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$A$ and $B$ are two points on a uniform ring of resistance $R$. The $\angle A C B=\theta$, where $C$ is the centre of the ring. The equivalent resistance between $A$ and $B$ is
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The correct answer is:
$\frac{R \theta(2 \pi-\theta)}{4 \pi^2}$
Resistance per unit length $=\rho=\frac{R}{2 \pi r}$
Lengths of sections $A P B$ and $A Q B$ are $r \theta$ and $r(2 \pi-\theta)$
Resistances of sections $A P B$ and $A Q B$ are
$\begin{aligned} & R_1=\rho r \theta=\frac{R}{2 \pi r} r \theta=\frac{R \theta}{2 \pi} \\ & \text { and } R_2=\frac{R}{2 \pi r} r(2 \pi-\theta)=\frac{R(2 \pi-\theta)}{2 \pi}\end{aligned}$
As $R_1$ and $R_2$ are in parallel between $A$ and $B$, their equivalent resistance is
$\begin{aligned} & R_{\mathrm{eq}}=\frac{R_1 R_2}{R_1+R_2} \\ & =\frac{\frac{R \theta}{2 \pi} \cdot \frac{R(2 \pi-\theta)}{2 \pi}}{\frac{R \theta}{2 \pi}+\frac{R(2 \pi-\theta)}{2 \pi}}=\frac{\frac{R^2 \theta(2 \pi-\theta)}{4 \pi^2}}{\frac{R}{2 \pi}[\theta+2 \pi-\theta]}=\frac{R(2 \pi-\theta) \theta}{4 \pi^2} \\ & \end{aligned}$
Lengths of sections $A P B$ and $A Q B$ are $r \theta$ and $r(2 \pi-\theta)$
Resistances of sections $A P B$ and $A Q B$ are
$\begin{aligned} & R_1=\rho r \theta=\frac{R}{2 \pi r} r \theta=\frac{R \theta}{2 \pi} \\ & \text { and } R_2=\frac{R}{2 \pi r} r(2 \pi-\theta)=\frac{R(2 \pi-\theta)}{2 \pi}\end{aligned}$
As $R_1$ and $R_2$ are in parallel between $A$ and $B$, their equivalent resistance is
$\begin{aligned} & R_{\mathrm{eq}}=\frac{R_1 R_2}{R_1+R_2} \\ & =\frac{\frac{R \theta}{2 \pi} \cdot \frac{R(2 \pi-\theta)}{2 \pi}}{\frac{R \theta}{2 \pi}+\frac{R(2 \pi-\theta)}{2 \pi}}=\frac{\frac{R^2 \theta(2 \pi-\theta)}{4 \pi^2}}{\frac{R}{2 \pi}[\theta+2 \pi-\theta]}=\frac{R(2 \pi-\theta) \theta}{4 \pi^2} \\ & \end{aligned}$
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