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(A) The total resistance of the wire is $R$. The circumference of the ring is $2\pi r$.Resistance per unit length is $\lambda = \frac{R}{2\pi r}$.The

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$\frac{R\alpha}{4\pi^2}(2\pi - \alpha)$

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(A) The total resistance of the wire is $R$. The circumference of the ring is $2\pi r$.Resistance per unit length is $\lambda = \frac{R}{2\pi r}$.The length of the arc $XWY$ is $l_1 = r\alpha$.Therefore,the resistance of the arc $XWY$ is $R_{XWY} = \lambda l_1 = \frac{R}{2\pi r} 	imes r\alpha = \frac{R\alpha}{2\pi}$.The length of the arc $XZY$ is $l_2 = r(2\pi - \alpha)$.Therefore,the resistance of the arc $XZY$ is $R_{XZY} = \lambda l_2 = \frac{R}{2\pi r} 	imes r(2\pi - \alpha) = \frac{R}{2\pi}(2\pi - \alpha)$.Since the two arcs are connected in parallel between points $X$ and $Y$,the equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R_{XWY} 	imes R_{XZY}}{R_{XWY} + R_{XZY}}$$R_{eq} = \frac{(\frac{R\alpha}{2\pi}) 	imes (\frac{R}{2\pi}(2\pi - \alpha))}{\frac{R\alpha}{2\pi} + \frac{R}{2\pi}(2\pi - \alpha)}$$R_{eq} = \frac{\frac{R^2\alpha}{4\pi^2}(2\pi - \alpha)}{\frac{R}{2\pi}(\alpha + 2\pi - \alpha)}$$R_{eq} = \frac{\frac{R^2\alpha}{4\pi^2}(2\pi - \alpha)}{\frac{R}{2\pi}(2\pi)}$$R_{eq} = \frac{R^2\alpha(2\pi - \alpha)}{4\pi^2} 	imes \frac{1}{R} = \frac{R\alpha}{4\pi^2}(2\pi - \alpha)$.

$A$ wire of resistance $R$ is bent to form a circular ring. Find the equivalent resistance between two points $X$ and $Y$ on its circumference. (The angle $XOY = \alpha$ is given.)

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