Question

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

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

Answer 1

The correct option is B $I_{AC}=I_{EF}$

Let the side of square be a.

We know that,
$I_{EF}=\frac{M{a}^2}{12}$
By perpendicular axes theorem, the moment of inertia of lamina about an axis passing through O and perpendicular to plane of lamina
$I_Z=\frac{M{a}^2}{12}+\frac{M{a}^2}{12}=\frac{M{a}^2}{6}.$
Again by perpendicular axes theorem,
$I_{AC}+I_{BD}=I_z$

Since $I_{AC}=I_{BD}$

$\Rightarrow I_{AC}=\frac{I_Z}{2}=\frac{M{a}^2}{12}$
$\therefore I_{AC}=I_{EF}$.