NEC Registration Exam (Geomatics) – Chapter 1 MCQs

AGeE01: Fundamentals of Surveying – Set 1 (60 Questions)

Each question has one correct option. An answer key with brief reasoning is provided immediately after
each question.

Q1. Surveying is best defined as:

A. The art of preparing topographic maps only

B. The science and art of determining relative positions of points on/near the Earth and representing
them on plans/maps
C. Measurement of elevations alone
D. Analysis of satellite images only
Answer: B
Explanation: Surveying deals with planimetric and altimetric positioning and depiction on plans/maps; not
restricted to topography alone.

Q2. Which ancient civilization is most associated with re■establishing field boundaries after floods using
rope stretchers?

A. Greek
B. Roman
C. Egyptian
D. Chinese
Answer: C
Explanation: Egyptians used rope stretching after Nile floods to reset agricultural boundaries.

Q3. In plane surveying, the Earth’s surface is assumed:

A. Spheroidal
B. Ellipsoidal
C. Flat over small areas
D. Irregular geoid everywhere
Answer: C
Explanation: For small areas, curvature is negligible and a plane surface is assumed.

Q4. Geodetic surveying is essential when:

A. Mapping a football field

B. Detailing a small building site
C. Establishing national control networks
D. Setting out a residential plot
Answer: C
Explanation: Geodetic surveys account for Earth curvature and are used for large areas and control
networks.

Q5. The fundamental principle 'work from whole to part' primarily aims to:

A. Save field time only

B. Reduce instrumental errors to zero
 C. Prevent accumulation/proliferation of errors in details
D. Avoid using triangles
Answer: C
Explanation: A strong control framework minimizes propagation of local errors.

Q6. A point should be located by at least two independent measurements to:

A. Make field notes lengthy

B. Allow a check on observations
C. Reduce need for experienced staff
D. Avoid booking angles
Answer: B
Explanation: Redundancy allows detection of blunders and improved reliability.

Q7. A well■conditioned triangle for triangulation has interior angles approximately:

A. 5°, 85°, 90°

B. 20°, 20°, 140°
C. 60°, 60°, 60°
D. 10°, 30°, 140°
Answer: C
Explanation: Equilateral or near■equilateral triangles minimize angular error effects.

Q8. Which is NOT a primary application of surveying?

A. Route location for highways

B. Boundary demarcation
C. Design of circuits in microchips
D. Hydrographic charting
Answer: C
Explanation: Microchip circuits belong to electronics; the others are surveying applications.
Q9. A numerical map scale of 1:25,000 means 1 cm on map equals:

A. 25 m on ground
B. 250 m on ground
C. 2.5 km on ground
D. 0.25 km on ground
Answer: B
Explanation: 1 cm × 25,000 = 25,000 cm = 250 m.

Q10. Which set lists only angular measurement instruments?

A. Chain, tape, EDM

B. Compass, theodolite, total station
C. Level, staff, tape
D. Ranging rod, cross■staff, prism
Answer: B
Explanation: Compass, theodolite, and total station measure directions/angles.

Q11. One radian equals approximately:

 A. 28.6479°
B. 45°
C. 57.2958°
D. 90°
Answer: C
Explanation: 1 rad = 180/π ≈ 57.2958°.

Q12. Which statement about grads (gons) is TRUE?

A. 100 grads = 360°

B. 400 grads = 360°
C. 200 grads = 360°
D. 90 grads = 360°
Answer: B
Explanation: The centesimal system divides a circle into 400 grads (gons).

Q13. Standardization in surveying primarily ensures:

A. Lower salaries
B. Uniformity and consistency of measurements
C. Faster GPS satellites
D. More rain■proof equipment
Answer: B
Explanation: Standards unify methods, units, instruments, and data exchange.

Q14. In chain surveying, the most suitable terrain is:

A. Open and fairly level ground

B. Dense forest and hilly terrain with obstacles
C. Urban areas with tall buildings
D. Swampy areas only
Answer: A
Explanation: Chain surveying requires clear lines and relatively level ground.

Q15. Tape correction for temperature is:

A. Always positive
B. Always negative
C. Depends on actual vs standard temperature
D. Zero for steel tapes
Answer: C
Explanation: Tape expands when hotter than standard (positive correction if measured too short) and
contracts when colder.

Q16. The main advantage of plane table surveying is:

A. Highest precision angular measurement

B. Direct plotting in the field reduces transcription errors
C. Unaffected by wind
D. Works well in rain
Answer: B
Explanation: Plane table allows on■site plotting and visualization; precision is limited.

Q17. Compass surveying is affected by:

A. Magnetic declination and local attraction

B. Refraction only
C. Coriolis effect strongly
D. Plate tectonics
Answer: A
Explanation: Magnetic influences change bearings; local attraction disturbs needle.

Q18. The line of sight in a theodolite must be:

A. Parallel to trunnion axis

B. Perpendicular to horizontal axis when collimation is correct
C. Parallel to vertical axis
D. Coincident with bubble axis
Answer: B
Explanation: For proper geometry, line of sight ■ trunnion (horizontal) axis.

Q19. Tacheometry determines distance primarily using:

A. EDM phase shift

B. Parallax constancy
C. Stadia intercepts and an anallactic lens
D. Micrometry of bubble tube
Answer: C
Explanation: Stadia method uses staff intercepts; anallactic lens removes additive constant.

Q20. Triangulation determines positions mainly from:

A. Distances only
B. Bearings only
C. Angles measured in a network of triangles with at least one known baseline
D. GPS only
Answer: C
Explanation: Angles from many stations with a measured baseline form the framework.
Q21. Trilateration uses primarily:

A. Distances
B. Angles
C. Directions only
D. Elevations only
Answer: A
Explanation: Trilateration computes positions from measured distances.

Q22. In traversing, the strength of figure improves when:

A. All angles are very acute

B. Traverse is long with few stations
C. Traverse is well distributed with checks (e.g., closing the traverse)
D. All lines pass through one point
Answer: C
Explanation: Well■shaped traverses with checks provide stronger geometry and error detection.

Q23. Which method adjusts a misclosure in an open traverse?

A. Transit rule or Bowditch rule (for closed), not applicable to truly open
B. Equal angle distribution only
C. Increase last bearing by misclosure
D. Use reciprocal leveling
Answer: A
Explanation: Transit/Bowditch apply to closed traverses; a truly open traverse cannot be proportionally
adjusted without additional control.

Q24. Bowditch (compass) rule distributes linear misclosure in proportion to:

A. Measured angles
B. Latitude only
C. Square of line length
D. Line length
Answer: D
Explanation: Corrections ∝ individual line lengths for Bowditch adjustment.

Q25. A well■conditioned triangulation baseline should be:

A. Very short compared to triangle sides

B. Measured precisely and located centrally in the network
C. Avoided to reduce work
D. Oriented due east■west only
Answer: B
Explanation: A central, precisely measured baseline provides robust scale to the network.

Q26. Direct spirit leveling determines:

A. Horizontal distances
B. Vertical angles only
C. Difference in elevation between points
 D. Magnetic declination
Answer: C
Explanation: Leveling measures elevation differences using a horizontal line of sight.

Q27. A two■peg test is used to:

A. Calibrate EDM constant

B. Check and adjust collimation error in a level
C. Determine magnetic declination
D. Compute geoid undulation
Answer: B
Explanation: Two■peg test detects line■of■sight not being truly horizontal.

Q28. Reciprocal leveling is most useful when:

A. No staff is available
B. A long line crosses an obstacle like a river/valley
C. Distances are very short
D. Temperature is stable
Answer: B
Explanation: Taking readings from both sides cancels curvature/refraction and collimation errors over
obstacles.

Q29. Curvature and refraction combined correction (per km of sight) is approximately:

A. +0.067 m
B. −0.067 m
C. +0.011 m
D. −0.011 m
Answer: B
Explanation: Net correction is roughly −0.067 d^2 (in meters for distance d in km), i.e., about −0.067 m
per km for long sights.
Q30. In precise leveling, the best practice is to:

A. Keep back■sight distances much longer than fore■sight

B. Balance back■sight and fore■sight distances
C. Sight lengths > 1 km
D. Readings only on top of staff
Answer: B
Explanation: Balancing distances minimizes systematic errors (collimation, curvature/refraction).

Q31. Trigonometric leveling derives elevation difference using:

A. Staff readings only

B. Vertical angle and slope distance with instrument and target heights
C. Magnetic bearings
D. EDM phase lag only
Answer: B
Explanation: Heights are computed from vertical angles, measured distances, and instrument/target
heights.
Q32. Contour lines:

A. Intersect each other frequently

B. Represent points of equal elevation
C. Always form straight lines
D. Are spaced equally regardless of slope
Answer: B
Explanation: Contours join equal elevation points; spacing varies with slope.

Q33. In steep terrain, contour lines appear:

A. Widely spaced
B. Random
C. Closely spaced
D. Always dashed
Answer: C
Explanation: Closer spacing indicates steeper slopes.

Q34. Index contours are:

A. Contours drawn lighter

B. Every 5th or so contour drawn thicker and labeled for readability
C. Only used at sea level
D. Used only in cadastral maps
Answer: B
Explanation: Index contours improve map readability and are labeled with elevation.

Q35. The primary objective of reconnaissance in topographic surveying is to:

A. Draw final map in office

B. Collect preliminary information and plan control/monumentation
C. Set out curves
D. Perform least squares adjustment
Answer: B
Explanation: Reconnaissance informs planning of control, logistics, and methods.

Q36. Detailing in topographic survey is the process of:

A. Creating DEMs from LiDAR

B. Observing and plotting features and spot levels based on established control
C. Selecting map projections
D. Printing maps
Answer: B
Explanation: Detailing captures terrain and feature information with respect to control.

Q37. Random errors are characterized by:

A. Large magnitude and one■sided bias

B. Unpredictable sign, follow statistical laws, reduce by averaging
C. Gross mistakes in booking
D. Instrument out of calibration only
Answer: B
Explanation: Random errors scatter around true value and diminish with repeated observations.

Q38. A blunder is best described as:

A. Systematic bias due to instrument

B. Random fluctuation
C. Human mistake like transposing digits
D. Refraction effect
Answer: C
Explanation: Blunders are human mistakes and must be detected/removed, not averaged.

Q39. Accuracy vs precision:

A. Accuracy = repeatability; Precision = closeness to truth

B. Accuracy = closeness to truth; Precision = repeatability
C. Both are identical
D. Neither is measurable in surveying
Answer: B
Explanation: Precision reflects consistency; accuracy reflects closeness to true value.

Q40. Least squares adjustment aims to:

A. Maximize sum of squared residuals

B. Minimize sum of squared weighted residuals
C. Equalize all residuals to zero
D. Ignore correlations
Answer: B
Explanation: LSQ finds parameter estimates that minimize weighted squared residuals.
Q41. Variance–covariance matrix in adjustments represents:

A. Instrument prices
B. Precision of observations and their correlations
C. Coordinate system definition
D. Map legend
Answer: B
Explanation: Diagonal terms indicate variances, off■diagonals indicate covariances (correlations).

Q42. Propagation of error (linearized) uses:

A. Jacobian (design) matrix to transfer variances to derived quantities

B. Magnetic declination charts
C. Only diagonal variances
D. No partial derivatives
Answer: A
Explanation: The law of covariance propagation uses partial derivatives (Jacobian).

Q43. In a traverse adjustment by transit rule, angular misclosure is distributed:

A. Proportional to line lengths

B. Equally to all angles
C. Proportional to the tangents of angles
D. To the last angle only
Answer: B
Explanation: Transit rule distributes angular misclosure equally among included angles.

Q44. Degrees of freedom in an adjustment problem roughly equal:

A. Number of parameters
B. Number of observations minus number of parameters (constraints considered)
C. Number of baselines
D. Always 1
Answer: B
Explanation: ν = m − u after considering constraints, where m obs., u unknowns.

Q45. Official length unit in Nepal for surveying is:

A. Foot
B. Meter (SI)
C. Yard
D. Mile
Answer: B
Explanation: Nepal officially uses SI units; local units are converted for records.

Q46. A commonly used horizontal datum for GPS work in Nepal is:

A. Tokyo Datum
B. WGS■84
C. NAD27
D. OSGB36
Answer: B
Explanation: WGS■84 is standard for GNSS; Everest (Nepal) also used historically for maps.

Q47. Traditional hill land unit 'Ropani' equals approximately:

A. 338.63 m²
B. 508.72 m²
C. 6772.63 m²
D. 31.80 m²
Answer: B
Explanation: 1 Ropani = 508.72 square meters.

Q48. A map at 1:5,000 shows two points 4 cm apart. Ground distance is:

A. 20 m
B. 200 m
C. 2,000 m
D. 400 m
Answer: B
Explanation: 4 cm × 5,000 = 20,000 cm = 200 m.

Q49. If a 30 m steel tape standardized at 20°C is used at 35°C, measured distances are:

A. Too long; apply negative (subtract) correction

B. Too short; apply positive (add) correction
C. Unaffected by temperature
D. Random
Answer: B
Explanation: Thermal expansion lengthens tape; measured value is shorter than true, so add correction.

Q50. Magnetic declination is the angle between:

A. True meridian and grid meridian

B. Magnetic meridian and true meridian
C. Astronomic azimuth and geodetic azimuth
D. Local attraction and dip
Answer: B
Explanation: Declination = Magnetic − True meridian angle.

Q51. Local attraction in compass work is checked by:

A. Comparing FB and BB bearings of the same line

B. Two■peg test
C. Reciprocal leveling
D. Measuring temperature
Answer: A
Explanation: Differences from 180° indicate local attraction at one/both stations.

Q52. Temporary adjustments of a level include:

A. Collimation correction (permanent)

 B. Centering, leveling up, and focusing
C. Tribrach calibration
D. Vertical circle indexing
Answer: B
Explanation: Temporary adjustments are done at each setup: centering (if needed), leveling, focusing.

Q53. Which is NOT a permanent adjustment?

A. Making line of sight horizontal in a dumpy level (collimation)

B. Index error removal in theodolite
C. Focusing the eyepiece on crosshairs at each setup
D. Making vertical axis truly vertical
Answer: C
Explanation: Eyepiece focusing is a temporary adjustment.

Q54. EDM measures distance by:

A. Counting steps
B. Measuring travel time/phase of electromagnetic waves
C. Optical parallax only
D. GPS ephemerides
Answer: B
Explanation: EDM uses phase/round■trip time of EM waves to compute distance.

Q55. In topographic mapping, selection of contour interval depends on:

A. Map color only

B. Relief, map scale, and purpose
C. Compass declination
D. Magnetic dip
Answer: B
Explanation: Steep relief and small scales need larger intervals; gentle relief and large scales use
smaller intervals.

Q56. A total station combines:

A. Compass and staff

B. EDM and electronic theodolite with data recorder
C. Level and plane table
D. GPS and sonar
Answer: B
Explanation: Total stations integrate angle/distance measurement and onboard computation.

Q57. Closing error in leveling over a loop should be:

A. Ignored
B. Within allowable tolerance (e.g., k√K mm) and adjusted by proportioning to line lengths
C. Doubled deliberately
D. Applied to first setup only
Answer: B
 Explanation: If within tolerance, distribute correction proportionally to sight lengths or sections.

Q58. The main cause of 'stadia additive constant' in tacheometry is:

A. Instrumental zero shift

B. Optical geometry between objective and stadia diaphragm
C. Atmospheric refraction
D. Magnetic field
Answer: B
Explanation: Additive constant arises from the separation between the objective and the instrument’s
center.

Q59. 'Back■bearing' of a line in whole■circle bearing system equals:

A. Fore■bearing + 180° or −180° (as needed)

B. Fore■bearing − 90°
C. Azimuth + declination
D. Always 0°
Answer: A
Explanation: Add/subtract 180° to get the back■bearing within 0–360°.

Q60. If the sum of interior angles in an n■sided traverse is Σθ, angular misclosure is:

A. Σθ − (n−2)×90°
B. Σθ − (n−2)×180°
C. Σθ − 360°
D. n×90° − Σθ
Answer: B
Explanation: Theoretical sum = (n−2)×180°; difference is the misclosure.
Q61. Orthometric height is:

A. Height above ellipsoid

B. Height along plumb line above geoid
C. Geoidal undulation
D. Dynamic height
Answer: B
Explanation: Orthometric height is measured along the direction of gravity above the geoid.

Q62. Bench Mark (BM) in Nepal vertical control traditionally refers to:

A. Ellipsoidal height point

B. Permanent mark with known elevation above MSL
C. Temporary staff point
D. Arbitrary datum only
Answer: B
Explanation: BMs have published elevations relative to MSL (historically Indian MSL Karachi).

Q63. Instrumental parallax in a telescope is eliminated by:

A. Focusing objective on target first

B. Focusing eyepiece on crosshairs first, then objective on target
C. Tightening tangent screws
D. Leveling up
Answer: B
Explanation: Eyepiece focus on crosshairs removes parallax between crosshair image and target image.

Q64. Which does NOT reduce random error effects?

A. Averaging repeated observations

B. Using least squares adjustment with proper weights
C. Using poorly conditioned triangles
D. Balancing sight lengths
Answer: C
Explanation: Poor geometry amplifies errors; the others mitigate them.

Q65. The geoid is best described as:

A. A mathematical sphere
B. An equipotential surface approximating mean sea level
C. The Earth’s crust
D. A topographic surface
Answer: B
Explanation: Geoid is an equipotential surface that closely matches MSL under continents.

Q66. Grid distance differs from ground distance because of:

A. Magnetic declination only

B. Projection scale factor and elevation (combined scale factor)
C. Eyepiece focusing
D. Two■peg test
Answer: B
Explanation: Projection scale factor + elevation factor cause grid/ground distance differences.

Q67. In Nepal, a common topographic map scale for hills is:

A. 1:5,000
B. 1:25,000
C. 1:100,000
D. 1:1,250
Answer: B
Explanation: Survey Department uses 1:25,000 for hills and 1:50,000 for Terai base maps.

Q68. A staff reading sequence BS=1.225 m, IS=1.870 m, FS=2.455 m on the same setup indicates:

A. The instrument is too low

B. An intermediate sight was taken between BS and FS; compute RL accordingly
C. Collimation error is zero
D. Refraction dominates
Answer: B
Explanation: Intermediate sight is a legitimate reading for a detail point; use HI method to compute RLs.

Q69. Spherical excess in large triangles is considered in:

A. Plane surveying
B. Geodetic surveying on ellipsoid/sphere
C. Cadastral plotting only
D. Leveling only
Answer: B
Explanation: Large triangles exceed 180° due to Earth’s curvature—handled in geodetic computations.

Q70. Which leveling method cancels staff graduation error best?

A. Single observation
B. Reciprocal leveling
C. Equalizing BS and FS distances and reversing staff
D. Using only one staff face
Answer: C
Explanation: Balancing distances and reading both faces can help cancel certain errors.

Q71. If a contour line closes within the map and values increase towards the center, it represents:

A. A depression
B. A summit/hill
C. A vertical cliff
D. A ridge only
Answer: B
Explanation: Increasing values inward indicate a hill; decreasing indicate a depression.

Q72. A resection is:

A. Determining position by observing to three or more known points

 B. Determining unknown heights from known baselines
C. Setting out a right angle
D. Computing tape corrections
Answer: A
Explanation: Resection computes the observer’s coordinates from observations to known control points.

Q73. Strength of figure in least squares is improved by:

A. Long skinny triangles

B. Well■distributed stations and intersection angles near 60°
C. All stations on a straight line
D. Fewer observations
Answer: B
Explanation: Good geometry (near■equilateral triangles, good crossing angles) improves precision.

Q74. Which correction is typically applied to long EDM slope distances to get horizontal distance?

A. Magnetic declination
B. Slope (vertical angle) correction and meteorological correction
C. Compass dip
D. Local attraction
Answer: B
Explanation: Reduce slope distance to horizontal using vertical angle; apply meteo (T, P, humidity) for
refractive index.

Q75. The plane table accessory used to transfer the location of the instrument onto the sheet is:

A. Plumb bob/ plumbing fork

B. Trough compass
C. Alidade
D. U■frame
Answer: A
Explanation: Plumbing fork with plumb bob centers the table over the ground point.

Q76. A staff intercept between the top and bottom stadia hairs is 1.500 m; multiplying constant K=100 and
additive constant C=0. If line of sight is horizontal, the horizontal distance is:

A. 1.5 m
B. 15 m
C. 150 m
D. 0.015 km
Answer: C
Explanation: D = K·s + C = 100×1.5 + 0 = 150 m.

Q77. Which is TRUE about mapping scale selection?

A. Smaller denominator → smaller scale map

B. Larger denominator → larger scale map
C. 1:500 is a larger scale than 1:5000
D. Scale choice does not affect contour interval
Answer: C
Explanation: 1:500 shows more detail (larger scale) than 1:5000; denominator inversely relates to scale
size.

Q78. A 'benchmark' used temporarily for a leveling day’s work is called:

A. GTS Bench Mark

B. Permanent Bench Mark
C. Temporary Bench Mark (TBM)
D. Arbitrary Bench Mark only
Answer: C
Explanation: TBM is established for short■term reference during a project.

Q79. Azimuth measured clockwise from true north is equivalent to:

A. Whole■circle bearing
B. Quadrantal bearing
C. Magnetic bearing
D. Grid convergence
Answer: A
Explanation: Whole■circle bearing is a 0°–360° azimuth from a reference meridian.

Q80. In Bowditch adjustment, the correction to latitude of a line equals:

A. Total linear misclosure × (line length / perimeter)

B. Angular misclosure / number of angles
C. Northing of line / total northing
D. Zero
Answer: A
Explanation: Both latitude and departure corrections are apportioned by line length relative to perimeter.
Q81. Which error source is MOST reduced by balancing BS and FS distances in leveling?

A. Random reading error

B. Curvature and refraction (systematic) and collimation error
C. Staff graduation error only
D. Magnetic declination
Answer: B
Explanation: Balanced sights cancel curvature/refraction and collimation effects approximately.

Q82. The plane that best approximates Earth’s mean sea level is:

A. Ellipsoid
B. Geoid
C. Topographic surface
D. Astronomical horizon
Answer: B
Explanation: Geoid is an equipotential surface approximating MSL.

Q83. A mapping project in Kathmandu municipality most appropriately uses:

A. Geodetic survey only

B. Plane surveying techniques with local control tied to national datum
C. Astronomical observations only
D. Open traverse without closure
Answer: B
Explanation: Municipal mapping is typically plane survey tied to national control for consistency.