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(B) The resistance of a wire is given by $R = \frac{\rho \ell}{A} = \frac{\rho \ell^2}{V}$,where $V$ is the volume of the wire. Since the volume remai

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$\frac{5}{3}\,\Omega$

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(B) The resistance of a wire is given by $R = \frac{ho \ell}{A} = \frac{ho \ell^2}{V}$,where $V$ is the volume of the wire. Since the volume remains constant during elongation,$R \propto \ell^2$.When the length is doubled,the new resistance $R' = R 	imes (2)^2 = 3 	imes 4 = 12\,\Omega$.This wire of $12\,\Omega$ is bent into a circle. The resistance of a part of the wire is proportional to the angle it subtends at the center.The circle is divided into two parts by the two points: one part subtends an angle of $60^o$ and the other subtends $360^o - 60^o = 300^o$.The resistance of the smaller arc $(R_1)$ is $R_1 = 12 	imes \frac{60}{360} = 2\,\Omega$.The resistance of the larger arc $(R_2)$ is $R_2 = 12 	imes \frac{300}{360} = 10\,\Omega$.These two resistances are in parallel between the two points.The equivalent resistance $R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{2 	imes 10}{2 + 10} = \frac{20}{12} = \frac{5}{3}\,\Omega$.

$A$ metal wire of resistance $3\,\Omega$ is elongated to make a uniform wire of double its previous length. The new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^o$ at the centre,the equivalent resistance between these two points will be

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