JEE Mains · Physics · STD 12 - 3. current electricity
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
- A \(\frac{{12}}{5}\,\Omega \)
- B \(\frac{{5}}{3}\,\Omega \)
- C \(\frac{{5}}{2}\,\Omega \)
- D \(\frac{{7}}{2}\,\Omega \)
Answer & Solution
Correct Answer
(B) \(\frac{{5}}{3}\,\Omega \)
Step-by-step Solution
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 \(=3 \times(\mathrm{B})^{2}=12 \,\Omega\)…
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