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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Answer & Solution

Correct Answer

(B) \(\frac{{5}}{3}\,\Omega \)

Detailed explanation

\(R=\frac{\rho \ell}{A}=\frac{\rho \ell}{(V / \ell)}=\frac{\rho \ell^{2}}{V} \quad(V \rightarrow \text { Volume of wire })\)

\(\Rightarrow\) Final resistance \(=3 \times(\mathrm{B})^{2}=12 \,\Omega\)

\(\begin{aligned} \mathrm{R}_{\mathrm{eq}} &=2\, \Omega \| 10\, \Omega \\ &=\frac{5}{3} \,\Omega \end{aligned}\)

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\((A)\) \(F_1\) - Layer	\((I)\) \(10\,km\)
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