
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
232.8k+ 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
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

