Question

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

• A. $\sqrt{2}{I}_{AC}=I_{EF}$
• B. $I_{AD}=3I_{EF}$
• C. $I_{AD}=4I_{EF}$
• D. $I_{AC}=\sqrt{2}{I}_{EF}$

Answer 1

Answer: (C)

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
$I_{AC}+I_{BD}=I_0$
or $2I_{AC}=I_0...(i)$
and $I_{EF}+I_{GH}=I_0$
or $2I_{EF}=I_0...(ii)$
From Eqs. (i) and (ii), we get $I_{AC}=I_{EF}$
$\therefore I_{AD}=I_{EF}+\frac{m{d}^2}{4}$
$=\frac{m{d}^2}{12}+\frac{m{d}^2}{4}\left(as{I}_{EF}=\frac{m{d}^2}{12}\right)$
So, $I_{AD}=\frac{m{d}^2}{3}=4I_{EF}$