The moment of inertia of a solid sphere about an axis passing through centre of gravity is (2MR²)/(5) where M is the mass and R is the radius. The radius of gyration about a parallel axis at a distance 2R from first axis is (a) 5R (b) √(22/5) R (c) 5R/2 (d) √(12/5) R

Answer: (b). Parallel axis theorem: I = 2/5MR² + M(2R)² = 2/5MR² + 4MR² = 22/5MR² K = √(I/M) = √(22/5) R So the answer is (b) √(22/5) R.

The moment of inertia of a solid sphere about an axis passing through… | MHT CET 2026 (14 May, Shift 1)

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Moment of Inertia & Radius of Gyration

The moment of inertia of a solid sphere about an axis passing through centre of gravity is $\frac{2MR^2}{5}$ where M is the mass and R is the radius. The radius of gyration about a parallel axis at a distance $2R$ from first axis is

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Answer: (b)

Parallel axis theorem: $I = \frac{2}{5}MR^2 + M(2R)^2 = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2$
$K = \sqrt{\frac{I}{M}} = \sqrt{\frac{22}{5}}\,R$
So the answer is (b) $\sqrt{\frac{22}{5}}\,R$ .

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Moment of Inertia & Radius of Gyration question from MHT CET 2026 (14 May, Shift 1)

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