A uniform thin bar of mass 6 m and length $12 L$ is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is :

• Correct answer
  A

  $20 \mathrm{~mL}^2$
• B

  $30 \mathrm{~mL}^2$
• C

  $\frac{12}{5} m L^2$
• D

  $6 \mathrm{~mL}^2$

Solution

Length of each side of hexagon $=2 L$ and mass of each side $=m$
Let $O$ be the centre of mass of hexagon. Therefore, perpendicular distance of $O$ from each side, $r=L \tan 60^{\circ}=L \sqrt{3}$ The desired moment of inertia of hexagon about $O$ is $\begin{aligned} I & =6\left[I_{\text {one side }}\right]=6\left[\frac{m(2 L)^2}{12}+m r^2\right] \\ & =6\left[\frac{m L^2}{3}+m(L \sqrt{3})^2\right] \\ & =20 m L^2\end{aligned}$

Hence, the correct answer is option: 'A'.