Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

• A

  $\alpha$
• Correct answer
  B

  $\alpha / 4$
• C

  $\alpha / 8$
• D

  $\alpha / 2$

Solution

$\alpha=\frac{\mathrm{M} \ell^2}{12}$
$\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4}\end{aligned}$
Correct option is (2)

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