Question

How many times resistance will increase if the length of a conducting wire becomes twice when stretched?

Answer 1

When a conducting wire is stretched, its length increases and its cross-sectional area decreases, which affects its resistance. Let's analyze how the resistance changes if the length of the wire is doubled.

1. **Resistance Formula**: The resistance \( R \) of a wire is given by:
   \[
   R = \rho \frac{L}{A}
   \]
   where:
   - \( \rho \) is the resistivity of the material (which remains constant if the material does not change),
   - \( L \) is the length of the wire,
   - \( A \) is the cross-sectional area of the wire.

2. **Initial Situation**: Let \( L_0 \) be the initial length and \( A_0 \) be the initial cross-sectional area. The initial resistance \( R_0 \) is:
   \[
   R_0 = \rho \frac{L_0}{A_0}
   \]

3. **After Stretching**: When the wire is stretched to twice its original length (\( L = 2L_0 \)), the volume of the wire remains constant. Therefore:
   \[
   L_0 \times A_0 = 2L_0 \times A
   \]
   where \( A \) is the new cross-sectional area. Solving for \( A \):
   \[
   A = \frac{A_0}{2}
   \]

4. **New Resistance**: The new resistance \( R \) with the new length and cross-sectional area is:
   \[
   R = \rho \frac{2L_0}{A} = \rho \frac{2L_0}{\frac{A_0}{2}} = \rho \frac{2L_0 \times 2}{A_0} = 4 \rho \frac{L_0}{A_0}
   \]
   So:
   \[
   R = 4 R_0
   \]

Therefore, if the length of a conducting wire is doubled by stretching, its resistance will increase by a factor of 4.

Answer 2

To determine how the resistance of a conducting wire changes when its length is doubled, we need to use the formula for resistance in a wire, which is:

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

where:
- \( R \) is the resistance of the wire,
- \( \rho \) is the resistivity of the material (a constant for a given material at a given temperature),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Initial Conditions

Let's denote the initial length of the wire as \( L_0 \) and its initial cross-sectional area as \( A_0 \). The initial resistance of the wire \( R_0 \) is:

\[ R_0 = \rho \frac{L_0}{A_0} \]

### When the Length is Doubled

When the wire is stretched so that its length becomes twice the original length, i.e., \( L = 2L_0 \), we need to determine how the cross-sectional area changes. Since the volume of the wire remains constant during stretching, we can use the fact that:

\[ L_0 A_0 = L A \]

Substituting \( L = 2L_0 \) into this equation:

\[ L_0 A_0 = 2L_0 A \]

Solving for \( A \):

\[ A = \frac{A_0}{2} \]

So, the cross-sectional area of the wire becomes half of its original area.

### New Resistance

Now, let’s calculate the new resistance \( R \) with the updated length and area:

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

Substituting \( L = 2L_0 \) and \( A = \frac{A_0}{2} \):

\[ R = \rho \frac{2L_0}{\frac{A_0}{2}} \]

Simplify the expression:

\[ R = \rho \frac{2L_0 \cdot 2}{A_0} = \rho \frac{4L_0}{A_0} \]

We know that \( R_0 = \rho \frac{L_0}{A_0} \). Therefore:

\[ R = 4R_0 \]

### Conclusion

The resistance of the wire increases by a factor of 4 when the length of the wire is doubled and its cross-sectional area is halved due to stretching.