Question

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

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

Answer 1

Answer: (C)

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 $I_{\text{ rod }}=\frac{m{l}^2}{12}+m{x}^2$
[From the theorem of parallel axes]
$=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}}{2}l\right)}^2=\frac{5m{l}^2}{6}$
Now the moment of inertia of system,
$I_{\text{system }}=6\times I_{\text{rod }}=6\times \frac{5m{l}^2}{6}=5m{l}^2$
$I_{\text{system }}=5(12I)=60I$ $[$ As $m{l}^2=12I]$