A conductor of resistance 3 $\mathrm{\Omega }$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is:

The resistance of a conductor of length l, cross sectional area A and made of a material of resistivity $\rho$ is given by

                    $R=\frac{\rho l}{A}=\frac{\rho l^2}{Al}=\left(\frac{\rho }{V}\right)l^2$

where V = Al is the volume of the conductor. Since $\rho$ is a constant and volume V cannot change if the conductor is stretched, it follows that R is proportional to l². Thus if l is doubled, R becomes four times. Hence the new resistance is 3 $\times$ 4 = 12 $\mathrm{\Omega }$. So, each side of the equilateral triangle has a resistance of 4 $\mathrm{\Omega }$. Therefore, the effective resistance between the ends of any side of the  triangle (such as side AB) is equal to the resistance of a parallel combination of R ₁ = 4 $\mathrm{\Omega }$ and R₂ = 4 + 4 = 8 $\mathrm{\Omega }$, which is given by 

R =$\frac{R_1\times R_2}{R_1+R_2}=\frac{4\times 8}{4+8}=\frac{8}{3}\Omega$