Ncert Solutions -

We know the formula for resistance:

[ R' = 9 \times 20 \ \Omega = 180 \ \Omega ] ncert solutions

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: We know the formula for resistance: [ R'

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 ] Assume the volume and resistivity remain constant

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

Original resistance: [ R = \rho \frac{L}{A} = 20 \ \Omega ]