There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is

Options

A. `+(1)/(2)`
B. `-(1)/(2)`
C. `+(1)/(4)`
D. `-(1)/(4)`

Answer

To solve the problem, we need to determine the Poisson's ratio of a wire that does not change in volume when stretched. The Poisson's ratio (ν) is defined as the negative ratio of lateral strain to longitudinal strain.

### Step-by-Step Solution:

1. **Understanding Poisson's Ratio**:
The Poisson's ratio (ν) is given by the formula:
\[ \nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} \]

2. **Volume of the Wire**:
Consider a cylindrical wire with radius \( R \) and length \( L \). The volume \( V \) of the wire can be expressed as:
\[ V = \pi R^2 L \]

3. **Change in Volume**:
When the wire is stretched, we denote the change in length as \( \Delta L \) and the change in radius as \( \Delta R \). The new length becomes \( L + \Delta L \) and the new radius becomes \( R - \Delta R \). The volume after stretching is:
\[ V' = \pi (R - \Delta R)^2 (L + \Delta L) \]

4. **Condition of No Change in Volume**:
Since the volume does not change, we have:
\[ V' = V \]
This leads to:
\[ \pi (R - \Delta R)^2 (L + \Delta L) = \pi R^2 L \]

5. **Expanding the Volume Equation**:
Expanding the left-hand side:
\[ (R - \Delta R)^2 = R^2 - 2R\Delta R + (\Delta R)^2 \]
Thus,
\[ V' = \pi (R^2 - 2R\Delta R + (\Delta R)^2)(L + \Delta L) \]

6. **Neglecting Higher Order Terms**:
For small deformations, we can neglect the term \( (\Delta R)^2 \). Therefore, we have:
\[ V' \approx \pi (R^2 - 2R\Delta R)(L + \Delta L) \]
Expanding this gives:
\[ V' \approx \pi R^2 L + \pi R^2 \Delta L - 2\pi R L \Delta R \]

7. **Setting the Change in Volume to Zero**:
Since \( V' = V \), we can set the change in volume to zero:
\[ \pi R^2 \Delta L - 2\pi R L \Delta R = 0 \]
Dividing through by \( \pi \):
\[ R^2 \Delta L = 2R L \Delta R \]

8. **Relating Strains**:
The longitudinal strain \( \epsilon_L \) is given by:
\[ \epsilon_L = \frac{\Delta L}{L} \]
The lateral strain \( \epsilon_T \) is given by:
\[ \epsilon_T = -\frac{\Delta R}{R} \]
Substituting these into the equation gives:
\[ R^2 \left(\epsilon_L L\right) = 2R L \left(-\epsilon_T R\right) \]
Simplifying leads to:
\[ \epsilon_L = -2\epsilon_T \]

9. **Finding Poisson's Ratio**:
Substituting into the Poisson's ratio formula:
\[ \nu = -\frac{\epsilon_T}{\epsilon_L} = -\frac{\epsilon_T}{-2\epsilon_T} = \frac{1}{2} \]

### Final Answer:
The Poisson's ratio of the material of the wire is:
\[ \nu = \frac{1}{2} \]