Question

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

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

Answer 1

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

The moment of inertia of a square lamina about an axis perpendicular to its plane, $I_z=\frac{M{L}^2}{6}$

Now, we know that diagonals $AC$ and $BD$ are perpendicular for a square.
Therefore, by perpendicular axis theorem,
$I_{AC}+I_{BD}=\frac{M{L}^2}{6}$, as $I_{AC}=I_{BD}$
$\Rightarrow I_{AC}=\frac{M{L}^2}{12}$​
Similarly, $EF$ and $GH$ are also perpendicular to each other, then by perpendicular axis theorem,
$2I_{EF}=\frac{M{L}^2}{6}\Rightarrow I_{EF}=\frac{M{L}^2}{12}$​
$\therefore I_{AC}=I_{EF}$