Theodolite, Tacheometry and Curves
JKSSB Junior Engineer (Civil) · Surveying · 13 practice questions with answers and explanations: 5 Easy, 6 Medium, 2 Hard.
Sample questions
Sample Easy question 1
The temporary adjustments of a theodolite include:
- Centring, levelling up and eliminating parallax
- Adjusting the plate bubble with the capstan screws
- Setting the vertical circle index
- Making the line of sight perpendicular to the trunnion axis
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Answer: A. Permanent adjustments are made by the manufacturer or an expert, and temporary ones at every set-up.
Sample Easy question 2
Transiting a theodolite means:
- Turning the telescope through 180° in the vertical plane
- Rotating the instrument about the vertical axis
- Changing the face of the vertical circle only
- Moving the instrument to a new station
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Answer: A. Swinging is a rotation about the vertical axis, and transiting is a rotation about the horizontal axis.
Sample Easy question 3
The least count of a vernier theodolite is the:
- Difference between two readings
- Value of the smallest main scale division divided by the number of vernier divisions
- Number of vernier divisions divided by the main scale reading
- Sum of the vernier divisions
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Answer: B. A common least count is 20 seconds.
Sample Medium question 4
A closed traverse of 1,000 m perimeter has a closing error of 0.2 m. The relative accuracy is:
- 1 in 500
- 1 in 5,000
- 1 in 50,000
- 1 in 200
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Answer: B. Accuracy = error ÷ perimeter = 0.2 ÷ 1,000 = 1 in 5,000.
Sample Medium question 5
A simple curve of radius 200 m has a deflection angle of 60° and the point of intersection is at chainage 1,000 m. The chainage of the end of the curve (PT) is about:
- 1,093.97 m
- 1,115.47 m
- 1,209.44 m
- 884.53 m
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Answer: A. T = R tan(Δ ÷ 2) = 115.47 m, so PC = 1,000 − T = 884.53 m, and PT = PC + L = 884.53 + 209.44 = 1,093.97 m.
Sample Medium question 6
The sum of the latitudes of a closed traverse is +0.12 m and the sum of the departures is −0.16 m. The closing error is:
- 0.04 m
- 0.2 m
- 0.28 m
- 0.4 m
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Answer: B. e = √(ΣL² + ΣD²) = √(0.12² + 0.16²) = 0.2 m.
Sample Hard question 7
The deflection angle to the end of a 20 m chord on a curve of radius 300 m, from δ = 1,718.9 c ÷ R minutes, is about:
- 115 minutes
- 57 minutes
- 15 minutes
- 229 minutes
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Answer: A. δ = 1,718.9 × 20 ÷ 300 = 114.6 minutes.
Sample Hard question 8
A staff intercept of 1.2 m is read on a sight inclined at 10° with k = 100 and C = 0. The horizontal distance D = k s cos²θ is about:
- 3.6 m
- 118.2 m
- 120 m
- 116.4 m
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Answer: D. D = 100 × 1.2 × cos²10° = 116.4 m.