Let `I` be the moment of interia of a uniform square plate about an axis `AB` that passses through its centre and is parallel to two its sides. `CD` is a line in the plane of the plate that passes through the centre of the plate and makes an angle `theta` with `AB`. The moment of inertia of the plate about the axis `CD` is then equal to-

Options

A. `I`
B. `I sin^(2) theta`
C. `I cos^(2) theta`
D. `I cos^(2) ((theta)/(2))`

Explanation

For the square plate, `I_(1)=I_(2)=(ma^(2))/(12)=I`
For an axis passing through the center `O` and `bot^(ar)` to the plane
`I_(1)+I_(2)=I_(z)` (`bot^(ar)` axes theorem)
`CD` and `EF` are `bot^(ar)`
`I_(CD)+I_(EF)=I_(z)`
By summetry `I_(CD)=I_(EF)=I_(1)=I_(2)=I`
In fact, `M.I.` about any axis passing through `O` (center) and lying in the plane of plane `=(ma^(2))/(12)=I`