Question

The moment of inertia of a rod of length $l$ about an axis passing through its centre of mass and perpendicular to the 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

• A. $16I$
• B. $40I$
• C. $60I$
• D. $80I$

Answer 1

The correct option is C $60I$

Moment of inertia of rod $AB$ about its centre and perpendicular to the length $I=\frac{m{l}^2}{12}$
$\Rightarrow m{l}^2=12I$
Now moment of inertia of the rod about the axis which is passing through $\text{O}$ and perpendicular to the plane of hexagon
$I_{\text{rod}}=\frac{m{l}^2}{12}+m{x}^2$
[From the theorem of parallel axes]


$=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}}{2}l\right)}^2=\frac{5m{l}^2}{6}$
So, total moment of inertia of system,
$I_{\text{system}}=6\times I_{\text{rod}}=6\times \frac{5m{l}^2}{6}=5m{l}^2$
$I_{\text{system}}=5\left(12I\right)=60I$
[As $m{l}^2=12I$]