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Radius of Gyration

The moment of...

Question

The moment of inertia of a circular disc of radius $2 m$ and mass $1 k g$ about an axis passing through its centre of mass and perpendicular to plane is $2 k g - m^{2}$ . Its moment of inertia about an axis parallel to this axis and passing through its edge in $k g - m^{2}$ is

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Solution

The correct option is C $6$
$\begin{matrix} B y a p p l y i n g t h e t h e o r e m o f p a r a l l e l a x e s m o m e n t o f i n e r t i a^{'} I^{'} o f a b o d y i s I = I g + M R^{2} w h e r e I g i s t h e m o m e n t o f i n e r t i a a n o u t a n a x i s p a s sin g t h r o u g h t h e c e n t r e o f m a s s b u t p e r p e n d i c u l a r t o t h e p l a n e o f t h e b o d y p l u s M R^{2} w h e r e M i s t h e m a s s o f t h e b o d y a n d R i s t h e d i s tan c e b e t w e e n t h e a x i s a n d t h e c e n t r e o f m a s s . I = 2 + 1 \times 2^{2} = 6 k g m^{2} \end{matrix}$
Hence,
option $(C)$ is correct answer.

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