For a circular curve of radius 250 m, the coefficient of lateral friction is 0.15 and the design speed is 35 kmph. The equilibrium superelevation would be

8.80%

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3.85%

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4.56%

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Solution

The correct option is B 3.85%
For equilibrium superelevation,
f = 0
$e = \frac{V^{2}}{127 R} = \frac{35^{2}}{127 \times 250}$

$= 0.0385 = 3.85 %$

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For a circular curve of radius 250 m, the coefficient of lateral friction is 0.15 and the design speed is 35 kmph. The equilibrium superelevation would be

The correct option is B 3.85%For equilibrium superelevation, f = 0 e=V2127R=352127×250 =0.0385=3.85%

The correct option is B 3.85%
For equilibrium superelevation,
f = 0
$e = \frac{V^{2}}{127 R} = \frac{35^{2}}{127 \times 250}$

$= 0.0385 = 3.85 %$