Question

When a metallic wire of resistance R is stretched N times its original length, then its new resistance will be?

Answer 1

When a metallic wire is stretched, its resistance changes due to changes in its length and cross-sectional area. To understand how the resistance changes, let’s break it down step-by-step.

### 1. Original Resistance

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

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

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 the wire is stretched \( N \) times its original length, the new length \( L' \) of the wire becomes:

\[ L' = N \cdot L \]

### 3. Volume Conservation

Assuming the wire is incompressible (no change in volume), the volume before stretching is equal to the volume after stretching. The volume of a wire is:

\[ \text{Volume} = L \cdot A \]

After stretching, the new volume should be:

\[ \text{Volume}' = L' \cdot A' \]

where \( A' \) is the new cross-sectional area. Since the volume remains constant:

\[ L \cdot A = L' \cdot A' \]

Substituting \( L' = N \cdot L \):

\[ L \cdot A = (N \cdot L) \cdot A' \]

Solving for \( A' \):

\[ A' = \frac{A}{N} \]

### 4. New Resistance

Now we need to find the new resistance \( R' \) of the stretched wire. Using the formula for resistance:

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

Substitute \( L' = N \cdot L \) and \( A' = \frac{A}{N} \):

\[ R' = \rho \frac{N \cdot L}{\frac{A}{N}} \]

Simplify:

\[ R' = \rho \frac{N^2 \cdot L}{A} \]

Since \( R = \rho \frac{L}{A} \), we get:

\[ R' = N^2 \cdot R \]

### Summary

When a metallic wire of resistance \( R \) is stretched \( N \) times its original length, its new resistance will be \( N^2 \cdot R \).

Answer 2

When a metallic wire of resistance \( R \) is stretched \( N \) times its original length, its new resistance can be determined using the relationship between resistance, length, and cross-sectional area of the wire.

Here's the step-by-step explanation:

1. **Original Wire:**
   - Let the original length of the wire be \( L \).
   - Let the original cross-sectional area be \( A \).
   - The resistance of the wire is given by:
     \[
     R = \rho \frac{L}{A}
     \]
     where \( \rho \) is the resistivity of the material.

2. **Stretched Wire:**
   - When the wire is stretched \( N \) times its original length, the new length \( L' \) is:
     \[
     L' = N \times L
     \]
   - The volume of the wire remains constant during stretching. Hence, the product of the original cross-sectional area \( A \) and the original length \( L \) is equal to the product of the new cross-sectional area \( A' \) and the new length \( L' \):
     \[
     A \times L = A' \times L'
     \]
     Solving for \( A' \):
     \[
     A' = \frac{A \times L}{L'} = \frac{A}{N}
     \]
     Thus, the new cross-sectional area \( A' \) is \( \frac{A}{N} \).

3. **New Resistance:**
   - The resistance of the stretched wire can be calculated as:
     \[
     R' = \rho \frac{L'}{A'}
     \]
     Substituting \( L' = N \times L \) and \( A' = \frac{A}{N} \):
     \[
     R' = \rho \frac{N \times L}{\frac{A}{N}} = \rho \frac{N^2 \times L}{A}
     \]
     Since \( R = \rho \frac{L}{A} \), we can substitute:
     \[
     R' = N^2 \times R
     \]

So, the new resistance \( R' \) of the wire when stretched \( N \) times its original length is \( N^2 \) times the original resistance \( R \).