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JEE Main 2026 (2 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2026 (2 Apr, Shift 2)

Q23JEE Main 2026 · 2 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Moment of inertia about an axis AB 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 AB axis with separation of 3 m as shown in figure below. The value of $R$ is given as $\sqrt{\frac{\alpha}{2}}$ . The value of $\alpha$ is ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

Rod about axis AB at its end: $I_{rod} = \frac{ML^2}{3} = \frac{40 \times 9}{3} = 120\ \text{kg m}^2$ .
Sphere about a parallel axis 3 m from its centre: $I = \frac{2}{5}mR^2 + md^2 = 4R^2 + 90$ .
$4R^2 + 90 = 120 \Rightarrow R^2 = 7.5 = \frac{15}{2}$
$R = \sqrt{\frac{15}{2}}$ , so $\alpha = 15$ .

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