1. 15 MR²/32 2. 13 MR²/32 3. 11 MR²/32 4. 9 MR²/32

Answer: 2. 13 MR²/32 Key idea: remaining part = full disc minus the cut-out disc, using the parallel axis theorem for the hole. Full disc: I₁=1/2MR² . Hole has radius R/2 , so its mass is m=M/4 (area is one-fourth). Hole about its own centre: 1/2 m(R/2)²=MR²/32 . Its centre is R/2 from the axis, so add m(R/2)²=MR²/16 . Total for hole: 3MR²/32 . Remaining: 16MR²/32-3MR²/32=13MR²/32 .

Topic: Moment of inertia, radius of gyration, MI of simple objects· Rotational Motion

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

1. \(15 MR^2/32\)

2. \(13 MR^2/32\)

3. \(11 MR^2/32\)

4. \(9 MR^2/32\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Answer: 2. \(13 MR^2/32\)

Key idea: remaining part = full disc minus the cut-out disc, using the parallel axis theorem for the hole.

Full disc: \(I_1=\tfrac12MR^2\).
Hole has radius \(R/2\), so its mass is \(m=M/4\) (area is one-fourth).
Hole about its own centre: \(\tfrac12 m(R/2)^2=\tfrac{MR^2}{32}\). Its centre is \(R/2\) from the axis, so add \(m(R/2)^2=\tfrac{MR^2}{16}\). Total for hole: \(\tfrac{3MR^2}{32}\).
Remaining: \(\tfrac{16MR^2}{32}-\tfrac{3MR^2}{32}=\tfrac{13MR^2}{32}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

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