The length of a metallic wire is increased by 20 % and its area of cross-section is reduced by 4 % . The percentage change in resistance of the metallic wire is _ _ _ _ _ _ _ _ .

Step-by-step solution

The given data is l ' = 1 . 2 l A ' = 0 . 96 A Using the formula R = ρ l A , the new resistance is R ' = 1 . 2 ρ l 0 . 96 A ⇒ R ' = 1 . 25 R ⇒ Δ R = 1 . 25 R - R = 0 . 25 R Hence, the percentage increase is = Δ R R × 100 = 0 . 25 R R × 100 = 25 %