Two identical thin rods of mass M kg and length L m are connected as shown in the figure below. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is $$\frac{x}{12}ML^{2}\text{kg m}^{2}$$. The value of x is ____ .

Moment of Inertia of Vertical Rod ($$I_1$$):

The axis passes through point $$P$$, which is the end of the vertical rod. For a rod about its end:

$$I_1 = \frac{1}{3} ML^2 = \frac{4}{12} ML^2$$

Moment of Inertia of Horizontal Rod ($$I_2$$):

The horizontal rod is at the bottom of the vertical rod. Its center of mass is at a distance $$d = L$$ from point $$P$$.
Using the Parallel Axis Theorem: 

$$I_2 = I_{cm} + Md^2$$

$$I_2 = \frac{1}{12} ML^2 + M(L)^2$$

$$I_2 = \frac{1}{12} ML^2 + \frac{12}{12} ML^2 = \frac{13}{12} ML^2$$

Total Moment of Inertia:

$$I_{total} = I_1 + I_2$$

$$I_{total} = \frac{4}{12} ML^2 + \frac{13}{12} ML^2$$

$$I_{total} = \frac{17}{12} ML^2$$

$$x = 17$$