A van is moving with a speed of 108 km / hr on level road where the coefficient of fraction between the tyres and the road is 0.5 for the safe driving of the van , the minimum radius of curvature of the road will be ( `g=10 m//s^(2)`)

Options

A. 80m
B. 180m
C. 40 m
D. 20 m

Explanation

To solve the problem, we need to find the minimum radius of curvature (r) for a van moving at a speed of 108 km/h on a level road with a coefficient of friction (μ) of 0.5. We'll use the relationship between centripetal force and frictional force to derive the formula for the radius of curvature.

### Step-by-Step Solution:

1. **Convert Speed from km/h to m/s:**
\[ \text{Speed} = 108 \text{ km/h} = \frac{108 \times 1000}{3600} \text{ m/s} = 30 \text{ m/s} \]

2. **Identify Given Values:**
- Speed (v) = 30 m/s
- Coefficient of friction (μ) = 0.5
- Acceleration due to gravity (g) = 10 m/s²

3. **Understand the Forces Involved:**
- The centripetal force required to keep the van moving in a circular path is provided by the frictional force. The centripetal force (F_c) is given by:
\[ F_c = \frac{mv^2}{r} \]
- The maximum frictional force (F_f) can be expressed as:
\[ F_f = \mu N \]
where N is the normal force. For a vehicle on a level road, the normal force is equal to the weight of the vehicle (N = mg).

4. **Set Up the Equation:**
- For safe driving, the centripetal force must not exceed the frictional force:
\[ \frac{mv^2}{r} \leq \mu mg \]
- Canceling mass (m) from both sides (assuming m ≠ 0):
\[ \frac{v^2}{r} \leq \mu g \]

5. **Rearranging for Radius (r):**
\[ r \geq \frac{v^2}{\mu g} \]

6. **Substituting the Values:**
\[ r \geq \frac{(30)^2}{0.5 \times 10} \]
\[ r \geq \frac{900}{5} = 180 \text{ m} \]

7. **Conclusion:**
The minimum radius of curvature for the safe driving of the van is **180 meters**.