
The roadway bridge over a canal is in the form of an arc of a circle of radius $20\,m$. What is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point?
A) $10\,m{s^{ - 1}}$
B) $12\,m{s^{ - 1}}$
C) $14\,m{s^{ - 1}}$
D) $16\,m{s^{ - 1}}$
Answer
135k+ views
Hint: When the object is moving in the circular path, then the acceleration due to gravity is given by the square of the velocity of the object divided by the radius of the circular path. By using this formula, the maximum velocity or the maximum speed of the car can be determined.
Useful formula:
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g$
Where, $V$ is the velocity of the car, $R$ is the radius of the circular path and $g$ is the acceleration due to gravity.
Complete step by step solution:
Given that,
The radius of the circular path is, $R = 20\,m$
The acceleration due to gravity is, $g = 9.81\,m{s^{ - 1}}$
Now,
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g\,.........................\left( 1 \right)$
By keeping the velocity in one side and the other terms in the other side, then the above equation (1) is written as,
${V^2} = R \times g$
By taking square root on the both sides, then the above equation is written as,
$V = \sqrt {R \times g} \,..............\left( 2 \right)$
By substituting the radius of the circular path and the acceleration due to gravity in the above equation (2), then the above equation (2) is written as,
$V = \sqrt {20 \times 9.8} $
On multiplying the above equation, then the above equation is written as,
$V = \sqrt {196} $
By taking the square root then the above equation is written as,
$V = 14\,m{s^{ - 1}}$
Thus, the above equation shows the maximum speed or velocity of the car.
Hence, the option (C) is the correct answer.
Note: From the equation (2), the velocity is directly proportional to the radius of the circular path. As the radius of the circular path increases, the velocity of the car will increase. In real time, imagine that we are driving a vehicle, if the circular path radius is less the speed also less. If the radius of the circular path is more the velocity is more.
Useful formula:
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g$
Where, $V$ is the velocity of the car, $R$ is the radius of the circular path and $g$ is the acceleration due to gravity.
Complete step by step solution:
Given that,
The radius of the circular path is, $R = 20\,m$
The acceleration due to gravity is, $g = 9.81\,m{s^{ - 1}}$
Now,
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g\,.........................\left( 1 \right)$
By keeping the velocity in one side and the other terms in the other side, then the above equation (1) is written as,
${V^2} = R \times g$
By taking square root on the both sides, then the above equation is written as,
$V = \sqrt {R \times g} \,..............\left( 2 \right)$
By substituting the radius of the circular path and the acceleration due to gravity in the above equation (2), then the above equation (2) is written as,
$V = \sqrt {20 \times 9.8} $
On multiplying the above equation, then the above equation is written as,
$V = \sqrt {196} $
By taking the square root then the above equation is written as,
$V = 14\,m{s^{ - 1}}$
Thus, the above equation shows the maximum speed or velocity of the car.
Hence, the option (C) is the correct answer.
Note: From the equation (2), the velocity is directly proportional to the radius of the circular path. As the radius of the circular path increases, the velocity of the car will increase. In real time, imagine that we are driving a vehicle, if the circular path radius is less the speed also less. If the radius of the circular path is more the velocity is more.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

A body is falling from a height h After it has fallen class 11 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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