Given,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 A and B is:
      

A. \[\dfrac{R\theta }{2\pi }\]
B. \[\dfrac{R(2\pi -\theta )}{4\pi }\]
C. \[R\left( 1-\dfrac{\theta }{2\pi } \right)\]
D. \[\dfrac{R}{4{{\pi }^{2}}}\left( 2\pi -\theta \right)\theta \]

Hint: In this question we have been asked to calculate the equivalent resistance between the given points. It is given that A and B are two points on the uniform circular ring. No, from the figure we can say that, the two arcs formed by the points A and B are connected in parallel to each other. Therefore, if we calculate the individual resistance of the bigger and smaller arc, we can calculate the equivalent resistance between points A and B.

Formula used:
\[l=r\theta \]
Where,
l is the length of the arc,
r is the radius of ring
\[\theta \] is the angle made by the arc
\[\dfrac{1}{{{R}_{AB}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}\]

Complete step by step solution:
From the given figure,
We know that the arc length is given by,
\[l=r\theta \]
Now, calculating the resistance of the smaller arc
We get,
\[{{R}_{1}}=\dfrac{r\theta }{2\pi r}R\]
On solving,
\[{{R}_{1}}=\dfrac{\theta }{2\pi }R\]…………….. (1)
Similarly, for bigger arc
We get,
\[{{R}_{2}}=\dfrac{2\pi -\theta }{2\pi }R\] …………………… (2)
Now,
From the figure we can say that the two arcs are connected in parallel combination with each other.Therefore, We know,
\[\dfrac{1}{{{R}_{AB}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}\]
After substituting values from (1) and (2)
We get,
\[\dfrac{1}{{{R}_{AB}}}=\dfrac{2\pi }{\theta R}+\dfrac{2\pi }{\left( 2\pi -\theta \right)R}\]
Therefore,
\[{{R}_{AB}}=\dfrac{R}{4{{\pi }^{2}}}(2\pi -\theta )\theta \]

Therefore, the correct answer is option D.

Note:
If two or more resistors in a circuit are connected to the same nodes, the resistors are said to be in parallel connection. The equivalent resistance in parallel connection is given by adding the reciprocals on individual resistance and then taking the reciprocal of the sum. The equivalent resistance is smaller than the resistor with smallest resistance.