Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 6 (Q.No. 40)
40.
For a circular curve of radius 200 m, the coefficient of lateral friction of 0.15 and the design speed is 40 kmph. The equilibrium superelevation (for equal pressure on inner and outer wheel) would be
Discussion:
10 comments Page 1 of 1.
Rahul verma said:
2 years ago
e+f = V2/127R.
For equilibrium superelevation f=0;
= 40 * 40/127 * 200,
= 0.0629 = 6.3%.
For equilibrium superelevation f=0;
= 40 * 40/127 * 200,
= 0.0629 = 6.3%.
Manjunatha R.N. said:
8 years ago
Equilibrium super elevation means f=0 then using, e=(V^2)/(127*R).
e=0.06299*100=6.3%.
e=0.06299*100=6.3%.
Onkar jadhav said:
6 years ago
Its e=v^2/gr
=123.43/(9.81 x 200)
=0.0629 x 100
=6.29
=123.43/(9.81 x 200)
=0.0629 x 100
=6.29
Rajesh Kumar Tripathy said:
4 years ago
@Priyanka
40/3.6=11.1111^2=123.4.
To convert the unit 1MPS-3.6 kmph.
40/3.6=11.1111^2=123.4.
To convert the unit 1MPS-3.6 kmph.
(1)
Prayag gosai said:
5 years ago
e = v^2/gr.
= (40/3.6)^2/(9.81 x 200),
= 0.0629x100,
= 6.30.
= (40/3.6)^2/(9.81 x 200),
= 0.0629x100,
= 6.30.
Vikram Syal said:
7 years ago
No, the answer is D and m/s should come instead of kmph.
Vishal kumar said:
4 years ago
e = V2/127R.
= 40 * 40/127 * 200 = 0.0629 = 6.3%.
= 40 * 40/127 * 200 = 0.0629 = 6.3%.
Ajeet said:
5 years ago
The answer should be 3.55% as e= V^2/255R.
Priyanka said:
5 years ago
@Onkar Jadhav.
How 123.43 comes?
How 123.43 comes?
Vikram syal said:
7 years ago
v2/gR.
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