$A$ uniform rod $AB$ of mass $2 \ kg$ and length $30 \ cm$ is at rest on a smooth horizontal surface. An impulse of force $0.2 \ Ns$ is applied to end $B$. The time taken by the rod to turn through a right angle will be $\frac{\pi}{X} \ s$,where $X = \text{ . . . . . . }$.

$A$ solid sphere $(A)$ of mass $5m$ and a spherical shell $(B)$ of mass $m$, both having the same radius $R$, are placed on a rough surface. When a force $F$ of the same magnitude is applied tangentially at the highest points of $A$ and $B$, they start rolling without slipping with accelerations $a_A$ and $a_B$, respectively. The ratio of $a_A$ to $a_B$ is . . . . . . .

$A$ uniform body of mass $M$ and radius $R$ has a small mass $m$ attached at its edge as shown in the figure. The system is placed on a perfectly rough horizontal surface such that mass $m$ is at the same horizontal level as the centre of the body. It is assumed that there is no slipping at point $A$. If $I_A$ is the moment of inertia of the combined system about the point of contact $A$,then the normal reaction at point $A$ just after the system is released from rest is ........ $N$. ($M = 6 \ kg$,$m = 2 \ kg$,$I_A = 4 \ kg \ m^2$,$R = 1 \ m$,$g = 10 \ m/s^2$)

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:

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:

$STATEMENT-1$ If there is no external torque on a body about its center of mass,then the velocity of the center of mass remains constant. because
$STATEMENT-2$ The linear momentum of an isolated system remains constant.