The mass of a bicycle rider along with the bicycle is 100 kg. he wants to cross over a circular turn of radius 100 m with a speed of `10 ms^(-1)`. If the coefficient of friction between the tyres and the road is 0.6, will the rider be able to cross the turn? Take `g = 10 ms^(-2)`.

Options

A. 300N
B. 600N
C. 1200N
D. 150N

Explanation

To determine if the bicycle rider can successfully navigate the circular turn, we need to compare the centripetal force required to maintain circular motion with the maximum frictional force available.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the rider and bicycle, \( m = 100 \, \text{kg} \)
- Radius of the circular turn, \( r = 100 \, \text{m} \)
- Speed of the bicycle, \( v = 10 \, \text{m/s} \)
- Coefficient of friction, \( \mu = 0.6 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Centripetal Force:**
The formula for centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \]
Substituting the values:
\[ F_c = \frac{100 \, \text{kg} \times (10 \, \text{m/s})^2}{100 \, \text{m}} = \frac{100 \times 100}{100} = 100 \, \text{N} \]

3. **Calculate the Maximum Frictional Force:**
The maximum frictional force \( F_f \) can be calculated using:
\[ F_f = \mu mg \]
Substituting the values:
\[ F_f = 0.6 \times 100 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.6 \times 1000 = 600 \, \text{N} \]

4. **Compare the Forces:**
Now, we compare the centripetal force and the maximum frictional force:
- Centripetal Force \( F_c = 100 \, \text{N} \)
- Maximum Frictional Force \( F_f = 600 \, \text{N} \)

Since \( F_f > F_c \) (i.e., \( 600 \, \text{N} > 100 \, \text{N} \)), the rider will be able to cross the turn.

5. **Conclusion:**
The rider can successfully navigate the turn because the frictional force is sufficient to provide the necessary centripetal force.