Question

A metallic wire of length 1m is stretched to double its length. Calculate the ratio of its initial and final resistance assuming that there is no change in its density on stretching

Answer 1

Originally the resistance is 'r',
length is 'l',
area of cross section = A
Resistance (r) = ρ× l /A
If it is stretched, length = 2l
As,there is no change in density,so volume will also not changed.Hence area remains constant.
New Resistance,r' = ρ×2l
​ A
= 2ρl
​ A
= 2r
Hence, new resistance r'= 2r
Ratio of initial resistance(r) to final resistance (r')=r/r'= 1/2