$A$ bar of length $L$ carrying a small mass $m$ at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the midpoint of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with the same $\omega$. The mass moves vertically up, comes back, and reaches the bar at the same point. At that place, the acceleration due to gravity is $g$.

According to the parallel axis theorem,$I = I_{cm} + Mx^2$. What will be the graph between $I$ and $x$?

$A$ uniform rod of length $l$ and density $\rho$ is revolving about a vertical axis passing through its one end. If $\omega$ is the angular velocity of the rod,then the centrifugal force per unit area of the rod is

$A$ 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 $(x=0, y=-h)$.

$A$ ring of mass $M$ and radius $R$ sliding with a velocity $v_0$ suddenly enters a rough surface where the coefficient of friction is $\mu$,as shown in the figure. Choose the correct statement$(s)$.

The total kinetic energy of a body of mass $10 \ kg$ and radius $0.5 \ m$ moving with a velocity of $2 \ m/s$ without slipping is $32.8 \ J$. The radius of gyration of the body is .......... $m$.