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 :

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Answer & Solution

Correct Answer

(B) \(\alpha / 4\)

Step-by-step Solution

Detailed explanation

\(\alpha=\frac{\mathrm{M} \ell^2}{12}\)

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