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 :

[JEE Main 2025, 2 Apr (Shift 1)]

Question

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 :

[JEE Main 2025, 2 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: b) $\alpha /4$. Moment of inertia of a uniform rod of mass M and length L about its center: $\alpha =\frac{M{L}^{2}}{12}$ Moment of inertia of each half of the rod about its own center: $I=\frac{(\frac{M}{2})(\frac{L}{2}){}^{2}}{12}\times 2$ Simplifying: $I=\frac{\alpha }{4}$

Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its lengt…

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