many instruments, consistent with typical f and i values that give f / i ≈ 100. Why Other Options Are Wrong: (b) inverts the ratio; (c) is the additive constant, not k; (d) and (e) are unrelated dimensional ratios. Common Pitfalls: Confusing multiplying and additive constants; forgetting that k is dimensionless and independent of staff distance. Final Answer: f / i

Question 8

Highway setting-out using contour maps: For locating a contour gradient (i.e., a line that cuts successive contour lines at equal vertical intervals) for a proposed highway on hilly terrain, from which feature and in which direction is it best set o…

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Introduction / Context: On contour maps, a contour gradient is a line that intersects contours at equal vertical intervals, giving a constant slope suitable for road or path location. Deciding where to begin setting out such a gradient on rugged ground (ridge, valley, or saddle) affects earthwork balance, drainage, and safety. This question checks practical route selection principles using contours. Given Data / Assumptions: Contours are reliable and drawn at a uniform contour interval. The desired highway grade is moderate and should minimize deep cutting or high embankment. Hydrologic safety (avoiding concentrated runoff) is preferred. Concept / Approach: Ridges are naturally elevated, convex features that shed water to both sides. Aligning a constant-grade line from a ridge down the hill generally avoids cross-drainage problems, reduces the need for high retaining works, and allows cuts on the uphill side to be modest while embankments on the downhill side remain manageable. In contrast, valleys focus drainage and can require frequent cross-drainage structures. Saddles are useful for passes, but a long alignment starting from a saddle toward the valley often descends into areas requiring heavier drainage works. Step-by-Step Solution: Identify ridge line on the map by U-shaped contours convex downslope.Trace a trial line that crosses contours at equal vertical steps, spacing them so the implied slope equals the design gradient.Adjust locally to avoid sharp bends or unstable ground while preserving near-constant contour spacing (uniform slope). Verification / Alternative check: Check volumes: ridge routes often balance cut and fill better on hillsides. Review hydrology: ridge routes place the carriageway above runoff, reducing flooding risk. Why Other Options Are Wrong: Saddle down the hill can lead into converging drainage; bottom-to-ridge or bottom-to-saddle tends to demand heavy, continuous climbing with deeper cut/fill and more drainage crossings; valley routes often require culverts and higher maintenance. Common Pitfalls: Confusing a valley line with a ridge on contour maps; forgetting that constant contour spacing corresponds to uniform slope but not necessarily minimum earthwork unless the line follows a ridge or shoulder. Final Answer: Ridge down the hill

Question 9

Reciprocal levelling between two pegs X and Y using two instrument positions: From near X, staff readings on X and Y are 1.755 m and 2.850 m. From near Y, staff readings on Y and X are 0.655 m and 1.560 m. If the reduced level (R.L.) of X is 105.500…

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Introduction / Context: Reciprocal levelling eliminates line-of-sight (collimation) error and the combined effect of curvature and refraction by observing both points from two positions near each point. This is essential when long sights or obstacles prevent equal sight lengths. The problem asks for the true R.L. of Y given four staff readings and the known R.L. of X. Given Data / Assumptions: From near X: r_X@X = 1.755 m, r_Y@X = 2.850 m. From near Y: r_Y@Y = 0.655 m, r_X@Y = 1.560 m. R.L.(X) = 105.500 m. Standard reciprocal-levelling relations apply. Concept / Approach: Let A = X and B = Y. The true difference in level (B relative to A) using reciprocal readings is obtained by averaging the two observed differences with appropriate sign. A compact relation is: (B − A) = 0.5 * [(r_A@A − r_B@A) + (r_A@B − r_B@B)] This cancels the collimation error term (proportional to sight length), leaving the true elevation difference unaffected by curvature/refraction as well. Step-by-Step Solution: Compute diff from near X: r_X@X − r_Y@X = 1.755 − 2.850 = −1.095 m.Compute diff from near Y: r_X@Y − r_Y@Y = 1.560 − 0.655 = +0.905 m.Average: {(−1.095) + (0.905)} / 2 = (−0.190) / 2 = −0.095 m.Thus R.L.(Y) − R.L.(X) = −0.095 m → R.L.(Y) = 105.500 − 0.095 = 105.405 m. Verification / Alternative check: Compute Y by each setup ignoring collimation, then average: From near X → Y = 105.500 + 1.755 − 2.850 = 104.405 m; From near Y → Y = 105.500 + (1.560 − 0.655) = 106.405 m; Average = 105.405 m, matching the reciprocal formula. Why Other Options Are Wrong: 105.000 m and 104.500 m underestimate; 105.800 m overestimates; 104.000 m is far from both independent estimates. Common Pitfalls: Mixing the sign in the averaging formula; averaging the four readings directly; forgetting that larger staff reading indicates lower ground (for the same H.I.). Final Answer: 105.405 m

Question 10

History of tacheometry: Stadia tacheometry (use of stadia lines and fixed constants for distance measurement) was introduced by James Watt in which year?

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Introduction / Context: Tacheometry is a rapid surveying technique for determining horizontal distance and elevation differences using angular measurements and a graduated staff. The stadia method, which employs fixed stadia hair constants, is historically associated with James Watt, who popularized fast optical distance estimation well before electronic distance measurement became available. Given Data / Assumptions: Question targets a historical milestone (not a formula). We assume the standard attribution of the stadia method to James Watt. Four century-spaced distractors are provided. Concept / Approach: James Watt lived from 1736 to 1819, with principal engineering activity in the mid-to-late 18th century. Stadia tacheometry is documented as a late-1700s development; thus 1770 is the historically consistent year among the options provided. Later dates (19th–20th century) correspond to the era of more advanced optical instruments but not the original introduction of stadia principles. Step-by-Step Solution: Locate James Watt’s period of innovation (mid/late 1700s).Match the stadia introduction to that period → circa 1770.Confirm that other dates either predate Watt’s career or postdate initial adoption by a century. Verification / Alternative check: Surveying histories agree that the stadia concept emerged in the latter half of the 18th century; 1770 is widely cited in traditional question banks as the answer linked to Watt. Why Other Options Are Wrong: 1670 predates Watt’s lifetime and the industrial context; 1870 and 1900 are too late for the method’s introduction; 1736 is his birth year, not the stadia milestone. Common Pitfalls: Confusing the advent of stadia with later refinements such as anallatic telescopes or EDM; memorizing only Watt’s birth/death dates. Final Answer: 1770

Question 11

Slope (dip) correction for chaining: What correction must be applied to each 30 m chain length measured along a slope of angle θ to obtain the horizontal distance?

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Introduction / Context: When a linear distance is measured along a slope, the recorded length exceeds the required horizontal distance. A slope (dip) correction converts the measured sloping length to its horizontal equivalent for plan computations and plotting. Given Data / Assumptions: Measured length along slope per segment = 30 m. Slope angle to the horizontal = θ. Standard trigonometric relations apply; small-angle approximations are optional but not required. Concept / Approach: For a segment measured along the slope (length L_s), the true horizontal length is L_h = L_s * cos θ. The correction to be applied to the measured length to get the horizontal value is Δ = L_h − L_s = L_s (cos θ − 1). This is a negative correction (to be subtracted), since cos θ ≤ 1 for θ ≥ 0. Step-by-Step Solution: Given L_s = 30 m.Compute L_h = 30 * cos θ.Slope correction per segment: Δ = 30 * (cos θ − 1) m (a negative quantity). Verification / Alternative check: For small θ (in radians), cos θ ≈ 1 − θ^2/2, so Δ ≈ 30 * (−θ^2/2), confirming a small negative correction that increases with the square of θ. Why Other Options Are Wrong: 30(sec θ − 1) and 30(tan θ − 1) are unrelated to the horizontal projection; 30(sin θ − 1) is dimensionally inconsistent; 30(1 − cos θ) gives the magnitude but with the opposite sign of the correction to be applied to the measured slope length. Common Pitfalls: Applying the magnitude with the wrong sign; mixing up whether the measured segment was horizontal or sloping; using sin instead of cos for horizontal projection. Final Answer: 30 (cos θ − 1) m

Question 12

Contours and terminology: The constant vertical distance between two adjacent contour lines on a map is known as the ______.

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Introduction / Context: Contours are lines joining points of equal elevation on a map. The spacing between successive contours represents slope, while the nominal vertical difference between them is a key map parameter used in design and volume computations. Correct terminology prevents confusion in communication and calculations. Given Data / Assumptions: Contours are at a fixed elevation step across the map. The map specifies a single value for the vertical step (e.g., 1 m, 5 m, 10 m). No variable-interval contouring is considered. Concept / Approach: The constant vertical difference between two successive contours is the contour interval. By contrast, the horizontal spacing between the same two contours on the ground (or map) depends on the ground slope and is sometimes called the horizontal equivalent or horizontal interval, which is not constant over the sheet unless the slope is uniform. Step-by-Step Solution: Identify what is constant: the vertical separation between contour elevations.Name of this constant: contour interval.Recognize that horizontal spacing varies with slope, not fixed by definition. Verification / Alternative check: On any standard topographic map, the legend lists the “contour interval” as the primary parameter describing vertical resolution. Why Other Options Are Wrong: Horizontal interval/equivalent refer to ground or map distance between contours and vary with slope; vertical equivalent is a nonstandard term; contour gradient is a line of constant slope, not the vertical spacing of contours. Common Pitfalls: Assuming wide spacing always means flat ground without checking the legend; confusing interval (vertical) with spacing (horizontal). Final Answer: Contour interval

Question 13

Line of sight error over distance: The combined effect of Earth’s curvature and atmospheric refraction on a levelling sight of length L kilometres produces an apparent error (on staff reading or line of collimation) approximately equal to:

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Introduction / Context: In precise levelling, the line of sight is not perfectly horizontal relative to the curved Earth, and atmospheric refraction bends light downward. The net effect causes a systematic apparent error that increases with the square of the sight length. Having a ready constant helps in checks and in two-peg test computations. Given Data / Assumptions: Sight length L is in kilometres. Error magnitude is requested in millimetres. Standard average refraction coefficient (~0.07) is assumed for typical conditions. Concept / Approach: Curvature alone would produce approximately 78.5 L^2 mm rise of the ground relative to a straight line of sight. Refraction (downward bending) partially compensates, often taken as about 1/7 of the curvature effect upward. The commonly used combined correction is about 67.3 L^2 mm (rounded as 67.2 L^2 mm) applied with appropriate sign in computations to remove the bias. Step-by-Step Solution: Curvature correction magnitude ≈ 78.5 L^2 mm.Refraction magnitude ≈ −11.2 L^2 mm (opposite sign, about 1/7 of curvature).Combined ≈ 78.5 − 11.2 ≈ 67.3 L^2 mm → rounded to 67.2 L^2 mm. Verification / Alternative check: Most textbooks and field tables list the combined effect between 67 and 68 L^2 mm depending on assumed refraction; 67.2 L^2 mm is the standard teaching constant for routine calculations. Why Other Options Are Wrong: 76.3 L^2 mm and 64.5 L^2 mm do not reflect the accepted mean combined value; 6.72 L mm is linear in L and dimensionally inconsistent with the well-known quadratic law. Common Pitfalls: Confusing sign conventions; forgetting that equal sight lengths cancel the effect; using metres instead of kilometres for L, causing large numerical errors. Final Answer: 67.2 L^2 mm

Question 14

Two-peg test with unequal sights: A level is set 25 m from peg A and 50 m from peg B. Staff readings (bubble centered) are 2.847 on A and 3.462 on B. Given R.L.(A) = 283.665 m and R.L.(B) = 284.295 m, determine the collimation error per 100 m (m/100…

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Introduction / Context: The two-peg test detects line-of-sight (collimation) error by comparing observed and true differences in level with unequal sight lengths. From the discrepancy, we compute the slope of the line of collimation and express it as an equivalent error per 100 m of sight length. Given Data / Assumptions: Distances: d_A = 25 m, d_B = 50 m. Readings: r_A = 2.847, r_B = 3.462 (bubble centered). R.L.(A) = 283.665 m; R.L.(B) = 284.295 m → true difference B−A = +0.630 m. Collimation error is assumed constant per unit length during the test. Concept / Approach: Let e be the collimation error per metre (positive if the line of sight rises per metre away from the instrument). The observed difference is r_B − r_A = 0.615 m. The relation is: Observed (B−A) = True (B−A) + e * (d_B − d_A) Solving for e yields the error per metre; multiply by 100 for per-100 m value. Step-by-Step Solution: Compute observed difference: 3.462 − 2.847 = 0.615 m.Set 0.615 = 0.630 + e * (50 − 25) → 0.615 = 0.630 + 25e.Thus e = (0.615 − 0.630) / 25 = −0.015 / 25 = −0.0006 m per m.Magnitude per 100 m = 0.0006 * 100 = 0.060 m per 100 m. Verification / Alternative check: With this e, recompute the expected observed difference: 0.630 + (−0.0006)*25 = 0.630 − 0.015 = 0.615 m, matching the observation. Why Other Options Are Wrong: 0.015, 0.030, and 0.045 m per 100 m are too small; 0.075 m is too large and inconsistent with the measured discrepancy of 0.015 m over a 25 m sight difference. Common Pitfalls: Using total distance (75 m) instead of sight difference (25 m); forgetting to convert to per-100 m units; mixing the sign—report the magnitude for specification compliance. Final Answer: 0.060 m

Question 15

Tacheometry by optical wedge (subtense wedge) system: The theodolites designed for this method have which distinctive feature?

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Introduction / Context: Tacheometric distance measurement can be achieved by several optical principles: fixed stadia hairs, anallatic lenses, and optical wedge (subtense wedge) systems. The wedge method introduces a fixed angular separation in the line of sight, allowing distance to be computed from the staff intercept formed by the wedge, without relying on traditional stadia hairs. Given Data / Assumptions: Instrument is a theodolite adapted for tacheometry. Optical wedge produces a constant small angular displacement. Standard staff with suitable markings is observed. Concept / Approach: In the optical wedge system, a glass wedge (or a pair forming a small prism angle) is fitted inside the telescope. The wedge splits or displaces the image to create a fixed angular subtense at the eye. The observed intercept on the staff, when multiplied by a constant, yields the horizontal distance. This differs from classical stadia where stadia hairs define the fixed subtense. Step-by-Step Solution: Recognize that a wedge provides constant angular separation.Relate staff intercept to distance via distance = K * intercept (constant depends on wedge angle).Note that no additional external attachment is required when the wedge is built inside the telescope. Verification / Alternative check: Manufacturer specifications of wedge tacheometers describe an internal optical wedge element and corresponding constants, confirming option (c). Why Other Options Are Wrong: (a) describes stadia-hair tacheometers; (d) implies an external wedge in front of the objective, uncommon in standard designs; (e) refers to a different micrometer-based technique; (b) is incomplete because although classical stadia hairs may be omitted, the defining feature is the internal wedge. Common Pitfalls: Confusing an anallatic lens (which corrects additive constant) with a wedge system; assuming the wedge is external rather than internal. Final Answer: A glass wedge fitted inside the telescope

Question 16

Reference surfaces in geodesy and levelling: The slightly irregular, curved surface that represents mean sea-level extended under the continents (zero potential surface) is called the ______.

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Introduction / Context: Accurate height systems require a physically meaningful reference. In geodesy and levelling, the geoid approximates mean sea level (MSL) globally, extended under land. It is an equipotential surface of Earth’s gravity field and provides the basis for orthometric heights and spirit levelling reductions. Given Data / Assumptions: The question seeks the name of the MSL-based zero-elevation surface. Minor local undulations are expected due to gravity variations. Surveying terminology distinguishes geometric, physical, and mathematical surfaces. Concept / Approach: The geoid is the equipotential surface that best fits global mean sea level. It is irregular because Earth’s mass distribution is not uniform. A level surface is any equipotential; the geoid is a particular one chosen as reference. A horizontal surface is tangent to a level surface at a point (a local plane). The reference ellipsoid is a smooth mathematical figure approximating Earth for map projections and geodetic computations, not the physical zero-height surface. Step-by-Step Solution: Identify required concept: MSL-extended, slightly irregular → geoid.Differentiate from generic level surfaces (many exist) and local horizontals (tangent planes).Exclude the ellipsoid (mathematical model) and plain MSL (a realization at coasts, not a continuous 3D surface under land). Verification / Alternative check: Modern geoid models (e.g., EGM family) represent this surface globally. Survey control networks use geoid undulations to convert between ellipsoidal and orthometric heights. Why Other Options Are Wrong: Mean sea level describes an observation, not the full global equipotential; a level surface is generic; a horizontal surface is local; the ellipsoid is a smooth reference used mainly for positions, not physical zero height. Common Pitfalls: Equating ellipsoidal heights (GNSS) directly with orthometric heights without applying geoid undulation; treating MSL at one tide gauge as the universal zero. Final Answer: Geoid surface

Question 17

Levelling logic check: if the back sight (BS) reading exceeds the fore sight (FS), what does it imply about relative elevations? In a single set of readings taken with the same instrument setup, if BS > FS, select the correct interpretation of the f…

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Introduction / Context: In differential levelling, each instrument setup produces a back sight (BS) and a fore sight (FS). Interpreting BS and FS correctly is essential for deciding whether the ground is rising or falling from the back point to the forward point. This question reinforces the core relationship used in both the Rise & Fall and Height of Instrument methods. Given Data / Assumptions: Both BS and FS are observed from the same instrument setup. Standard sign convention: Rise = BS − FS; Fall = FS − BS. Readings are taken on a vertical staff; instrument is properly levelled. Concept / Approach: The Height of Instrument (HI) equals RL(back point) + BS. The reduced level (RL) of the forward point equals HI − FS. If BS is greater than FS, the computed RL at the forward point becomes larger than the back point's RL, indicating a rise. The same conclusion follows from the Rise & Fall method where Rise = BS − FS; if positive, the forward point is higher. Step-by-Step Solution: Let RL_back be the reduced level of the back point.Compute HI: HI = RL_back + BS.Compute RL_forward: RL_forward = HI − FS = RL_back + (BS − FS).If BS > FS, then (BS − FS) > 0 → RL_forward > RL_back → forward point higher (rise). Verification / Alternative check: Using the Rise & Fall method directly: Rise = BS − FS. If the numerical result is positive, there is a rise from back to fore—identical conclusion. Why Other Options Are Wrong: Lower elevation (fall) corresponds to FS > BS, not BS > FS. Indeterminate is incorrect; a single BS–FS pair from one setup is sufficient to infer rise or fall. Equal elevation would require BS = FS. Common Pitfalls: Reversing the sign convention; mixing readings from different setups; forgetting that larger staff reading implies lower point only for a fixed HI, whereas the BS–FS comparison already embeds HI consistently. Final Answer: The forward staff point is at a higher elevation (there is a rise from back to fore).

Question 18

Cartography of very flat terrain: which method best represents subtle relief? For mapping a region with extremely low relief (near-flat topography), select the most effective technique for representing general topography on a plan.

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Introduction / Context: On nearly level terrain, the relief variations are subtle. If contour intervals are too large, many intermediate undulations will be missed and the map will appear deceptively flat. Choosing an appropriate contour interval is therefore central to faithful representation of the ground surface in topographic mapping. Given Data / Assumptions: Terrain is very flat with small elevation differences. Goal is to depict general topography, not only isolated points. Standard planimetric accuracy and contouring practices apply. Concept / Approach: Contours connect points of equal elevation. The contour interval controls vertical resolution of the map. For flat lands, a small contour interval captures gentle rises and depressions. Conversely, a large interval would skip these variations. Spot levels are helpful supplements, but they cannot substitute for continuous depiction across the whole area when one seeks to visualize general relief patterns. Step-by-Step Solution: Identify the need for high vertical resolution in flat terrain.Select small contour interval so minor relief becomes visible as multiple lines.Use spot levels as additional checks, not the sole method. Verification / Alternative check: Surveying standards recommend reducing the contour interval as the terrain becomes flatter; e.g., 0.2–0.5 m intervals are common for low-relief urban layouts. Why Other Options Are Wrong: Large interval (A) masks subtle relief. Sparse spot levels (C) and only salient spots (D) do not portray continuous topography. Hachures alone (E) are outdated and unsuitable for accurate quantitative elevation representation. Common Pitfalls: Keeping a uniform interval across very different terrains; failing to densify control to support small intervals. Final Answer: Drawing contours at a small interval

Question 19

Earthwork by sections at equal spacing: identify the correct Trapezoidal Rule for volume An embankment is divided into n cross-sections at equal spacing D (metres). If A1, A2, …, An are the respective end-area cross-sectional areas (m^2), which expr…

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Introduction / Context: For linear earthworks such as roads and canals, volumes are commonly computed from a sequence of cross-sections spaced at equal distances. Two widely used numerical methods are the Trapezoidal Rule and Simpson’s Rule. Correct identification of each formula avoids systematic bias in estimated quantities and costs. Given Data / Assumptions: n sections at equal spacing D. Areas A1 (first) to An (last) represent cross-sections perpendicular to the centerline. Ground varies smoothly between sections. Concept / Approach: The Trapezoidal Rule treats each adjacent pair of sections as forming a prismoid whose end areas are averaged, then multiplied by spacing D. Summing across all intervals leads to the compact weighted sum with coefficients 1 for the ends and 2 for all interior areas. Simpson’s Rule, in contrast, uses alternating 4 and 2 coefficients and requires an odd number of segments (even number of intervals) with D/3 as the multiplier. Step-by-Step Solution: Start with one interval: V_i ≈ D * (A_i + A_{i+1})/2.Sum over i = 1 to n−1: V = Σ V_i = (D/2) * [A1 + An + 2*(A2 + ... + A_{n-1})].Hence, select the expression with D/2 and weight 2 on all interior areas. Verification / Alternative check: Dimensional check: Areas in m^2 times D in m gives m^3. Weighting matches the trapezoidal composite integration rule. Why Other Options Are Wrong: Option B/C are Simpson-type formulas (require D/3 and alternating 4,2 weights), not trapezoidal. Option D ignores interior areas beyond the ends and is incomplete. Option E lacks proper weighting and overcounts interior sections. Common Pitfalls: Confusing trapezoidal and Simpson weights; using D instead of D/2; forgetting to include all interior areas with coefficient 2. Final Answer: V = (D/2) * [A1 + An + 2*(A2 + A3 + ... + A_{n-1})]

Question 20

Fundamentals of contouring: choose the correct set of properties Select the option that correctly states key properties of contour lines on topographic maps.

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Introduction / Context: Contours are the backbone of representing three-dimensional terrain on a two-dimensional map. Understanding their geometric and topological properties helps interpret slopes, ridges, valleys, and depressions accurately. This item reviews three fundamental properties every surveyor must know. Given Data / Assumptions: Contours represent loci of equal elevation. Standard mapping conventions apply (no overhangs shown in a single-valued surface representation). Continuous, differentiable ground surface assumed for interpretation. Concept / Approach: (a) On a single-valued surface representation, contours of the same elevation do not cross; two distinct same-elevation contours will not merge to form a longer line unless they are in fact the same contour closing elsewhere. (b) A contour is a closed curve; if it appears to end, it continues off the sheet. (c) The gradient vector is normal to the contour; hence, steepest slope is at right angles to the contour line at a point. Step-by-Step Solution: Recall definition: contour = set of points of equal RL.Property 1: same-elevation lines cannot unite or split arbitrarily on a single-valued surface.Property 2: contours close on themselves or outside map limits.Property 3: steepest slope is perpendicular to contours. Verification / Alternative check: These properties are consistent with level curves in calculus and standard cartographic rules. Why Other Options Are Wrong: Any subset of the properties omits essential facts; hence (d) is the most complete and correct choice. Common Pitfalls: Assuming contours can terminate; misinterpreting steepness by spacing along, rather than perpendicular to, contours. Final Answer: All of the above

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