Ncert Solutions -

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

[ \boxed{180 \ \Omega} ] If you have a specific NCERT chapter, class, and subject (e.g., Class 10 Science Ch 12 Electricity, or Class 12 Physics Ch 1 Electric Charges), I can create a complete, accurate, and original set of solutions for all the exercises. Just tell me the details. ncert solutions

If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy: [ R' = 9 \times 20 \ \Omega

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 ] and subject (e.g.

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