If a force is applicable to an elastic wire of the material of Poisson's ratio 0.2 there is a decrease of the cross-sectional area by 1 % . The percentage increase in its length is :

Options

A. `1%`
B. `5%`
C. `2.5%`
D. `1.5%`

Answer

To solve the problem, we need to find the percentage increase in the length of an elastic wire when the cross-sectional area decreases by 1%. We will use the concept of Poisson's ratio, which relates lateral strain to linear strain.

### Step-by-Step Solution:

1. **Understanding Poisson's Ratio**:
Poisson's ratio (σ) is defined as the ratio of lateral strain to linear strain. Mathematically, it is given by:
\[ \sigma = -\frac{\text{Lateral Strain}}{\text{Linear Strain}} = -\frac{\Delta R / R}{\Delta L / L} \]
where:
- \( \Delta R \) is the change in radius,
- \( R \) is the original radius,
- \( \Delta L \) is the change in length,
- \( L \) is the original length.

2. **Given Information**:
- Poisson's ratio \( \sigma = 0.2 \)
- The decrease in cross-sectional area \( \Delta A / A = -1\% = -0.01 \)

3. **Relating Area to Radius**:
The cross-sectional area \( A \) of a circular wire is given by:
\[ A = \pi R^2 \]
The relative change in area can be expressed in terms of the relative change in radius:
\[ \frac{\Delta A}{A} = 2 \frac{\Delta R}{R} \]
Therefore, we can write:
\[ -0.01 = 2 \frac{\Delta R}{R} \]

4. **Calculating Change in Radius**:
Rearranging the equation gives:
\[ \frac{\Delta R}{R} = -0.005 \]
This indicates a decrease of 0.5% in the radius.

5. **Using Poisson's Ratio**:
Now, substituting \( \frac{\Delta R}{R} \) into the Poisson's ratio equation:
\[ 0.2 = -\frac{-0.005}{\Delta L / L} \]
Rearranging gives:
\[ \Delta L / L = \frac{0.005}{0.2} \]

6. **Calculating Change in Length**:
\[ \Delta L / L = 0.025 \]
This indicates a percentage increase in length:
\[ \Delta L / L \times 100 = 2.5\% \]

### Final Answer:
The percentage increase in the length of the wire is **2.5%**.