The breaking stress for a wire of radius r of given material is F `N//m^(2)`. The breaking stress for the wire of same material of radius 2r is:

Options

A. F/4
B. F/2
C. F
D. 2F

Answer

To solve the problem, we need to understand the concept of breaking stress and how it relates to the radius of a wire. Breaking stress is defined as the maximum force (or weight) that a wire can withstand per unit area before it breaks.

### Step-by-Step Solution:

1. **Understanding Breaking Stress**:
- Breaking stress (σ) is defined as the force (F) applied divided by the cross-sectional area (A) of the wire.
- Mathematically, this can be expressed as:
\[ \sigma = \frac{F}{A} \]

2. **Cross-Sectional Area of the Wire**:
- For a circular wire of radius \( r \), the cross-sectional area \( A \) is given by:
\[ A = \pi r^2 \]

3. **Breaking Stress for Wire of Radius \( r \)**:
- Given that the breaking stress for the wire of radius \( r \) is \( F \) N/m², we can write:
\[ F = \frac{W}{\pi r^2} \]
- Here, \( W \) is the maximum weight that can be hung from the wire.

4. **Cross-Sectional Area for Wire of Radius \( 2r \)**:
- Now, consider a wire of radius \( 2r \). The cross-sectional area \( A' \) for this wire is:
\[ A' = \pi (2r)^2 = \pi \cdot 4r^2 = 4\pi r^2 \]

5. **Breaking Stress for Wire of Radius \( 2r \)**:
- The breaking stress \( F' \) for this wire can be expressed as:
\[ F' = \frac{W}{A'} = \frac{W}{4\pi r^2} \]

6. **Relating the Breaking Stresses**:
- Now, we can relate the breaking stresses \( F \) and \( F' \):
\[ F' = \frac{W}{4\pi r^2} \]
- From the earlier expression for \( F \):
\[ F = \frac{W}{\pi r^2} \]
- Dividing the two equations:
\[ \frac{F'}{F} = \frac{\frac{W}{4\pi r^2}}{\frac{W}{\pi r^2}} = \frac{1}{4} \]
- Thus, we find:
\[ F' = \frac{F}{4} \]

7. **Conclusion**:
- Therefore, the breaking stress for the wire of radius \( 2r \) is:
\[ F' = \frac{F}{4} \]

### Final Answer:
The breaking stress for the wire of radius \( 2r \) is \( \frac{F}{4} \) N/m².