The moment of inertia of a thin square plate ABCD, as shown A in the figure, of uniform thickness about an axis passing through the centre O and perpendicular to the plane of

Options

A. `l_(1)+l_(2)`
B. `l_(3)+l_(4)`
C. `l_(1)+l_(3)`
D. `l_(1)+l_(2)+l_(3)+l_(4)` ltbRgt where `l_(1), l_(2), l_(3)` and `l_(4)` are respectively the moments of inertia about axis 1, 2, 3 and 4 which are in the plane of the plate.

Answer

A, B, C

Explanation

Since ABCD is a square plate, by symmetry, `I_(3) =I_(4)` and `I_(1) =I_(2)`.
Let `I_(0)` = Moment of inertia about an axis perpendicular to square plate and passing through the centre O.
`thereforeI_(0)=I_(1)+I_(2)`, by theorem of perpendicular axes.
Also `I_(0)=I_(3)+I_(4)`, by theorem of perpendicular axes.
or `I_(0)+I_(0)=(I_(1)+I_(2))+(I_(3)+I_(4))`
or `2I_(0)=I_(1)+I_(1)+I_(3)+I_(3)=2(I_(1)+I_(3))`
or `I_(0)=I_(1)+I_(3)`.
Hence options (a), (b) and (c) are correct.