A bob of mass `10 kg` is attached to a wire `0.3 m` long. Its breaking stress is `4.8xx10^(7) N//m^(2)`. Then area of cross-section of the wire is `10^(-6) m^(2)`. What is the maximum angular velocity with which it can be rotated in a horizontal circle?

Options

A. `8 rad//s`
B. `4 rad//s`
C. `2 rad//s`
D. `1rad//s`

Explanation

To find the maximum angular velocity with which the bob can be rotated in a horizontal circle, we can follow these steps:

### Step 1: Understand the forces acting on the bob
When the bob is rotated in a horizontal circle, the tension in the wire provides the necessary centripetal force. The tension (T) in the wire can be expressed in terms of the mass (m), angular velocity (ω), and the length of the wire (l).

### Step 2: Write the equation for centripetal force
The centripetal force required to keep the bob moving in a circle is given by:
\[ T = m \cdot a_c \]
where \( a_c = \omega^2 \cdot l \) is the centripetal acceleration. Therefore, we can write:
\[ T = m \cdot \omega^2 \cdot l \]

### Step 3: Relate tension to stress
The stress (σ) in the wire is defined as the force (in this case, tension) per unit area (A):
\[ \sigma = \frac{T}{A} \]
Substituting the expression for tension, we have:
\[ \sigma = \frac{m \cdot \omega^2 \cdot l}{A} \]

### Step 4: Rearranging for angular velocity
To find the angular velocity (ω), we rearrange the equation:
\[ T = \sigma \cdot A \]
Substituting for T:
\[ m \cdot \omega^2 \cdot l = \sigma \cdot A \]
Thus,
\[ \omega^2 = \frac{\sigma \cdot A}{m \cdot l} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{\sigma \cdot A}{m \cdot l}} \]

### Step 5: Substitute the known values
Now we can substitute the values provided in the question:
- Mass, \( m = 10 \, \text{kg} \)
- Length, \( l = 0.3 \, \text{m} \)
- Breaking stress, \( \sigma = 4.8 \times 10^7 \, \text{N/m}^2 \)
- Area of cross-section, \( A = 10^{-6} \, \text{m}^2 \)

Substituting these values into the equation:
\[ \omega = \sqrt{\frac{(4.8 \times 10^7) \cdot (10^{-6})}{10 \cdot 0.3}} \]

### Step 6: Calculate the maximum angular velocity
Calculating the expression inside the square root:
\[ \omega = \sqrt{\frac{4.8 \times 10^1}{3}} \]
\[ \omega = \sqrt{16} \]
\[ \omega = 4 \, \text{radians/second} \]

### Final Answer
The maximum angular velocity with which the bob can be rotated in a horizontal circle is:
\[ \omega_{\text{max}} = 4 \, \text{radians/second} \]

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