A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

The correct option is v²b/Rg.

To prevent a vehicle from skidding outwards on a curved road due to centrifugal force, the road is banked. Banking of roads involves raising the outer edge of the road relative to its inner edge. Let us derive the expression for the required difference in elevation between these edges.

Let:
$v$ = velocity of the vehicle,
$R$ = radius of curvature of the curved road,
$b$ = width of the road,
$h$ = difference in elevation between the outer and inner edges of the road (height of banking), and
$g$ = acceleration due to gravity.

Let the angle of banking of the road with the horizontal be $\theta$.

For a vehicle to safely navigate a turn of radius $R$ at a speed $v$ without relying on friction, the banking angle $\theta$ must satisfy the relation:

$\tan \left(\theta \right)=\frac{v^2}{Rg}$

From the geometry of the banked road, if $b$ is the width of the road (the hypotenuse of the right-angled triangle formed by the incline) and $h$ is the elevation difference (the opposite side):

$\sin \left(\theta \right)=\frac{h}{b}$

For small angles of banking, $\theta$ is very small, which allows us to use the approximation:

$\tan \left(\theta \right)\approx \sin \left(\theta \right)$

Equating the two expressions, we get:

$\frac{h}{b}=\frac{v^2}{Rg}$

Solving for the difference in elevation ($h$):

$h=\frac{v^{2}b}{Rg}$