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Question

Four thin rods of same mass $$M$$ and same length $$l$$, form a square as shown in the figure. Moment of inertia of this system about an axis passing through centre $$O$$ and perpendicular to plane is :
670135_69fea811d4b64504807468559ca51475.PNG


A
43MI2
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B
MI23
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C
MI26
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D
23MI2
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Solution

The correct option is A $$\cfrac { 4 }{ 3 } M{ I }^{ 2 }$$
Moment of inertia of rod $$AB$$ about
$$P=\cfrac { 1 }{ 12 } M{ I }^{ 2 }\quad $$
$$MI$$ of rod $$AB$$ about
$$O=\cfrac { M{ I }^{ 2 } }{ 12 } +M{ \left( \cfrac { I }{ 2 }  \right)  }^{ 2 }=\cfrac { 1 }{ 3 } M{ I }^{ 2 }$$ (from the theorem of parallel axis)
and the system consists of $$4$$ rods of similar type so by the symmetry
$${ I }_{ system }=\cfrac { 4 }{ 3 } M{ I }^{ 2 }$$

Physics

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