Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of 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. `Isin^(2)theta`
C. `Icos^(2)theta`
D. `Icos^(2)(theta/2)`

Answer

A

Explanation

`A'B'_|_ AB` and `C''D'_|_CD`
From symmetry `I_(AB)=I_(A'B)` and `I_(CD)=I_(CD)`
From theorem of perpendicular axis

`I_(zz)=I_(AB)I_(A'B')=I_(CD)+I_(CD')`
`implies 2I_(AB)=2II_(CD)`
`I_(AB)=I_(CD)`
Alternative method:
The relation between `I_(AB)` and `I_(CD)` should be true for all values of `theta`
At `theta=I_(CD)=I_(AB)`
Similarly at `theta=pi/2`
`I_(CD)=I_(AB)` (By symmetry)
Keeping these things in mind, option a is correct.