If the volume of a wire remains constant when subjected to tensile stress, the value of poisson's ratio of material of the wire is

Options

A. 0.1
B. 0.2
C. 0.4
D. 0.5

Explanation

To find the value of Poisson's ratio when the volume of a wire remains constant under tensile stress, we can follow these steps:

### Step 1: Understand the Definition of Poisson's Ratio
Poisson's ratio (ν) is defined as the negative ratio of lateral strain to longitudinal strain. Mathematically, it is expressed as:
\[ \nu = -\frac{\Delta r / r}{\Delta l / l} \]
where:
- \( \Delta r \) is the change in radius,
- \( r \) is the original radius,
- \( \Delta l \) is the change in length,
- \( l \) is the original length.

### Step 2: Consider the Volume of the Wire
The volume \( V \) of the wire can be expressed as:
\[ V = \pi r^2 l \]
Given that the volume remains constant, we have:
\[ \frac{dV}{dt} = 0 \]
This implies that any change in radius and length must satisfy the condition of constant volume.

### Step 3: Differentiate the Volume Expression
Taking the differential of the volume expression, we have:
\[ dV = \pi (2r \, dr \cdot l + r^2 \, dl) = 0 \]
Setting \( dV = 0 \) gives us:
\[ 2r \, dr \cdot l + r^2 \, dl = 0 \]

### Step 4: Relate Changes in Radius and Length
Rearranging the above equation, we can express the relationship between the changes in radius and length:
\[ 2r \, dr = -r^2 \, dl \]
Dividing both sides by \( r^2 \) and simplifying, we obtain:
\[ \frac{2 \, dr}{r} = -\frac{dl}{l} \]

### Step 5: Substitute into Poisson's Ratio Formula
Now we can substitute this relationship into the Poisson's ratio formula:
\[ \nu = -\frac{\Delta r / r}{\Delta l / l} \]
Substituting \( \Delta r / r = -\frac{1}{2} \Delta l / l \):
\[ \nu = -\left(-\frac{1}{2}\right) = \frac{1}{2} \]

### Conclusion
Thus, the value of Poisson's ratio (ν) when the volume of the wire remains constant under tensile stress is:
\[ \nu = 0.5 \]