A body of mass 500 g is fastened to one end of a steel wire of length 2m and area of cross-section `2m m^(2)`. If the breaking stress of the wire is `1.25 xx 10^(7) N//m^(2)`, then the maximum angular velocity with which the body can be rotated in a horizontal circle is

Options

A. 2 rad/s
B. 3 rad/s
C. 4 rad/s
D. 5 rad/s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass from grams to kilograms
Given:
- Mass, \( m = 500 \, \text{g} \)

To convert grams to kilograms:
\[ m = 500 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.5 \, \text{kg} \]

### Step 2: Convert the area of cross-section from mm² to m²
Given:
- Area, \( A = 2 \, \text{mm}^2 \)

To convert mm² to m²:
\[ A = 2 \, \text{mm}^2 \times \left( \frac{1 \, \text{m}}{1000 \, \text{mm}} \right)^2 = 2 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the maximum tension using breaking stress
Given:
- Breaking stress, \( \sigma = 1.25 \times 10^7 \, \text{N/m}^2 \)

The maximum tension \( T \) in the wire can be calculated using:
\[ T = \sigma \times A \]
Substituting the values:
\[ T = 1.25 \times 10^7 \, \text{N/m}^2 \times 2 \times 10^{-6} \, \text{m}^2 = 25 \, \text{N} \]

### Step 4: Use the formula for centripetal force to find angular velocity
The centripetal force required to keep the mass moving in a circle is given by:
\[ T = m \cdot r \cdot \omega^2 \]
Where:
- \( r \) is the radius (length of the wire), \( r = 2 \, \text{m} \)
- \( \omega \) is the angular velocity

Rearranging the formula to solve for \( \omega \):
\[ \omega^2 = \frac{T}{m \cdot r} \]
Substituting the known values:
\[ \omega^2 = \frac{25 \, \text{N}}{0.5 \, \text{kg} \cdot 2 \, \text{m}} = \frac{25}{1} = 25 \]
Taking the square root:
\[ \omega = \sqrt{25} = 5 \, \text{rad/s} \]

### Final Answer
The maximum angular velocity with which the body can be rotated in a horizontal circle is:
\[ \omega = 5 \, \text{rad/s} \]

---