Question

What is the new resistance of a wire of resistance R is stretched to thrice its length?

Answer 1

When a wire is stretched, its resistance changes due to changes in both its length and cross-sectional area. Let's break down how to determine the new resistance of a wire when its length is increased.

### Basic Concept

1. **Original Resistance**: The resistance \( R \) of a wire can be described by the formula:
   \[
   R = \rho \frac{L}{A}
   \]
   where:
   - \( \rho \) is the resistivity of the material,
   - \( L \) is the original length of the wire,
   - \( A \) is the original cross-sectional area of the wire.

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

3. **Volume Conservation**: The volume of the wire remains constant during stretching. Therefore, the original volume \( V \) of the wire, given by:
   \[
   V = A \cdot L
   \]
   must equal the volume after stretching. The new cross-sectional area \( A' \) and new length \( L' \) give:
   \[
   V = A' \cdot L'
   \]
   Substituting \( L' = 3L \):
   \[
   A \cdot L = A' \cdot 3L
   \]
   Solving for \( A' \):
   \[
   A' = \frac{A}{3}
   \]

4. **New Resistance**: The new resistance \( R' \) can be calculated using the new length \( L' \) and the new cross-sectional area \( A' \):
   \[
   R' = \rho \frac{L'}{A'}
   \]
   Substitute \( L' = 3L \) and \( A' = \frac{A}{3} \):
   \[
   R' = \rho \frac{3L}{\frac{A}{3}} = \rho \frac{3L \cdot 3}{A} = 9 \left(\rho \frac{L}{A}\right)
   \]
   \[
   R' = 9R
   \]

### Summary

When a wire is stretched to three times its original length, its resistance increases by a factor of 9. Thus, the new resistance of the wire is \( 9R \), where \( R \) is the original resistance.

Answer 2

When a wire is stretched to three times its original length, its resistance increases. The resistance \( R \) of a wire is given by the formula \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

When the wire is stretched to three times its original length, the new length \( L' \) is \( 3L \). The volume of the wire remains constant, so the new cross-sectional area \( A' \) can be found using the fact that \( A \cdot L = A' \cdot L' \). Substituting \( L' = 3L \) gives \( A' = \frac{A}{3} \).

The new resistance \( R' \) is then calculated as:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{3L}{\frac{A}{3}} = 9 \rho \frac{L}{A} = 9R \]

So, the new resistance is 9 times the original resistance.