JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Rotational Mechanics

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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 '

α

'. 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) $α$
(B) $α /4$ ✓ Correct
(C) $α /8$
(D) $α /2$

Answer: (B)

Solution

The full rod has

α = \frac{M L ^{2}}{12}

about its centre. Each half carries mass

\frac{M}{2}

and length

\frac{L}{2}

, so about its own centre it contributes

\frac{1}{12} \cdot \frac{M}{2} \cdot (\frac{L}{2})^{2} = \frac{M L ^{2}}{96}

. The two arms cross at the shared centre, and for an axis perpendicular to the plane this is the same as the perpendicular-to-length axis, so they simply add:

α^{'} = 2 \times \frac{M L ^{2}}{96} = \frac{M L ^{2}}{48} = \frac{α}{4}

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Rotational Mechanics

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