Let the total mass of the rod be $M$ and its total length be $L$. When it is bent at its midpoint, it forms two halves, each of mass $m = \frac{M}{2}$ and length $l = \frac{L}{2}$. The axis passes through the bending point (which is one end of each half) and is perpendicular to the plane containing both halves. For a uniform rod of mass $m$ and length $l$, the moment of inertia about an axis passing through its end and perpendicular to its length is given by $I = \frac{ml^2}{3}$. The moment of inertia of each half about the given axis is: $I_1 = I_2 = \frac{1}{3} m l^2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{ML^2}{24}$ The total moment of inertia is the scalar sum of the moments of inertia of both halves: $I_{total} = I_1 + I_2 = \frac{ML^2}{24} + \frac{ML^2}{24} = \frac{2ML^2}{24} = \frac{ML^2}{12}$