Question

A wire of resistance 160 Ω is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

• A. 10 Ω
• B. 16 Ω
• C. 40 Ω
• D. 640 Ω

Answer 1

The Correct Option is A

Given that the volume of the wire remains constant, we use the relation: \[ A_1 L_1 = A_2 L_2 \] Where: - \( A_1 \) and \( L_1 \) are the area and length of the original wire, - \( A_2 \) and \( L_2 \) are the area and length of the new wire. The volume of the wire is constant, so the area of cross-section \( A_2 \) and the length \( L_2 \) change according to the new dimensions. Since the length of the new wire is one-fourth of the original, we have: \[ A_1 L_1 = A_2 L_2 \quad \Rightarrow \quad A_2 = 4 A_1 \] For resistance \( R \), we know: \[ R = \rho \frac{L}{A} \] Thus, for the new wire: \[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{L/4}{4A} = \rho \frac{L}{16A} = \frac{1}{16} R_1 \] Substituting \( R_1 = 160 \Omega \): \[ R_2 = \frac{1}{16} \times 160 = 10 \Omega \] Thus, the new resistance is \( \boxed{10} \Omega \).