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NEET 1993 (AIPMT) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from NEET 1993 (AIPMT)

Q203NEET 1993 · AIPMTEasy

Moment of Inertia & Radius of Gyration

In a rectangle ABCDABCD (BC=2ABBC = 2AB). The moment of inertia is minimum along axis through

Figure

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Answer: (d)

  1. MI is smallest when the mass is closest to the axis, i.e. about a central axis parallel to the longer side.
  2. AD=BCAD = BC is the longer side, and EGEG is the central axis parallel to it: IEG=M(AB)212I_{EG} = \frac{M(AB)^2}{12}.
  3. IHF=M(BC)212=4IEGI_{HF} = \frac{M(BC)^2}{12} = 4I_{EG}; BCBC is an edge axis and BDBD (diagonal) also has more mass far away.
  4. So the answer is (d) EGEG.
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