Question

Resistance of a wire is \( R \). If the length of the wire is stretched by \( n \) times the original length, then what will be the new resistance of the wire?

Hint:

Volume of the wire stays constant on stretching, so R is proportional to length squared; increasing length n times raises resistance by n squared.

Answer 1

Step 1: Write the resistance formula.
For a wire of resistivity \( \rho \), length \( L \), and area of cross-section \( A \),
\[ R = \rho\frac{L}{A} \]

Step 2: Apply the constant-volume condition.
Stretching only changes the shape, not the amount of material, so the volume stays constant. Original volume \( = LA \).
New length \( L' = nL \). If new area is \( A' \), then
\[ L'A' = LA \;\Rightarrow\; (nL)A' = LA \;\Rightarrow\; A' = \frac{A}{n} \]

Step 3: New resistance.
\[ R' = \rho\frac{L'}{A'} = \rho\frac{nL}{A/n} = \rho\frac{n^2 L}{A} \]

Step 4: Express in terms of R.
Since \( R = \rho L/A \),
\[ R' = n^2 \left(\rho\frac{L}{A}\right) = n^2 R \]

So stretching the wire to \( n \) times its length makes its resistance \( n^2 \) times the original.
\[\boxed{R' = n^2 R}\]