
- A$I_{AC}$ =$\sqrt 2 $$I_{EF}$
- B$\sqrt 2 $$I_{AC}=I_{EF}$
- C$I_{AO}=3I_{EF}$
- ✓$I_{AC}=I_{EF}$

${I_z} = {I_x} + {I_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or,\,\,{I_z} = 2{I_y}$
$\begin{array}{l}
\left( {\,{I_x} = {I_y}\,by\,symmetry\,of\,the\,figure} \right)\\
\therefore \,{I_{EF}} = \frac{{{I_z}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
\end{array}$
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 ${I_{EF}} = {I_{AC}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The speed of the block when it reaches the point $Q$ is :
$(A)$ $5 \ ms ^{-1}$ $(B)$ $10 \ ms ^{-1}$$(C)$ $10 \sqrt{3} \ ms ^{-1}$ $(D)$ $20 \ ms ^{-1}$
$2.$ The magnitude of the normal reaction that acts on the block at the point $Q$ is :
$(A)$ $7.5 \ N$ $(B)$ $8.6 \ N$ $(C)$ $11.5 \ N$ $(D)$ $22.5 \ N$
Give the answer question $1$ and $2.$

