Q 11-06-253JEE MainJEE Main 2025 (2 Apr, Shift 1)Easy
The 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. The moment of inertia of the cross about an axis passing through its center and normal to the plane containing the cross is:
Answer: (B) $\alpha/4$
$\alpha = \dfrac{ML^2}{12}$.
Each half has mass $M/2$ and length $L/2$, and in the cross each half is centred on the axis. So each contributes
$$\frac{(M/2)(L/2)^2}{12} = \frac{ML^2}{96}$$
$$I_{\text{cross}} = 2\times\frac{ML^2}{96} = \frac{ML^2}{48} = \frac{\alpha}{4}$$
Solution by Sreeraj P, M.Sc Physics