A steel wire can support a maximum load of W before reaching its elastic limit. How much load can another wire, made out of identical steel, but with a radius one half the radius of the first wire, support before reaching its elastic limit?

Options

A. W
B. `(W)/(4)`
C. `(W)/(4)`
D. `4W`

Explanation

To solve the problem, we need to determine how much load a second wire, made of identical steel but with a radius that is half the radius of the first wire, can support before reaching its elastic limit.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The first wire can support a maximum load \( W \) before reaching its elastic limit.
- The radius of the first wire is \( r \).
- The radius of the second wire is \( r_2 = \frac{1}{2}r \).

2. **Calculate the Cross-Sectional Area**:
- The cross-sectional area \( A_1 \) of the first wire is given by:
\[ A_1 = \pi r^2 \]
- The cross-sectional area \( A_2 \) of the second wire is:
\[ A_2 = \pi r_2^2 = \pi \left(\frac{1}{2}r\right)^2 = \pi \left(\frac{1}{4}r^2\right) = \frac{\pi r^2}{4} \]

3. **Determine the Stress in Each Wire**:
- The stress \( \sigma_1 \) in the first wire when the load \( W \) is applied is:
\[ \sigma_1 = \frac{W}{A_1} = \frac{W}{\pi r^2} \]
- For the second wire, let the maximum load it can support be \( W' \). The stress \( \sigma_2 \) in the second wire is:
\[ \sigma_2 = \frac{W'}{A_2} = \frac{W'}{\frac{\pi r^2}{4}} = \frac{4W'}{\pi r^2} \]

4. **Equate the Stresses**:
- Since both wires are made of identical steel and are subjected to the same elastic limit, the stresses must be equal:
\[ \sigma_1 = \sigma_2 \]
\[ \frac{W}{\pi r^2} = \frac{4W'}{\pi r^2} \]

5. **Solve for \( W' \)**:
- Cancel \( \pi r^2 \) from both sides:
\[ W = 4W' \]
- Rearranging gives:
\[ W' = \frac{W}{4} \]

6. **Conclusion**:
- The second wire can support a maximum load of \( \frac{W}{4} \) before reaching its elastic limit.

### Final Answer:
The second wire can support a maximum load of \( \frac{W}{4} \).