A wire of length 25 m and cross-sectional ar | vedclass

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(D) Given: Length $L = 25 \ m$,Area $A = 5 \ mm^2 = 5 	imes 10^{-6} \ m^2$,Resistivity $ho = 2 	imes 10^{-6} \ \Omega \ m$. First,calculate the to

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$2.5$

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(D) Given: Length $L = 25 \ m$,Area $A = 5 \ mm^2 = 5 	imes 10^{-6} \ m^2$,Resistivity $ho = 2 	imes 10^{-6} \ \Omega \ m$. First,calculate the total resistance of the wire: $R = \frac{ho L}{A} = \frac{2 	imes 10^{-6} 	imes 25}{5 	imes 10^{-6}} = 10 \ \Omega$. When the wire is bent into a circle,the two diametrically opposite points divide the wire into two equal semicircular parts,each having a resistance of $R' = \frac{R}{2} = \frac{10}{2} = 5 \ \Omega$. These two parts are connected in parallel between the diametrically opposite points. Therefore,the equivalent resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} = \frac{2}{5} \implies R_{eq} = \frac{5}{2} = 2.5 \ \Omega$.

$A$ wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be (in $\Omega$)

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