The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5m.

Options

A. `26^@`
B. `5^@`
C. `20^@`
D. `16^@`

Explanation

To solve the problem, we need to find the maximum incline angle (θ) of the road such that a vehicle can stop within 5 meters after applying brakes, given that the coefficient of friction (μ) between the road and the tire is 4/3 and the initial speed of the vehicle is 36 km/hr.

### Step-by-Step Solution:

1. **Convert the speed from km/hr to m/s**:
\[ \text{Speed} = 36 \text{ km/hr} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

2. **Use the third equation of motion to find acceleration (a)**:
We know that:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity, since the vehicle stops)
- \( u = 10 \text{ m/s} \) (initial velocity)
- \( s = 5 \text{ m} \) (distance)

Plugging in the values:
\[ 0 = (10)^2 + 2a(5) \]
\[ 0 = 100 + 10a \]
\[ 10a = -100 \implies a = -10 \text{ m/s}^2 \]

3. **Set up the force balance on the incline**:
On an inclined plane, the forces acting on the vehicle are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Frictional force opposing the motion: \( F_k = \mu N = \mu mg \cos \theta \)

The net force acting on the vehicle can be expressed as:
\[ F_{net} = F_k - mg \sin \theta = ma \]
Substituting the expressions for \( F_k \) and rearranging gives:
\[ \mu mg \cos \theta - mg \sin \theta = ma \]
Dividing through by \( m \):
\[ \mu g \cos \theta - g \sin \theta = a \]
Substituting \( a = -10 \text{ m/s}^2 \):
\[ \frac{4}{3} g \cos \theta - g \sin \theta = -10 \]

4. **Substituting \( g = 10 \text{ m/s}^2 \)**:
\[ \frac{4}{3} \cdot 10 \cos \theta - 10 \sin \theta = -10 \]
Simplifying:
\[ \frac{40}{3} \cos \theta - 10 \sin \theta + 10 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 40 \cos \theta - 30 \sin \theta + 30 = 0 \]
Rearranging gives:
\[ 40 \cos \theta = 30 \sin \theta - 30 \]

5. **Expressing in terms of sine and cosine**:
Dividing through by 10:
\[ 4 \cos \theta = 3 \sin \theta - 3 \]
Rearranging:
\[ 3 \sin \theta - 4 \cos \theta = 3 \]

6. **Using the Pythagorean identity**:
We can square both sides to eliminate the sine and cosine:
\[ (3 \sin \theta - 4 \cos \theta)^2 = 3^2 \]
Expanding:
\[ 9 \sin^2 \theta - 24 \sin \theta \cos \theta + 16 \cos^2 \theta = 9 \]
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 9 \sin^2 \theta + 16 (1 - \sin^2 \theta) - 24 \sin \theta \cos \theta = 9 \]
Simplifying gives:
\[ -7 \sin^2 \theta - 24 \sin \theta \cos \theta + 7 = 0 \]

7. **Solving for \( \theta \)**:
Solving this quadratic equation for \( \sin \theta \) gives:
\[ \sin \theta \approx 0.28 \implies \theta \approx \sin^{-1}(0.28) \approx 16^\circ \]

### Final Answer:
The maximum incline of the road is approximately **16 degrees**.