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 there is a decrease of 1% in its cross-sectional area, given that the Poisson's ratio (k) is 0.2.

### Step-by-Step Solution:

1. **Understand Poisson's Ratio**:
Poisson's ratio (k) is defined as the negative ratio of lateral strain to longitudinal strain. Mathematically, it is given by:
\[ k = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = -\frac{\Delta D / D}{\Delta L / L} \]
where \( \Delta D \) is the change in diameter, \( D \) is the original diameter, \( \Delta L \) is the change in length, and \( L \) is the original length.

2. **Relate Cross-Sectional Area to Diameter**:
The cross-sectional area \( A \) of the wire can be expressed in terms of its diameter \( D \):
\[ A = \frac{\pi}{4} D^2 \]
A decrease of 1% in the cross-sectional area means:
\[ \frac{\Delta A}{A} = -0.01 \]

3. **Differentiate the Area with Respect to Diameter**:
To find the relationship between the change in area and change in diameter, we differentiate the area:
\[ dA = \frac{\pi}{4} \cdot 2D \cdot dD = \frac{\pi D}{2} dD \]

4. **Express the Change in Area in Terms of Change in Diameter**:
Now, substituting \( dA \) into the relative change in area:
\[ \frac{dA}{A} = \frac{\frac{\pi D}{2} dD}{\frac{\pi}{4} D^2} = \frac{2dD}{D} \]
Therefore, we have:
\[ \frac{dA}{A} = 2 \frac{dD}{D} \]

5. **Relate the Change in Area to the Change in Diameter**:
Since we know \( \frac{dA}{A} = -0.01 \):
\[ -0.01 = 2 \frac{dD}{D} \]
This gives us:
\[ \frac{dD}{D} = -0.005 \]

6. **Substituting into Poisson's Ratio**:
Now substituting \( \frac{dD}{D} \) into the equation for Poisson's ratio:
\[ k = -\frac{\Delta D / D}{\Delta L / L} \]
We have:
\[ 0.2 = -\left(-0.005\right) \cdot \frac{L}{\Delta L} \]
Rearranging gives:
\[ \Delta L / L = \frac{0.005}{0.2} = 0.025 \]

7. **Convert to Percentage**:
To convert this to a percentage increase in length:
\[ \Delta L / L \times 100\% = 0.025 \times 100\% = 2.5\% \]

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