Question

How will the resistivity of a wire change if it is stretched to double its original length without loss of mass?

Answer 1

When a wire is stretched to double its original length without a loss of mass, its **resistivity** remains the same. However, the **resistance** of the wire changes. Let's break it down:

### Resistivity (ρ):
Resistivity is an intrinsic property of the material and does not depend on the wire's dimensions (length or cross-sectional area). It only depends on the material itself, temperature, and other intrinsic factors. Hence, when the wire is stretched to double its length, the **resistivity (ρ)** remains constant.

### Resistance (R):
The resistance of the wire is given by the formula:

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

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

If the wire is stretched to **double its length**:
1. **Length (L):** The new length becomes \( 2L \).
2. **Cross-sectional area (A):** The volume of the wire remains constant (since there is no loss of mass). The volume \( V \) is given by \( V = A \times L \). When the length doubles, the cross-sectional area decreases to maintain the same volume. The new cross-sectional area \( A' \) becomes \( \frac{A}{2} \).

Now, substituting these into the resistance formula:

\[
R' = \rho \frac{2L}{A/2} = 4 \times \left( \rho \frac{L}{A} \right) = 4R
\]

### Conclusion:
- **Resistivity** of the wire remains unchanged.
- **Resistance** of the wire increases by a factor of 4.

Answer 2

If a wire is stretched to double its original length without losing any mass, its resistivity will remain unchanged.

Here's why: Resistivity (\(\rho\)) is an intrinsic property of a material and depends only on the material's composition and temperature, not on its dimensions or shape. However, stretching the wire affects its dimensions:

- **Original length** = \(L\)
- **Original cross-sectional area** = \(A\)

When stretched to double its length (\(2L\)) while keeping the mass constant, the volume of the wire remains the same. Since volume \(V\) is given by \(V = A \times L\), stretching the wire to double its length means the new cross-sectional area \(A'\) must decrease to maintain the same volume. Specifically:

\[ \text{Volume before stretching} = A \times L \]
\[ \text{Volume after stretching} = A' \times (2L) \]
\[ A \times L = A' \times (2L) \]
\[ A' = \frac{A}{2} \]

The resistivity \(\rho\) of the wire is given by:

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

where \(R\) is the resistance. When stretched, the resistance \(R'\) of the wire becomes:

\[ R' = \rho \times \frac{2L}{A'} \]
\[ R' = \rho \times \frac{2L}{\frac{A}{2}} \]
\[ R' = \rho \times \frac{4L}{A} \]

The resistance quadruples because the length doubles and the cross-sectional area halves.

However, the resistivity itself, being a material property, does not change with the dimensions of the wire.