$I_{\text {ring}}=m a^{2}$
$I_{s q \text { lamina }}=\frac{m(2 a)^{2}}{6}=\frac{2}{3} m a^{2}$
$I_{r o d}=$
$\left.I_{r o d}=I_{1}+I_{2} \text { [perpendicular axis theorem }\right]$
$I_{1}=\left(\frac{m}{4}\right) a^{2}+0 \rightarrow$ axis passing through centre of mass and parallel to axis
$I_{2}=\left(\frac{m}{4}\right) \frac{(2 a)^{2}}{12}$
$I_{r o d}=\left(\frac{m}{4}\right) a^{2}\left[1+\frac{1}{3}\right]=\frac{m a^{2}}{3}$
$I_{4 r o d}=4 I_{r o d}=\frac{4 m a^{2}}{3}$
largest moment of inertia is from rods forming a square of side $2 a .$
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$A$. In uniform circular motion $\vec{\omega}, \vec{v}$ and $\vec{a}$ are always mutually perpendicular
$B$. In non-uniform circular motion, $\vec{\omega}, \vec{v}$ and $\vec{a}$ are always mutually perpendicular
