Let R1 be the initial resistance of the wire in the left gap. R2 be the resistance in the right gap l1 be the initial balancing length ( 40 cm ) Let l2 be the remaining length ( 100−40=60cm ) The initial ratio is ‌
R1
R2
=‌
40
60
=‌
2
3
When the wire is stretched to double its length, its resistance becomes 4R1 Let l1 " be the new balancing length l2 'be the new remaining length (100- l1 ) The new ratio is ‌