1.

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

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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