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 :
- 10 Ω
- 5 Ω
- 40 Ω
- 20 Ω
The Correct Option is A
Solution and Explanation
Let R0 be initial resistance = 10 Ω. When stretched to length 2l, the resistance becomes R1 = $\frac{2l}{l}$R0 = 20 Ω. When bent into a circle, the total resistance remains 20 Ω.
The resistance of the semi-circle along any diameter is given by 20/2 = 10 Ω.
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