A car has to take a safe circular turn on horizontal flat rough road of radius 200 m. If the coefficient of friction between the tyres and road is 0.1 ,then maximum speed of car would be(All are in SI units)

Options

A. `sqrt(20g)`
B. `sqrt(10g)`
C. `sqrt(g)`
D. `g/sqrt(20)`

Answer

To find the maximum speed of a car taking a circular turn on a horizontal flat rough road, we can use the formula derived from the balance of forces acting on the car. The centripetal force required to keep the car moving in a circle is provided by the frictional force between the tires and the road.

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the circular turn (R) = 200 m
- Coefficient of friction (μ) = 0.1
- Acceleration due to gravity (g) = 9.81 m/s² (standard value)

2. **Understand the relationship between friction and centripetal force:**
The maximum frictional force (F_friction) that can act on the car is given by:
\[ F_{\text{friction}} = \mu \cdot N \]
where N is the normal force. On a flat surface, the normal force is equal to the weight of the car (mg), so:
\[ F_{\text{friction}} = \mu \cdot mg \]

3. **Set the frictional force equal to the centripetal force:**
The centripetal force (F_c) required to keep the car moving in a circle is given by:
\[ F_c = \frac{mv^2}{R} \]
Setting the two forces equal gives:
\[ \mu mg = \frac{mv^2}{R} \]

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

5. **Rearrange the equation to solve for v (maximum speed):**
\[ v^2 = \mu g R \]
\[ v = \sqrt{\mu g R} \]

6. **Substitute the known values into the equation:**
\[ v = \sqrt{0.1 \cdot 9.81 \cdot 200} \]

7. **Calculate the value:**
- First, calculate \(0.1 \cdot 9.81 = 0.981\)
- Then, calculate \(0.981 \cdot 200 = 196.2\)
- Finally, take the square root:
\[ v = \sqrt{196.2} \approx 14.0 \, \text{m/s} \]

### Conclusion:
The maximum speed of the car while taking the circular turn is approximately **14.0 m/s**.