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A solid cube of wood of side \( 2 a \) and mass \( M \) is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis \( A B \). A bullet of mass \( m(< < M) \) and speed \( v \) is shot horizontally at the face opposite to \( A B C D \) at a height of \( \frac{4 a}{3} \) from the surface to impart the cube and angular speed \( \omega \). It strikes the face and embeds in the cube. Then \( \omega \) is close to (note: the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is \( \frac{2 M a^{2}}{3} . \) )

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The correct answer is:
\( \frac{m v}{2 M a} \)
Applying the parallel axis theorem, , here is the moment of inertia passing through centre of mass, is the distance between two axes.
The moment of inertia of solid cube about is
The cube is free to rotate about axis. Thus, the angular momentum should be conserved.
Angular momentum , here .
So,
.
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