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

The moment of inertia of a thin square plate ABCD, fig, of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is
where `l_1, l_2, l_3 and l_4` are respectively the moments of intertial about axis 1,2,3 and 4 which are in the plane of the plate.

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 the the plane of the plane we use perpendicular axis theorem. If we consider `ABCD` to be in the `X-Y`-plane, then we know that

`I_(zz')=I_("xx"')+I_(yy')`
`:. I_(zz')=I_(1)+I_(2)`.........i
Also `I_(zz')=I_(3)+I_(4)`.........ii
Adding eqn i and ii we get
`2I_(zz')=I_(1)+I_(2)+I_(3)+I_(4)`
But `I_(1)=I_(2)` and `I_(3)=I_(4)` (by symmetry)
`:. 2I_(zz')=I_(1)+I_(1)+I_(3)+I_(3)`
`=2I_(I)1+2I_(3)`
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