A certain wire has length L and radius r, and a resistance of R. That wire is melted down and the same volume of the same metal is used to produce a wire with length 2L. What is the resistance of the resulting wire?

• A. $R/4$
• B. $R/2$
• C. 8R
• D. R
• E. 4R
• F. 2R

Answer: (F)

Verified step by step guidance

1. Start by recalling the formula for the resistance of a wire: \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
2. Since the wire is melted and reformed, the volume of the wire remains constant. The volume \( V \) of the original wire is \( V = A \cdot L = \pi r^2 L \).
3. The new wire has a length of \( 2L \). Let the new radius be \( r' \). The volume of the new wire is \( V = \pi (r')^2 (2L) \). Set the volumes equal: \( \pi r^2 L = \pi (r')^2 (2L) \).
4. Solve for the new radius \( r' \): \( r' = \frac{r}{\sqrt{2}} \).
5. Substitute the new length and area into the resistance formula: \( R' = \frac{\rho (2L)}{\pi (r')^2} = \frac{\rho (2L)}{\pi \left(\frac{r}{\sqrt{2}}\right)^2} = \frac{\rho (2L)}{\pi \frac{r^2}{2}} = \frac{2 \cdot \rho L}{\pi r^2} \cdot 2 = 2R \). Thus, the resistance of the new wire is \( 2R \).