Question # 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 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:

# 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 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:

(A) ( frac{7}{2} Omega )

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

(C) ( frac{12}{5} Omega )

(D) ( frac{5}{2} Omega )

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_{e q}=frac{10 times 2}{10+2}=frac{5}{3} Omega )