Q16. Stadia interval factor is:
Difficulty: Easy
A. Ratio of stadia interval to the focal length of the objective
B. Sum of focal length of the objective and distance between vertical axis of the instrument and objective
C. Ratio of multiplying constant to additive constant
D. Ratio of focal length of the objective to stadia interval
Correct Answer: D. Ratio of focal length of the objective to stadia intervalSolution:Stadia interval factor = multiplying constant K = f/i. Option B describes the additive constant; the ratio must be f/i, not i/f.
Q17. Consider the following statements and select the correct option.
i. The stadia method is based on the principle that the ratio of the perpendicular to the base is constant (K) in similar isosceles triangles.
ii. If angle = 34° 22′ 26.4″, then the constant K is 100.
Difficulty: Hard
A. Statement i is wrong, but ii is correct
B. Both the statements are wrong
C. Statement i is correct, but ii is wrong
D. Both the statements are correct
Correct Answer: D. Both the statements are correctSolution:The similar-triangles principle underlies the stadia method (i, true). Using tan(θ/2)=1/(2K), an angle of 34°22’26.4″ corresponds exactly to K=100 (ii, true).
Q18. The following statements pertain to theodolite used in surveying. Identify the incorrect statement.
Difficulty: Medium
A. The line of sight should be parallel to the horizontal axis
B. The horizontal axis should be perpendicular to the vertical axis
C. The vertical cross hair should be perpendicular to the horizontal axis
D. The axis of the plate bubble should be in a plane perpendicular to the vertical axis
Correct Answer: A. The line of sight should be parallel to the horizontal axisSolution:The line of sight (collimation) must be perpendicular, not parallel, to the horizontal axis — so the telescope sweeps a true vertical plane when tilted. B, C, and D are all correct conditions.
Q19. If the latitude and departure of a line AB are 40 m and 30 m respectively, estimate the length of the line AB.
Difficulty: Easy
A. 35 m
B. 70 m
C. 50 m
D. 10 m
Correct Answer: C. 50 mSolution:Latitude and departure are perpendicular vector components. Length = √(40²+30²) = √2500 = 50 m, using the Pythagorean theorem.
Q20. Following are some errors in total station survey work: Vertical collimation error, Centering error, Horizontal collimation error, Eccentricity error. Categorize them as temporary (T) or permanent (P) adjustment errors.
Difficulty: Hard
A. T: Vertical & Horizontal collimation | P: Centering, Eccentricity
B. T: Centering, Horizontal collimation, Eccentricity | P: Vertical collimation
C. T: Eccentricity, Horizontal collimation | P: Centering, Vertical collimation
D. T: Centering | P: Vertical collimation, Horizontal collimation, Eccentricity
Correct Answer: D. T: Centering | P: Vertical collimation, Horizontal collimation, EccentricitySolution:Centring is redone at every setup (temporary). Collimation and eccentricity errors are fixed properties of the instrument’s internal calibration (permanent).
Q21. Which of the following actions is NOT a permanent adjustment in case of transit theodolite?
Difficulty: Medium
A. Axis of the telescope level adjusted parallel to the line of collimation
B. Axis of the plate level adjusted perpendicular to vertical axis
C. Bringing the vertical axis of the theodolite exactly over the station mark
D. Horizontal axis is perpendicular to vertical axis when the instrument is levelled
Correct Answer: C. Bringing the vertical axis of the theodolite exactly over the station markSolution:This describes centring — a temporary adjustment redone at every setup, not a permanent (internal axis) adjustment.
Q22. Bowditch rule is also termed as:
Difficulty: Easy
A. Compass rule
B. Transit rule
C. Graphical rule
D. Axis rule
Correct Answer: A. Compass ruleSolution:Bowditch’s rule is universally also called the compass rule — it distributes error proportional to line length, used when angular and linear precision are comparable.
Q23. A smart station is used to indicate:
Difficulty: Easy
A. A total station with an integrated GPS module
B. A total station attached to a computer
C. Total station with electromagnetic distance measuring equipment
D. A total station with software to calculate and display quantities
Correct Answer: A. A total station with an integrated GPS moduleSolution:A smart station combines a total station’s precision angle/distance measurement with a built-in GPS/GNSS receiver in one hybrid unit.
Q24. Select the incorrect statement from the following.
Difficulty: Medium
A. In the total station, angles and distances are recorded digitally
B. The total station cannot measure horizontal distance less than 2 km
C. The total station has all facilities of tacheometer operated electronically
D. The total station is operated through the control panel
Correct Answer: B. The total station cannot measure horizontal distance less than 2 kmSolution:Total stations measure short distances (a few metres to hundreds of metres) with millimetre precision — the 2 km figure is an upper range limit, not a lower one.
Q25. While using a theodolite, how to change the reading on the horizontal circle while measuring a horizontal angle?
Difficulty: Easy
A. Both upper and lower clamp are tightened
B. Upper clamp is tightened and lower clamp is loosened
C. Upper clamp is loosened and lower clamp is tightened
D. Both upper and lower clamp are loosened
Correct Answer: C. Upper clamp is loosened and lower clamp is tightenedSolution:With the lower clamp tight (circle fixed) and upper clamp loose, the vernier rotates freely relative to the fixed graduated circle, changing the reading.
Q26. Which type of error is represented by a closed traverse, if the algebraic sum of latitude of all the lines is zero?
Difficulty: Easy
A. Compensating error
B. Negative error
C. No error
D. Positive error
Correct Answer: C. No errorSolution:For a perfectly closed traverse, north and south latitudes must exactly balance — an algebraic sum of zero means there is no closing error in that component.
Q27. If in a closed traverse, the sum of north latitudes exceeds south latitudes, and the sum of west departures exceeds east departures, the bearing of the closing line is in the:
Difficulty: Medium
A. SE quadrant
B. NE quadrant
C. NW quadrant
D. SW quadrant
Correct Answer: C. NW quadrantSolution:Excess north latitude gives a positive (N) latitude error; excess west departure gives a negative (W) departure error. Positive latitude + negative departure = NW quadrant.
Q28. The process of turning the telescope in the theodolite in vertical plane through 180 degrees about the trunnion axis, is called:
Difficulty: Easy
A. Swinging
B. Centring
C. Transiting
D. Levelling
Correct Answer: C. TransitingSolution:Transiting (plunging/reversing) is the 180° vertical-plane rotation of the telescope about the trunnion axis, used for face-left/face-right observations.
Q29. Calculate the intersect angle (degree) for the anallactic telescope:
Difficulty: Hard
A. 0.467
B. 0.573
C. 0.592
D. 0.598
Correct Answer: B. 0.573Solution:For K=100: tan(θ/2)=1/200, so θ/2=0.2865°, giving θ=0.573°.
Q30. The least count of a theodolite is:
Difficulty: Easy
A. 1 degree
B. 2 minutes
C. 2 seconds
D. 20 seconds
Correct Answer: D. 20 secondsSolution:A standard vernier transit theodolite has a least count of 20 seconds, much coarser than the 1-second resolution of an electronic theodolite (Q9).
