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(D) The diameter of the circle is $d = 2 \, m$,so the radius is $r = 1 \, m$.The circumference of the circle is $C = \pi d = 2\pi \, m$.The wire is di

Vedclass · Accepted Answer

$0.88 	imes 10^{-6} \, \Omega$

Vedclass · Answer

(D) The diameter of the circle is $d = 2 \, m$,so the radius is $r = 1 \, m$.The circumference of the circle is $C = \pi d = 2\pi \, m$.The wire is divided into two semicircular arcs,each of length $L_{arc} = \pi r = \pi 	imes 1 = \pi \, m$.The resistance of each semicircular arc is $R_1 = R_3 = ho_L 	imes L_{arc} = 10^{-6} 	imes \pi = \pi 	imes 10^{-6} \, \Omega$.The wire connected across the diameter has length $L_{diameter} = d = 2 \, m$.The resistance of this wire is $R_2 = ho_L 	imes L_{diameter} = 10^{-6} 	imes 2 = 2 	imes 10^{-6} \, \Omega$.These three resistances $(R_1, R_2, R_3)$ are connected in parallel between points $A$ and $B$.The equivalent resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$\frac{1}{R_{AB}} = \frac{1}{\pi 	imes 10^{-6}} + \frac{1}{2 	imes 10^{-6}} + \frac{1}{\pi 	imes 10^{-6}}$$\frac{1}{R_{AB}} = \frac{2}{\pi 	imes 10^{-6}} + \frac{1}{2 	imes 10^{-6}} = \frac{1}{10^{-6}} \left( \frac{2}{\pi} + \frac{1}{2} ight) = \frac{1}{10^{-6}} \left( \frac{4 + \pi}{2\pi} ight)$$R_{AB} = 10^{-6} 	imes \frac{2\pi}{4 + \pi} \approx 10^{-6} 	imes \frac{6.28}{7.14} \approx 0.88 	imes 10^{-6} \, \Omega$.

$A$ wire of resistance per unit length $\rho_L = 10^{-6} \, \Omega/m$ is bent into a circle of diameter $2 \, m$. $A$ piece of wire of the same material is connected across the diameter $AB$. Find the resistance between $A$ and $B$.

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