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MHT CET 2026 (18 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (18 Apr, Shift 2)

Q35MHT CET 2026 · 18 APR · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

Given identical rings are arranged in a hexagonal plane pattern so as to touch each neighbouring ring as shown in figure. Each ring has mass M and radius R. The moment of inertia of the system of seven rings about an axis passing through the centre of central ring and normal to plane of all rings is

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Answer: (a)

Central ring: $MR^2$ .
Each outer ring centre is $2R$ away: $MR^2 + M(2R)^2 = 5MR^2$ .
$I = MR^2 + 6 \times 5MR^2 = 31MR^2$
So the answer is (a).

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