Q. A wire of resistance 4Ω is stretched to twice its original length. The resistance of stretched wire would be -
A) 2 Ω
B) 4 Ω
C) 8 Ω
D) 16 Ω
Explanation
[Solution from source] Stretching keeps the volume constant, so doubling the length halves the cross-section. R = ρl/A becomes ρ(2l)/(A/2) = 4R = 4 × 4 = 16 Ω. 8 Ω only accounts for the change in length. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.