Question

When A wire is stretched and its radius becomes R^2?

Answer 1

When a wire is stretched, its dimensions change, including its radius. The relationship between the original and final dimensions of the wire can be understood by considering the conservation of volume and how stretching affects the physical properties of the wire. Here's a detailed breakdown:

### Assumptions and Basic Principles

1. **Volume Conservation**: Assuming the wire is made of a material that does not change its volume when stretched, the volume of the wire before and after stretching will remain constant.

2. **Wire Geometry**: Let's assume the wire is cylindrical. The volume of a cylindrical wire is given by:
   \[
   V = \pi r^2 L
   \]
   where \( r \) is the radius of the wire, and \( L \) is its length.

### Stretching the Wire

1. **Initial Dimensions**:
   - Radius: \( r_0 \)
   - Length: \( L_0 \)
   - Volume: \( V_0 = \pi r_0^2 L_0 \)

2. **Stretched Dimensions**:
   - New radius: \( r \)
   - New length: \( L \)
   - Volume: \( V = \pi r^2 L \)

Since volume is conserved:
   \[
   \pi r_0^2 L_0 = \pi r^2 L
   \]
   \[
   r_0^2 L_0 = r^2 L
   \]

3. **Radius Becomes \( R^2 \)**:
   If the radius of the wire changes to \( R^2 \) (where \( R \) is a new variable representing some proportional change), you would substitute \( r = R^2 \) into the volume conservation equation:

   \[
   r_0^2 L_0 = (R^2)^2 L
   \]
   \[
   r_0^2 L_0 = R^4 L
   \]
   Solving for the new length \( L \):
   \[
   L = \frac{r_0^2 L_0}{R^4}
   \]

### Interpretation

- **Radius Increase**: If the radius increases, the length of the wire must decrease to keep the volume constant.
- **Radius as \( R^2 \)**: If \( r \) becomes \( R^2 \), the length must be adjusted accordingly based on the initial radius and length of the wire.

In summary, when the wire’s radius changes to \( R^2 \), the length of the wire changes inversely with the fourth power of \( R \). The exact new length \( L \) can be calculated using the formula derived above.

Answer 2

When a wire is stretched, its physical dimensions change, which can affect its properties. If we specifically focus on the radius of the wire and how it changes when the wire is stretched, here's what happens:

1. **Initial Assumptions**:
   - **Original Radius**: Let's denote the initial radius of the wire as \( r_0 \).
   - **Original Length**: Assume the initial length of the wire is \( L_0 \).
   - **New Radius**: When stretched, the new radius becomes \( r \), which is given to be \( r = r_0^2 \).
   - **New Length**: Let the new length of the wire be \( L \).

2. **Volume Conservation**:
   - For an incompressible material, the volume of the wire is conserved. Therefore, the initial volume (\( V_0 \)) is equal to the final volume (\( V \)).
   - The volume of the wire is given by \( V = \pi r^2 L \).
   - Initially: \( V_0 = \pi r_0^2 L_0 \).
   - After stretching: \( V = \pi r^2 L \).

   Since \( V_0 = V \):
   \[
   \pi r_0^2 L_0 = \pi r^2 L
   \]
   Simplifying, we get:
   \[
   r_0^2 L_0 = r^2 L
   \]

3. **New Length Calculation**:
   - Given \( r = r_0^2 \), substitute this into the equation:
   \[
   r_0^2 L_0 = (r_0^2)^2 L
   \]
   \[
   r_0^2 L_0 = r_0^4 L
   \]
   Solving for \( L \):
   \[
   L = \frac{r_0^2 L_0}{r_0^4} = \frac{L_0}{r_0^2}
   \]

4. **Summary**:
   - When the wire is stretched and its radius becomes \( r_0^2 \), the new length of the wire \( L \) is \( \frac{L_0}{r_0^2} \).

This result shows that stretching the wire and changing its radius affects its length in such a way that the volume remains constant, assuming the wire is incompressible.