Question

What is the change in resistance when a wire is stretched 3 times its length?

Answer 1

When a wire is stretched to three times its original length, its resistance increases. This is due to changes in both the length and the cross-sectional area of the wire.

Here's a step-by-step explanation:

1. **Initial Resistance Formula:**
   The resistance \( R \) of a wire is given by:
   \[
   R = \frac{\rho L}{A}
   \]
   where:
   - \( \rho \) is the resistivity of the material
   - \( L \) is the length of the wire
   - \( A \) is the cross-sectional area of the wire

2. **Stretching the Wire:**
   When you stretch the wire to three times its original length, the new length \( L' \) is:
   \[
   L' = 3L
   \]

3. **Change in Cross-Sectional Area:**
   Stretching a wire not only increases its length but also decreases its cross-sectional area. If the wire is stretched uniformly, the volume of the wire remains constant. Therefore:
   \[
   A \times L = A' \times L'
   \]
   where \( A' \) is the new cross-sectional area. Substituting \( L' = 3L \):
   \[
   A \times L = A' \times 3L
   \]
   Solving for \( A' \):
   \[
   A' = \frac{A}{3}
   \]

4. **New Resistance:**
   The new resistance \( R' \) with the stretched wire is:
   \[
   R' = \frac{\rho L'}{A'}
   \]
   Substituting \( L' = 3L \) and \( A' = \frac{A}{3} \):
   \[
   R' = \frac{\rho \times 3L}{\frac{A}{3}}
   \]
   Simplify this:
   \[
   R' = \frac{3 \rho L}{\frac{A}{3}} = \frac{3 \times 3 \rho L}{A} = 9 \frac{\rho L}{A}
   \]
   So:
   \[
   R' = 9R
   \]

Thus, when the wire is stretched to three times its length, its resistance increases by a factor of 9.

Answer 2

When a wire is stretched to three times its original length, the resistance of the wire changes significantly due to the relationship between resistance, length, and cross-sectional area.

The resistance \(R\) of a wire is given by the formula:

\[
R = \rho \frac{L}{A}
\]

Where:
- \(R\) is the resistance,
- \(\rho\) 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 of the wire.

### Stretching the Wire:
1. **Length**: If the wire is stretched to three times its original length, then the new length \(L_{\text{new}} = 3L\).
  
2. **Cross-sectional Area**: When a wire is stretched, its volume remains constant (assuming no material is lost or added). The volume of the wire is given by:

   \[
   V = L \times A
   \]

   So, for the stretched wire:

   \[
   V_{\text{new}} = L_{\text{new}} \times A_{\text{new}} = \text{constant}
   \]

   Since \(L_{\text{new}} = 3L\), the new cross-sectional area \(A_{\text{new}}\) becomes:

   \[
   A_{\text{new}} = \frac{A}{3}
   \]

3. **Resistance Change**: The new resistance \(R_{\text{new}}\) after stretching becomes:

   \[
   R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}} = \rho \frac{3L}{\frac{A}{3}} = \rho \frac{3L \times 3}{A} = 9 \times \rho \frac{L}{A}
   \]

   Therefore, the new resistance is **9 times the original resistance**.

### Conclusion:
When a wire is stretched to three times its original length, its resistance increases by a factor of 9.