Question

The resistance of a wire is 16 ohms. By melting it, the wire is stretched to half of its original length. What will be the resistance of the new wire?

Hint:

When the length of a wire is changed, its resistance changes quadratically due to the change in area and length.

Answer 1

When the wire is stretched, its length is halved. The resistance \( R \) of a wire is given by: \[ R = \rho \frac{L}{A}, \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire. - When the wire is stretched, the volume remains constant, so \( L_1 A_1 = L_2 A_2 \), where the subscript 1 and 2 represent the initial and final conditions. The final resistance is given by: \[ R_2 = R_1 \left( \frac{L_2}{L_1} \right)^2. \] Since the length is halved, the final resistance will be: \[ R_2 = 16 \left( \frac{1}{2} \right)^2 = 16 \times \frac{1}{4} = 4 \, \Omega. \]