Question

What happens when the wire is now stretched to twice its original length by a process that keeps its volume constant?

Answer 1

When a wire is stretched to twice its original length while keeping its volume constant, several physical changes occur, primarily related to stress, strain, and the material's properties. Let’s break down the process step-by-step to understand what happens:

### 1. Understanding the Basics: Stress and Strain

**Stress** (\( \sigma \)) is defined as the force (\( F \)) applied per unit area (\( A \)) of the cross-section of the wire:
\[
\sigma = \frac{F}{A}
\]

**Strain** (\( \epsilon \)) is defined as the change in length (\( \Delta L \)) divided by the original length (\( L_0 \)):
\[
\epsilon = \frac{\Delta L}{L_0}
\]

### 2. Initial Conditions

Let’s consider a wire with:
- **Original Length (\( L_0 \))**: \( L \)
- **Original Cross-sectional Area (\( A \))**
- **Original Volume (\( V_0 \))**: \( V_0 = L \times A \)

### 3. Stretching the Wire

When the wire is stretched to **twice its original length**, the new length (\( L_f \)) becomes:
\[
L_f = 2L
\]

Given that the volume is constant, we can derive the new cross-sectional area (\( A_f \)) using the formula for volume:

\[
V_0 = L_0 \times A_0 = L_f \times A_f
\]

Since \( V_0 \) remains constant:

\[
L \times A = (2L) \times A_f
\]

### 4. Finding the New Cross-Sectional Area

Rearranging the above equation to solve for \( A_f \):

\[
A_f = \frac{A}{2}
\]

Thus, the new cross-sectional area is half of the original area.

### 5. Implications of Stretching

#### a. Change in Stress

The stress in the wire after it has been stretched can be calculated using the new area:

\[
\sigma_f = \frac{F}{A_f} = \frac{F}{\frac{A}{2}} = \frac{2F}{A}
\]

This means that the stress in the wire has doubled because the cross-sectional area has decreased while the force applied remains constant.

#### b. Change in Strain

The strain, as noted earlier, is given by:

\[
\epsilon = \frac{\Delta L}{L_0}
\]

Here, the change in length (\( \Delta L \)) is \( L_f - L_0 = 2L - L = L \):

\[
\epsilon = \frac{L}{L} = 1 \quad \text{(which is 100% strain)}
\]

This indicates that the wire has undergone significant deformation.

### 6. Material Properties

The behavior of the wire will depend on its material properties, particularly its **Young's Modulus** (\( E \)), which is defined as the ratio of stress to strain in the elastic region of the material:

\[
E = \frac{\sigma}{\epsilon}
\]

Since both stress and strain have changed significantly, the new Young's Modulus can be expressed as:

\[
E_f = \frac{\sigma_f}{\epsilon_f} = \frac{2\sigma}{1} = 2E
\]

This means that the effective stiffness of the material has increased as the wire is stretched and its cross-sectional area decreases.

### 7. Conclusion

To summarize, stretching a wire to twice its original length while maintaining constant volume results in:
- **Doubling of stress** due to the reduction in cross-sectional area.
- **100% strain**, indicating that the wire has been elongated to twice its length.
- An increase in **effective Young's Modulus**, which signifies a change in how the material responds to further stress.

This entire process illustrates fundamental concepts in material mechanics and elasticity, demonstrating how changes in dimensions under load affect the physical properties of materials.

Answer 2

When a wire is stretched to twice its original length while keeping its volume constant, several physical properties of the wire will change. Here’s a detailed explanation of what happens:

### Original Conditions

Let's start by defining some initial parameters for the wire:
- **Original Length**: \( L_0 \)
- **Original Cross-Sectional Area**: \( A_0 \)
- **Original Volume**: \( V_0 = L_0 \times A_0 \)

### Stretching the Wire

When the wire is stretched to twice its original length, the new length \( L \) becomes:
\[ L = 2L_0 \]

Since the volume is constant, the volume after stretching \( V \) remains equal to the original volume \( V_0 \). The volume of the wire is given by:
\[ V = L \times A \]
where \( A \) is the new cross-sectional area. Because the volume is unchanged:
\[ L_0 \times A_0 = L \times A \]

Substituting \( L = 2L_0 \) into this equation:
\[ L_0 \times A_0 = 2L_0 \times A \]

Solving for the new cross-sectional area \( A \):
\[ A = \frac{A_0}{2} \]

### Effects on Physical Properties

1. **Cross-Sectional Area**:
   - The cross-sectional area of the wire is halved. Originally, it was \( A_0 \), and now it is \( \frac{A_0}{2} \).

2. **Volume**:
   - As stipulated, the volume remains constant. Both before and after stretching, the volume is \( L_0 \times A_0 \).

3. **Young's Modulus**:
   - Young's modulus \( E \) of the wire is a material property and does not change with stretching. It is defined as the ratio of stress to strain:
     \[ E = \frac{\text{Stress}}{\text{Strain}} \]
     Stress is the force applied per unit area, and strain is the relative change in length.

4. **Resistance**:
   - If the wire is a conductor and you’re considering electrical resistance, it will change. The resistance \( R \) of a wire is given by:
     \[ R = \rho \frac{L}{A} \]
     where \( \rho \) is the resistivity of the material.
   - After stretching, \( L = 2L_0 \) and \( A = \frac{A_0}{2} \), so the new resistance \( R' \) is:
     \[ R' = \rho \frac{2L_0}{\frac{A_0}{2}} = 4 \rho \frac{L_0}{A_0} = 4R \]
     Thus, the resistance of the wire becomes four times its original resistance.

5. **Tensile Stress and Strain**:
   - **Strain** is the ratio of the change in length to the original length:
     \[ \text{Strain} = \frac{\Delta L}{L_0} \]
     Since the length increased by a factor of 2:
     \[ \text{Strain} = \frac{2L_0 - L_0}{L_0} = 1 \]
     This indicates a 100% increase in length.
   - **Stress** is the force applied per unit area. If the wire is stretched uniformly, the stress is inversely proportional to the cross-sectional area. Since the area is halved, the stress for the same force applied will be doubled.

### Summary

When the wire is stretched to twice its original length with constant volume:
- The cross-sectional area is halved.
- The volume remains the same.
- The resistance of the wire increases by a factor of four.
- Young’s modulus remains constant as it is a material property.
- The strain is 1 (or 100% increase in length), and the stress doubles if the applied force remains constant.

These changes illustrate the interplay between length, cross-sectional area, and volume in a wire under stretching conditions.