The breaking force for a wire of diameter D of a material is F. The breaking force for a wire of the same material of radius D is

Options

A. `F`
B. `2F`
C. `(F)/(4)`
D. `4F`

Explanation

To solve the problem, we need to determine the breaking force for a wire with a radius equal to the diameter of another wire, given that the breaking force for the first wire (with diameter D) is F.

### Step-by-step Solution:

1. **Understand the Given Information**:
- The diameter of the first wire is D.
- The breaking force for this wire is F.
- The radius of the second wire is equal to D, which means the radius \( R_2 = D \).

2. **Relate Diameter and Radius**:
- The radius \( R_1 \) of the first wire can be expressed as:
\[ R_1 = \frac{D}{2} \]

3. **Define Breaking Stress**:
- Breaking stress (σ) is defined as the breaking force (BF) per unit area (A):
\[ \sigma = \frac{BF}{A} \]
- For both wires made of the same material, the breaking stress will be the same.

4. **Calculate the Area of Each Wire**:
- The area \( A_1 \) for the first wire (with radius \( R_1 \)) is:
\[ A_1 = \pi R_1^2 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]
- The area \( A_2 \) for the second wire (with radius \( R_2 = D \)) is:
\[ A_2 = \pi R_2^2 = \pi D^2 \]

5. **Set Up the Relationship Using Breaking Stress**:
- Since the breaking stress is the same for both wires:
\[ \frac{F}{A_1} = \frac{BF_2}{A_2} \]
- Substituting the areas:
\[ \frac{F}{\frac{\pi D^2}{4}} = \frac{BF_2}{\pi D^2} \]

6. **Cross-Multiply to Solve for BF_2**:
- Cross-multiplying gives:
\[ F \cdot \pi D^2 = BF_2 \cdot \frac{\pi D^2}{4} \]
- Simplifying:
\[ BF_2 = 4F \]

7. **Conclusion**:
- The breaking force for the wire with radius D is:
\[ BF_2 = 4F \]

### Final Answer:
The breaking force for a wire of the same material with radius D is \( 4F \).