A racing car of mass `10^2` kg goes around a circular track (horizontal) of radius 10cm. The maximum thrust that track can withstand is `10^5`N. The maximum speed with which car can go around is

Options

A. 10`m//s`
B. 100`m//s`
C. 50`m//s`
D. 20`m//s`

Answer

To solve the problem of finding the maximum speed with which a racing car can go around a circular track, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the car, \( m = 10^2 \, \text{kg} = 100 \, \text{kg} \)
- Radius of the circular track, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) (convert to meters)
- Maximum thrust (which acts as the maximum centripetal force), \( F_{\text{max}} = 10^5 \, \text{N} \)

2. **Understand the Relationship Between Thrust and Centripetal Force:**
- The maximum thrust that the track can withstand is equal to the centripetal force required to keep the car moving in a circular path. The formula for centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
- Here, \( F_c \) is the centripetal force, \( m \) is the mass of the car, \( v \) is the velocity of the car, and \( r \) is the radius of the circular path.

3. **Set the Maximum Thrust Equal to the Centripetal Force:**
- Since the maximum thrust is equal to the centripetal force, we can write:
\[ F_{\text{max}} = \frac{mv^2}{r} \]
- Substituting the known values:
\[ 10^5 = \frac{100 \cdot v^2}{0.1} \]

4. **Rearranging the Equation to Solve for \( v^2 \):**
- Multiply both sides by \( 0.1 \):
\[ 10^5 \cdot 0.1 = 100 \cdot v^2 \]
\[ 10^4 = 100 \cdot v^2 \]
- Now, divide both sides by \( 100 \):
\[ v^2 = \frac{10^4}{100} = 10^2 \]

5. **Taking the Square Root to Find \( v \):**
- Finally, take the square root of both sides to find \( v \):
\[ v = \sqrt{10^2} = 10 \, \text{m/s} \]

### Final Answer:
The maximum speed with which the car can go around the track is \( 10 \, \text{m/s} \).