A wire of length l = 3m and area of cross section `10^(-2) cm^2` and breaking stress `4.8xx10^(-7) N/m^2` is attached with block of mass 10kg. Find the maximum possible value of angular velocity with which block can be moved in circle with string fixed at one end.

Explanation

To find the maximum possible value of angular velocity with which a block can be moved in a circle using a wire, we need to consider the forces acting on the block and the properties of the wire.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the wire, \( l = 3 \, \text{m} \)
- Area of cross-section, \( A = 10^{-2} \, \text{cm}^2 = 10^{-2} \times 10^{-4} \, \text{m}^2 = 10^{-6} \, \text{m}^2 \)
- Breaking stress, \( \sigma = 4.8 \times 10^{-7} \, \text{N/m}^2 \)
- Mass of the block, \( m = 10 \, \text{kg} \)

2. **Calculate the Breaking Force:**
The breaking force \( F_b \) can be calculated using the formula:
\[ F_b = \sigma \times A \]
Substituting the values:
\[ F_b = 4.8 \times 10^{-7} \, \text{N/m}^2 \times 10^{-6} \, \text{m}^2 = 4.8 \times 10^{-13} \, \text{N} \]

3. **Relate Breaking Force to Centripetal Force:**
For circular motion, the centripetal force \( F_c \) required to keep the block moving in a circle is given by:
\[ F_c = m \cdot \frac{v^2}{r} \]
where \( v \) is the linear velocity and \( r \) is the radius of the circle (which is equal to the length of the wire, \( l \)).
Since \( v = r \cdot \omega \) (where \( \omega \) is the angular velocity), we can rewrite the centripetal force as:
\[ F_c = m \cdot \frac{(l \cdot \omega)^2}{l} = m \cdot l \cdot \omega^2 \]

4. **Set Breaking Force Equal to Centripetal Force:**
To find the maximum angular velocity before the wire breaks, set the breaking force equal to the centripetal force:
\[ F_b = m \cdot l \cdot \omega^2 \]
Substituting the values:
\[ 4.8 \times 10^{-13} = 10 \cdot 3 \cdot \omega^2 \]

5. **Solve for Angular Velocity \( \omega \):**
Rearranging the equation gives:
\[ \omega^2 = \frac{4.8 \times 10^{-13}}{30} \]
\[ \omega^2 = 1.6 \times 10^{-14} \]
Taking the square root:
\[ \omega = \sqrt{1.6 \times 10^{-14}} \approx 1.26 \times 10^{-7} \, \text{rad/s} \]

### Final Answer:
The maximum possible value of angular velocity \( \omega \) with which the block can be moved in a circle is approximately:
\[ \omega \approx 1.26 \times 10^{-7} \, \text{rad/s} \]