JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 5)
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 :
$\alpha / 4$
$\alpha / 8$
$\alpha$
$\alpha / 2$
Explanation
_2nd_April_Morning_Shift_en_5_1.png)
$$\alpha=\frac{\mathrm{M} \ell^2}{12}\quad\text{..... (1)}$$
_2nd_April_Morning_Shift_en_5_2.png)
$$\begin{aligned} &\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4} \end{aligned}\\ &\text { Correct option is (2) } \end{aligned}$$
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