A road has a horizontal curve of 400 m radius on which a super-elevation of 0.07 is provided. The coefficient of lateral friction mobilised on the curve when a vehicle is travelling at 100 kmph is

0.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.13

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.07

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.13
$e + f = \frac{V^{2}}{127 R}$

$\Rightarrow f = \frac{V^{2}}{127 R} - e = \frac{(100)^{2}}{127 \times 400} - 0.07$

$= 0.13$

Suggest Corrections

Similar questions

A road is passing through a hilly area with a horizontal curve. If the superelevation provided is 1 in 10 and the coefficient of lateral friction is 0.16, minimum radius of the horizontal curve which the vehicles can negotiate safely at a speed of 40 kmph in metres is

Q. A super- elevation e is provided on a circular horizontal curve such that a vehicle can be stopped on the curve without sliding. Assuming a design speed v and maximum coefficient of side friction

f_{m a x}

, which one of the following criteria should be satisfied ?

Q. A road is being designed for a speed of 110 km/hr on a horizontal curve with a super elevation of 8%. If the coefficient of side friction is 0.10, the minimum radius of the curve (in m) required for safe vehicular movement is

Q. 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

Q. Full amount of super-elevation on a horizontal curve is provided at the

Join BYJU'S Learning Program

A road has a horizontal curve of 400 m radius on which a super-elevation of 0.07 is provided. The coefficient of lateral friction mobilised on the curve when a vehicle is travelling at 100 kmph is

The correct option is C 0.13e+f=V2127R ⇒f=V2127R−e=(100)2127×400−0.07 =0.13

The correct option is C 0.13
$e + f = \frac{V^{2}}{127 R}$

$\Rightarrow f = \frac{V^{2}}{127 R} - e = \frac{(100)^{2}}{127 \times 400} - 0.07$

$= 0.13$