$A$ wheel of moment of inertia $2 \ kg \ m^2$ is rotating about an axis passing through its centre and perpendicular to its plane at a speed of $60 \ rad \ s^{-1}$. Due to friction, it comes to rest in $5$ minutes. The angular momentum of the wheel three minutes before it stops rotating is:

In the following problems,indicate the correct direction of the friction force acting on the cylinder,which is pulled on a rough surface by a constant force $F$. $A$ cylinder is pulled horizontally by a force $F$ acting at a point above the centre of mass of the cylinder,as shown in the figure. The friction force can be given by which of the following diagrams?

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:

Fill in the blanks:
$(1)$ If $|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}$,then the angle between $\vec{A}$ and $\vec{B}$ is ............ .
$(2)$ The angle between the angular momentum and linear momentum of a particle in rotational motion is ............ .
$(3)$ $A$ force $F\hat{k}$ acts on a particle having position vector $(2\hat{i} + \hat{j})$. The torque acting on the particle is ............ .

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$,perpendicular to the rod in the horizontal direction. Find the speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $t = \frac{\pi m l}{12 J}$.

$A$ ring and a disc have the same mass and the same radius. The ratio of the moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is: (in $: 1$)