two rods of equal length(l) and equal mass M are kept along x and y axis respectively such that their centre of mass lie at origin.the moment of inertia about an line y=x,is

Moment of inertia of a rod of about an axis passing through centre and perpendicular to length:

             $I=\frac{1}{12}M{L}^2$

Moment of inertia of the rod about an axis inclined an angle theta to the original axis:
          $I\text{'}=I {\sin }^2\theta$

The line x=y makes an angle of 45 degrees x and y axes.
Hence, 
            $$\begin{gathered} I\text{'} = I {\sin }^{2}45 \\ =\frac{I}{2} \end{gathered}$$

Moment of inertia of both rods about the line passing through x=y:

          $$\begin{gathered} I_{two rods}=2 I\text{'} \\ =2\left(\frac{I}{2}\right) \\ =I \\ =\frac{1}{12}M{L}^2 \end{gathered}$$