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Q.
When a wire of uniform cross-section $ a $ , length $ l $ and resistance $R$ is bent into a complete circle, resistance between two of diametrically opposite points will be
When wire is bent to form a complete circle then
$ 2\,\pi r=R $
$ \Rightarrow r=\frac{R}{2\pi } $
Resistance of each semicircle
$=\pi r=\frac{\pi R}{2 \pi}=\frac{R}{2}$
Thus, net resistance in parallel combination of two semicircular resistances
$R'=\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}=\frac{\frac{R^{2}}{4}}{R}=\frac{R}{4}$
