If Poission's ratio `sigma` is `-(1)/(2)` for a material, then the material is

Options

A. elastic fatigue
B. incompressible
C. compressible
D. None of these

Explanation

To determine the nature of the material given that Poisson's ratio \( \sigma \) is \( -\frac{1}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Poisson's Ratio**:
Poisson's ratio \( \sigma \) is defined as the negative ratio of transverse strain to axial strain. It indicates how much a material deforms in the lateral direction when subjected to axial stress.

2. **Given Value**:
We are given that \( \sigma = -\frac{1}{2} \).

3. **Relation Between Poisson's Ratio and Volume Change**:
The change in volume \( \frac{dV}{V} \) can be expressed in terms of Poisson's ratio as:
\[ \frac{dV}{V} = 1 + 2\sigma \frac{dL}{L} \]
where \( \frac{dL}{L} \) is the strain.

4. **Substituting the Given Value**:
Substitute \( \sigma = -\frac{1}{2} \) into the equation:
\[ \frac{dV}{V} = 1 + 2 \left(-\frac{1}{2}\right) \frac{dL}{L} \]
This simplifies to:
\[ \frac{dV}{V} = 1 - \frac{dL}{L} \]

5. **Analyzing the Expression**:
If we assume that the material is undergoing a uniform deformation, we can set \( \frac{dL}{L} = 1 \) (which would imply maximum elongation):
\[ \frac{dV}{V} = 1 - 1 = 0 \]
This indicates that the volume change \( dV \) is zero, meaning the volume remains constant.

6. **Bulk Modulus Relation**:
The bulk modulus \( K \) is defined as:
\[ K = -\frac{p}{\frac{dV}{V}} \]
Since \( \frac{dV}{V} = 0 \), this implies that:
\[ K \rightarrow \infty \]
Therefore, the material is incompressible.

7. **Conclusion**:
Since the bulk modulus is infinite, we conclude that the material is incompressible.

### Final Answer:
The material is **incompressible**.

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