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A thin uniform rectangular plate of mass $2 \mathrm{~kg}$ is placed in $x y$-plane as shown in figure. The moment of inertial about $x$-axis is $I_{x}=0.2 \mathrm{kgm}^{2}$ and the moment of inertia about $Y$-axis is $I_{y}=0.3 \mathrm{kgm}^{2}$. The radius of gyration of the plate about the axis passing through $O$ and perpendicular to the plane of the plate is
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The correct answer is:
$50 \mathrm{~cm}$
Given, mass, $M=2 \mathrm{~kg}$
$\begin{aligned}
&I_{x}=0.2 \mathrm{~kg}-\mathrm{m}^{2} \\
&I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}$
$\text { and } \quad I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}$
According to the perpendicular axis theorem, the moment of inertia of the rectangular plate about an axis passing through $O$ and perpendicular to the plane of plate is
$I=I_{x}+I_{y}=0.2+0.3$
$=0.5 \mathrm{~kg} \mathrm{~m}{ }^{2}$
Also, $I=M k^{2}$
where, $k=$ radius of gyration.
$\Rightarrow \quad k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.5}{2}}=\sqrt{0.25}=0.5 \mathrm{~m}$ or $50 \mathrm{~cm}$
$\begin{aligned}
&I_{x}=0.2 \mathrm{~kg}-\mathrm{m}^{2} \\
&I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}$
$\text { and } \quad I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}$
According to the perpendicular axis theorem, the moment of inertia of the rectangular plate about an axis passing through $O$ and perpendicular to the plane of plate is
$I=I_{x}+I_{y}=0.2+0.3$
$=0.5 \mathrm{~kg} \mathrm{~m}{ }^{2}$
Also, $I=M k^{2}$
where, $k=$ radius of gyration.
$\Rightarrow \quad k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.5}{2}}=\sqrt{0.25}=0.5 \mathrm{~m}$ or $50 \mathrm{~cm}$
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