A and B are two points on a uniform ring of radius r. The resistance of the ring is R. \(\angle AOB =\theta\) as shown in the figure. The equivalent resistance between points A & B is _________.

(D)

From figure,
\(l_1=r \theta\)
and \(l_2=r(2 \pi-\theta)\)
From \(R =\frac{\rho l}{A}\),
\(R=\frac{\rho(2 \pi r)}{A}\) (Total Resistance)
\(R _1=\frac{\rho l_1}{A}\)
\(R _1=\frac{\rho r \theta}{A}\) (Resistance of minor arc)
and \(R_2=\frac{\rho l_2}{A}\)
\(R_2=\frac{\rho r(2 \pi-\theta)}{A}\)
(Resistance of major arc)
Now \(R_{A B}=\frac{R_1 R_2}{R_1+R_2}\)
\(\begin{array}{l} R _{ AB }=\frac{\frac{\rho r \theta}{A} \cdot \frac{\rho r(2 \pi-\theta)}{ A }}{\frac{\rho r \theta}{A}+\frac{\rho r(2 \pi-\theta)}{ A }} \\ R _{ AB }=\frac{\frac{\rho r \theta}{A} \cdot \frac{\rho r(2 \pi-\theta)}{ A }}{\frac{\rho r \theta+2 \pi \rho r-\rho r \theta}{A}}\end{array}\)
\(\begin{array}{ll}\therefore & R_{A B}=\frac{\rho r(2 \pi-\theta) \theta}{2 \pi A} \times \frac{2 \pi}{2 \pi} \\ \therefore & R_{A B}=\frac{R \theta}{4 \pi^2}(2 \pi-\theta)\end{array}\)