A conductor of resistance 3 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

Step-by-step solution

Resistance R l^2 aligned R₁ R₂ & = ( l₁ l₂ )^2 3 R₂ & = ( l 2 l )^2 R₂ & =12 aligned Stretched wire is bent in the form of an equilateral triangle. Resistance of each side = 12 3 =4 R₁=R₂=R₃=4 Equivalent resistance of R₁ and R₂ (in series) aligned R^ & =R₁+R₂ & =4+4=8 aligned Now, R^ and R₃ are in parallel so total resistance aligned R & = R^ R₃ R^ +R₃ = 8 4 8+4 & = 32 12 = 8 3 aligned