When a wire of uniform cross-section a, length land resistance R is bent into a complete circle, resistance between two of dimetrically opposite points will be

$\frac{R}{4}$

$\frac{R}{8}$

4R

$\frac{R}{2}$

Solution

$\frac{R}{4}$

When wire is bent to form a complete circle then

$2 πr = R \Rightarrow r = \frac{R}{2 π}$

Thus, net resistance in parallel combination of two semicircular resistance

$R' = \frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2} + \frac{R}{2}} = \frac{\frac{R^{2}}{4}}{R} R' = \frac{R}{4}$

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