The radius of the curved road on a national highway is `R`. The width of the road is `b`. The outer edge of the road is raised by `h` with respect to the inner edge so that a car with velocity `v` can pass safe over it. The value of `h` is

Options

A. `(v^(2)b)/(Rg)`
B. `(v)/(Rgb)`
C. `(v^(2)R)/(bg)`
D. `(v^(2)b)/(R)`

Explanation

To solve the problem, we need to determine the height \( h \) that the outer edge of the curved road should be raised to ensure that a car can safely navigate the curve at a velocity \( v \).

### Step-by-Step Solution:

1. **Understanding the Forces**: When a car is turning on a curved road, it experiences a centripetal force that keeps it moving in a circular path. This force is provided by the friction between the tires and the road. The road is banked to help provide this force without relying solely on friction.

2. **Banking Angle**: The banking of the road creates an angle \( \theta \) with the horizontal. The relationship between the angle of banking, the velocity of the car, the radius of the curve, and the acceleration due to gravity is given by the equation:
\[ \tan(\theta) = \frac{v^2}{gR} \]
where \( g \) is the acceleration due to gravity.

3. **Relating Height and Width**: The height \( h \) of the outer edge of the road can be related to the width \( b \) of the road and the angle \( \theta \) using the tangent function:
\[ \tan(\theta) = \frac{h}{b} \]

4. **Equating the Two Expressions for \( \tan(\theta) \)**: Since both expressions represent \( \tan(\theta) \), we can set them equal to each other:
\[ \frac{h}{b} = \frac{v^2}{gR} \]

5. **Solving for Height \( h \)**: Rearranging the equation to solve for \( h \):
\[ h = b \cdot \frac{v^2}{gR} \]

6. **Final Expression**: Thus, the height \( h \) that the outer edge of the road should be raised is given by:
\[ h = \frac{bv^2}{gR} \]

### Conclusion:
The value of \( h \) is \( \frac{bv^2}{gR} \).