Q. 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 :
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 \times2}{10+2} = \frac{5}{3} \Omega$
