College

An 8.00-meter-long length of wire has a resistance of 4.00 Ω. The wire is uniformly stretched to a length of 16.0 meters. Find the resistance of the wire after it has been stretched.

Answer :

Final answer:

The resistance of an 8.00-m-long wire that is stretched to 16.0 m increases by a factor of four, resulting in a new resistance of 16.00 Ω.

Explanation:

When an 8.00-m-long wire with a resistance of 4.00 Ω is uniformly stretched to a length of 16.0 m, we can find the new resistance by recognizing that the resistance R of a wire is proportional to its length L and inversely proportional to its cross-sectional area A, given by the formula R = ρ(L/A), where ρ is the resistivity of the material. Since volume remains constant when stretching the wire, and the volume V is equal to A × L, as the wire is stretched to twice its length (L' = 2L), its cross-sectional area (A') must become one-half the original (A/2). Therefore, the new resistance R' will be

R' = ρ(L'/A') = ρ(2L/(A/2)) = ρ(4L/A) = 4R

Thus, the resistance increases by a factor of four. In this case, the resistance of the wire after being stretched becomes 16.00 Ω.