A cylindrical metallic wire of initial resistance 5 is drawn so that its length becomes three times its original length. This elongated wire is then folded to form a closed equilateral triangle. What will be the equivalent resistance between any two vertices of this triangle?

Step-by-step solution

Let the initial length of the wire be L₀ and the initial area of cross-section be A₀ . The initial resistance is given by R₀ = L₀ A₀ = 5 . When the wire is drawn to three times its original length, the new length is L = 3L₀ . Since the volume of the wire remains constant, A₀ L₀ = A L , which gives the new area of cross-section as A = A₀ 3 . The new resistance of the elongated wire is R = L A = 3L₀ A₀ 3 = 9 L₀ A₀ = 9R₀ . Substituting R₀ = 5 , we get R = 9 5 = 45 . The wire is then folded to form a closed equilateral