The maximum load that a wire can sustain is W. If the wire is cut to half its value, the maximum load it can sustain is

Options

A. `W `
B. `W/2`
C. `W/4`
D. `2W`

Explanation

To solve the problem, we need to understand how the maximum load a wire can sustain is related to its physical properties, particularly its length and cross-sectional area.

### Step-by-Step Solution:

1. **Understanding the Maximum Load**:
The maximum load \( W \) that a wire can sustain is determined by the formula:
\[ F_{\text{max}} = \sigma \cdot A \]
where \( \sigma \) is the stress (which is the maximum load per unit area), and \( A \) is the cross-sectional area of the wire.

2. **Cutting the Wire**:
When the wire is cut to half its length, the new length of the wire becomes \( L/2 \). The cross-sectional area \( A \) remains unchanged.

3. **Stress and Maximum Load**:
The stress \( \sigma \) in the wire is defined as:
\[ \sigma = \frac{F_{\text{max}}}{A} \]
Since the wire is made of the same material and has the same cross-sectional area, the maximum stress that the wire can handle remains the same.

4. **Maximum Load After Cutting**:
If we denote the maximum load after cutting the wire as \( W' \), we can express it as:
\[ W' = \sigma \cdot A \]
Since \( \sigma \) does not change and \( A \) remains the same, we can conclude that:
\[ W' = W \]

5. **Conclusion**:
Therefore, even after cutting the wire to half its length, the maximum load it can sustain remains the same, which is \( W \).

### Final Answer:
The maximum load that the wire can sustain after being cut to half its length is \( W \).

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