[ A \times L = A' \times L' ] [ A \times L = A' \times (3L) ] [ A' = \frac{A}{3} ]
We know the formula for resistance:
New resistance: [ R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} ] [ R' = \rho \frac{3L \times 3}{A} = \rho \frac{9L}{A} ] [ R' = 9 \left( \rho \frac{L}{A} \right) ] [ R' = 9 \times R ]
If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy:
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:
Ncert Solutions Verified May 2026
[ A \times L = A' \times L' ] [ A \times L = A' \times (3L) ] [ A' = \frac{A}{3} ]
We know the formula for resistance:
New resistance: [ R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} ] [ R' = \rho \frac{3L \times 3}{A} = \rho \frac{9L}{A} ] [ R' = 9 \left( \rho \frac{L}{A} \right) ] [ R' = 9 \times R ] ncert solutions
If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy: [ 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: Assume the volume and resistivity remain constant
