Final answer:
The percentage change in resistance when an aluminum wire is stretched to make its length 0.4% larger is 1.2%. This increase is due to the combined effects of the increased length and the decreased cross-sectional area. When stretched to four times its length, the resistance increases by a factor of 16.
Step-by-step explanation:
The question involves examining the relationship between the change in the length of a wire and the resultant change in its electrical resistance. When an aluminum wire is stretched, its length increases while its cross-sectional area decreases. According to the formula for resistance (R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area), both a change in length (L) and a change in cross-sectional area (A) affect the overall resistance of the wire.
As a wire is stretched and its length increases by a certain percentage, its cross-sectional area must decrease to conserve volume (assuming constant volume). If a wire's length increases by 0.4%, not only does the resistance increase due to the longer length but it also increases further due to the decreased cross-sectional area. For a small percentage stretch, the percentage change in resistance is approximately the sum of the percent change in length and twice the percentage change in the area (since the area is proportional to the square of the diameter), making the answer 1.2% (choice c).
If the wire is stretched to four times its original length, the resistance increase can be found by considering the same principles. The volume remains constant, so the cross-sectional area must decrease. If the length becomes four times greater, the cross-sectional area is one-fourth of the original area. Resistance is directly proportional to the length and inversely proportional to the cross-sectional area. Therefore, when the wire is stretched to four times its length, its resistance increases by a factor of 16 (four times due to length and four times due to area change).