question_answer

A metal wire of resistance \[3\Omega \]is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle \[60{}^\circ \] at the centre, the equivalent resistance between these two points will be :-          [JEE Main 9-4-2019 Afternoon]

A) \[\frac{12}{5}\Omega \]                    

B) \[\frac{5}{3}\Omega \]

C) \[\frac{5}{2}\Omega \]                                  

D) \[\frac{7}{2}\Omega \]

Correct Answer: B

Solution :

\[R=\frac{\rho {{\ell }^{2}}}{A\ell D}d=\frac{\rho d{{\ell }^{2}}}{m}\] \[R\propto {{\ell }^{2}}\] \[R=12\Omega \](new resistance of wire) \[{{R}_{1}}=2\Omega \,\,\,\,{{R}_{2}}=10\Omega \] \[{{R}_{eq}}=\frac{10\times 2}{10+2}=\frac{5}{3}\Omega .\]