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The moment of inertia of a thin square p...

The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through its center and perpendicular to its plane will

A

`I_(1) = I_(2)`

B

`I_(3)+I_(4)`

C

`I_(1) + I_(3)`

D

`I_(1) + I_(2) + I_(3) + I_(4)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
To find the moment of inertia of ABCD about an axis passing through centre O and perpendicular to the plane of the plate, we use perpendicular axis theorem. If we consider ABCD to be in the X-Y plane, then we know that,
`I_(Z Z') = I_(XX') + I_(YY')`
`therefore I_(XX') = I_(1) + I_(2)`.......(i)
Also, `I_(Z Z') = I_(3) + I_(4)`.......(iii)
Adding Eqs. (i) and (ii), we get

`2I_(Z Z') =I_(1) + I_(2) + I_(3) + I_(4)`
But `I_(1) =I_(2)` and `I_(3) = I_(4)` (by symmetry)
`therefore 2I_(Z Z') = I_(1) + I_(1) + I_(3) + I_(3) = 2I_(1) + 2I_(3)`
`I_(Z Z') = I_(1) + I_(3)`
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