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When a wire of uniform cross-section $a$, length $l$ and resistance R is bent into a complete circle, resistance between any two diametrically opposite points will be -
A
$\dfrac{R}{2}$
B
$\dfrac{R}{4}$
C
$\dfrac{R}{8}$
D
$4R$
Detailed Solution
When the wire is bent into a circle, two diametrically opposite points A and B divide it into two equal semicircles.
Each semicircle has half the length of the wire, so its resistance is $\dfrac{R}{2}$.
Between A and B the two semicircles are in parallel.
$R_{AB} = \dfrac{(R/2)(R/2)}{(R/2) + (R/2)}$
$R_{AB} = \dfrac{R^2/4}{R} = \dfrac{R}{4}$
Each semicircle has half the length of the wire, so its resistance is $\dfrac{R}{2}$.
Between A and B the two semicircles are in parallel.
$R_{AB} = \dfrac{(R/2)(R/2)}{(R/2) + (R/2)}$
$R_{AB} = \dfrac{R^2/4}{R} = \dfrac{R}{4}$