The road way bridge over a canal is in the form of an arc of a circle of radius 20m . What is the minimum speed with which a car can cross the bridge without leaving contact with the ground at the highest point ( 9.8 m / s)

The correct option is 14 m /s.

Underlying Concept and Force Analysis:
When a car crosses a roadway bridge that is in the form of a convex circular arc of radius $R$, it performs circular motion in a vertical plane. At the highest point of the bridge, two primary vertical forces act on the car:

1. The weight of the car, $mg$, acting vertically downwards towards the center of the circular path.
2. The normal reaction force, $N$, exerted by the road surface acting vertically upwards.

The net force towards the center of the circle provides the necessary centripetal force for the circular path. Therefore, we can write the equation of motion at the highest point as:

$mg-N=\frac{m{v}^2}{R}$

where:
- $m$ is the mass of the car,
- $v$ is the speed of the car,
- $R$ is the radius of the circular arc, and
- $g$ is the acceleration due to gravity.

Condition for Maintaining Contact:
For the car to maintain contact with the bridge, the normal reaction force $N$ must be greater than or equal to zero ($N\ge 0$). If the speed increases to the point where the normal force becomes zero, the car is on the verge of losing contact with the road.

Rearranging the centripetal force equation for $N$:

$N=mg-\frac{m{v}^2}{R}$

Setting $N\ge 0$ to find the threshold speed:

$mg-\frac{m{v}^2}{R}\ge 0$

$g\ge \frac{v^2}{R}$

$v^2\le gR$

$v\le \sqrt{gR}$

Thus, the maximum speed with which the car can safely cross the highest point without losing contact is:

$v_{\text{limit}}=\sqrt{gR}$

Calculation:
Given parameters:
- Radius of the circular path, $R=20\text{m}$
- Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Substituting these values into the formula:

$v=\sqrt{9.8\times 20}$

$v=\sqrt{196}$

$v=14\text{m/s}$

Therefore, the threshold speed is 14 m/s.