The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is `("Take g"=10ms^(-2))`

Options

A. `sqrt15 "ms"^(-1)`
B. `sqrt3 "ms"^(-1)`
C. `sqrt30 "ms"^(-1)`
D. `sqrt10 "ms"^(-1)`

Explanation

To find the maximum speed with which a cyclist can take a circular turn without skidding, we can use the formula derived from the concepts of circular motion and friction.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Coefficient of friction (μ) = 0.1
- Radius of the circular turn (r) = 3 m
- Acceleration due to gravity (g) = 10 m/s²

2. **Use the Formula for Maximum Speed**:
The formula to calculate the maximum speed (V_max) at which a cyclist can take a turn without skidding is:
\[ V_{\text{max}} = \sqrt{\mu \cdot r \cdot g} \]

3. **Substitute the Values into the Formula**:
Now, substitute the values of μ, r, and g into the formula:
\[ V_{\text{max}} = \sqrt{0.1 \cdot 3 \cdot 10} \]

4. **Calculate the Value Inside the Square Root**:
First, calculate the product:
\[ 0.1 \cdot 3 \cdot 10 = 3 \]

5. **Take the Square Root**:
Now, take the square root of the result:
\[ V_{\text{max}} = \sqrt{3} \text{ m/s} \]

6. **Final Result**:
Therefore, the maximum speed with which the cyclist can take the turn without skidding is:
\[ V_{\text{max}} = \sqrt{3} \text{ m/s} \]

### Conclusion:
The maximum speed with which a cyclist can take a circular turn of radius 3 m without skidding is \(\sqrt{3} \text{ m/s}\).

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