A car is taking a turn of radius 400 m on a horizontal circle . Coefficient of friction between the ground and tyres of car is 0.2 . Find the maximum speed with a safe turn can be taken

Options

A. 80 m/s
B. `40 sqrt(2) m/s`
C. `20 sqrt(2) m/s`
D. None of these

Answer

To find the maximum speed at which a car can safely take a turn on a horizontal circle, we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of the turn, \( R = 400 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (standard value)

2. **Understand the Forces Acting on the Car:**
- The car experiences a centripetal force required to keep it moving in a circular path, which is provided by the frictional force between the tires and the road.
- The maximum frictional force can be expressed as:
\[ F_f^{\text{max}} = \mu \cdot N \]
- Here, \( N \) (normal force) is equal to the weight of the car, \( mg \). Therefore:
\[ F_f^{\text{max}} = \mu \cdot mg \]

3. **Centripetal Force Requirement:**
- The centripetal force required to keep the car moving in a circle is given by:
\[ F_c = \frac{mv^2}{R} \]
- For the car to negotiate the turn safely, the maximum frictional force must be equal to or greater than the required centripetal force:
\[ F_f^{\text{max}} \geq F_c \]
- Thus, we can set up the equation:
\[ \mu mg = \frac{mv^2}{R} \]

4. **Cancel the Mass:**
- Since mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{v^2}{R} \]

5. **Rearranging for Maximum Speed \( v_{\text{max}} \):**
- Rearranging the equation gives:
\[ v^2 = \mu g R \]
- Taking the square root of both sides, we find:
\[ v_{\text{max}} = \sqrt{\mu g R} \]

6. **Substituting the Values:**
- Now, substituting the known values into the equation:
\[ v_{\text{max}} = \sqrt{0.2 \cdot 10 \cdot 400} \]
- Simplifying this:
\[ v_{\text{max}} = \sqrt{0.2 \cdot 4000} = \sqrt{800} = \sqrt{400 \cdot 2} = 20\sqrt{2} \, \text{m/s} \]

7. **Final Answer:**
- The maximum speed with which the car can safely take the turn is:
\[ v_{\text{max}} = 20\sqrt{2} \, \text{m/s} \]

### Conclusion:
The correct option is \( 20\sqrt{2} \, \text{m/s} \).