Question

For the given uniform square lamina ABCD, whose centre is O. Then ( Here $I$ denotes moment of inertia)

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

Answer 1

The correct option is C $I_{AD}=4I_{EF}$


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