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.

Options

A. `I_(1)+I_(2)`
B. `I_(3)+I_(4)`
C. `I_(1)+I_(3)`
D. `I_(1)+I_(2)+I_(3)+I_(4)`

Answer

A, B, C

Explanation

According to the theorem of perpendicular axis:
Moment of inertia of the plate about an axis passing through centre `O` and perpendicular to the plate is `I_(0)=I_(1)+I_(2)=I_(3)+I_(4)`.
because diagonals `1` and `2` as well as `3` and `4` are mutully perpendicular).
Now according to symmetry,
`I_(1)=I_(2)=I` (say)
and `I_(3)=I_(4)=I'` (say)
For a square plate.
`I_(0)=(M(L^(2)+L^(2)))/12=(ML^(2))/6`
Hence `I=I'=(ML^(2))/6`
Hence `I=I'=(ML^(2))/12`
i.e. `I_(1)=I_(2)=I_(3)=I_(4)=(ML^(2))/12`
Hence `I_(0)=I_(1)+I_(2)=I_(3)+I_(4)=I_(1)+I_(3)`
i.e. options `a` and `b`, `c` are correct.