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A uniform metal wire of length l has 10 Ω resistance. Now this wire is stretched to a length 2l and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

Asked in Re-NEET 2024 · Wires bent into shapes

Answer: (1) 10 Ω

Step-by-step solution

Stretching to twice the length at constant volume halves the area, so R∝ l²: R'=4×10=40 Ω.

Across a diameter the ring is two 20 Ω halves in parallel.

R_AB=(20×20)/(20+20)=10 Ω.

Why the other options are wrong

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