# Civil Engineering - GATE Exam Questions - Discussion

### Discussion :: GATE Exam Questions - Section 3 (Q.No.46)

46.

Design rate of super elevation for horizontal highway curve of radius 450 m for a mixed traffic condition, having a speed of 125 km/hour is

 [A]. 1 [B]. 0.05 [C]. 0.07 [D]. 0.154

Explanation:

No answer description available for this question.

 Koushik Pal said: (Mar 28, 2014) e = v^2/225R. So, e = 125^2/225*450. = 0.154.

 Berhane said: (Sep 5, 2014) For mixed traffic the supper elevation is calculated as follows: e = v^2/225*R that means 125^2/225*450 = 15625/101250 = 0.15432098.

 Shubham said: (Jul 18, 2016) But maximum superelevation should be 7%. It can't be more than 7%. So the answer should be 0.07.

 Aashish Ahirwar said: (Jan 11, 2017) V^2/225R = (125)^2/225 * 450 = 0.154.

 Vinay Singh said: (Aug 29, 2017) The answer should be .07. because it is max super elevation.

 Sadulla said: (Nov 14, 2017) Yes @Vinay. It should be 0.07.

 Priya said: (Dec 25, 2017) It's should be v2/225R.

 Shubham Ramdasrao Umak said: (Jan 30, 2018) According to me, its answer is = 0.07. Because whenever we are designing super elevation the maximum elevation is restricted to 0.07.

 Ramyasree said: (Aug 21, 2018) The answer is 0.154 only because in the question he didn't asked as per IRC. So as per IRC, it's is in between 7-10%.

 Thonduru Radhika said: (Dec 8, 2018) For mixed traffic, super elevation can be calculated. It should be V^2 / 225R. Since this value is less than 0.07, the super elevation of 0.154 may be adopted.

 Ankit said: (Dec 19, 2019) e = .75v^2÷gR. = .75x34.72^2÷(9.81x450). = 0.07 answer.