
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) ${0.6^ \circ }$
Answer
145.8k+ 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.
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
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Ideal and Non-Ideal Solutions Raoult's Law - JEE

At which height is gravity zero class 11 physics JEE_Main

Charging and Discharging of Capacitor

Which of the following is used as a coolant in an automobile class 11 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Other Pages
Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

JEE Main Chemistry Question Paper with Answer Keys and Solutions

NCERT Solutions for Class 11 Physics In Hindi Chapter 1 Physical World

The mass of a lead ball is M It falls down in a viscous class 11 physics JEE_Main

JEE Advanced Chemistry Notes 2025

The breaking stress of a material is 109 pascal If class 11 physics JEE_Main
