Answer

Verified

87k+ views

**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.

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

The thickness of the depletion layer is approximately class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main