The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to plate is

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

To find the moment of inertia of ABCD about an axis passing through centre O and perpendicular to the plane of the plate, we use perpendicular axis theorem. If we consider ABCD to be in the X-Y plane, then we know that,
`I_(Z Z') = I_(XX') + I_(YY')`
`therefore I_(XX') = I_(1) + I_(2)`.......(i)
Also, `I_(Z Z') = I_(3) + I_(4)`.......(iii)
Adding Eqs. (i) and (ii), we get

`2I_(Z Z') =I_(1) + I_(2) + I_(3) + I_(4)`
But `I_(1) =I_(2)` and `I_(3) = I_(4)` (by symmetry)
`therefore 2I_(Z Z') = I_(1) + I_(1) + I_(3) + I_(3) = 2I_(1) + 2I_(3)`
`I_(Z Z') = I_(1) + I_(3)`