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

Options

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

Explanation

To solve the problem, we need to determine the value of Poisson's ratio for a wire whose volume remains constant when subjected to tensile stress. Here’s a step-by-step solution:

### Step 1: Understand Poisson's Ratio
Poisson's ratio (ν) is defined as the negative ratio of lateral strain to longitudinal strain. Mathematically, it is expressed as:
\[ \nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta R / R}{\Delta L / L} \]
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: Volume of the Wire
The volume \( V \) of a cylindrical wire is given by:
\[ V = \pi R^2 L \]
Since the volume remains constant, we can differentiate this equation to find the relationship between changes in radius and length.

### Step 3: Differentiate the Volume
Differentiating the volume with respect to changes in radius and length gives:
\[ \Delta V = \pi (2R \Delta R L + R^2 \Delta L) = 0 \]
Since the volume remains constant, \( \Delta V = 0 \).

### Step 4: Set Up the Equation
From the differentiated volume equation, we have:
\[ 0 = 2R \Delta R L + R^2 \Delta L \]
Rearranging this gives:
\[ R^2 \Delta L = -2R L \Delta R \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ R \Delta L = -2L \Delta R \]

### Step 5: Express Lateral Strain
Now, we can express the lateral strain \( \frac{\Delta R}{R} \) in terms of the longitudinal strain \( \frac{\Delta L}{L} \):
\[ \frac{\Delta R}{R} = -\frac{1}{2} \frac{\Delta L}{L} \]

### Step 6: Substitute into Poisson's Ratio
Substituting this expression into the formula for Poisson's ratio:
\[ \nu = -\frac{\Delta R / R}{\Delta L / L} = -\left(-\frac{1}{2}\right) = \frac{1}{2} \]

### Conclusion
Thus, the value of Poisson's ratio for the material of the wire is:
\[ \nu = 0.5 \]

### Final Answer
The value of Poisson's ratio of the material of the wire is **0.5**.

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