Q 11-06-017JEE MainAIEEE 2007Medium
For the given uniform square lamina ABCD, whose centre is O, E and F are the midpoints of AB and DC. Then
Answer: (C) $I_{AC} = I_{EF}$
All axes here lie in the plane of the lamina and pass through O (except AD).
By the perpendicular axis theorem, for any two perpendicular in-plane axes through O: $I_1 + I_2 = I_z$ (the axis perpendicular to the plane).
EF and the line GH through O parallel to AB are perpendicular and, by symmetry, equal: $2I_{EF} = I_z$.
The diagonals AC and BD are also perpendicular and equal: $2I_{AC} = I_z$.
So $I_{AC} = I_{EF}$.
(For reference, $I_{AD} = I_{EF} + m\left(\dfrac{a}{2}\right)^2 = \dfrac{ma^2}{12} + \dfrac{ma^2}{4} = 4I_{EF}$.)
Solution by Sreeraj P, M.Sc Physics