By stretching if the radius of the wire is 1/3 of the initial. Find the ratio of new and old resistance
Answers
Answer:
Designate the original end-end resistance of the wire as R. Assuming the wire is uniformly stretched to twice its original length, then in order to maintain constant volume (which for a uniform cylinder, is equal to length times cross-sectional area, the cross sectional area of the wire will decrease by a factor of two (i.e., its final cross-sectional area will be one-half of its original value). Assuming uniform chemical composition is maintained, the wire’s end-end resistance will be proportional to cross-sectional area, then the resistance will increase by a factor of two. However, the stretching has also increased the length by 2x, and resistance varies directly with length. Thus the combined effect of stretching the wire to 2x its original length, while simultaneously decreasing its cross-sectional area by 2x, will cause its final resistance to become R٠2٠2 = 4R ohms.
Explanation: