A body of mass m = 10 kg is attached to a wire of length 0.3m. The maximum angular velocity with which it can be rotated in a horizontal circle is (Given, breaking stress of wire `= 4.8 xx 10^(7) Nm^(-2)` and area of cross-section of a wire `= 10^(-2) m^(2)`)

Options

A. `4 rad s^(-1)`
B. `8 rad s^(-1)`
C. `1 rad s^(-1)`
D. 2 rad

Explanation

To solve the problem of finding the maximum angular velocity with which a body can be rotated in a horizontal circle without breaking the wire, we can follow these steps:

### Step 1: Identify the given values
- Mass of the body, \( m = 10 \, \text{kg} \)
- Length of the wire (which acts as the radius of the circular motion), \( r = 0.3 \, \text{m} \)
- Breaking stress of the wire, \( \sigma = 4.8 \times 10^7 \, \text{N/m}^2 \)
- Area of cross-section of the wire, \( A = 10^{-6} \, \text{m}^2 \)

### Step 2: Calculate the maximum tension in the wire
The maximum tension \( T_{\text{max}} \) that the wire can withstand before breaking can be calculated using the formula:
\[ T_{\text{max}} = \sigma \times A \]
Substituting the given values:
\[ T_{\text{max}} = (4.8 \times 10^7 \, \text{N/m}^2) \times (10^{-6} \, \text{m}^2) = 48 \, \text{N} \]

### Step 3: Relate tension to centripetal force
In circular motion, the tension in the wire provides the centripetal force required to keep the body moving in a circle. The centripetal force \( F_c \) is given by:
\[ F_c = m \omega^2 r \]
Where \( \omega \) is the angular velocity. Setting the maximum tension equal to the centripetal force gives:
\[ T_{\text{max}} = m \omega_{\text{max}}^2 r \]

### Step 4: Rearranging to find maximum angular velocity
Rearranging the equation to solve for \( \omega_{\text{max}} \):
\[ \omega_{\text{max}}^2 = \frac{T_{\text{max}}}{m r} \]
Taking the square root to find \( \omega_{\text{max}} \):
\[ \omega_{\text{max}} = \sqrt{\frac{T_{\text{max}}}{m r}} \]

### Step 5: Substitute the values
Now substituting the values we have:
\[ \omega_{\text{max}} = \sqrt{\frac{48 \, \text{N}}{10 \, \text{kg} \times 0.3 \, \text{m}}} \]
Calculating the denominator:
\[ 10 \times 0.3 = 3 \]
Thus:
\[ \omega_{\text{max}} = \sqrt{\frac{48}{3}} = \sqrt{16} = 4 \, \text{radians/second} \]

### Conclusion
The maximum angular velocity with which the body can be rotated in a horizontal circle without breaking the wire is:
\[ \omega_{\text{max}} = 4 \, \text{radians/second} \]

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