The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes:

$A$ non-uniform cylinder of mass $m$,length $l$,and radius $r$ has its center of mass at a distance $l/4$ from the geometric center,lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density $\rho$. The moment of inertia of the cylinder about its center of mass is $I$. The angular acceleration of the cylinder just after it is released from the horizontal position shown in the figure is:

$A$ block of mass $m = 1 \, kg$ slides with velocity $v = 6 \, m/s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision,making an angle $\theta$ before momentarily coming to rest. If the rod has mass $M = 2 \, kg$ and length $l = 1 \, m$,the value of $\theta$ is approximately (Take $g = 10 \, m/s^2$) (in $^{\circ}$)

$A$ rod of length $1\,m$ is standing vertically. When its upper end is released and it falls such that the lower end touches the ground without slipping,the speed of the upper end when it hits the ground is:

The position of an object having mass $0.1 \text{ kg}$ as a function of time $t$ is given as $\vec{r} = (10t^2\hat{i} + 5t^3\hat{j}) \text{ m}$. At $t = 1 \text{ s}$, which of the following statements are correct?
$A$. The linear momentum $\vec{p} = (2\hat{i} + 1.5\hat{j}) \text{ kg} \cdot \text{m/s}$.
$B$. The force acting on the object $\vec{F} = (2\hat{i} + 3\hat{j}) \text{ N}$.
$C$. The angular momentum of the object about its origin $\vec{L} = 15\hat{k} \text{ J} \cdot \text{s}$.
$D$. The torque acting on the object about its origin $\vec{\tau} = 20\hat{k} \text{ N} \cdot \text{m}$.
Choose the correct answer from the options given below:

$A$ uniform rod of length $l$ and density $\rho$ is revolving about a vertical axis passing through its one end. If $\omega$ is the angular velocity of the rod,then the centrifugal force per unit area of the rod is