The Poisson's ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is

Options

A. `3%`
B. `2.5%`
C. `1%`
D. `0.5%`

Answer

To solve the problem, we will use the concept of Poisson's ratio and the relationship between lateral strain and longitudinal strain.

### Step-by-Step Solution:

1. **Understanding Poisson's Ratio:**
Poisson's ratio (σ) is defined as the ratio of lateral strain to longitudinal strain. Mathematically, it is expressed as:
\[ \sigma = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta A/A}{\Delta L/L} \]
where:
- \(\Delta A\) is the change in cross-sectional area,
- \(A\) is the original cross-sectional area,
- \(\Delta L\) is the change in length,
- \(L\) is the original length.

2. **Given Values:**
- Poisson's ratio, \(σ = 0.4\)
- Decrease in cross-sectional area, \(\Delta A/A = -2\%\) (which is -0.02 in decimal form).

3. **Relating Change in Area to Change in Radius:**
The area \(A\) of a circular wire can be expressed as:
\[ A = \pi r^2 \]
The change in area can be related to the change in radius:
\[ \frac{\Delta A}{A} = 2 \frac{\Delta r}{r} \]
Thus, if \(\Delta A/A = -0.02\), we can write:
\[ -0.02 = 2 \frac{\Delta r}{r} \]
From this, we can find \(\Delta r/r\):
\[ \frac{\Delta r}{r} = -0.01 \]

4. **Using Poisson's Ratio to Find Longitudinal Strain:**
Now, we can use the Poisson's ratio to find the longitudinal strain:
\[ \sigma = -\frac{\Delta r/r}{\Delta L/L} \]
Rearranging gives:
\[ \Delta L/L = -\frac{\Delta r/r}{\sigma} \]
Substituting the values:
\[ \Delta L/L = -\frac{-0.01}{0.4} = 0.025 \]

5. **Converting to Percentage:**
To express this as 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%**.

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