A wire of resistance 9 ohm is bent in a form of equilateral triangle. the equivalent resistance between any two of its vertices will be

Let ABC be the equilateral triangle formed. AB = BC = CA = L (say)

The original length of the wire = 3L, whose resistance is 9 Ω.

Since, resistance is directly proportional to the length, so each side of the triangle have equal resistance.

That is, R_AB = R_AC = R_CA = 3 Ω

Let us find the resistance across A and B.

R_AC and R_BC are in series, so, they have a combined resistance of 3+3 = 6 Ω

This 6 Ω is in parallel with R_AB. So the equivalent resistance across A and B is, R_eq = (6×3)/(6+3) = 2 Ω.