$A$ rotating body has angular momentum $L$. If its frequency is doubled and its kinetic energy is halved,what will be its new angular momentum?

$STATEMENT-1$ 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.

$A$ thin rod of length '$L$' lies along the $x$-axis with its ends at $x = 0$ and $x = L$. Its linear mass density $\lambda$ varies with $x$ as $\lambda = k{\left( {\frac{x}{L}} \right)^n}$,where $n$ is a non-negative constant. If the position $x_{CM}$ of the center of mass of the rod is plotted against '$n$',which of the following graphs best approximates the dependence of $x_{CM}$ on $n$?

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:

On a solid sphere lying on a horizontal surface,a force $F$ is applied at a height of $R/2$ from the centre of mass. The initial acceleration of a point at the top of the sphere is (there is no slipping at any point).

$A$ solid sphere of mass $m$,radius $R$,having moment of inertia about an axis passing through its center of mass as $I$ is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through the rim (edge) and perpendicular to its plane remains $I$. Then the radius of the disc is: