$A$ solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown below. The cube is free to rotate about the fixed axis $AB$. $A$ bullet of mass $m (< M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height $h$ above the surface to impart the cube an angular speed $\omega_{c}$,so that the cube just topples over. Then,$\omega_{c}$ is (Note: the moment of inertia of the cube about an axis passing through the centre of mass and parallel to the edge is $2Ma^{2}/3$)

The figure shows a system consisting of $(i)$ a ring of outer radius $3R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and $(ii)$ an inner disc of radius $2R$ rotating anti-clockwise with angular speed $\omega/2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at distance $R$ from the origin,where $OP$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,
$(A)$ the point $O$ has linear velocity $3R\omega\hat{i}$.
$(B)$ the point $P$ has a linear velocity $\frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(C)$ the point $P$ has linear velocity $\frac{13}{4}R\omega\hat{i} - \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(D)$ The point $P$ has a linear velocity $(3 - \frac{\sqrt{3}}{4})R\omega\hat{i} + \frac{1}{4}R\omega\hat{k}$.

$A$ particle of mass $M=0.2 \ kg$ is initially at rest in the $xy$-plane at a point $(x=-l, y=-h)$,where $l=10 \ m$ and $h=1 \ m$. The particle is accelerated at time $t=0$ with a constant acceleration $a=10 \ m/s^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin,in $SI$ units,are represented by $\vec{L}$ and $\vec{\tau}$,respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x, y$ and $z$-directions,respectively. If $\hat{k}=\hat{i} \times \hat{j}$,then which of the following statement$(s)$ is(are) correct?
$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t=2 \ s$.
$(B)$ $\vec{\tau}=2 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(C)$ $\vec{L}=4 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(D)$ $\vec{\tau}=\hat{k}$ when the particle passes through the point $(x=0, y=-h)$.

Three solid spheres each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side length $d$. The ratio $I_{0} / I_{A}$ of the moment of inertia $I_{0}$ of the system about an axis passing through the centroid and perpendicular to the plane of the triangle,to the moment of inertia $I_{A}$ about an axis passing through the center of any one of the spheres and perpendicular to the plane of the triangle,is:

$A$ $2 \, kg$ steel rod of length $0.6 \, m$ is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end,the speed of the free end of the rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \, ms^{-1}$. (Take $g = 10 \, ms^{-2}$)

$A$ body with a moment of inertia of $3 \ kg \cdot m^2$ rotating with an angular speed of $2 \ rad/s$ has the same kinetic energy as a mass of $12 \ kg$ moving with a speed of ......... $m/s$.