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The moment of inertia of a circular disc of radius $2 \mathrm{~m}$ and mass $1 \mathrm{~kg}$ about an axis passing through the centre of mass but perpendicular to the plane of the disc is $2 \mathrm{~kg} \mathrm{~m}^{2}$. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is (see the given figure).
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Verified Answer
The correct answer is:
$6 \mathrm{~kg} \mathrm{~m}^{2}$
According to parallel axis theorem
$$
\begin{aligned}
\mathrm{I}_{X^{\prime} Y^{\prime}} &=I_{X Y}+M^{2} \\
&=2+(1)(2)^{2} \\
&=6 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{I}_{X^{\prime} Y^{\prime}} &=I_{X Y}+M^{2} \\
&=2+(1)(2)^{2} \\
&=6 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}
$$
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