$A$ rod of mass $m$ and length $L$,pivoted at one of its ends,is hanging vertically. $A$ bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x=x_M$. Then
$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$

$A$ $2 \, kg$ steel rod of length $0.6 \, m$ is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end,the speed of the free end of the rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \, ms^{-1}$. (Take $g = 10 \, ms^{-2}$)

$A$ uniform thin rod of length $2L$ and mass $m$ lies on a horizontal table. $A$ horizontal impulse $J$ is given to the rod at one end. There is no friction. The total kinetic energy of the rod just after the impulse will be

$A$ man is sitting in a smooth groove on a horizontal circular table at the edge by holding a rope joined to the centre. The moment of inertia of the table is $I$. The mass of the man is $M$. The man now pulls the rope so that he comes to the centre. The angular velocity of the table:

The position vector $\vec{r}$ of a particle of mass $m$ is given by the following equation:
$\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}$
where $\alpha = 10/3 \ m \ s^{-3}$,$\beta = 5 \ m \ s^{-2}$ and $m = 0.1 \ kg$. At $t = 1 \ s$,which of the following statement$(s)$ is(are) true about the particle?
$(A)$ The velocity $\vec{v}$ is given by $\vec{v} = (10 \hat{i} + 10 \hat{j}) \ m \ s^{-1}$
$(B)$ The angular momentum $\vec{L}$ with respect to the origin is given by $\vec{L} = -(5/3) \hat{k} \ N \ m \ s$
$(C)$ The force $\vec{F}$ is given by $\vec{F} = (2 \hat{i} + 1 \hat{j}) \ N$
$(D)$ The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau} = -(20/3) \hat{k} \ N \ m$

$A$ square lamina $OABC$ of side length $10 \ cm$ is pivoted at $O$. Forces act on the lamina as shown in the figure. If the lamina remains stationary,then the magnitude of $F$ is: