$A$ uniform disc is rolling on a horizontal surface. At a certain instant,$B$ is the point of contact and $A$ is the topmost point of the disc,where $R$ is the radius of the disc.

$A$ yo-yo is resting on a rough horizontal table. Forces $F_1, F_2$ and $F_3$ are applied separately as shown. The correct statement is

$A$ thin ring of mass $2 \ kg$ has a radius of $0.5 \ m$. It is rolling without slipping on a horizontal plane with a velocity of $1 \ m/s$. $A$ small ball of mass $0.1 \ kg$ moving in the opposite direction with a velocity of $20 \ m/s$ hits the ring at a height of $0.75 \ m$ and moves vertically upward with a velocity of $10 \ m/s$ after the collision. Immediately after the collision:

$A$ 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 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:

The angular momentum of a rigid body of mass $m$ about an axis is $n$ times the linear momentum $(P)$ of the body. The total kinetic energy of the rigid body is: