JEE Main 2026 Apr 2 Shift 2, Physics Q25: Moment of Inertia

JEE Main2026Apr 2, Shift 2Physics

Moment of inertia about an axis $A B$ for a rod of mass $40$ kg and length $3$ m is same as that of a solid sphere of mass of $10$ kg and radius $R$ about an axis parallel to $A B$ axis with separation of $3$ m as shown in figure below. The value of $R$ is given as $\frac{α}{2}$ . The value of $α$ is______.

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Solution

Moment of inertia of the rod about $A B$ (axis through one end, perpendicular to rod): $I_{rod} = \frac{M L ^{2}}{3} = \frac{40 \cdot 9}{3} = 120$ kg $\cdot$ m $^{2}$ .

Moment of inertia of the sphere about a parallel axis at distance $3$ m (parallel-axis theorem): $I_{sphere} = \frac{2}{5} m R^{2} + m d^{2} = \frac{2}{5} (10) R^{2} + 10 (9) = 4 R^{2} + 90$ .

Setting them equal: $4 R^{2} + 90 = 120$ , so $4 R^{2} = 30$ and $R^{2} = \frac{15}{2}$ .

Thus $R = \frac{15}{2}$ , giving $α = 15$ .

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JEE Main 2026 Apr 2 Shift 2, Physics Q25: Moment of Inertia

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