Question

A wire of resistance \(R\) is stretched to double its original length. What is its new resistance?

• A. \(2R\)
• B. \(4R\)
• C. \(\frac{R}{2}\)
• D. \(R\)

Hint:

If a wire is stretched to \(n\) times its original length while volume remains constant, the new resistance becomes \(n^2 R\).

Answer 1

The Correct Option is B

Concept: The resistance of a wire is given by the relation \[ R = \rho \frac{L}{A} \] where
• \( \rho \) = resistivity of the material (constant for a given material),
• \( L \) = length of the wire,
• \( A \) = cross-sectional area. When a wire is stretched, its volume remains constant (assuming uniform stretching). Since \[ \text{Volume} = A \times L \] if the length increases, the cross-sectional area must decrease proportionally.

Step 1: Apply the constant volume condition. Let the original length be \(L\) and the original area be \(A\). If the wire is stretched to double its length, then \[ L' = 2L \] Since the volume remains constant, \[ AL = A' L' \] \[ AL = A'(2L) \] \[ A' = \frac{A}{2} \] Thus, the new cross-sectional area becomes half.

Step 2: Substitute into the resistance formula. The new resistance is \[ R' = \rho \frac{L'}{A'} \] Substituting \(L' = 2L\) and \(A' = \frac{A}{2}\): \[ R' = \rho \frac{2L}{A/2} \] \[ R' = \rho \times \frac{2L \times 2}{A} \] \[ R' = 4\rho \frac{L}{A} \] \[ R' = 4R \] Thus, the new resistance becomes \[ \boxed{4R} \]