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in Physics by (84.3k points)
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The moment of inertia of thin square plate `ABCD` of uniform thickness about an axis passing through the center `O` and perpendicular to the plane of the plate is
image
(`i`) `I_(1)+I_(2)`
(`ii`) `I_(2)+I_(4)`
(`iii`) `I_(1)+I_(3)`
(`iv`) `I_(1)+I_(2)+I_(3)+I_(4)`
where `I_(1)`, `I_(2)`, `I_(3)` and `I_(4)` are repectively moments of inertia about axes `1`, `2`, `3` and `4` which are in the plane of the plane
A. (`i`),(`ii`)
B. (`i`), (`ii`), (`iii`)
C. (`ii`), (`iii`)
D. (`i`), (`iii`)

1 Answer

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by (82.1k points)
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Best answer
As explained in the previous problem
`I_(1)=I_(2)=I_(3)=I_(4)`
`I_(0)=I_(1)+I_(2)=I_(2)+I_(4)` (`bot^(ar)` axes theorem)
`=I_(1)+I_(3)`

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