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The moment of inertia of a thin circular disc about an axis passing through its centre and perpendicular to its plane is $I$. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is
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The correct answer is:
$\frac{5}{2} I$
Moment of inertia of a circular disc about an axis passing through centre of gravity and perpendicular to its plane
$$
I=\frac{1}{2} M R^2
$$
From Eq. (i) $\quad M R^2=2 I$
Then, moment of inertia of disc about tangent
$$
\begin{aligned}
\text { in a plane }=\frac{5}{4} & M R^2 \\
& =\frac{5}{4}(2 I) \\
& =\frac{5}{2} I
\end{aligned}
$$
$$
I=\frac{1}{2} M R^2
$$
From Eq. (i) $\quad M R^2=2 I$
Then, moment of inertia of disc about tangent
$$
\begin{aligned}
\text { in a plane }=\frac{5}{4} & M R^2 \\
& =\frac{5}{4}(2 I) \\
& =\frac{5}{2} I
\end{aligned}
$$
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