The given data is
l′=1.2lA′=0.96A
Using the formula R=Aρl, the new resistance is
R′=0.96A1.2ρl
⇒R′=1.25R
⇒ΔR=1.25R−R=0.25R
Hence, the percentage increase is
=RΔR×100=R0.25R×100=25
JEE Main 2023 — Physics Mechanics
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 ________.
Held on 6 Apr 2023 · Verified 6 Jul 2026.
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