MCQ
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$I_{AO}=3I_{EF}$
- ✓$I_{AC}=I_{EF}$
✓
Answer
Correct option: D.
$I_{AC}=I_{EF}$
d
By the theorem of perpendicular axes,
By the theorem of perpendicular axes,
${I_z} = {I_x} + {I_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or,\,\,{I_z} = 2{I_y}$
$\begin{array}{l}
\left( {\,{I_x} = {I_y}\,by\,symmetry\,of\,the\,figure} \right)\\
\therefore \,{I_{EF}} = \frac{{{I_z}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
\end{array}$
Again, by the same theorem
$\begin{array}{l}
{I_z} = {I_{AC}} + {I_{BD}} = 2{I_{AC}}\\
\left( {\therefore \,{I_{AC}} = {I_{BD}}\,by\,symmetry\,of\,the\,figure} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,{I_{AC}} = \frac{{{I_z}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
\end{array}$
From $(i)$ and $(ii)$, we get ${I_{EF}} = {I_{AC}}$
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