When a wire is stretched 4 times its length?
2 Answers
When a wire is stretched to 4 times its original length, several properties of the wire change due to the stretching. Here’s a detailed look at what happens:
### 1. **Change in Diameter**
When a wire is stretched, its volume remains constant if it’s incompressible. As the length increases, the diameter of the wire decreases to maintain the same volume. Specifically, if a wire is stretched to \( L_{\text{new}} \) which is 4 times its original length \( L_{\text{original}} \), then the new diameter \( d_{\text{new}} \) can be calculated using the relationship of volumes:
\[
\text{Volume}_{\text{original}} = \text{Volume}_{\text{new}}
\]
\[
A_{\text{original}} \cdot L_{\text{original}} = A_{\text{new}} \cdot L_{\text{new}}
\]
Where \( A \) is the cross-sectional area. Since the cross-sectional area \( A \) of a wire is proportional to the square of the diameter:
\[
A = \frac{\pi d^2}{4}
\]
Substituting and simplifying:
\[
\frac{\pi d_{\text{original}}^2}{4} \cdot L_{\text{original}} = \frac{\pi d_{\text{new}}^2}{4} \cdot L_{\text{new}}
\]
\[
d_{\text{new}} = \frac{d_{\text{original}}}{\sqrt{4}} = \frac{d_{\text{original}}}{2}
\]
So, the diameter is reduced to half of its original value.
### 2. **Change in Cross-Sectional Area**
As the diameter reduces to half, the cross-sectional area reduces to:
\[
A_{\text{new}} = \left(\frac{d_{\text{original}}}{2}\right)^2 \cdot \frac{\pi}{4} = \frac{A_{\text{original}}}{4}
\]
Thus, the cross-sectional area is reduced to one-fourth of its original value.
### 3. **Change in Resistance**
The electrical resistance \( R \) of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
Where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. For the stretched wire:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
\[
R_{\text{new}} = \rho \frac{4 L_{\text{original}}}{\frac{A_{\text{original}}}{4}} = 16 \rho \frac{L_{\text{original}}}{A_{\text{original}}} = 16 R_{\text{original}}
\]
So, the resistance of the wire increases by a factor of 16.
### 4. **Mechanical Stress and Strain**
- **Strain**: Strain is defined as the change in length divided by the original length. For a wire stretched to 4 times its length:
\[
\text{Strain} = \frac{\Delta L}{L_{\text{original}}} = \frac{L_{\text{new}} - L_{\text{original}}}{L_{\text{original}}} = \frac{4L_{\text{original}} - L_{\text{original}}}{L_{\text{original}}} = 3
\]
So, the strain is 3 (or 300%).
- **Stress**: Stress is the force applied divided by the cross-sectional area. As the cross-sectional area reduces, the stress in the wire increases if the same force is applied.
### Summary
When a wire is stretched to 4 times its original length:
- The diameter of the wire reduces to half of its original diameter.
- The cross-sectional area reduces to one-fourth of its original area.
- The electrical resistance increases to 16 times its original value.
- The strain is 300%, and stress increases due to reduced cross-sectional area.
These changes are important in both electrical and mechanical contexts.
### 1. **Change in Diameter**
When a wire is stretched, its volume remains constant if it’s incompressible. As the length increases, the diameter of the wire decreases to maintain the same volume. Specifically, if a wire is stretched to \( L_{\text{new}} \) which is 4 times its original length \( L_{\text{original}} \), then the new diameter \( d_{\text{new}} \) can be calculated using the relationship of volumes:
\[
\text{Volume}_{\text{original}} = \text{Volume}_{\text{new}}
\]
\[
A_{\text{original}} \cdot L_{\text{original}} = A_{\text{new}} \cdot L_{\text{new}}
\]
Where \( A \) is the cross-sectional area. Since the cross-sectional area \( A \) of a wire is proportional to the square of the diameter:
\[
A = \frac{\pi d^2}{4}
\]
Substituting and simplifying:
\[
\frac{\pi d_{\text{original}}^2}{4} \cdot L_{\text{original}} = \frac{\pi d_{\text{new}}^2}{4} \cdot L_{\text{new}}
\]
\[
d_{\text{new}} = \frac{d_{\text{original}}}{\sqrt{4}} = \frac{d_{\text{original}}}{2}
\]
So, the diameter is reduced to half of its original value.
### 2. **Change in Cross-Sectional Area**
As the diameter reduces to half, the cross-sectional area reduces to:
\[
A_{\text{new}} = \left(\frac{d_{\text{original}}}{2}\right)^2 \cdot \frac{\pi}{4} = \frac{A_{\text{original}}}{4}
\]
Thus, the cross-sectional area is reduced to one-fourth of its original value.
### 3. **Change in Resistance**
The electrical resistance \( R \) of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
Where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. For the stretched wire:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
\[
R_{\text{new}} = \rho \frac{4 L_{\text{original}}}{\frac{A_{\text{original}}}{4}} = 16 \rho \frac{L_{\text{original}}}{A_{\text{original}}} = 16 R_{\text{original}}
\]
So, the resistance of the wire increases by a factor of 16.
### 4. **Mechanical Stress and Strain**
- **Strain**: Strain is defined as the change in length divided by the original length. For a wire stretched to 4 times its length:
\[
\text{Strain} = \frac{\Delta L}{L_{\text{original}}} = \frac{L_{\text{new}} - L_{\text{original}}}{L_{\text{original}}} = \frac{4L_{\text{original}} - L_{\text{original}}}{L_{\text{original}}} = 3
\]
So, the strain is 3 (or 300%).
- **Stress**: Stress is the force applied divided by the cross-sectional area. As the cross-sectional area reduces, the stress in the wire increases if the same force is applied.
### Summary
When a wire is stretched to 4 times its original length:
- The diameter of the wire reduces to half of its original diameter.
- The cross-sectional area reduces to one-fourth of its original area.
- The electrical resistance increases to 16 times its original value.
- The strain is 300%, and stress increases due to reduced cross-sectional area.
These changes are important in both electrical and mechanical contexts.
When a wire is stretched to four times its original length, several physical properties of the wire are affected, particularly its cross-sectional area, volume, and stress-strain characteristics. Here’s a detailed explanation:
### 1. **Change in Cross-Sectional Area**
Assuming the wire is stretched uniformly and is made of a material that remains homogeneous, the volume of the wire must remain constant. Let's denote:
- The original length of the wire as \( L_0 \).
- The original cross-sectional area of the wire as \( A_0 \).
- The original volume of the wire as \( V_0 = A_0 \times L_0 \).
When the wire is stretched to four times its original length, the new length \( L \) is \( 4L_0 \). To maintain the same volume, the cross-sectional area must change accordingly. The new volume \( V \) must be equal to the original volume:
\[ V = A \times L = A_0 \times L_0 \]
Substitute \( L = 4L_0 \):
\[ A \times 4L_0 = A_0 \times L_0 \]
Solving for the new cross-sectional area \( A \):
\[ A = \frac{A_0}{4} \]
Thus, the cross-sectional area of the wire after stretching is one-fourth of the original cross-sectional area.
### 2. **Stress and Strain**
- **Strain** is defined as the change in length divided by the original length. For the wire stretched to four times its length:
\[
\text{Strain} = \frac{\text{Change in Length}}{\text{Original Length}} = \frac{4L_0 - L_0}{L_0} = \frac{3L_0}{L_0} = 3
\]
So, the strain is 3, meaning the wire is under a strain of 300%.
- **Stress** is the force applied per unit area. The relationship between stress (\( \sigma \)) and strain (\( \epsilon \)) in a material is given by Hooke's Law for elastic deformation:
\[
\sigma = E \cdot \epsilon
\]
where \( E \) is the Young's Modulus of the material.
Because the cross-sectional area has decreased, the stress in the wire will be affected. If the original force applied was \( F_0 \), the new stress \( \sigma \) after stretching can be calculated using:
\[
\sigma = \frac{F}{A} = \frac{F_0}{\frac{A_0}{4}} = 4 \cdot \frac{F_0}{A_0}
\]
So, the stress will increase by a factor of 4.
### 3. **Mechanical Properties**
- **Young’s Modulus** (\( E \)) is a measure of the stiffness of the material and remains constant for a given material regardless of the wire's dimensions. It relates stress and strain.
- **Elastic Limit**: If the wire is stretched beyond its elastic limit, it may not return to its original length when the force is removed, leading to permanent deformation.
### Summary
When a wire is stretched to four times its original length, the cross-sectional area decreases to one-fourth of its original value. The strain is 3 (or 300%), and the stress increases by a factor of 4 if the same force is applied. These changes affect how the wire behaves under load and how it will return to its original shape after the load is removed.
### 1. **Change in Cross-Sectional Area**
Assuming the wire is stretched uniformly and is made of a material that remains homogeneous, the volume of the wire must remain constant. Let's denote:
- The original length of the wire as \( L_0 \).
- The original cross-sectional area of the wire as \( A_0 \).
- The original volume of the wire as \( V_0 = A_0 \times L_0 \).
When the wire is stretched to four times its original length, the new length \( L \) is \( 4L_0 \). To maintain the same volume, the cross-sectional area must change accordingly. The new volume \( V \) must be equal to the original volume:
\[ V = A \times L = A_0 \times L_0 \]
Substitute \( L = 4L_0 \):
\[ A \times 4L_0 = A_0 \times L_0 \]
Solving for the new cross-sectional area \( A \):
\[ A = \frac{A_0}{4} \]
Thus, the cross-sectional area of the wire after stretching is one-fourth of the original cross-sectional area.
### 2. **Stress and Strain**
- **Strain** is defined as the change in length divided by the original length. For the wire stretched to four times its length:
\[
\text{Strain} = \frac{\text{Change in Length}}{\text{Original Length}} = \frac{4L_0 - L_0}{L_0} = \frac{3L_0}{L_0} = 3
\]
So, the strain is 3, meaning the wire is under a strain of 300%.
- **Stress** is the force applied per unit area. The relationship between stress (\( \sigma \)) and strain (\( \epsilon \)) in a material is given by Hooke's Law for elastic deformation:
\[
\sigma = E \cdot \epsilon
\]
where \( E \) is the Young's Modulus of the material.
Because the cross-sectional area has decreased, the stress in the wire will be affected. If the original force applied was \( F_0 \), the new stress \( \sigma \) after stretching can be calculated using:
\[
\sigma = \frac{F}{A} = \frac{F_0}{\frac{A_0}{4}} = 4 \cdot \frac{F_0}{A_0}
\]
So, the stress will increase by a factor of 4.
### 3. **Mechanical Properties**
- **Young’s Modulus** (\( E \)) is a measure of the stiffness of the material and remains constant for a given material regardless of the wire's dimensions. It relates stress and strain.
- **Elastic Limit**: If the wire is stretched beyond its elastic limit, it may not return to its original length when the force is removed, leading to permanent deformation.
### Summary
When a wire is stretched to four times its original length, the cross-sectional area decreases to one-fourth of its original value. The strain is 3 (or 300%), and the stress increases by a factor of 4 if the same force is applied. These changes affect how the wire behaves under load and how it will return to its original shape after the load is removed.