Question

A wire of length 2 m and resistance 8 \(\Omega\) is stretched to double its original length, keeping the volume constant. What is the new resistance of the wire?

• A. \( 16 \, \Omega \)
• B. \( 32 \, \Omega \)
• C. \( 8 \, \Omega \)
• D. \( 4 \, \Omega \)

Hint:

When a wire is stretched with constant volume, the resistance changes by the square of the factor by which the length changes. If the length is doubled, the resistance increases by \( 2^2 = 4 \) times.

Answer 1

The Correct Option is B

The resistance of a wire is given by the formula: \[ R = \rho \frac{l}{A} \] Where: - \( \rho \) is the resistivity of the material, - \( l \) is the length of the wire, - \( A \) is the cross-sectional area. Initially, the wire has: - Length \( l = 2 \, \text{m} \), - Resistance \( R = 8 \, \Omega \). When the wire is stretched to double its original length, the new length is: \[ l' = 2 \times 2 = 4 \, \text{m} \] Since the volume of the wire remains constant during stretching, the volume before and after stretching is equal: \[ V = A \cdot l = A' \cdot l' \] \[ A \cdot 2 = A' \cdot 4 \] \[ A' = \frac{A}{2} \] The new cross-sectional area is half the original area. The new resistance \( R' \) is given by: \[ R' = \rho \frac{l'}{A'} \] Substitute \( l' = 4 \, \text{m} \) and \( A' = \frac{A}{2} \): \[ R' = \rho \frac{4}{\frac{A}{2}} = \rho \frac{4 \cdot 2}{A} = \rho \frac{8}{A} \] Since the original resistance is \( R = \rho \frac{2}{A} = 8 \, \Omega \), we can express the new resistance in terms of the original resistance: \[ R' = \rho \frac{8}{A} = 4 \cdot \rho \frac{2}{A} = 4 \cdot R \] \[ R' = 4 \cdot 8 = 32 \, \Omega \] Alternatively, since resistance is proportional to \( \frac{l}{A} \), and \( l \) doubles while \( A \) halves, the resistance changes by: \[ R' = R \cdot \frac{l'}{l} \cdot \frac{A}{A'} = 8 \cdot \frac{4}{2} \cdot \frac{A}{\frac{A}{2}} = 8 \cdot 2 \cdot 2 = 32 \, \Omega \] Thus, the new resistance of the wire is \( 32 \, \Omega \).