The superelevation needed for a vehicle travelling at a speed of 60 kmph on a curve of radius 128 m on a surface with a coefficient of friction of 0.15 is

Zero

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0.071

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0.022

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0.015

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Solution

The correct option is B 0.071
Superelevation is given by the relation

$e + f = \frac{v^{2}}{g R}$

if V is in km/hr, then

$e + f = \frac{V^{2}}{127 R}$

$\Rightarrow e + 0.15 = \frac{(60)^{2}}{127 \times 128}$

$\Rightarrow e = 0.071$

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The superelevation needed for a vehicle travelling at a speed of 60 kmph on a curve of radius 128 m on a surface with a coefficient of friction of 0.15 is

The correct option is B 0.071Superelevation is given by the relation e+f=v2gR if V is in km/hr, then e+f=V2127R ⇒e+0.15=(60)2127×128 ⇒e=0.071

The correct option is B 0.071
Superelevation is given by the relation

$e + f = \frac{v^{2}}{g R}$

if V is in km/hr, then

$e + f = \frac{V^{2}}{127 R}$

$\Rightarrow e + 0.15 = \frac{(60)^{2}}{127 \times 128}$

$\Rightarrow e = 0.071$