[ R' = 9 \times 20 \ \Omega = 180 \ \Omega ]
[ A \times L = A' \times L' ] [ A \times L = A' \times (3L) ] [ A' = \frac{A}{3} ] ncert solutions
We know the formula for resistance:
[ \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. [ R' = 9 \times 20 \ \Omega
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: A wire of resistance 20 Ω is stretched
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 ]
