Find the moment of inertia of system formed using two identical rods about the given axis of rotation.

The correct option is 17/12 ML².

Let us find the total moment of inertia of the T-shaped system formed by two identical rods, each of mass $M$ and length $L$, about the axis passing through point $O$ perpendicular to the plane of the figure.

We can divide the system into two individual parts:
1. A vertical rod of mass $M$ and length $L$ rotating about one of its ends.
2. A horizontal rod of mass $M$ and length $L$ whose center of mass is at a distance of $L$ from the axis of rotation.

Step 1: Moment of inertia of the vertical rod ($I_1$)
The axis of rotation passes through the end point $O$ of the vertical rod and is perpendicular to its length. The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its end is given by:

$I_1=\frac{1}{3}M{L}^2$

Step 2: Moment of inertia of the horizontal rod ($I_2$)
The moment of inertia of the horizontal rod about an axis passing through its own center of mass (perpendicular to its length) is:

$I_{\text{com}}=\frac{1}{12}M{L}^2$

Since the horizontal rod is attached to the bottom of the vertical rod of length $L$, its center of mass is at a perpendicular distance $d=L$ from the axis of rotation passing through $O$.
Using the parallel axis theorem:

$I_2=I_{\text{com}}+M{d}^2$

Substituting $I_{\text{com}}$ and $d=L$ into the equation, we get:

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

Step 3: Total moment of inertia of the system ($I_{\text{total}}$)
The total moment of inertia of the combined system is the sum of the moments of inertia of the individual rods:

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

$I_{\text{total}}=\frac{1}{3}M{L}^2+\frac{13}{12}M{L}^2$

To add the fractions, convert $\frac{1}{3}$ to have a common denominator of 12:

$I_{\text{total}}=\frac{4}{12}M{L}^2+\frac{13}{12}M{L}^2=\frac{17}{12}M{L}^2$