In a rectangle ABCD ( BC = 2AB). The moment of inertia is minimum… | NEET 1993 (AIPMT)

Q203NEET 1993 · AIPMTEasy

Moment of Inertia & Radius of Gyration

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

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Answer: (d)

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

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