Question

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

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

Answer 1

The Correct Option is C

Let the each side of square lamina is d. So, $\hspace5mm I_{EF}=I_{GH} \hspace10mm (due \, to \, symmetry) $ and $\hspace5mm I_{AC}=I_{BD} \hspace10mm (due \, to \, symmetry) $ Now, according to theorem of perpendicular axis $\hspace10mm I_{AC}+I_{BD}=I_0$ or $\hspace10mm 2I_{AC}=I_0 \hspace5mm ...(i) $ and $\hspace10mm I_{EF}+I_{GH}=I_0$ or $\hspace10mm 2I_{EF}=I_0 \hspace5mm ...(ii) $ From Eqs. (i) and (ii), we get $I_{AC}=I_{EF}$ $\therefore \hspace5mm I_{AD}=I_{EF}+ \frac {md^2}{4}$ $\hspace5mm =\frac {md^2}{12}+ \frac {md^2}{4} \hspace5mm \bigg (as \, I_{EF}= \frac {md^2}{12}\bigg ) $ So, $\hspace5mm I_{AD}= \frac {md^2}{3}=4I_{EF}$