What will be maximum speed of a car on a curved road of radius 30 m , If the coefficient of friction between the tyres and the road is 0.4?
`(g=9.8 m//s^(2))`

Options

A. `10.84 m//s`
B. `9.84 m//s`
C. `8.84 m//s`
D. `6.84 m//s`

Explanation

To find the maximum speed of a car on a curved road, we can use the relationship between the frictional force and the centripetal force required to keep the car moving in a circle.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The car experiences a centripetal force that is required to keep it moving in a circular path. This force is provided by the friction between the tires and the road.
- The centripetal force \( F_c \) is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
- The frictional force \( F_f \) that prevents the car from slipping is given by:
\[ F_f = \mu mg \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity.

2. **Set the Forces Equal**:
- For the maximum speed before slipping occurs, the frictional force will equal the centripetal force:
\[ \mu mg = \frac{mv^2}{r} \]

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

4. **Rearrange for Speed**:
- Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \mu g r \]

5. **Substitute Known Values**:
- Now, substitute the known values into the equation. Given:
- \( \mu = 0.4 \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( r = 30 \, \text{m} \)
- Substitute these values into the equation:
\[ v^2 = 0.4 \times 9.8 \times 30 \]

6. **Calculate**:
- First, calculate \( 0.4 \times 9.8 \):
\[ 0.4 \times 9.8 = 3.92 \]
- Then, multiply by \( 30 \):
\[ 3.92 \times 30 = 117.6 \]
- Therefore:
\[ v^2 = 117.6 \]

7. **Find the Square Root**:
- Finally, take the square root to find \( v \):
\[ v = \sqrt{117.6} \approx 10.84 \, \text{m/s} \]

### Final Answer:
The maximum speed of the car on the curved road is approximately \( 10.84 \, \text{m/s} \).

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