Two identical rings $A$ and $B$ of same mass $M$ and radius $R$ are revolving. Ring $A$ revolves around its own diameter, and ring $B$ revolves about a tangential axis in its own plane. Both rings $A$ and $B$ have the same rotational kinetic energy. The ratio of the angular velocity of ring $B$ $(\omega_B)$ to that of ring $A$ $(\omega_A)$ is:

State whether the following statements are true or false:
$(1)$ In the law of conservation of momentum, the center of mass velocity remains constant.
$(2)$ "If the resultant of internal forces on a system is zero, then the total linear momentum of the system remains constant." This is the statement of the law of conservation of linear momentum.
$(3)$ The position of the center of mass of a rigid body must always be inside the rigid body.
$(4)$ All particles of a rigid body undergoing rotational motion have the same linear velocity.

The position vectors of two $1 \ kg$ particles,$(A)$ and $(B),$ are given by $\overrightarrow{r}_{A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 \hat{k}) \ m$ and $\vec{r}_B = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k}) \ m$,respectively. Given $\alpha_1 = 1 \ m/s^2, \alpha_2 = 3n \ m/s, \alpha_3 = 2 \ m, \beta_1 = 2 \ m/s, \beta_2 = -1 \ m/s^2, \beta_3 = 4p \ m/s$,where $t$ is time,$n$ and $p$ are constants. At $t = 1 \ s$,$|\overrightarrow{V}_{A}| = |\overrightarrow{V}_{B}|$ and the velocities $\overrightarrow{V}_{A}$ and $\overrightarrow{V}_{B}$ are orthogonal. At $t = 1 \ s$,the magnitude of angular momentum of particle $(A)$ with respect to particle $(B)$ is $\sqrt{L} \ kg \ m^2/s$. The value of $L$ is:

$A$ 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 $g$.

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of [where $g$ is gravitational acceleration]:

Find the increase in the moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.