A string of length `0.1` m cannot bear a tension more than 100 N. It is lied to a body of mass 100 g and rotated in a horizontal circle. The maximum angular velocity can be

Options

A. `100"rad s"^(-1)`
B. `1000"rad s"^(-1)`
C. `10000s^(-1)`
D. `0.1"rad s"^(-1)`

Explanation

To solve the problem, we need to find the maximum angular velocity (\( \omega \)) at which a body can be rotated in a horizontal circle without exceeding the maximum tension that the string can bear.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Length of the string (radius of the circle, \( r \)) = 0.1 m
- Maximum tension (\( T_{\text{max}} \)) = 100 N
- Mass of the body (\( m \)) = 100 g = 0.1 kg

2. **Understand the Relationship**:
- The tension in the string provides the centripetal force required to keep the body moving in a circular path. The formula for centripetal force (\( F_c \)) is given by:
\[ F_c = m \omega^2 r \]
- In this case, the maximum tension equals the centripetal force:
\[ T_{\text{max}} = m \omega^2 r \]

3. **Substitute the Known Values**:
- Substitute the values of \( T_{\text{max}} \), \( m \), and \( r \) into the equation:
\[ 100 = 0.1 \cdot \omega^2 \cdot 0.1 \]

4. **Simplify the Equation**:
- Rearranging the equation gives:
\[ 100 = 0.01 \cdot \omega^2 \]
- To isolate \( \omega^2 \), divide both sides by 0.01:
\[ \omega^2 = \frac{100}{0.01} = 10000 \]

5. **Calculate \( \omega \)**:
- Take the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{10000} = 100 \text{ rad/s} \]

### Final Answer:
The maximum angular velocity (\( \omega \)) is **100 rad/s**.

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