Question

What is a wire of resistance R is melted and stretched to N times?

Answer 1

When a wire of resistance \( R \) is melted and stretched to \( N \) times its original length, its resistance changes due to both its increased length and decreased cross-sectional area.

### Step-by-step Explanation:

1. **Original Resistance**:
   The resistance 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,
   - \( L \) is the length of the wire,
   - \( A \) is the cross-sectional area of the wire.

2. **Stretching the Wire**:
   When the wire is stretched to \( N \) times its original length, its volume remains constant. Since the volume of the wire is given by \( V = L \cdot A \), stretching it means the new length \( L' = N \cdot L \).

   To conserve the volume, the new cross-sectional area \( A' \) will be reduced. Since volume is constant:

   \[
   L \cdot A = L' \cdot A' \quad \Rightarrow \quad A' = \frac{A}{N}
   \]

3. **New Resistance**:
   The resistance of the stretched wire is now:

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

   Substituting \( L' = N \cdot L \) and \( A' = \frac{A}{N} \), we get:

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

### Final Result:
The new resistance \( R' \) after the wire is stretched to \( N \) times its original length is:

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

So, the resistance increases by a factor of \( N^2 \) when the wire is stretched \( N \) times its original length.