Questions and answers for exam Preparation

MCQs

Total Questions : 132 | Page 2 of 14 pages

Question 11. The angle at point of intersection of tangents indicate____________

Radius of the arc
Angle of the arc
Curvature angle
Deflection angle

Discuss Question

Answer: Option D. -> Deflection angle
Answer: (d).Deflection angle

Question 12. The angle at point of intersection of tangents indicate____________Which of the following curves helps in avoiding overturning of vehicles?The tangent distance of a long curve is given as____________

Radius of the arcSimple curve\(T = t_l – (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)
Angle of the arcTransition curve\(T = t_l + (t_s – t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)
Curvature angleCompound curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)
Deflection angleReverse curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)

Discuss Question

Answer: Option D. -> Deflection angleReverse curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)
Answer: Deflection angleTransition curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)

Question 13. Find the value of the long curve tangent distance, if the tangent length of short and long curves were given as 23.21m and 65.87m. The total deflection is 67˚54ꞌ and the deflection angle at short curve is given as 28˚43ꞌ.

112.06m
121.06m
211.06m
121.68m

Discuss Question

Answer: Option A. -> 112.06m
Answer: (a).112.06m

Question 14. Determine the value of chainage of point of the compound curve, if the chainage at T1 is given as 226.43m and the curve length as 23.64m.

205.07
250.07
207.7
202.79

Discuss Question

Answer: Option B. -> 250.07
Answer: (b).250.07

Question 15. The angle at point of intersection of tangents indicate____________Which of the following curves helps in avoiding overturning of vehicles?The tangent distance of a long curve is given as____________What would be the short curve length of tangent if the radius of curvature is given as 43.21m and deflection of about 76˚54ꞌ?

Radius of the arcSimple curve\(T = t_l – (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)34.13m
Angle of the arcTransition curve\(T = t_l + (t_s – t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)43.13m
Curvature angleCompound curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)43.31m
Deflection angleReverse curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)34.31m

Discuss Question

Answer: Option B. -> Angle of the arcTransition curve\(T = t_l + (t_s – t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)43.13m
Answer: Deflection angleTransition curve\(T = t_l + (t_s + t_l) \frac{sin⁡(Δ1)}{sin⁡Δ}\)34.31m

Question 16. If the radii of the curves in a reverse curve are equal, calculate the distance between the tangent points T1 and T2. Assume R = 98.54m with deflection angle 54˚31ꞌ.