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# A curved road of diameter $1.8\,km$ is banked so that no friction is required at a speed of $30\,m{s^{ - 1}}$. What is the banking angle?(A) ${6^ \circ }$(B) ${16^ \circ }$(C) ${26^ \circ }$(D) ${0.6^ \circ }$

Last updated date: 20th Jun 2024
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Hint: A banked turn or banking angle is the turn or the change of the direction of the vehicle where the vehicle banks or inclines with some angle, which is usually seen in the turn of the vehicle. The banking angle in roads is the phenomenon in which the edge of the road is raised in the curve roads about the inner edge of the road and provides necessary centripetal force for the safe turn.

Useful formula:
Banking angle, $\tan \theta = \dfrac{{{v^2}}}{{rg}}$
Where,
$\theta$ is the banking angle of the road
$v$ is the velocity of the object
$r$ is the radius of the curved road
$g$ is the acceleration due to gravity

Complete step by step solution:
Radius of the curved road, $r = \dfrac{D}{2}$
Substituting the value of $D$ in the above equation,
$\Rightarrow$ $r = \dfrac{{1.8 \times {{10}^3}\,m}}{2}$
By solving the above equation,
$\Rightarrow$ $r = \dfrac{{1.8 \times 1000\,m}}{2}$
Multiplying the values in numerator,
$\Rightarrow$ $r = \dfrac{{1800\,m}}{2}$
By cancelling the numerator by denominator,
$\Rightarrow$ $r = 900\,m$
Substituting the values of $v$, $r$ and $g$ in the Banking angle formula, as we known already the value of acceleration due to gravity, $g = 9.8\,m{s^{ - 2}}$,
$\Rightarrow$ $\tan \theta = \dfrac{{{{30}^2}}}{{900 \times 9.8}}$
Squaring the numerator in the above equation,
$\Rightarrow$ $\tan \theta = \dfrac{{900}}{{900 \times 9.8}}$
By cancelling the term in the above equation,
$\Rightarrow$ $\tan \theta = \dfrac{1}{{9.8}}$
By solving the above equation,
$\Rightarrow$ $\tan \theta = 0.1$
Taking $\theta$ in one side and other terms in another side,
$\theta = {\tan ^{ - 1}}\left( {0.1} \right) \\ \theta = {6^ \circ } \\$

$\therefore$ Hence, the option (A) is correct.

Note: If the object is moving in the flat surface, the weight of the object acts downward and the reaction force acts upwards. Both the force's vertical component, there is no horizontal component of force. If the object takes the turn in the curved road, then the weight of the object gives some horizontal force. So, there is a possibility of skidding of the object. To avoid this problem, the banking angle is used to maintain the proper centre of gravity and maintain the vertical force.