The resistance of a wire is 10^{-6} , Omega p | vedclass

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(C) The circumference of the circle is $C = \pi d = 2\pi \, m$. The total resistance of the circular wire is $R_{total} = (2\pi \, m) 	imes (10^{-6}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The circumference of the circle is $C = \pi d = 2\pi \, m$. The total resistance of the circular wire is $R_{total} = (2\pi \, m) 	imes (10^{-6} \, \Omega/m) = 2\pi 	imes 10^{-6} \, \Omega$.When bent into a circle, the wire is divided into two equal semicircular arcs, each with resistance $R = \frac{R_{total}}{2} = \pi 	imes 10^{-6} \, \Omega$.A wire of the same material is connected across the diameter $AB$. The length of this wire is equal to the diameter $d = 2 \, m$. Its resistance is $R_1 = 2 	imes 10^{-6} \, \Omega$.The two semicircular arcs are in parallel with each other, and this combination is in parallel with the diameter wire $R_1$. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R_1} = \frac{2}{R} + \frac{1}{R_1} = \frac{2}{\pi 	imes 10^{-6}} + \frac{1}{2 	imes 10^{-6}}$$\frac{1}{R_{eq}} = \frac{1}{10^{-6}} \left( \frac{2}{\pi} + 0.5 ight) = \frac{1}{10^{-6}} \left( \frac{4 + \pi}{2\pi} ight)$$R_{eq} = \frac{2\pi}{4 + \pi} 	imes 10^{-6} \approx \frac{6.28}{7.14} 	imes 10^{-6} \approx 0.88 	imes 10^{-6} \, \Omega$.

The resistance of a wire is $10^{-6} \, \Omega$ per meter. It is bent in the form of a circle of diameter $2 \, m$. A wire of the same material is connected across its diameter. The total resistance across its diameter $AB$ will be

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