Question

A metal wire having length of 2.0 m is elongated by 2 mm under a load. The initial diameter of the wire is 2 mm. Find the change in diameter of wire after elongation if poisons ratio is 0.2.

• A. 4 x 10$^{-5}$ m
• B. 8 x 10$^{-3}$ m
• C. 3.2 x 10$^{-4}$ m
• D. 4 x 10$^{-7}$ m

Hint:

To solve dimensional change problems, use the relation: $\Delta d = -\mu \cdot \frac{\Delta L}{L} \cdot d$. Ensure all length and diameter values are converted to meters before performing the calculation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a material is subjected to an axial tensile force, it elongates in the direction of the load (longitudinal strain) and contracts in the lateral directions (lateral strain).
The ratio of this lateral strain to the longitudinal strain is a material property known as Poisson's ratio ($\mu$ or $\nu$).
We can use Poisson's ratio to calculate changes in cross-sectional dimensions during deformation.
Key Formula or Approach:
The formula for Poisson's ratio ($\mu$) is:
\[ \mu = -\frac{\text{Lateral Strain } (\epsilon_{\text{lat}})}{\text{Longitudinal Strain } (\epsilon_{\text{long}})} \]
Where:
Longitudinal strain is:
\[ \epsilon_{\text{long}} = \frac{\Delta L}{L} \]
Lateral strain is:
\[ \epsilon_{\text{lat}} = \frac{\Delta d}{d} \]
Therefore, the change in diameter ($\Delta d$) can be calculated as:
\[ \Delta d = -\mu \cdot \epsilon_{\text{long}} \cdot d \]

Step 2: Detailed Explanation:
Let us identify the parameters from the problem statement:
The initial length of the wire is:
\[ L = 2.0 \text{ m} \]
The elongation (change in length) is:
\[ \Delta L = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \]
The initial diameter of the wire is:
\[ d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \]
The Poisson's ratio is:
\[ \mu = 0.2 \]
First, we calculate the longitudinal strain ($\epsilon_{\text{long}}$):
\[ \epsilon_{\text{long}} = \frac{2 \times 10^{-3} \text{ m}}{2.0 \text{ m}} = 1 \times 10^{-3} \]
Next, we calculate the lateral strain ($\epsilon_{\text{lat}}$) using Poisson's ratio:
\[ \epsilon_{\text{lat}} = -\mu \cdot \epsilon_{\text{long}} = -0.2 \cdot (1 \times 10^{-3}) = -2 \times 10^{-4} \]
Now, we find the change in diameter ($\Delta d$):
\[ \Delta d = \epsilon_{\text{lat}} \cdot d \]
\[ \Delta d = (-2 \times 10^{-4}) \cdot (2 \times 10^{-3} \text{ m}) \]
\[ \Delta d = -4 \times 10^{-7} \text{ m} \]
The negative sign indicates a reduction in diameter as the wire is stretched.
The magnitude of the change in diameter is $4 \times 10^{-7}$ m.

Step 3: Final Answer
The change in the diameter of the wire after elongation is $4 \times 10^{-7}$ m.