A thin uniform rod of length 2 m cross-sectional area 'A' and density… | MHT CET 2026 (13 Apr, Shift 2)

Q28MHT CET 2026 · 13 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A thin uniform rod of length $2$ m cross-sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$ . If the value of $\omega$ in terms of its rotational kinetic energy $E$ is $(\alpha E/Ad)^{1/2}$ , then the value of $\alpha$ is

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Answer: (b)

Mass $M = A \cdot 2 \cdot d = 2Ad$ , $I = \frac{ML^2}{12} = \frac{2Ad \cdot 4}{12} = \frac{2Ad}{3}$ .
$E = \frac{1}{2}I\omega^2 = \frac{Ad\,\omega^2}{3}$
$\omega = \left(\frac{3E}{Ad}\right)^{1/2}$ , so $\alpha = 3$ .
So the answer is (b) 3.

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Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 Apr, Shift 2)

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