Ncert Solutions (2025)

[ A \times L = A' \times L' ] [ A \times L = A' \times (3L) ] [ A' = \frac{A}{3} ]

If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy: ncert solutions

[ R' = 9 \times 20 \ \Omega = 180 \ \Omega ] [ A \times L = A' \times L'

I have structured it exactly like an official NCERT Solutions answer: step-by-step, using the correct formulas, and ending with the final answer boxed. A wire of resistance 20 Ω is stretched to three times its original length. Calculate its new resistance. Assume the volume and resistivity remain constant. Solution: then to keep volume constant