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: Note how R depends on dimensions.
Resistance is \(R = \rho L/A\). When a wire is stretched, both \(L\) (increases) and \(A\) (decreases) change, while resistivity \(\rho\) and the volume of metal stay fixed.

Step 2: Write R using volume instead of area.
Since volume \(V = A L\), we have \(A = V/L\). Substituting,
\(R = \rho \dfrac{L}{V/L} = \dfrac{\rho L^2}{V}\).
This shows that at constant volume, \(R \propto L^2\).

Step 3: Apply the length change.
The new length is \(L' = nL\), so the new resistance is
\(\dfrac{R'}{R} = \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{nL}{L}\right)^2 = n^2\).

Step 4: Result.
Therefore \(R' = n^2 R\); for example, doubling the length (\(n=2\)) makes the resistance four times as large.
\[\boxed{R' = n^2 R}\]