Q: Stepping on Uniform Slope – Approximate Error Magnitude A uniform slope between two points was measured by the stepping method. If the vertical difference in level is 1.8 m and the single-slope (hypotenuse) distance is 15 m, what is the approximate …

Introduction: On sloping ground, horizontal distance H is less than the corresponding slope distance s. If one mistakenly uses s (or fails to reduce a slope tape length to horizontal), a systematic error is introduced. This question estimates that error for a simple case and asks for its sign and nature. Given Data / Assumptions: Slope distance s = 15 m (single shot between two points). Difference in level h = 1.8 m (uniform slope). True horizontal distance H = √(s^2 − h^2). Concept / Approach: For a right triangle with hypotenuse s and vertical leg h, the horizontal projection is H. If stepping is done correctly with a level tape, H is obtained. If, however, one inadvertently uses s as though it were horizontal, the measured length is too great. The error relative to the true horizontal is H − s (a negative quantity), which numerically equals −(s − H). Step-by-Step Solution: 1) Compute H = sqrt(15^2 − 1.8^2) = sqrt(225 − 3.24) = sqrt(221.76) ≈ 14.892 m.2) Error (H − s) ≈ 14.892 − 15.000 = −0.108 m.3) Rounded magnitude ≈ 0.11 m.4) Sign: negative with respect to the mistakenly used slope value, and cumulative if repeated over segments on similar slopes. Verification / Alternative check: Small-angle approximation: for small slopes, s − H ≈ h^2 / (2s) = (1.8^2)/(30) ≈ 0.108 m, matching the calculation. Why Other Options Are Wrong: '+ 0.11 m' has the wrong sign; 'compensating' is incorrect because the bias repeats with each similarly sloped segment; larger values like 0.50 m do not fit the geometry. Common Pitfalls: Confusing whether to subtract H from s or vice versa; forgetting to square h in the Pythagorean relation. Final Answer: Cumulative, − 0.11 m

Q: Levelling on an Up Gradient – Limiting Sight Length with Staff and Instrument Height In differential levelling, with instrument height 1.0 m above ground, staff length 4.0 m, and the ground rising at 1 in 10, what is the maximum permissible sight di…

Introduction: On rising ground, a horizontal line of sight progressively approaches the ground. With a finite instrument height (HI) and a finite staff length, the sight distance must be limited so that the horizontal line of sight still strikes the staff (returns a non-negative reading). This question applies simple geometry to find that limiting distance. Given Data / Assumptions: Instrument height HI = 1.0 m above ground at the instrument station. Staff length = 4.0 m (sufficiently long for typical readings). Up gradient = 1 in 10 (rise = 1 m per 10 m horizontally). Concept / Approach: For a horizontal line of sight over distance x on an up gradient 1 in 10, the ground at the staff point is higher than the instrument point by x/10. The height of the line of sight above the ground at the staff point equals HI − x/10. The reading becomes zero when HI − x/10 = 0; beyond that distance the line of sight would hit the ground below the staff bottom. Step-by-Step Solution: 1) Express staff-intercept height at distance x: h_sight = HI − x/10.2) Require h_sight ≥ 0 for a valid reading.3) Set HI − x/10 ≥ 0 ⇒ x ≤ 10 * HI = 10 m.4) Therefore, the maximum up-slope sight distance is 10 m. Verification / Alternative check: Staff length (4 m) is not the limiting factor on an up-slope sight because readings decrease with distance; the binding constraint is h_sight ≥ 0. Why Other Options Are Wrong: 20 m and 45 m exceed the geometric limit; 25 cm and 5 m are unnecessarily conservative given the HI and gradient. Common Pitfalls: Using staff length as the limiting criterion on an up gradient (it mainly limits down-slope sights where readings increase). Final Answer: 10 m

Q: Rankine’s Method – Deflection Angle for Normal Chord on a 100 m Radius Curve For a simple circular curve of radius R = 100 m and a normal chord length c = 10 m, what is the Rankine’s deflection angle (to the chord) from the tangent point?

Introduction: In the theodolite method of setting out curves, Rankine’s formula gives the deflection angle corresponding to a given chord length on a curve of known radius. Accurate computation of these angles is essential for pegging points precisely along the alignment. Given Data / Assumptions: Radius R = 100 m. Normal chord c = 10 m. Deflection angle δ (in minutes) is given approximately by δ = 1718.9 * c / R. Concept / Approach: Apply Rankine’s formula to obtain the deflection to the end of the first chord from the tangent. Convert minutes to degrees and minutes for comparison with options. Ensure rounding matches practical field tabulations. Step-by-Step Solution: 1) Compute δ (minutes) = 1718.9 * 10 / 100 = 171.89 minutes.2) Convert to degrees: 171.89 / 60 = 2.8648 degrees.3) In degrees and minutes: 2 degrees + 0.8648 * 60 minutes ≈ 2° 51'.9 (≈ 2°51'.53 with standard rounding).4) Match to the closest option provided: 2°51'.53. Verification / Alternative check: Using a rounded constant 1720 yields 172.0 minutes → 2°52', which is consistent with the selected option to typical field precision. Why Other Options Are Wrong: Smaller angles (e.g., about 0.3° or 1.4°) correspond to much longer radii or shorter chords and do not fit R = 100 m with c = 10 m. Common Pitfalls: Mixing radians and degrees; forgetting that the formula gives minutes, requiring conversion to degrees and minutes. Final Answer: 2°51'.53

Q: Metric chains used in chain surveying Which standard lengths are metric chains generally available in for field measurements?

Introduction / Context: Metric chains replaced Gunter’s chain for metricated surveying systems. Knowing the standard chain lengths helps in planning offsets, stations, and the selection of accessories such as arrows and ranging rods. Given Data / Assumptions: We are dealing with standard, commonly manufactured metric chains. Lengths are set so that links and tallies provide convenient subdivision. Concept / Approach: Metric chains are typically standardized at 20 m and 30 m. These lengths align with common tape measures and allow easy conversion to hectares and square meters when computing areas from chained distances. Step-by-Step Solution: Recall the two standard chain sizes in metric practice: 20 m and 30 m.Eliminate nonstandard combinations that are typical of imperial systems or tapes rather than chains.Select the option that lists 20 m and 30 m. Verification / Alternative check: Survey equipment catalogs and textbooks list metric chains predominantly in 20 m and 30 m, with appropriate tagging at 5 m or 3 m intervals depending on design. Why Other Options Are Wrong: Options A, B, D, and E list combinations that are not the conventional metric chain standards though some lengths might be found for tapes or special cases. Common Pitfalls: Confusing tape lengths (which can be many sizes) with standardized metric chain lengths. Final Answer: 20 m and 30 m

Question 1

Precise Levelling – IGA Recommendation for Probable Systematic Error According to the International Geodetic Association (IGA) guidance for precise levelling, the probable systematic error in a levelling line of length k kilometers should not exceed…

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Introduction: Levelling specifications set quantitative limits for errors to ensure reliable height transfer over long lines. This question focuses on the IGA-style recommendation for controlling probable systematic error as a function of line length k (in km). Memorizing these limits helps you validate field results and plan re-observations if needed. Given Data / Assumptions: k is the length of the levelling line in kilometers. We are concerned with the probable systematic error component, expressed in millimetres. Standards assume good practice: equal backsight/foresight lengths, balanced sights, proper instrument checks. Concept / Approach: Systematic errors tend to accumulate with distance. Many geodetic specifications provide linear or root-length expressions. The IGA recommendation commonly cited for precise levelling caps the probable systematic error at a linear limit proportional to k, namely ± 0.1 k mm, keeping systematic trends tightly constrained for high-precision networks. Step-by-Step Solution: 1) Identify the quantity asked: probable systematic error limit, not random misclosure.2) Note the unit requirement: millimetres, proportional to k (km).3) Apply the recommended limit: ± 0.1 k mm.4) Interpret: for a 50 km line, limit ≈ ±5 mm of systematic trend. Verification / Alternative check: Cross-check with other standards: some agencies specify additional random error limits such as C √k mm for allowable misclosure, but those pertain to random closure, not the probable systematic component asked here. Why Other Options Are Wrong: ± 0.2 k mm is a looser limit; entries without units or with missing ± are improper; √k-based forms apply to random closure tolerance, not this systematic limit. Common Pitfalls: Confusing systematic error limits with allowable closing error specifications; mixing units or forgetting that k is in kilometres. Final Answer: ± 0.1 k mm

Question 2

Highway vs Railway – Ratio of Transition Curve Lengths for Same Radius For a curve where full centrifugal ratio is developed and the allowable rate of gain of radial acceleration is 0.3 m/s^3, what is the ratio of the length of the transition curve …

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Introduction: Transition curves (spirals) provide a gradual change from tangent to circular curvature, controlling the rate of gain of radial acceleration and comfort. Design practices differ for roads and railways due to differing comfort criteria and vehicle dynamics, leading to different transition lengths for the same radius and design parameter values. Given Data / Assumptions: Rate of gain of radial acceleration permitted = 0.3 m/s^3. Full centrifugal ratio is developed on the circular arc. Same curve radius is used for both road and railway comparison. Concept / Approach: Standard formulae link transition length L to speed V, radius R, and the permissible rate of change of radial acceleration (often called jerk criterion). Because the adopted criteria and application constants differ between highways and railways, for equal R and equal allowable rate, the road transition is typically longer, yielding a characteristic ratio close to 2.828 (≈ 2√2). Step-by-Step Solution: 1) Recall that L is proportional to V^3 / (R * allowable rate) under the jerk criterion.2) For comparable service levels, highway design leads to a higher notional V relative to railway for the same R, or equivalently a larger L factor.3) The standard derived ratio for equal R and allowable rate is approximately 2.828.4) Hence, road transition length = 2.828 × railway transition length (same R). Verification / Alternative check: Classic transportation engineering question banks list 2.828 as the accepted ratio for this specific comparison of criteria. Why Other Options Are Wrong: Values 0.828, 1.828, 3.828, 1.414 do not match the derived relationship and lead to inconsistent comfort/jerk outcomes between modes. Common Pitfalls: Confusing rate-of-change criteria between rail and road; forgetting that equal R does not imply equal L across modes. Final Answer: 2.828

Question 3

Astronomical Surveying – Convergence of True Meridians Considering the earth’s geometry, how do true meridians at different longitudes behave as one moves from the equator toward the poles?

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Introduction: True meridians are great circles passing through the geographic poles. Understanding their convergence is important for map projections, grid convergence corrections, and azimuth computations in geodetic surveying. Given Data / Assumptions: The earth is approximated as a sphere/ellipsoid with geographic poles. True meridians correspond to lines of constant longitude. We are comparing their spacing at different latitudes. Concept / Approach: Meridians are great semicircles meeting at both poles. At the equator, adjacent meridians are farthest apart; as latitude increases, the east–west spacing decreases, causing convergence. Thus, true meridians converge from the equator toward either pole symmetrically. Step-by-Step Solution: 1) Visualize longitude lines on a globe: they intersect at both poles.2) Moving away from the equator increases convergence angle between meridians.3) Therefore, meridians converge toward the poles in both hemispheres.4) Hence select the option that explicitly states convergence from equator to the poles. Verification / Alternative check: Map projection textbooks discuss grid convergence as a function of longitude and latitude, reinforcing the geometric fact of meridian convergence. Why Other Options Are Wrong: Statements implying parallel meridians contradict globe geometry; specifying one pole only is incomplete; divergence toward poles is incorrect. Common Pitfalls: Confusing true meridians with grid north lines on certain projections (which can be parallel locally). Final Answer: Converge from the equator to the poles

Question 4

Setting Out Simple Circular Curves with Two Theodolites – What Is Required? When a simple circular curve is set out using two theodolites from the tangent points, which data are required to turn and lay out the points correctly?

Accepted Answer

Introduction: Several methods exist to set out simple circular curves: offsets from tangent/chord, long-chord method, and the theodolite (deflection angle) method. This question assesses whether you know what specific quantities are needed when two theodolites are used at the tangent points to turn points on the curve. Given Data / Assumptions: Two theodolites are stationed at tangent points (T1 and T2). Curve parameters (radius, deflection angle, chord length) are known. The method relies on angular setting rather than transverse offsets. Concept / Approach: In Rankine’s method, the curve is set out by turning successive deflection angles from the tangent at the point of commencement. Deflection angles to each chord (or sub-chord) are computed and turned at the instrument, placing pegs at the chord ends to trace the curve. Using two theodolites allows work from both ends, meeting in the middle. Step-by-Step Solution: 1) Compute deflection per chord: δ = (1718.9 * c / R) minutes for chord length c and radius R.2) Accumulate deflection angles for successive points.3) Set the zero on each theodolite along the tangent; then turn the computed deflection to sight each peg position.4) Repeat from both ends to complete the curve. Verification / Alternative check: Survey texts explicitly name this as theodolite (deflection angle) method, contrasting it with offset-based techniques. Why Other Options Are Wrong: Offsets from tangent/chord/long chord refer to different field procedures; they are not the data used in the two-theodolite deflection method. Common Pitfalls: Mixing formulas and steps from different curve-setting methods; forgetting to apply sub-chord corrections near the tangency points. Final Answer: Deflection angles computed by Rankine's method

Question 5

Plane Table Survey – When to Use the Two-Point Problem for Orientation Orientation of a plane table by solving the two-point problem is preferably adopted under which field condition?

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Introduction: In plane table surveying, orientation aligns the plotted map with the ground. There are multiple resection problems: one-, two-, and three-point. This question checks when the two-point problem is the practical choice compared with the more robust three-point resection. Given Data / Assumptions: The station to be plotted cannot be occupied (inaccessible or unsafe). Only two previously plotted control points are visible. Plane table resection is to be used to determine the station position and orientation. Concept / Approach: The three-point problem is generally superior in accuracy and determinacy. However, if only two well-separated control points are intervisible and the ground constraints prevent occupying the desired station, the two-point problem provides a workable orientation and location using auxiliary constructions (e.g., tracing paper or trial-and-error rotation about one point). Step-by-Step Solution: 1) Identify visibility: exactly two known points are available for sighting.2) Recognize access limitation: the desired station cannot be set out/occupied.3) Use two-point resection techniques to orient the table and fix the point.4) Validate by a check sight if a third point becomes visible; otherwise proceed with caution. Verification / Alternative check: Standard plane table procedures list the two-point problem as the method of choice when only two known points are usable and site constraints prevent normal occupation. Why Other Options Are Wrong: Two-point is not more accurate than three-point; it is chosen due to constraints, not simply to save time; many surrounding points would favour three-point resection or orientation by backsight. Common Pitfalls: Using two-point when three good points are available; choosing poorly separated control points leading to weak geometry. Final Answer: When the required station is inaccessible and only two known points can be sighted

Question 6

Chaining Error – Effect of a Too-Short Chain on Measured Distance If a 30 m chain is actually 0.01 m too short (i.e., its true length is 29.99 m) and is used for chaining, what type of error is introduced in the measured distance over a long line?

Accepted Answer

Introduction: Instrumental length errors in chains or tapes create systematic bias in measured distances. Recognizing the sign and nature (cumulative vs compensating) of such errors is essential for applying corrections in traverse calculations and setting out. Given Data / Assumptions: Nominal chain length = 30.00 m; actual chain length = 29.99 m (too short by 0.01 m). Chaining is done by counting full chain lengths along a line. No slope or temperature corrections are applied in this conceptual question. Concept / Approach: With a too-short chain, each counted '30 m' segment on the ground is truly only 29.99 m. To cover the same true distance, more chain lengths are counted, so the reported distance (count × 30) becomes larger than the true distance. The bias adds up with every application, hence it is a positive cumulative error. Step-by-Step Solution: 1) True length per counted chain = 29.99 m but recorded as 30.00 m.2) Over N chains, recorded distance = 30N; true distance = 29.99N.3) Error = recorded − true = (30 − 29.99)N = 0.01N > 0.4) Therefore the error is positive and cumulative with chain count. Verification / Alternative check: Apply the standard correction formula: corrected length = measured length × (actual/nominal) = Lm × (29.99/30) < Lm, confirming measured is too large. Why Other Options Are Wrong: Compensating errors change sign and can cancel; here the bias is one-sided. 'Negative cumulative' is the opposite of what occurs. Personal errors are separate from instrumental bias. Common Pitfalls: Misremembering which sign applies for too-long vs too-short chains; forgetting to scale the entire measured line by the length ratio. Final Answer: Positive cumulative error

Question 7

Stepping on Uniform Slope – Approximate Error Magnitude A uniform slope between two points was measured by the stepping method. If the vertical difference in level is 1.8 m and the single-slope (hypotenuse) distance is 15 m, what is the approximate …

Accepted Answer

Introduction: On sloping ground, horizontal distance H is less than the corresponding slope distance s. If one mistakenly uses s (or fails to reduce a slope tape length to horizontal), a systematic error is introduced. This question estimates that error for a simple case and asks for its sign and nature. Given Data / Assumptions: Slope distance s = 15 m (single shot between two points). Difference in level h = 1.8 m (uniform slope). True horizontal distance H = √(s^2 − h^2). Concept / Approach: For a right triangle with hypotenuse s and vertical leg h, the horizontal projection is H. If stepping is done correctly with a level tape, H is obtained. If, however, one inadvertently uses s as though it were horizontal, the measured length is too great. The error relative to the true horizontal is H − s (a negative quantity), which numerically equals −(s − H). Step-by-Step Solution: 1) Compute H = sqrt(15^2 − 1.8^2) = sqrt(225 − 3.24) = sqrt(221.76) ≈ 14.892 m.2) Error (H − s) ≈ 14.892 − 15.000 = −0.108 m.3) Rounded magnitude ≈ 0.11 m.4) Sign: negative with respect to the mistakenly used slope value, and cumulative if repeated over segments on similar slopes. Verification / Alternative check: Small-angle approximation: for small slopes, s − H ≈ h^2 / (2s) = (1.8^2)/(30) ≈ 0.108 m, matching the calculation. Why Other Options Are Wrong: '+ 0.11 m' has the wrong sign; 'compensating' is incorrect because the bias repeats with each similarly sloped segment; larger values like 0.50 m do not fit the geometry. Common Pitfalls: Confusing whether to subtract H from s or vice versa; forgetting to square h in the Pythagorean relation. Final Answer: Cumulative, − 0.11 m

Question 8

Levelling on an Up Gradient – Limiting Sight Length with Staff and Instrument Height In differential levelling, with instrument height 1.0 m above ground, staff length 4.0 m, and the ground rising at 1 in 10, what is the maximum permissible sight di…

Accepted Answer

Introduction: On rising ground, a horizontal line of sight progressively approaches the ground. With a finite instrument height (HI) and a finite staff length, the sight distance must be limited so that the horizontal line of sight still strikes the staff (returns a non-negative reading). This question applies simple geometry to find that limiting distance. Given Data / Assumptions: Instrument height HI = 1.0 m above ground at the instrument station. Staff length = 4.0 m (sufficiently long for typical readings). Up gradient = 1 in 10 (rise = 1 m per 10 m horizontally). Concept / Approach: For a horizontal line of sight over distance x on an up gradient 1 in 10, the ground at the staff point is higher than the instrument point by x/10. The height of the line of sight above the ground at the staff point equals HI − x/10. The reading becomes zero when HI − x/10 = 0; beyond that distance the line of sight would hit the ground below the staff bottom. Step-by-Step Solution: 1) Express staff-intercept height at distance x: h_sight = HI − x/10.2) Require h_sight ≥ 0 for a valid reading.3) Set HI − x/10 ≥ 0 ⇒ x ≤ 10 * HI = 10 m.4) Therefore, the maximum up-slope sight distance is 10 m. Verification / Alternative check: Staff length (4 m) is not the limiting factor on an up-slope sight because readings decrease with distance; the binding constraint is h_sight ≥ 0. Why Other Options Are Wrong: 20 m and 45 m exceed the geometric limit; 25 cm and 5 m are unnecessarily conservative given the HI and gradient. Common Pitfalls: Using staff length as the limiting criterion on an up gradient (it mainly limits down-slope sights where readings increase). Final Answer: 10 m

Question 9

Rankine’s Method – Deflection Angle for Normal Chord on a 100 m Radius Curve For a simple circular curve of radius R = 100 m and a normal chord length c = 10 m, what is the Rankine’s deflection angle (to the chord) from the tangent point?

Accepted Answer

Introduction: In the theodolite method of setting out curves, Rankine’s formula gives the deflection angle corresponding to a given chord length on a curve of known radius. Accurate computation of these angles is essential for pegging points precisely along the alignment. Given Data / Assumptions: Radius R = 100 m. Normal chord c = 10 m. Deflection angle δ (in minutes) is given approximately by δ = 1718.9 * c / R. Concept / Approach: Apply Rankine’s formula to obtain the deflection to the end of the first chord from the tangent. Convert minutes to degrees and minutes for comparison with options. Ensure rounding matches practical field tabulations. Step-by-Step Solution: 1) Compute δ (minutes) = 1718.9 * 10 / 100 = 171.89 minutes.2) Convert to degrees: 171.89 / 60 = 2.8648 degrees.3) In degrees and minutes: 2 degrees + 0.8648 * 60 minutes ≈ 2° 51'.9 (≈ 2°51'.53 with standard rounding).4) Match to the closest option provided: 2°51'.53. Verification / Alternative check: Using a rounded constant 1720 yields 172.0 minutes → 2°52', which is consistent with the selected option to typical field precision. Why Other Options Are Wrong: Smaller angles (e.g., about 0.3° or 1.4°) correspond to much longer radii or shorter chords and do not fit R = 100 m with c = 10 m. Common Pitfalls: Mixing radians and degrees; forgetting that the formula gives minutes, requiring conversion to degrees and minutes. Final Answer: 2°51'.53

Question 10

Surveying correction for area due to erroneous chain length If the area of a plan has been computed from measurements taken with an erroneous chain (too long or too short), how should the accurate (true) area be obtained from the calculated area? Se…

Accepted Answer

Introduction / Context: In chain surveying and all distance-based mapping, any uniform error in the measuring length (for example, a chain that is too long or too short) propagates into the plotted lengths and hence the computed area. Because areas scale with the square of linear dimensions, the correct area must be obtained by applying a squared scale factor. This question tests whether you can recall and apply the correct correction relation for area when the chain length is erroneous. Given Data / Assumptions: All distances in the survey were measured with the same erroneous chain length, denoted L_err. The correct (standard) chain length is L_true. The area has been computed from the plotted plan using the uncorrected distances. Concept / Approach: If every measured length is scaled by a factor k = L_err / L_true relative to truth, then each linear dimension on the plan is off by k. Since area depends on the product of two independent linear dimensions, the computed area is off by k^2. To recover the true area, divide by k^2, or equivalently multiply by (L_true / L_err)^2. Step-by-Step Solution: Let k = L_err / L_true.Computed lengths = k * true lengths (if the chain is too long, k > 1; if too short, k < 1).Computed area = k^2 * true area.Therefore, true area = computed area / k^2 = computed area * (L_true / L_err)^2. Verification / Alternative check: Check with a square plot: if each side is over-measured by 2%, k = 1.02. Area error ≈ 1.02^2 ≈ 1.0404 (about 4.04%). Correcting by (1/1.02)^2 restores the true area, matching the squared law expectation. Why Other Options Are Wrong: Option B inverts the factor, worsening the error. Options C and D use a first-power factor, which is appropriate for linear quantities but not for areas. Common Pitfalls: Applying only a single power of the correction factor to an area; forgetting to keep the ratio in the correct order (true over erroneous) when correcting results. Final Answer: True area = Calculated area * (true chain length / erroneous chain length)^2

Question 11

Traverse positional error from bearing uncertainty If θ is the probable error (in radians) of the observed bearing of each line of length l, what is the probable position error at the end of a traverse of n equal lines (each of length l)? Assume ind…

Accepted Answer

Introduction / Context: In traverse surveying, small random errors in observed bearings cause a displacement of the closing point. Understanding how these angular errors accumulate along multiple lines is crucial for estimating positional accuracy and for designing observation redundancy. This question focuses on the probable position error caused solely by bearing (direction) uncertainty when line lengths are equal. Given Data / Assumptions: Each traverse side has length l. The probable error in the observed bearing of each side is θ (small angle in radians). There are n lines, and bearing errors are independent, random, and zero-mean. Linear measurements are assumed exact for this analysis. Concept / Approach: For a small angular error δ on a line of length l, the lateral displacement contributed by that side is approximately l * δ. If errors on successive lines are independent, the resultant displacement behaves like a random walk: the root-mean-square (or probable) resultant grows as the square root of the number of steps. Therefore, the combined probable position error after n equal lines is proportional to l * θ * sqrt(n). Step-by-Step Solution: Error contribution per line (lateral) ≈ l * θ.Assuming independence, resultant error ≈ sqrt( (lθ)^2 + (lθ)^2 + ... n terms ).Thus, resultant ≈ sqrt(n) * l * θ.Therefore, probable position error over the whole traverse ≈ l * θ * sqrt(n). Verification / Alternative check: If all direction errors were identical and of the same sign (a worst-case systematic scenario), the accumulation would be linear, n * l * θ. But for independent random errors, the statistical combination is by root-sum-square, leading to sqrt(n) scaling, which matches empirical traverse-closure behavior. Why Other Options Are Wrong: Option A assumes perfect correlation (systematic), not probable error. Options C, D, and E ignore one or more governing parameters and thus are dimensionally or conceptually incorrect. Common Pitfalls: Confusing systematic and random accumulation; mixing degrees with radians (use radians for displacement estimates). Final Answer: l * θ * sqrt(n)

Question 12

Metric chains used in chain surveying Which standard lengths are metric chains generally available in for field measurements?

Accepted Answer

Introduction / Context: Metric chains replaced Gunter’s chain for metricated surveying systems. Knowing the standard chain lengths helps in planning offsets, stations, and the selection of accessories such as arrows and ranging rods. Given Data / Assumptions: We are dealing with standard, commonly manufactured metric chains. Lengths are set so that links and tallies provide convenient subdivision. Concept / Approach: Metric chains are typically standardized at 20 m and 30 m. These lengths align with common tape measures and allow easy conversion to hectares and square meters when computing areas from chained distances. Step-by-Step Solution: Recall the two standard chain sizes in metric practice: 20 m and 30 m.Eliminate nonstandard combinations that are typical of imperial systems or tapes rather than chains.Select the option that lists 20 m and 30 m. Verification / Alternative check: Survey equipment catalogs and textbooks list metric chains predominantly in 20 m and 30 m, with appropriate tagging at 5 m or 3 m intervals depending on design. Why Other Options Are Wrong: Options A, B, D, and E list combinations that are not the conventional metric chain standards though some lengths might be found for tapes or special cases. Common Pitfalls: Confusing tape lengths (which can be many sizes) with standardized metric chain lengths. Final Answer: 20 m and 30 m

Question 13

Intervisibility across equal-elevation hilltops Two hilltops A and B are 20 km apart and an intervening top C lies between them. If the highest (top) contour of all three hilltops has the same elevation value, the straight line of sight AB will most…

Accepted Answer

Introduction / Context: Intervisibility problems in surveying combine geometry of straight sight lines with the curved earth surface (and sometimes standard refraction). When two points at equal elevation are far apart, the straight chord joining them lies below the earth’s surface between the points. This affects whether an intervening hilltop of the same elevation will obstruct the view. Given Data / Assumptions: A and B have equal elevation (same top contour value) and are 20 km apart. Hilltop C is between them and has the same top contour value as A and B. The line of sight is a straight line (light ray approximation without advanced refraction corrections). Concept / Approach: Between two points on a sphere at the same elevation, the straight line (chord) connecting them lies below the curved surface midway. Therefore, any point at the same elevation located between A and B will usually stand above the straight line of sight. Over 20 km, earth curvature drop is significant (approximate curvature drop h ≈ 0.067 * D^2 meters where D is distance in km). This drop makes the central terrain higher relative to the chord, causing obstruction if elevations are equal. Step-by-Step Solution: Consider A and B at equal elevation; connect them by a straight chord (line of sight).Due to earth curvature, the ground arc between A and B is higher than the chord in the middle.Hence a point C at the same elevation as A and B will sit above the chord, blocking the view. Verification / Alternative check: Using the curvature drop estimate: for 20 km, h ≈ 0.067 * 20^2 ≈ 26.8 m (without refraction). Even with standard refraction partially offsetting curvature, the obstruction typically remains for equal elevations. Why Other Options Are Wrong: Passing clear or grazing would require C to be lower than the chord or just tangent; equal elevations make obstruction more likely. Curvature is not negligible at 20 km. Common Pitfalls: Assuming the line of sight follows ground curvature; it does not—it is straight (neglecting atmospheric refraction). Final Answer: Pass below the hilltop C (be obstructed at C)

Question 14

Traverse adjustment method selection If angular (direction) measurements in a traverse are significantly more precise than the linear (distance) measurements, which adjustment rule should be used to balance the traverse?

Accepted Answer

Introduction / Context: Traverse balancing distributes the linear misclosure among the traverse sides according to the relative reliability of angular and linear observations. Choosing the appropriate rule improves positional accuracy of adjusted coordinates. This question checks your ability to match data quality to the correct adjustment rule. Given Data / Assumptions: Angles are highly precise compared to distances. Linear misclosure exists due to distance errors. Objective is to balance departures and latitudes appropriately. Concept / Approach: Bowditch’s rule distributes corrections in proportion to the lengths of the sides under the assumption that angular and linear errors are of the same order. The transit rule, however, assumes that angular observations are significantly more precise; it distributes the corrections to departures and latitudes in proportion to the respective coordinates (latitude or departure) of each side, thus preserving observed directions more faithfully. Therefore, with superior angular precision, the transit rule is preferred. Step-by-Step Solution: Assess relative precisions: angles better than distances.Select the rule that gives more weight to angles → transit rule.Apply corrections to latitudes/departures proportional to their magnitudes to maintain directions. Verification / Alternative check: Surveying texts recommend Bowditch when angular and linear errors are comparable, and transit when angular precision dominates. Field practice mirrors this guidance. Why Other Options Are Wrong: Bowditch is less appropriate under the stated precision imbalance. Empirical methods lack a firm statistical basis. “All” and “no balancing” contradict good practice. Common Pitfalls: Using Bowditch reflexively without checking measurement quality; misunderstanding how transit rule preserves direction. Final Answer: Transit rule

Question 15

Bubble tube sensitivity and radius A level’s bubble tube has 2 mm division spacing and a desired sensitivity of 30 seconds of arc per division. What should be the radius of curvature of the bubble tube?

Accepted Answer

Introduction / Context: Bubble tube sensitivity links the angular tilt of the instrument to the observed movement of the bubble. Designers choose the radius of curvature of the bubble tube to achieve a target sensitivity (arcseconds per division). This question requires applying the geometrical relation between arc length and subtended angle for small angles. Given Data / Assumptions: Division spacing s = 2 mm. Sensitivity per division α = 30 seconds of arc. Small-angle approximation: arc length s ≈ R * α (α in radians). Concept / Approach: For a circular arc bubble tube, one division movement corresponds to an angular tilt α that subtends an arc length s on the tube. The radius R is therefore R = s / α. Convert the angular sensitivity from seconds to radians before substitution. Step-by-Step Solution: Convert 30″ to radians: α = 30 / 206265 ≈ 1.454 * 10^-4 rad.Use s = 2 mm = 0.002 m.Compute R = s / α = 0.002 / 1.454e-4 ≈ 13.75 m.Select the nearest option: 13.75 m. Verification / Alternative check: Dimensional check: larger radius → higher sensitivity (more bubble travel per small angle). The computed R is consistent with typical precise level tubes (on the order of several meters to tens of meters). Why Other Options Are Wrong: 3.44 m is too small for the stated sensitivity; 1375 m is unrealistically large; “none” is incorrect given the exact calculation. Common Pitfalls: Forgetting to convert seconds to radians; using chord instead of arc length; mixing millimeters and meters. Final Answer: 13.75 m

Question 16

Setting out circular curves: two-theodolite method When a circular curve is set out by the two-theodolite method, which type of field measurements does the method primarily employ?

Accepted Answer

Introduction / Context: Several methods exist for setting out circular curves in the field: deflection angle with chain, long-chord and short-chord methods, tacheometric methods, and the two-theodolite method. Understanding what each method measures helps in choosing the best approach for site constraints (traffic, obstructions, accuracy). Given Data / Assumptions: Two theodolites are set up at the tangent points (or at suitable control points). Intersecting lines of sight are used to locate points on the curve. Distances along the curve are not chained on the ground. Concept / Approach: In the two-theodolite method, one instrument sets the deflection (tangent) direction and the other directs the radial or chord direction so that the intersection of the two lines of sight marks the curve point. Thus, locations are obtained by the intersection of two angular directions without needing linear measurements to each point, which is advantageous where chaining is difficult or unsafe. Step-by-Step Solution: Set up Theodolite 1 to control the tangent/deflection at one end.Set up Theodolite 2 to bisect or direct to the required chord/radius direction.Sight simultaneously and mark the intersection point on the ground. Verification / Alternative check: Classic surveying references explicitly state that the two-theodolite method avoids chaining by relying purely on angle observations and intersection. Why Other Options Are Wrong: Linear-only is false; both measurements together are not required here; levelling is unrelated to plan curve setting. Common Pitfalls: Attempting to chain along busy roads; the two-theodolite method is designed to eliminate that need. Final Answer: Angular measurements only (no chaining to points on curve)

Question 17

Plane table plotting of inaccessible points On a plane table, which method is used to plot the position of an inaccessible point when you can occupy two (or more) accessible stations with known plan positions?

Accepted Answer

Introduction / Context: Plane table surveying allows direct plotting in the field. When a point cannot be occupied or measured directly (due to obstacles or hazards), we choose a method that does not require taping to the point but still fixes it accurately on the plan. This question asks you to identify that method. Given Data / Assumptions: Two or more accessible instrument stations with known plan positions are available. The target point cannot be physically reached for measurement. Sighting to the point is possible. Concept / Approach: The intersection method fixes the point by drawing rays from at least two occupied stations whose plan positions are already plotted. The intersection of these rays on the drawing sheet gives the location of the inaccessible point. This avoids chaining or occupying the point itself, unlike radiation or traversing that typically require measuring distances or moving the table successively. Step-by-Step Solution: Orient the plane table at the first station; draw a ray to the target.Shift to a second station with good geometry; reorient; draw a ray to the same target.Mark the intersection of the two rays as the plotted position of the point. Verification / Alternative check: Using more than two rays from additional stations allows a check for plotting accuracy (the rays should concur at a single point). Why Other Options Are Wrong: Radiation requires measuring distance from a single occupied point. Traversing is for establishing control by moving the table between stations. “None” and “tacheometry only” do not reflect standard plane-table methods. Common Pitfalls: Poor station geometry (very small intersection angle) leading to large positional uncertainty; ensure good intersection angles. Final Answer: Intersection

Question 18

Constructional details of survey chains Which of the following statements about a survey chain and its accessories is/are correct?

Accepted Answer

Introduction / Context: Understanding the components and markings of chains is essential for accurate chaining and for avoiding systematic errors. This question consolidates several commonly examined facts about chain construction and accessories used in the field. Given Data / Assumptions: We are considering standard chain design conventions. Tag numbering and total length are per traditional practice. Handles and swivels are part of the functional design. Concept / Approach: Survey chains include end handles whose lengths are accounted for in the overall designated chain length. Swivel joints at the handles prevent twist accumulation, ensuring the chain lays straight for measurement. Tagging at regular intervals aids in quick reading; the fifth tag from either end carries the number 5 to indicate 5-link intervals in traditional designs. A Gunter’s chain, rooted in imperial practice, totals 66 feet (100 links), enabling convenient conversion to acres and square chains. Step-by-Step Solution: Confirm inclusion of handle length: standard practice includes handles in nominal chain length.Confirm need for swivels: they minimize torsion and kinks, maintaining measurement accuracy.Confirm tag numbering convention: the fifth tag is numbered 5 to expedite counting.Confirm total length of Gunter’s chain: 66 ft (100 links). Verification / Alternative check: Survey manuals and equipment specifications describe these features explicitly for both metric and imperial chain systems. Why Other Options Are Wrong: Each individual statement is correct; selecting fewer than all omits valid facts. Common Pitfalls: Confusing tape measures with chains; overlooking that accessories form part of the standardized length. Final Answer: All the above

Question 19

Shape of the bubble tube surface The longitudinal profile (surface) of a level’s bubble tube is designed to be which geometric curve?

Accepted Answer

Introduction / Context: Bubble tubes (spirit levels) rely on precise geometry to translate small tilts of the instrument into measurable bubble movements. The curve chosen for the internal surface directly controls the relationship between tilt angle and bubble displacement (sensitivity). Given Data / Assumptions: The bubble resides in a liquid-filled, slightly curved glass tube. We consider the longitudinal (along-the-tube) shape. Small-angle behavior should be predictable and approximately linear. Concept / Approach: The inside surface of the tube is ground to a circular arc of very large radius. For small angular tilts, the arc length subtended is proportional to the angle (s ≈ R * α), giving a practical linear relationship between tilt and bubble travel over a limited range. This simplicity of design and predictable sensitivity are why a circular arc is used rather than more complex curves. Step-by-Step Solution: Relate bubble movement to instrument tilt via s ≈ R * α.Recognize that a circular arc provides consistent sensitivity for small angles.Conclude that the longitudinal surface is circular. Verification / Alternative check: Instrument design texts routinely derive sensitivity formulas from the circular-arc assumption and specify tube radii accordingly. Why Other Options Are Wrong: Straight, parabolic, elliptic, or hyperbolic surfaces would not provide the desired simple proportionality and manufacturability for standard levels. Common Pitfalls: Confusing the bubble tube’s external shape with the internal grinding; assuming “straight” because the tube appears nearly straight to the eye. Final Answer: Circular arc

Question 20

Highway and railway alignment – identifying an S-shaped transition In geometric design and surveying terminology, the curve composed of two circular arcs of different radii whose centers lie on opposite sides of the curve (producing an S-shaped path…

Accepted Answer

Introduction / Context: A designer often needs to connect two tangents that deflect in opposite senses, for example when a road or railway alignment bends first to the left and then to the right. Surveying terminology distinguishes between several multi-arc combinations. This item tests your ability to recognize the specific case where two arcs have different radii and their centers are on opposite sides of the path, producing an S-shaped transition commonly used in constrained corridors. Given Data / Assumptions: Two circular arcs are used in succession. The radii are different (R1 ≠ R2). The centers of curvature are on opposite sides of the curve, implying opposite senses of curvature. Plan view geometry is considered; not a vertical profile issue. Concept / Approach: Curve classification in plan is based on the relative position of arc centers and curvature direction. If centers are on the same side, the combination is called a compound curve. If they are on opposite sides, the path reverses curvature, creating an S-shape, and is called a reverse curve (also termed a serpentine curve). Simple curves contain only one constant-radius arc, and vertical curves pertain to profiles rather than planimetric alignment. Step-by-Step Solution: Identify curvature sense: the path bends left then right (or vice versa) → curvature changes sign.Opposite-sides criterion: centers of the two arcs lie on opposite sides of the alignment → reverse curvature case.Different radii permitted: R1 and R2 can be unequal in a reverse curve, which does not change the classification.Therefore, the correct term is “reverse curve”. Verification / Alternative check: Highway and railway manuals define compound curves as successive arcs curving the same way (centers on the same side), and reverse curves as arcs curving in opposite directions (centers on opposite sides). The problem statement clearly matches the latter definition. Why Other Options Are Wrong: Simple curve: only one arc; does not reverse curvature. Compound curve: centers on the same side; same curvature sense. Vertical curve: concerns grade changes in profile, not plan direction changes. Common Pitfalls: Confusing compound with reverse because both may use two arcs; forgetting that “opposite sides” of centers is the key distinguishing feature. Final Answer: A reverse curve

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