In a wire of length L, the increase in its length is `DeltaL`. If the length is reduced to half, then the increase in its length will be

Options

A. `DeltaL`
B. `2DeltaL`
C. `(DeltaL)/(2)`
D. none

Explanation

To solve the problem, we need to understand the relationship between the increase in length of a wire (ΔL) and its original length (L). The increase in length is directly proportional to the original length of the wire when the force and cross-sectional area remain constant.

### Step-by-Step Solution:

1. **Understanding the Proportionality**:
The increase in length (ΔL) of a wire is directly proportional to its original length (L) when the force (F) applied and the cross-sectional area (A) are constant. This can be expressed mathematically as:
\[ \Delta L \propto L \]

2. **Setting Up the Relationship**:
If we denote the original length of the wire as \( L_1 \) and the increase in length as \( \Delta L_1 \), we can write:
\[ \Delta L_1 = k \cdot L_1 \]
where \( k \) is a proportionality constant that depends on the material properties and the applied force.

3. **Reducing the Length**:
Now, if the length of the wire is reduced to half, we have:
\[ L_2 = \frac{L_1}{2} \]
The increase in length for this new length (ΔL2) can be expressed as:
\[ \Delta L_2 = k \cdot L_2 = k \cdot \left(\frac{L_1}{2}\right) \]

4. **Substituting the Values**:
Substituting \( L_2 \) into the equation for ΔL2 gives:
\[ \Delta L_2 = \frac{k \cdot L_1}{2} = \frac{\Delta L_1}{2} \]

5. **Conclusion**:
Therefore, the increase in length when the original length is reduced to half is:
\[ \Delta L_2 = \frac{\Delta L_1}{2} \]

### Final Answer:
If the length of the wire is reduced to half, the increase in its length will be half of the original increase in length:
\[ \Delta L_2 = \frac{\Delta L_1}{2} \]