Question

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

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

Answer 1

Answer: (D)

Let $I$ be the moment of inertia of the square lamina
about an axis through $O$ and perpendicular to its plane. Then by
perpendicular axes theorem, $I_{AC}+I_{BD}=I$
$\text{ or }{I}_{AC}=\frac{I}{2}\left[\because I_{AC}=I_{BD}\right]$
Again by perpendicular axes theorem,
$I_{EF}+I_{GH}=I$
or $I_{EF}=\frac{I}{2}\left[\because I_{EF}=I_{GH}\right]$
Hence $I_{AC}=I_{FF}$