If a copper wire is stretched to increase its length by 20 % . The percentage increase in resistance of the wire is _____%.

Step-by-step solution

Before heating, resistance of wire is R 0 = ρ l 0 A 0 After heating, resistance is R = ρ l A As volume = constant Therefore, A 0 l 0 = 1 . 2 l 0 A ⇒ A = A 0 1 . 2 So, R = ρ l A = ρ × 1 . 2 l 0 A 0 1 . 2 = ρ l 0 A 0 × 1 . 44 = 1 . 44   R 0 Percentage increase in R = R - R 0 R 0 × 100 = 44 %