A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is `4.8 xx 10^(7) N//m^(2)`. The area of cross section of the wire is `10^(-6) m^(2)`. The maximum angular velocity with which it can be rotated in a horizontal circle

Answer

To find the maximum angular velocity with which a bob of mass 10 kg can be rotated in a horizontal circle using a wire of length 0.3 m, we will follow these steps:

### Step 1: Calculate the Maximum Tension in the Wire
The breaking stress (σ) of the wire is given as \(4.8 \times 10^{7} \, \text{N/m}^2\) and the area of cross-section (A) is \(10^{-6} \, \text{m}^2\). The maximum tension (T_max) that the wire can withstand is calculated using the formula:

\[ T_{\text{max}} = \sigma \times A \]

Substituting the values:

\[ T_{\text{max}} = (4.8 \times 10^{7} \, \text{N/m}^2) \times (10^{-6} \, \text{m}^2) = 48 \, \text{N} \]

### Step 2: Relate Tension to Centripetal Force
When the bob is rotating in a circle, the tension in the wire provides the necessary centripetal force. The centripetal force (F_c) can be expressed as:

\[ F_c = m \cdot \omega^2 \cdot r \]

Where:
- \(m\) is the mass of the bob (10 kg),
- \(\omega\) is the angular velocity,
- \(r\) is the radius of the circular path, which is equal to the length of the wire (0.3 m).

Thus, we can write:

\[ T = m \cdot \omega^2 \cdot l \]

Substituting the values:

\[ T = 10 \cdot \omega^2 \cdot 0.3 = 3 \cdot \omega^2 \]

### Step 3: Set the Maximum Tension Equal to the Tension from Centripetal Force
Since the maximum tension the wire can withstand is 48 N, we set up the inequality:

\[ 3 \cdot \omega^2 \leq 48 \]

### Step 4: Solve for Angular Velocity
Now, we solve for \(\omega^2\):

\[ \omega^2 \leq \frac{48}{3} = 16 \]

Taking the square root of both sides gives:

\[ \omega \leq 4 \, \text{rad/s} \]

### Conclusion
The maximum angular velocity with which the bob can be rotated in a horizontal circle is:

\[ \omega_{\text{max}} = 4 \, \text{rad/s} \]