A road is 10 m wide. Its radius of curvature is 50 m . The outer edge is above the lower edge by a distance of 1.5 m . This road is most suited for the velocity

Options

A. `8.5 ms^(-1)`
B. `6.5 ms^(-1)`
C. `5.5 ms^(-1)`
D. none of these

Answer

To solve the problem, we need to determine the most suited velocity for a vehicle on a banked road given the width of the road, the radius of curvature, and the height difference between the outer and inner edges.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Width of the road (w) = 10 m
- Radius of curvature (R) = 50 m
- Height difference (h) = 1.5 m

2. **Calculate the Banking Angle (θ)**:
The banking angle can be found using the relationship between the height difference and the width of the road:
\[ \tan(\theta) = \frac{h}{w} = \frac{1.5}{10} = 0.15 \]
To find θ, we take the arctangent:
\[ \theta = \tan^{-1}(0.15) \]

3. **Establish the Forces**:
For a vehicle on a banked curve, the forces acting on it are:
- Normal force (N)
- Gravitational force (mg)
- Centripetal force required for circular motion (\( \frac{mv^2}{R} \))

The normal force can be resolved into two components:
- Vertical component: \( N \cos(\theta) = mg \)
- Horizontal component: \( N \sin(\theta) = \frac{mv^2}{R} \)

4. **Set Up the Equations**:
From the vertical force balance:
\[ N \cos(\theta) = mg \quad \text{(1)} \]
From the horizontal force balance:
\[ N \sin(\theta) = \frac{mv^2}{R} \quad \text{(2)} \]

5. **Divide Equation (1) by Equation (2)**:
This gives us:
\[ \frac{N \cos(\theta)}{N \sin(\theta)} = \frac{mg}{\frac{mv^2}{R}} \]
Simplifying, we find:
\[ \cot(\theta) = \frac{gR}{v^2} \]
Rearranging gives:
\[ v^2 = gR \cot(\theta) \]

6. **Substituting for Cotangent**:
Since \( \cot(\theta) = \frac{1}{\tan(\theta)} \):
\[ v^2 = gR \frac{1}{\tan(\theta)} \]

7. **Substituting Known Values**:
Using \( g = 10 \, \text{m/s}^2 \), \( R = 50 \, \text{m} \), and \( \tan(\theta) = 0.15 \):
\[ v^2 = 10 \times 50 \times \frac{1}{0.15} \]
\[ v^2 = 500 \times \frac{1}{0.15} = 500 \times \frac{100}{15} = \frac{50000}{15} \approx 3333.33 \]
Taking the square root:
\[ v \approx \sqrt{3333.33} \approx 57.74 \, \text{m/s} \]

8. **Final Result**:
The most suited velocity for the vehicle on the banked road is approximately \( 5\sqrt{3} \, \text{m/s} \).