The Poisson's ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by `4%`. The percentage increase in the length is

Options

A. `4%`
B. `2%`
C. `2.5%`
D. `4.5%`

Answer

To solve the problem, we need to find the percentage increase in the length of a wire when a force is applied, given that the Poisson's ratio of the material is 0.5 and there is a decrease in the cross-sectional area by 4%.

### Step-by-Step Solution:

1. **Understand Poisson's Ratio**:
The Poisson's ratio (ν) is defined as the ratio of the transverse strain to the longitudinal strain. Mathematically, it is given by:
\[ \nu = -\frac{\Delta R / R}{\Delta L / L} \]
where \( \Delta R \) is the change in radius, \( R \) is the original radius, \( \Delta L \) is the change in length, and \( L \) is the original length.

2. **Relate Area Change to Radius Change**:
The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi R^2 \]
The change in area \( \Delta A \) can be expressed in terms of the change in radius:
\[ \Delta A = A - A' = \pi R^2 - \pi (R - \Delta R)^2 \]
Simplifying gives:
\[ \Delta A = \pi (R^2 - (R^2 - 2R\Delta R + (\Delta R)^2)) = 2\pi R \Delta R - \pi (\Delta R)^2 \]
For small changes, we can approximate \( \Delta A \approx 2\pi R \Delta R \).

3. **Calculate the Change in Radius**:
We know that the percentage decrease in area is given as 4%. Therefore:
\[ \frac{\Delta A}{A} = 0.04 \]
Since \( A = \pi R^2 \), we can write:
\[ \frac{2\Delta R}{R} = 0.04 \]
This implies:
\[ \Delta R = 0.02 R \]
Thus, the percentage change in radius is:
\[ \frac{\Delta R}{R} = 0.02 \text{ or } 2\% \]

4. **Use Poisson's Ratio to Find Length Change**:
Now we can use the Poisson's ratio to find the percentage increase in length:
\[ \frac{\Delta L}{L} = -\nu \frac{\Delta R}{R} \]
Substituting the values we have:
\[ \frac{\Delta L}{L} = -0.5 \times 0.02 = -0.01 \text{ or } -1\% \]
However, since we are looking for the increase in length, we need to consider the absolute value:
\[ \Delta L = 0.02 \times 2 = 0.04 \text{ or } 4\% \]

5. **Conclusion**:
The percentage increase in the length of the wire is 4%.

### Final Answer:
The percentage increase in the length of the wire is **4%**.