When a wire is stretched 4 times its original length?
2 Answers
When a wire is stretched to 4 times its original length, several physical properties and behaviors of the wire change. Here’s a detailed explanation:
### 1. **Change in Cross-Sectional Area:**
When a wire is stretched, its length increases, but its volume remains constant (assuming incompressibility). If the original length of the wire is \( L_0 \) and its original cross-sectional area is \( A_0 \), then the volume \( V \) of the wire before stretching is:
\[
V = A_0 \times L_0
\]
After stretching to 4 times its original length (\( 4L_0 \)), the new cross-sectional area \( A_f \) can be calculated using the constant volume:
\[
A_f = \frac{V}{4L_0} = \frac{A_0 \times L_0}{4L_0} = \frac{A_0}{4}
\]
Thus, the cross-sectional area becomes one-fourth of its original area.
### 2. **Change in Volume:**
As noted, the volume of the wire remains the same before and after stretching if we assume no volume loss or gain due to stretching.
### 3. **Change in Electrical Resistance:**
The electrical resistance \( R \) of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
- **Initial Resistance:** \( R_0 = \rho \frac{L_0}{A_0} \)
- **Final Resistance after Stretching:** \( R_f = \rho \frac{4L_0}{A_f} \)
Substituting \( A_f = \frac{A_0}{4} \):
\[
R_f = \rho \frac{4L_0}{\frac{A_0}{4}} = \rho \frac{16L_0}{A_0} = 16 \times R_0
\]
Therefore, the resistance of the wire increases by a factor of 16.
### 4. **Change in Young’s Modulus:**
The Young’s modulus \( E \) of a material is a measure of its stiffness and is given by:
\[
E = \frac{\text{Stress}}{\text{Strain}}
\]
where Stress \( \sigma = \frac{F}{A} \) and Strain \( \epsilon = \frac{\Delta L}{L_0} \).
When the wire is stretched, the strain increases. For a wire stretched to 4 times its length, the strain becomes 3 (since \( \frac{4L_0 - L_0}{L_0} = 3 \)).
The Young’s modulus \( E \) remains constant for a given material under elastic deformation, so the material's stiffness doesn't change with stretching. However, beyond the elastic limit, this could change.
In summary, stretching a wire to 4 times its original length results in a reduction in the cross-sectional area to one-fourth of its original value and an increase in electrical resistance by a factor of 16. The volume remains constant, and the Young's modulus remains unchanged within the elastic limit.
### 1. **Change in Cross-Sectional Area:**
When a wire is stretched, its length increases, but its volume remains constant (assuming incompressibility). If the original length of the wire is \( L_0 \) and its original cross-sectional area is \( A_0 \), then the volume \( V \) of the wire before stretching is:
\[
V = A_0 \times L_0
\]
After stretching to 4 times its original length (\( 4L_0 \)), the new cross-sectional area \( A_f \) can be calculated using the constant volume:
\[
A_f = \frac{V}{4L_0} = \frac{A_0 \times L_0}{4L_0} = \frac{A_0}{4}
\]
Thus, the cross-sectional area becomes one-fourth of its original area.
### 2. **Change in Volume:**
As noted, the volume of the wire remains the same before and after stretching if we assume no volume loss or gain due to stretching.
### 3. **Change in Electrical Resistance:**
The electrical resistance \( R \) of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
- **Initial Resistance:** \( R_0 = \rho \frac{L_0}{A_0} \)
- **Final Resistance after Stretching:** \( R_f = \rho \frac{4L_0}{A_f} \)
Substituting \( A_f = \frac{A_0}{4} \):
\[
R_f = \rho \frac{4L_0}{\frac{A_0}{4}} = \rho \frac{16L_0}{A_0} = 16 \times R_0
\]
Therefore, the resistance of the wire increases by a factor of 16.
### 4. **Change in Young’s Modulus:**
The Young’s modulus \( E \) of a material is a measure of its stiffness and is given by:
\[
E = \frac{\text{Stress}}{\text{Strain}}
\]
where Stress \( \sigma = \frac{F}{A} \) and Strain \( \epsilon = \frac{\Delta L}{L_0} \).
When the wire is stretched, the strain increases. For a wire stretched to 4 times its length, the strain becomes 3 (since \( \frac{4L_0 - L_0}{L_0} = 3 \)).
The Young’s modulus \( E \) remains constant for a given material under elastic deformation, so the material's stiffness doesn't change with stretching. However, beyond the elastic limit, this could change.
In summary, stretching a wire to 4 times its original length results in a reduction in the cross-sectional area to one-fourth of its original value and an increase in electrical resistance by a factor of 16. The volume remains constant, and the Young's modulus remains unchanged within the elastic limit.
When a wire is stretched to 4 times its original length, several important physical properties are affected, especially in the context of mechanical deformation, electrical resistance, and material strength. Let’s break this down in a detailed way:
### 1. **Effect on Length and Cross-Sectional Area**
- **Initial Length**: Let’s assume the original length of the wire is \( L_0 \).
- **Final Length**: When stretched, the final length becomes \( L_f = 4L_0 \).
Stretching a wire doesn’t just increase its length. It also affects its cross-sectional area, assuming that the volume of the wire remains constant (this holds in many practical cases where the wire is deformed but not broken apart or chemically altered). As the wire elongates, its diameter or cross-sectional area must decrease to conserve the volume.
If the initial cross-sectional area is \( A_0 \) and the volume \( V_0 \) is conserved, then:
\[
V_0 = L_0 \times A_0 = L_f \times A_f
\]
\[
L_0 \times A_0 = 4L_0 \times A_f
\]
\[
A_f = \frac{A_0}{4}
\]
Thus, the cross-sectional area decreases by a factor of 4 when the length increases by a factor of 4.
### 2. **Effect on Resistance**
Electrical resistance (\( R \)) of a wire depends on three factors:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) is the resistivity of the material (a constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.
Initially, the resistance is:
\[
R_0 = \rho \frac{L_0}{A_0}
\]
After stretching, the resistance becomes:
\[
R_f = \rho \frac{4L_0}{A_f} = \rho \frac{4L_0}{\frac{A_0}{4}} = 16 \times R_0
\]
So, the resistance of the wire increases by a factor of 16 when the wire is stretched to 4 times its original length.
### 3. **Effect on Mechanical Properties**
When a wire is stretched, it undergoes **tensile stress** and **strain**.
- **Tensile stress** is the force per unit area acting on the wire:
\[
\text{Stress} = \frac{\text{Force}}{\text{Cross-sectional Area}}
\]
Since the cross-sectional area decreases by a factor of 4, the stress on the wire increases significantly for the same applied force.
- **Tensile strain** is the ratio of the change in length to the original length:
\[
\text{Strain} = \frac{\Delta L}{L_0} = \frac{L_f - L_0}{L_0} = \frac{4L_0 - L_0}{L_0} = 3
\]
So, the strain is 3 (or 300% of the original length), indicating substantial deformation.
For many materials, especially metals, there is a limit to how much they can be stretched before they reach the **yield point** (beyond which they permanently deform) or the **ultimate tensile strength** (beyond which they break).
### 4. **Effect on Conductivity**
While the resistivity \( \rho \) of the material itself remains constant (assuming the wire isn’t heated significantly), the **conductivity** (inverse of resistance) decreases sharply due to the large increase in resistance. After stretching the wire 4 times its original length, the conductivity decreases by a factor of 16 because resistance has increased 16 times.
### Summary
- The final length of the wire is 4 times the original length.
- The cross-sectional area reduces by a factor of 4.
- The resistance increases by a factor of 16.
- The tensile stress increases because of the reduced cross-sectional area.
- The strain is significant, and the wire may approach or exceed its mechanical limits depending on the material.
In electrical and mechanical systems, significant stretching like this could result in failure if the wire is not designed to withstand such deformation.
### 1. **Effect on Length and Cross-Sectional Area**
- **Initial Length**: Let’s assume the original length of the wire is \( L_0 \).
- **Final Length**: When stretched, the final length becomes \( L_f = 4L_0 \).
Stretching a wire doesn’t just increase its length. It also affects its cross-sectional area, assuming that the volume of the wire remains constant (this holds in many practical cases where the wire is deformed but not broken apart or chemically altered). As the wire elongates, its diameter or cross-sectional area must decrease to conserve the volume.
If the initial cross-sectional area is \( A_0 \) and the volume \( V_0 \) is conserved, then:
\[
V_0 = L_0 \times A_0 = L_f \times A_f
\]
\[
L_0 \times A_0 = 4L_0 \times A_f
\]
\[
A_f = \frac{A_0}{4}
\]
Thus, the cross-sectional area decreases by a factor of 4 when the length increases by a factor of 4.
### 2. **Effect on Resistance**
Electrical resistance (\( R \)) of a wire depends on three factors:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) is the resistivity of the material (a constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.
Initially, the resistance is:
\[
R_0 = \rho \frac{L_0}{A_0}
\]
After stretching, the resistance becomes:
\[
R_f = \rho \frac{4L_0}{A_f} = \rho \frac{4L_0}{\frac{A_0}{4}} = 16 \times R_0
\]
So, the resistance of the wire increases by a factor of 16 when the wire is stretched to 4 times its original length.
### 3. **Effect on Mechanical Properties**
When a wire is stretched, it undergoes **tensile stress** and **strain**.
- **Tensile stress** is the force per unit area acting on the wire:
\[
\text{Stress} = \frac{\text{Force}}{\text{Cross-sectional Area}}
\]
Since the cross-sectional area decreases by a factor of 4, the stress on the wire increases significantly for the same applied force.
- **Tensile strain** is the ratio of the change in length to the original length:
\[
\text{Strain} = \frac{\Delta L}{L_0} = \frac{L_f - L_0}{L_0} = \frac{4L_0 - L_0}{L_0} = 3
\]
So, the strain is 3 (or 300% of the original length), indicating substantial deformation.
For many materials, especially metals, there is a limit to how much they can be stretched before they reach the **yield point** (beyond which they permanently deform) or the **ultimate tensile strength** (beyond which they break).
### 4. **Effect on Conductivity**
While the resistivity \( \rho \) of the material itself remains constant (assuming the wire isn’t heated significantly), the **conductivity** (inverse of resistance) decreases sharply due to the large increase in resistance. After stretching the wire 4 times its original length, the conductivity decreases by a factor of 16 because resistance has increased 16 times.
### Summary
- The final length of the wire is 4 times the original length.
- The cross-sectional area reduces by a factor of 4.
- The resistance increases by a factor of 16.
- The tensile stress increases because of the reduced cross-sectional area.
- The strain is significant, and the wire may approach or exceed its mechanical limits depending on the material.
In electrical and mechanical systems, significant stretching like this could result in failure if the wire is not designed to withstand such deformation.