Question

For the given uniform square lamina $ABCD$ of side length $l$, whose centre is $\text{O}$, identify the correct option.

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

Answer 1

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

The moment of inertia of a square lamina of side $l$ about an axis perpendicular to its plane and passing through its centre $\text{O}$ is
$I_O=\frac{m{l}^2}{6}$

The axis $GH$ and $EF$ are perpendicular, hence by perpendicular axis theorem,
$I_O=I_{EF}+I_{GH}$
$\because I_{EF}=I_{GH}$ due to symmetry of square plate
$2I_{EF}=\frac{m{l}^2}{6}$
$\therefore I_{EF}=I_{GH}=\frac{m{l}^2}{12}...\left(i\right)$

The diagonal axis $BD$ and $AC$ are also perpendicular and axis through $\text{O}$ is $\perp$ to the plane of the plate, hence by perpendicular axis theorem,
$I_O=I_{AC}+I_{BD}$
$\because I_{AC}=I_{BD}$ due to symmetry of square plate.
$\Rightarrow 2I_{AC}=I_O$
$\therefore I_{AC}=I_{BD}=\frac{m{l}^2}{12}...\left(ii\right)$

From Eq. $\left(i\right)$ and $\left(ii\right)$
$I_{EF}=I_{AC}$