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.
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%.
Vikram Syal said:
7 years ago
No, the answer is D and m/s should come instead of kmph.
Vikram syal said:
7 years ago
v2/gR.
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
Priyanka said:
5 years ago
@Onkar Jadhav.
How 123.43 comes?
How 123.43 comes?
Ajeet said:
5 years ago
The answer should be 3.55% as e= V^2/255R.
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.
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%.
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)
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%.
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