Q1. A simple circular curve of radius 600 m is to be set out on field. Calculate the value of versed sine for the curve if the deflection angle (Δ) is 120°.
Difficulty: Medium
A. 120 m
B. 300 m
C. 453.33 m
D. 333.33 m
Correct Answer: B. 300 mSolution:Versed sine M = R(1−cos(Δ/2)) = 600×(1−cos60°) = 600×0.5 = 300 m.
Q2. In a plane survey, the length of an arc 12 km long lying on the earth’s surface is _____ greater than the subtended chord.
Difficulty: Medium
A. 0.1 cm
B. 1 cm
C. 100 cm
D. 10 cm
Correct Answer: B. 1 cmSolution:Arc−Chord ≈ Arc³/(24R²). For a 12 km arc on the earth (R≈6370 km), this works out to about 1 cm — the basis for plane surveying’s flat-earth assumption.
Q3. Offsets from the tangents method of curve setting can be adopted if:
Difficulty: Easy
A. Both the deflection angle and the radius of curvature are large
B. The deflection angle is large and the radius of curvature is small
C. Both the deflection angle and the radius of curvature are small
D. The deflection angle is small and the radius of curvature is large
Correct Answer: C. Both the deflection angle and the radius of curvature are smallSolution:This linear (tape-only) method suits sharp curves only — small radius keeps offsets short and measurable, small deflection keeps the curve’s extent manageable.
Q4. _____ is the method of locating an offset point from 2 different points on a chain line in such a way that all the three points form a near-equilateral triangle.
Difficulty: Easy
A. Method of ties
B. Swing offset method
C. Oblique offset method
D. Perpendicular offset method
Correct Answer: A. Method of tiesSolution:Locating a point via two tie distances from two chain-line points, forming a near-equilateral triangle, minimises the sensitivity of the plotted position to small measurement errors.
Q5. The angle of intersection of two straights is 120°. Find the ratio of the length of long chord to the tangent length.
Difficulty: Hard
A. √3
B. 1
C. 1/3
D. 0.866
Correct Answer: A. √3Solution:Deflection angle Δ=180−120=60°. L/T = 2Rsin(30°)/[Rtan(30°)] = 1/(1/√3) = √3.
Q6. The length of the horizontal transition curve based on the allowable rate change of centrifugal acceleration is 48 m, based on the rate of introduction of super-elevation is 40 m, and based on the empirical formula recommended by IRC is 45 m. What will be the shift in the transition curve? Take radius = 200 m.
Difficulty: Hard
A. 0.42 m
B. 0.48 m
C. 0.33 m
D. 3.68 m
Correct Answer: B. 0.48 mSolution:Design length L = max(48,40,45) = 48 m. Shift S = L²/24R = 48²/(24×200) = 0.48 m.
Q7. The length of the first sub chord is 10 m and the radius of the curve is 200 m. Find the first tangential angle for setting out a circular curve.
Difficulty: Hard
A. 1° 25′ 57″
B. 100°
C. 85.95°
D. 100
Correct Answer: A. 1° 25′ 57″Solution:δ (minutes) = 1718.9×c/R = 1718.9×10/200 = 85.945′ = 1°25’57”. Note: the raw formula result is in MINUTES, not degrees.
Q8. A 3-degree curve is situated on a ruling gradient of 1 in 250 on a Broad Gauge track. What should be the actual ruling gradient considering the grade compensation of curvature?
Difficulty: Hard
A. 1 in 280
B. 1 in 300
C. 0.12%
D. 0.28%
Correct Answer: D. 0.28%Solution:Ruling gradient = 1in250 = 0.4%. BG compensation = 0.04%×3 = 0.12%. Actual = 0.4−0.12 = 0.28%.
Q9. The following is data recorded during curve setting using the two-theodolite method, with tangential angles 0°21’25” for P1 and 1°12’05” for P2. The values of A, B and C (deflection angles and POT reading) are respectively…
Difficulty: Hard
A. A = 0°21’25”; B = 1°33’30”; C = 357°14’25”
B. A = 0°21’25”; B = 1°33’30”; C = 177°14’25”
C. A = 0°21’25”; B = 1°33’30”; C = 182°45’35”
D. A = 0°21’25”; B = 1°33’30”; C = 182°45’35”
Correct Answer: A. A = 0°21’25”; B = 1°33’30”; C = 357°14’25”Solution:Deflection angles are cumulative: A = 0°21’25”; B = A+1°12’05” = 1°33’30”. The closing reading C = 360° − 2°45’35” = 357°14’25”.
Q10. Calculate the length of a simple circular curve, if its deflection angle is 45° and radius 400 m.
Difficulty: Easy
A. 314.16 m
B. 326.19 m
C. 298.78 m
D. 227.89 m
Correct Answer: A. 314.16 mSolution:l = πRΔ/180° = (π×400×45)/180 = 314.16 m.
Q11. The degree of the curve is an angle subtended at the centre by a chord of length _____ and the degree of a curve with radius 688 m will be equal to _____.
Difficulty: Medium
A. 30 m; 2.5”
B. 30 m; 2.5°
C. 30 m; 3”
D. 20 m; 2°
Correct Answer: B. 30 m; 2.5°Solution:Degree of curve uses a standard 30 m chord; D = 1720/R = 1720/688 = 2.5°.
Q12. If g₁ = +1.2% and g₂ = −0.8% and rate of change of grade = 0.1% per 20 m chain, then the length of the vertical curve is:
Difficulty: Medium
A. 10 chains
B. 20 chains
C. 30 chains
D. 15 chains
Correct Answer: B. 20 chainsSolution:Total grade change = 1.2−(−0.8) = 2.0%. Length = 2.0/0.1 = 20 chains.
Q13. In a highway an ascending gradient of 1 in 100 meets with a descending gradient of 1 in 80, determine the deviation angle.
Difficulty: Medium
A. 9/400
B. 9/800
C. 1/800
D. 1/400
Correct Answer: A. 9/400Solution:N = g₁−g₂ = 1/100−(−1/80) = 1/100+1/80 = 18/800 = 9/400.
