For the given uniform square lamina ABCD, whose centre is O,

Options

A. `I_(AC) = I_(EF)`
B. `sqrt(2) I_(AC) = I_(EF)`
C. `I_(AD) = 3 I_(EF)`
D. `I_(AB) = sqrt(2) I_(EF)`

Answer

A

Explanation

(a) Let the each side of square lamina is `d`.
So, `I_(EF) = I_(GH)` (due to symmetry)
and `I_(AC) = I_(BD)` (due to symmetry)
Now, according to theorem of perpendicular axis,
`I_(AC) + I_(BD) = I_0`
`rArr 2 I_(AC) = I_0` …(i)
and `I_(EF) + I_(GH) = I_0`
`rArr 2 I_(EF) = I_0` ...(ii)
From Eqs. (i) and (ii), we get `I_(AC) = I_(EF)`
:. `I_(AD) = I_(EF) + (md^2)/(4)`
=`(md^2)/(12) + (md^2)/(4) (as I_(EF) = (md^2)/(12))`
So, `I_(AD) = (md^2)/(3) = 4 I_(EF)`.
.