A square-shaped wire with a resistance of each side 3 $$\Omega$$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in a unit of $$\Omega$$ will is _________.

We start by noting that the wire originally forms a square, and along each of its four sides the resistance is given as $$3\;\Omega$$. Resistance of a conductor is directly proportional to its length (for the same material and cross-section). Therefore the total resistance around the complete square is obtained by simply adding the resistances of the four equal sides:

$$R_{\text{square}} = 3\;\Omega + 3\;\Omega + 3\;\Omega + 3\;\Omega = 12\;\Omega.$$

Now the square is reshaped into a perfect circle. The material, length and cross-section of the wire all remain unchanged, so its total resistance also remains unchanged. Hence the resistance of the whole circumference of the new circle is still

$$R_{\text{circle}} = 12\;\Omega.$$

We are asked to find the resistance between two diametrically opposite points on this circle. Choosing those opposite points divides the circle into two equal semicircular arcs. Because the wire is uniform, each semicircle owns exactly half the total length and therefore half the total resistance.

So, the resistance of each semicircle is

$$R_{\text{semi}} = \dfrac{R_{\text{circle}}}{2} = \dfrac{12\;\Omega}{2} = 6\;\Omega.$$

Between the chosen diametrically opposite points, current can travel through either the upper semicircle or the lower semicircle; these two paths connect the same pair of nodes, so they are effectively in parallel.

For two resistors $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance is found using the parallel-combination formula

$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}.$$

Substituting $$R_1 = R_2 = 6\;\Omega$$ gives

$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{6\;\Omega} + \dfrac{1}{6\;\Omega} = \dfrac{2}{6\;\Omega} = \dfrac{1}{3\;\Omega}.$$

Taking the reciprocal yields the equivalent resistance:

$$R_{\text{eq}} = 3\;\Omega.$$

So, the answer is $$3\;\Omega$$.