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(A) When a wire of resistance $R$ is bent into a circle,it is divided into two equal semicircular parts.Each part has a resistance of $R' = \frac{R}{2

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$\frac{R}{4}$

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(A) When a wire of resistance $R$ is bent into a circle,it is divided into two equal semicircular parts.Each part has a resistance of $R' = \frac{R}{2}$.These two semicircular parts are connected in parallel between the two diametrically opposite points.The equivalent resistance $R_{eq}$ between these points is given by the parallel combination formula:$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$.Therefore,$R_{eq} = \frac{R}{4}$.

When a wire of uniform cross-section $a$,length $l$ and resistance $R$ is bent into a complete circle,what is the resistance between any two diametrically opposite points?

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