The moment of inertia of a rod of length L about an axis passing through its center 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 center of mass and perpendicular to its plane will be $nI$. Find the value of n.

Moment of inertia of rod AB about its centre and perpendicular to the length $=\frac{m{l}^2}{12}=I\Rightarrow m{l}^2=12I$

Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of hexagon ${\mathrm{I}}_{\mathrm{rod}}=\frac{m{l}^2}{12}+m{x}^2$

[From the theorem of parallel axes]

$=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}}{2}l\right)}^2=\frac{5m{l}^2}{6}$

Now the moment of inertia of system,

$\begin{array}{r}{I}_{\text{system }}=6\times I_{\mathrm{rod}}=6\times \frac{5m{l}^2}{6}=5m{l}^2\end{array}$

$I_{\text{system }}=5\left(12I\right)=60I \left[\text{ As }m{l}^2=12I\right]$