The moment of inertia of a rod of length l about an axis passing through its centre of mass and perpendicular to rod is I. The moment of inertia of hexagonal shape formed by six such rods, about an axis passing through its centre of mass and perpendicular to its plane will be

The correct option is 60 I.

Let us understand the step-by-step derivation to find the total moment of inertia of the hexagonal shape formed by six identical rods.

Step 1: Moment of inertia of a single rod
Let each rod have a mass $m$ and length $l$.
The moment of inertia of a single rod about an axis passing through its center of mass and perpendicular to its length is given by:
$I=\frac{1}{12}m{l}^2$
From this, we can express the term $m{l}^2$ in terms of $I$:
$m{l}^2=12I$

Step 2: Distance from the center of the hexagon to the center of mass of each rod
A regular hexagon is formed by six such rods of length $l$.
Let $d$ be the perpendicular distance from the center of the hexagon (which is also the center of mass of the hexagon) to the midpoint (center of mass) of any one of the rods.
Using basic trigonometry for a regular hexagon:
$d=lcos(30°)=l\frac{\sqrt{3}}{2}$
Squaring both sides gives:
$d^2=\frac{3}{4}{l}^2$

Step 3: Moment of inertia of one rod about the central axis of the hexagon
Using the parallel axis theorem, the moment of inertia $I_{\text{axis}}$ of one rod about the axis passing through the center of the hexagon and perpendicular to its plane is:
$I_{\text{axis}}=I+m{d}^2$
Substitute the value of $d^2$ into the equation:
$I_{\text{axis}}=I+m\left(\frac{3}{4}\right)$
Substitute $m{l}^2=12I$ into the equation:
$I_{\text{axis}}=I+\frac{3}{4}\left(12I\right)$
$I_{\text{axis}}=I+9I=10I$

Step 4: Total moment of inertia of the hexagonal shape
Since the hexagonal frame is made up of six identical rods, the total moment of inertia $I_{\text{total}}$ is the sum of the moments of inertia of all six rods:
$I_{\text{total}}=6\times I_{\text{axis}}$
$I_{\text{total}}=6\times 10I=60I$