ABC is a triangular plate of uniform thicknes | vedclass

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(C) The moment of inertia $I$ of a body about an axis is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.For a planar obje

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$I_{BC} > I_{AB}$

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(C) The moment of inertia $I$ of a body about an axis is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.For a planar object,the moment of inertia about an axis is proportional to the square of the average distance of the mass from the axis.In the given right-angled triangle $ABC$ with sides $AB=4$,$BC=3$,and $AC=5$:$1$. The axis $BC$ is at a distance from the mass distributed along $AB$. The average distance of the mass from axis $BC$ is larger than the average distance of the mass from axis $AB$.$2$. Specifically,the mass is more spread out relative to the axis $BC$ compared to axis $AB$.$3$. Therefore,the radius of gyration $k_{BC} > k_{AB}$.$4$. Consequently,$I_{BC} > I_{AB}$.$5$. The axis $CA$ (the hypotenuse) is closest to the center of mass of the triangle,resulting in the smallest moment of inertia.Thus,the correct relation is $I_{BC} > I_{AB} > I_{CA}$.

$ABC$ is a triangular plate of uniform thickness. Its sides are as shown in the figure. If $I_{AB}$,$I_{BC}$,and $I_{CA}$ are the moments of inertia of the plate about axes $AB$,$BC$,and $CA$ respectively,then which of the following relations is correct?

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