In the given figure,about which axis will the | vedclass

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(A) The moment of inertia $I$ of a body about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the mass element $

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$AB$

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(A) The moment of inertia $I$ of a body about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the mass element $m_i$ from the axis of rotation.For a given mass distribution,the moment of inertia is larger if the mass is distributed farther from the axis of rotation.In the given triangular frame,the axis $AC$ is the longest side (hypotenuse). When rotating about $AC$,the mass of the frame is,on average,closer to the axis compared to the other two sides.Conversely,the axis $AB$ is the shortest side. When rotating about $AB$,the mass of the frame (specifically the side $BC$) is distributed at a larger average distance from the axis $AB$.Since $AB Therefore,the moment of inertia is maximum about the axis $AB$.

In the given figure,about which axis will the moment of inertia of the triangular frame be maximum? [Given: $AB < BC < AC$]

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