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A solid cube of wood of side $2 \mathrm{a}$ and mass $\mathrm{M}$ is resting on a horizontal surface as shown in the figure. The cube is free to rotate about the fixed axis $\mathrm{AB}$. A bullet of mass $\mathrm{m}(< < \mathrm{M})$ and speed $\mathrm{v}$ is shot horizontally at the face opposite to $\mathrm{ABCD}$ at a height ' $\mathrm{h}$ ' above the surface to impart the cube an angular speed $\omega_{\mathrm{c}}$ so that the cube just topples over. Then $\omega_{\mathrm{c}}$ is (note : the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is $2 \mathrm{Ma}^{3} / 3$ )
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The correct answer is:
$\sqrt{3 g(\sqrt{2}-1) / 4 a}$
conservation of energy
$$
\begin{array}{l}
\frac{1}{2} \mathrm{I}_{\mathrm{A}} \omega_{\mathrm{c}}^{2}=\mathrm{Mg}(\mathrm{a} \sqrt{2}-\mathrm{a}) \\
\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Ma}^{2}=\frac{2}{3} \mathrm{Ma}^{2}+\mathrm{M}(\mathrm{a} \sqrt{2})^{2} \\
\mathrm{I}_{\mathrm{A}}=\frac{8}{9} \mathrm{Ma}^{2} \\
\frac{1}{2} \frac{8}{3} \mathrm{Ma}^{2} \omega_{\mathrm{c}}^{2}=\operatorname{Mg} \mathrm{a}(\sqrt{2}-1) \\
\omega_{\mathrm{c}}=\sqrt{3 \mathrm{~g}(\sqrt{2}-1) / 4 \mathrm{a}}
\end{array}
$$
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