When a wire is stretched and its radius becomes R^2?
2 Answers
When a wire is stretched and its radius changes, its volume remains constant. This is because the stretching is assumed to be purely mechanical, and no material is added or removed. The relationship between the initial and final dimensions of the wire can be explained using the concept of conservation of volume.
Let’s analyze the situation step by step:
### 1. **Initial State of the Wire:**
- Let the original radius of the wire be \( R_0 \).
- Let the original length of the wire be \( L_0 \).
- The volume of the wire in the initial state is given by:
\[
V_0 = \pi R_0^2 L_0
\]
### 2. **Stretched State of the Wire:**
- After the wire is stretched, its new radius is \( R_f = R_0^2 \) (as given in the problem).
- Let the new length of the wire be \( L_f \).
- The volume of the wire in the stretched state is:
\[
V_f = \pi R_f^2 L_f
\]
### 3. **Conservation of Volume:**
Since the volume remains the same before and after stretching, we can set the initial and final volumes equal to each other:
\[
V_0 = V_f
\]
Substituting the volume expressions, we get:
\[
\pi R_0^2 L_0 = \pi R_f^2 L_f
\]
Simplifying the equation:
\[
R_0^2 L_0 = R_f^2 L_f
\]
### 4. **Substituting \( R_f = R_0^2 \):**
Now, substitute \( R_f = R_0^2 \) into the equation:
\[
R_0^2 L_0 = (R_0^2)^2 L_f
\]
Simplifying further:
\[
R_0^2 L_0 = R_0^4 L_f
\]
### 5. **Solving for \( L_f \):**
Divide both sides by \( R_0^4 \):
\[
L_f = \frac{L_0}{R_0^2}
\]
### Conclusion:
When the wire is stretched such that its radius becomes \( R_0^2 \), the new length of the wire \( L_f \) will be \( \frac{L_0}{R_0^2} \).
In simpler terms:
- The radius decreases significantly (from \( R_0 \) to \( R_0^2 \)).
- The length of the wire increases correspondingly to conserve volume, becoming much longer by a factor of \( \frac{1}{R_0^2} \).
Let’s analyze the situation step by step:
### 1. **Initial State of the Wire:**
- Let the original radius of the wire be \( R_0 \).
- Let the original length of the wire be \( L_0 \).
- The volume of the wire in the initial state is given by:
\[
V_0 = \pi R_0^2 L_0
\]
### 2. **Stretched State of the Wire:**
- After the wire is stretched, its new radius is \( R_f = R_0^2 \) (as given in the problem).
- Let the new length of the wire be \( L_f \).
- The volume of the wire in the stretched state is:
\[
V_f = \pi R_f^2 L_f
\]
### 3. **Conservation of Volume:**
Since the volume remains the same before and after stretching, we can set the initial and final volumes equal to each other:
\[
V_0 = V_f
\]
Substituting the volume expressions, we get:
\[
\pi R_0^2 L_0 = \pi R_f^2 L_f
\]
Simplifying the equation:
\[
R_0^2 L_0 = R_f^2 L_f
\]
### 4. **Substituting \( R_f = R_0^2 \):**
Now, substitute \( R_f = R_0^2 \) into the equation:
\[
R_0^2 L_0 = (R_0^2)^2 L_f
\]
Simplifying further:
\[
R_0^2 L_0 = R_0^4 L_f
\]
### 5. **Solving for \( L_f \):**
Divide both sides by \( R_0^4 \):
\[
L_f = \frac{L_0}{R_0^2}
\]
### Conclusion:
When the wire is stretched such that its radius becomes \( R_0^2 \), the new length of the wire \( L_f \) will be \( \frac{L_0}{R_0^2} \).
In simpler terms:
- The radius decreases significantly (from \( R_0 \) to \( R_0^2 \)).
- The length of the wire increases correspondingly to conserve volume, becoming much longer by a factor of \( \frac{1}{R_0^2} \).
When a wire is stretched, its physical properties change in a way that's governed by several principles from materials science and mechanics. If the radius of a wire becomes \( R^2 \) when it is stretched, this indicates a transformation in the cross-sectional area of the wire, which in turn affects its mechanical and electrical properties.
Here’s a detailed breakdown of what happens when a wire is stretched and its radius changes:
### 1. **Understanding the Stretching Process**
When a wire is stretched, its length increases, and its cross-sectional area typically decreases. This is due to the conservation of volume, assuming the wire’s material is incompressible. The fundamental relationship for this process can be understood using the concept of volume conservation:
- **Initial Volume (V₀):** For a wire with initial radius \( r_0 \) and initial length \( L_0 \), the initial volume is given by:
\[
V_0 = \pi r_0^2 L_0
\]
- **Stretched Volume (V):** When the wire is stretched to a new length \( L \) and its new radius is \( r \), the new volume is:
\[
V = \pi r^2 L
\]
Since the volume is conserved (assuming no volume loss), we have:
\[
\pi r_0^2 L_0 = \pi r^2 L
\]
Simplifying, we get:
\[
r^2 = r_0^2 \frac{L_0}{L}
\]
### 2. **Implications of Changing Radius**
Given that the radius of the wire becomes \( R^2 \), it implies that the final radius \( r \) is such that:
\[
r = R^2
\]
From the volume conservation equation, substituting \( r = R^2 \), we have:
\[
(R^2)^2 = r_0^2 \frac{L_0}{L}
\]
which simplifies to:
\[
R^4 = r_0^2 \frac{L_0}{L}
\]
This equation can be used to determine how the length \( L \) relates to the initial dimensions and the radius transformation.
### 3. **Mechanical Properties**
When the wire is stretched, the following mechanical properties are affected:
- **Stress and Strain:** The tensile stress \( \sigma \) in the wire is given by:
\[
\sigma = \frac{F}{A}
\]
where \( F \) is the applied force, and \( A \) is the cross-sectional area of the wire. As the cross-sectional area decreases (with \( A \propto r^2 \)), the stress in the wire increases.
- **Young’s Modulus (E):** This is a measure of the stiffness of the material and is given by:
\[
E = \frac{\sigma}{\epsilon}
\]
where \( \epsilon \) is the strain (relative change in length). The relationship between stress and strain remains linear for elastic deformation according to Hooke’s Law, provided the material remains within its elastic limit.
### 4. **Electrical Properties**
For electrical applications, stretching a wire changes its resistance \( R \), which 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.
As the wire stretches and its radius becomes \( R^2 \), the new cross-sectional area \( A \) is \( \pi (R^2)^2 \). This means:
\[
A = \pi R^4
\]
Substituting this into the resistance formula:
\[
R = \rho \frac{L}{\pi R^4}
\]
Thus, stretching the wire increases its resistance if the length increases and the cross-sectional area decreases.
### Summary
- When a wire is stretched and its radius becomes \( R^2 \), its volume is conserved if the material is incompressible.
- The new cross-sectional area of the wire becomes \( \pi (R^2)^2 \).
- Mechanical properties like stress and Young’s Modulus are affected.
- Electrical properties such as resistance increase as the cross-sectional area decreases.
Understanding these changes is crucial for designing and analyzing materials in various applications, from mechanical engineering to electrical circuit design.
Here’s a detailed breakdown of what happens when a wire is stretched and its radius changes:
### 1. **Understanding the Stretching Process**
When a wire is stretched, its length increases, and its cross-sectional area typically decreases. This is due to the conservation of volume, assuming the wire’s material is incompressible. The fundamental relationship for this process can be understood using the concept of volume conservation:
- **Initial Volume (V₀):** For a wire with initial radius \( r_0 \) and initial length \( L_0 \), the initial volume is given by:
\[
V_0 = \pi r_0^2 L_0
\]
- **Stretched Volume (V):** When the wire is stretched to a new length \( L \) and its new radius is \( r \), the new volume is:
\[
V = \pi r^2 L
\]
Since the volume is conserved (assuming no volume loss), we have:
\[
\pi r_0^2 L_0 = \pi r^2 L
\]
Simplifying, we get:
\[
r^2 = r_0^2 \frac{L_0}{L}
\]
### 2. **Implications of Changing Radius**
Given that the radius of the wire becomes \( R^2 \), it implies that the final radius \( r \) is such that:
\[
r = R^2
\]
From the volume conservation equation, substituting \( r = R^2 \), we have:
\[
(R^2)^2 = r_0^2 \frac{L_0}{L}
\]
which simplifies to:
\[
R^4 = r_0^2 \frac{L_0}{L}
\]
This equation can be used to determine how the length \( L \) relates to the initial dimensions and the radius transformation.
### 3. **Mechanical Properties**
When the wire is stretched, the following mechanical properties are affected:
- **Stress and Strain:** The tensile stress \( \sigma \) in the wire is given by:
\[
\sigma = \frac{F}{A}
\]
where \( F \) is the applied force, and \( A \) is the cross-sectional area of the wire. As the cross-sectional area decreases (with \( A \propto r^2 \)), the stress in the wire increases.
- **Young’s Modulus (E):** This is a measure of the stiffness of the material and is given by:
\[
E = \frac{\sigma}{\epsilon}
\]
where \( \epsilon \) is the strain (relative change in length). The relationship between stress and strain remains linear for elastic deformation according to Hooke’s Law, provided the material remains within its elastic limit.
### 4. **Electrical Properties**
For electrical applications, stretching a wire changes its resistance \( R \), which 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.
As the wire stretches and its radius becomes \( R^2 \), the new cross-sectional area \( A \) is \( \pi (R^2)^2 \). This means:
\[
A = \pi R^4
\]
Substituting this into the resistance formula:
\[
R = \rho \frac{L}{\pi R^4}
\]
Thus, stretching the wire increases its resistance if the length increases and the cross-sectional area decreases.
### Summary
- When a wire is stretched and its radius becomes \( R^2 \), its volume is conserved if the material is incompressible.
- The new cross-sectional area of the wire becomes \( \pi (R^2)^2 \).
- Mechanical properties like stress and Young’s Modulus are affected.
- Electrical properties such as resistance increase as the cross-sectional area decreases.
Understanding these changes is crucial for designing and analyzing materials in various applications, from mechanical engineering to electrical circuit design.