Question

A given wire of resistance 1Ω is stretched to double its length.What will be its new resistance??

Answer 1

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Answer 2

Answer:

R = ρ X L / A - [1]

WHEN THE WIRE is stretched its volume is constant

since volume is constant and question is based on LENGTH multiply both numerator and denominator with L so the equation becomes

R = [ρ X L^2 ]  /  V

SO R is directly proportional to L^2

R1 / R2 = L1^2 / L2^2     where R1 = R     ; L1 = L ;      L2 = 2L   R2 = ?

R / R2 = L^2 / 4L^2

BY CALCULATING R2 = 4R = 4 x 1  = 4 ohm