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MHT CET 2025 (25 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2025 (25 Apr, Shift 2)

Q44MHT CET 2025 · 25 APR · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

The moment of inertia of a solid sphere of mass ' $m$ ' and radius ' $R$ ' about its diametric axis is ' $I$ '. Its moment of inertia about a tangent in the plane is

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Answer: (c)

$I = \frac{2}{5}mR^2$ about a diameter.
Tangent is parallel to a diameter at distance $R$ : $I_t = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$
$\frac{I_t}{I} = \frac{7/5}{2/5} = 3.5$
So the answer is (c) $3.5\,I$ .

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