Surveying MCQ With Answers for JKSSB Junior Engineer (Civil)

Technical · 2026-10-27 · 6 min read

Surveying carried 10 marks in the board's earlier Junior Engineer (Civil) syllabus, which lists chain, compass and plane table surveying, levelling, tacheometry, the theodolite, traversing, contouring and curves. The board's 2025 paper had 120 one-mark questions in 120 minutes, and about one in three questions needed a calculation, so practise numerical problems as well as definitions. Our guide to the 2026 notice shows where this section fits. This page gives a revision list, worked problems and solved MCQs.

Quick revision list

  • Chain surveying: ranging, offsets, tape corrections, and the optical square for setting out right angles
  • Compass surveying: whole circle and quadrantal bearings, back bearings, magnetic declination and local attraction
  • Levelling: height of instrument and rise-and-fall methods, the arithmetical check, reciprocal levelling, and the curvature and refraction correction
  • Theodolite: temporary adjustments, face left and face right, swinging and transiting, and the closing error of a traverse
  • Tacheometry: stadia formula D = ks + C, with k usually 100 and C usually zero
  • Plane table: radiation, intersection, traversing and resection
  • Contours: contour interval, steep and gentle slopes, and which contour lines may or may not meet
  • Curves: simple circular curve with tangent length T = R tan(Δ ÷ 2), length L = πRΔ ÷ 180 and long chord 2R sin(Δ ÷ 2)

Worked numerical problems

  • Bearing conversion: a whole circle bearing of 215°30' lies in the third quadrant, so the quadrantal bearing is S (215°30' − 180°) W = S 35°30' W
  • Back bearing: the back bearing of a line with a forward bearing of 125° is 125° + 180° = 305°
  • Levelling: a point has an RL of 100.00 m, the backsight is 1.50 m and the foresight to the next point is 2.30 m. The height of instrument is 101.50 m, so the next RL is 101.50 − 2.30 = 99.20 m
  • Curvature and refraction: over 2 km the combined correction is 0.0673 × 2² = 0.269 m
  • Stadia: a staff intercept of 1.50 m with k = 100 and C = 0 gives a horizontal distance of 150 m
  • Tape correction: a 30 m tape that is 0.02 m too long measures 150 m on the ground. The true length is 150 × 30.02 ÷ 30 = 150.1 m
  • Simple curve: R = 300 m and a deflection angle of 40° give T = 300 × tan 20° = 109.19 m and L = π × 300 × 40 ÷ 180 = 209.44 m

Levelling in brief

In the height of instrument method the height of instrument is the reduced level of the benchmark plus the backsight, and the reduced level of each point is the height of instrument minus the foresight to it. The rise-and-fall method compares each reading with the one before. Both methods must pass the same check: the sum of the backsights minus the sum of the foresights equals the sum of the rises minus the sum of the falls, and equals the difference between the last and the first reduced levels. The height of instrument method is quicker for many intermediate points, and the rise-and-fall method is the better check on each reading.

Common mistakes

  • Forgetting to add or subtract 180° correctly when the bearing is in the third or fourth quadrant
  • Using the combined correction 0.0673 d² with d in metres, when it needs kilometres
  • Calling a tape that is too long a source of an under-measurement: a long tape records fewer tape lengths, so the true length is more than the measured length
  • Mixing up the deflection angle and the angle of intersection of a curve

Traversing in brief

In a closed traverse the angles are measured with the theodolite and the lengths with a chain or tape, or by tacheometry. The angular misclosure is checked against (2n − 4) × 90° and shared equally among the angles. The linear misclosure shows up as a gap in the sum of the latitudes and departures, and it is spread among the sides by Bowditch's rule, in proportion to the side lengths. The accuracy of the traverse is the closing error divided by the perimeter.

Practise

Work the solved questions below, then read the Junior Engineer (Civil) guide for the dates and the marks for every subject. The other subject pages in this series cover fluid mechanics, mechanics and structures, RCC and construction management.

By Shivam · Written with AI assistance and fact-checked · Questions on this page are original · Editorial policy · Report an error

Practice questions

Q1. The quadrantal bearing of a line with a whole circle bearing of 215°30' is:

  1. S 54°30' W
  2. S 35°30' W
  3. N 35°30' W
  4. N 54°30' E
Show answer

S 35°30' W — A bearing between 180° and 270° lies in the third quadrant, so it is S (WCB − 180°) W.

Q2. The back bearing of a line whose forward bearing is 125° is:

  1. 305°
  2. 55°
  3. 235°
  4. 215°
Show answer

305° — Back bearing = forward bearing + 180° when the forward bearing is less than 180°.

Q3. Local attraction at a station in compass surveying is detected by comparing:

  1. The length of the line with the chain
  2. The level at both ends
  3. The time of the observation
  4. The fore bearing and the back bearing of a line
Show answer

The fore bearing and the back bearing of a line — Without local attraction the two bearings differ by exactly 180°.

Q4. Magnetic declination is the angle between:

  1. The true meridian and the grid north
  2. The compass needle and the horizontal
  3. The magnetic meridian and the true meridian
  4. The line of sight and the horizontal
Show answer

The magnetic meridian and the true meridian — It varies from place to place and changes with time.

Q5. The combined correction for curvature and refraction over a distance of d km is, in metres:

  1. 0.0785 d²
  2. 0.0673 d²
  3. 0.673 d
  4. 0.0673 d
Show answer

0.0673 d² — Curvature alone is 0.0785 d², and refraction reduces it by one-seventh to 0.0673 d².

Q6. The combined curvature and refraction correction over a sight of 2 km is about:

  1. 0.269 m
  2. 0.135 m
  3. 0.538 m
  4. 1.346 m
Show answer

0.269 m — 0.0673 × 2² = 0.269 m.

Q7. Reciprocal levelling eliminates:

  1. Only the staff reading error
  2. Only the personal error
  3. Only the temperature error
  4. The errors of collimation, curvature and refraction
Show answer

The errors of collimation, curvature and refraction — Taking readings from both banks makes these errors equal and opposite.

Q8. The arithmetical check for the rise-and-fall method is:

  1. ΣBS + ΣFS = Σrise + Σfall
  2. ΣBS = ΣFS
  3. ΣBS − ΣFS = Σrise − Σfall = last RL − first RL
  4. Σrise = ΣFS
Show answer

ΣBS − ΣFS = Σrise − Σfall = last RL − first RL — All three differences must agree.

Q9. The reduced level of a point is 100.00 m, the backsight is 1.50 m and the foresight to the next point is 2.30 m. The reduced level of the next point is:

  1. 100.80 m
  2. 99.20 m
  3. 101.50 m
  4. 98.50 m
Show answer

99.20 m — HI = 100.00 + 1.50 = 101.50 m and RL = 101.50 − 2.30 = 99.20 m.

Q10. Which statement about contours is correct?

  1. Contours of different elevations cannot cross except at an overhanging cliff or a cave
  2. Contours of different elevations always cross each other
  3. Contours never close on themselves
  4. Contours are always straight lines
Show answer

Contours of different elevations cannot cross except at an overhanging cliff or a cave — Where they meet on a vertical face, several contours coincide.

Q11. Closely spaced contours indicate:

  1. A gentle slope
  2. A flat plain
  3. A water body
  4. A steep slope
Show answer

A steep slope — A small horizontal gap for the same vertical interval means a steeper slope.

Q12. The vertical distance between two consecutive contours is called the:

  1. Horizontal equivalent
  2. Datum
  3. Contour interval
  4. Gradient
Show answer

Contour interval — The horizontal distance between them is the horizontal equivalent.

Q13. The temporary adjustments of a theodolite include:

  1. Adjusting the plate bubble with the capstan screws
  2. Centring, levelling up and eliminating parallax
  3. Making the line of sight perpendicular to the trunnion axis
  4. Setting the vertical circle index
Show answer

Centring, levelling up and eliminating parallax — Permanent adjustments are made by the manufacturer or an expert, and temporary ones at every set-up.

Q14. Transiting a theodolite means:

  1. Turning the telescope through 180° in the vertical plane
  2. Rotating the instrument about the vertical axis
  3. Changing the face of the vertical circle only
  4. Moving the instrument to a new station
Show answer

Turning the telescope through 180° in the vertical plane — Swinging is a rotation about the vertical axis, and transiting is a rotation about the horizontal axis.

Q15. In tacheometry with k = 100 and C = 0, the staff intercept is 1.50 m. The horizontal distance is:

  1. 15 m
  2. 100 m
  3. 1.5 m
  4. 150 m
Show answer

150 m — D = ks + C = 100 × 1.50 = 150 m.

Q16. A point that cannot be reached is best located on a plane table by:

  1. Radiation from a single station only
  2. Traversing
  3. Intersection
  4. Resection
Show answer

Intersection — Rays drawn from two stations meet at the point.

Q17. Resection in plane table surveying is used to:

  1. Locate an inaccessible point
  2. Locate the table station by sighting known points
  3. Measure the height of a point
  4. Draw a contour
Show answer

Locate the table station by sighting known points — The three-point problem is solved by Bessel's method or trial and error.

Q18. A 30 m tape that is 0.02 m too long measures 150 m. The true length is:

  1. 150.1 m
  2. 149.9 m
  3. 150.02 m
  4. 148.0 m
Show answer

150.1 m — True length = 150 × (30.02 ÷ 30) = 150.1 m.

Q19. The sum of the interior angles of a closed traverse of 5 sides is:

  1. 360°
  2. 450°
  3. 720°
  4. 540°
Show answer

540° — (2n − 4) × 90° = (10 − 4) × 90° = 540°.

Q20. The tangent length of a simple circular curve of radius 300 m and deflection angle 40° is about:

  1. 209.4 m
  2. 205.2 m
  3. 109.2 m
  4. 300 m
Show answer

109.2 m — T = R tan(Δ ÷ 2) = 300 × tan 20° = 109.2 m.

Related articles

More JKSSB Junior Engineer (Civil) guides

Study guides

Back to the blog