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The length of a given cylindrical wire is increased by 100 %.Due to the consequent decrease in diameter the change inthe resistance of the wire will be |
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Answer» The total volume remains the same before and after stretching. Therefore A xl = A’ x l’ Here l’ = 2 l A’ = A x l / l’ = A x l / 2 l = A/2 Percentage change in resistance = Rf– Ri/ Rlx 100 = ? l’/A’ - l/A / l ? l/A x 100 = [ (l’/A’ X A/l) – 1 ] x 100 [(2 l/A/2 x A / l)-1 ] x 100 = 300% |
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