$A$ solid sphere of mass $M$,radius $R$ and having moment of inertia about an axis passing through the centre of mass as $I$,is recast into a disc of thickness $t$,whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. Then,the radius of the disc will be

$A$ thin rod $AB$ of length $L$ and mass $m$ is held horizontally so that it can freely rotate in a vertical plane about the end $A$ as shown in the figure. The potential energy of the rod when it hangs vertically is taken to be zero. The end $B$ of the rod is released from rest from a horizontal position. At the instant the rod makes an angle $\theta$ with the horizontal:

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 sphere of radius $R$ is placed on a rough horizontal surface and given a linear velocity $v_0$ and angular velocity $\omega_0$ as shown. The sphere comes to rest after moving some distance to the right. It follows that:

Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
$1.$ The ratio of $x_1/x_2$ is
$(A)$ $2$ $(B)$ $1/2$ $(C)$ $\sqrt{2}$ $(D)$ $1/\sqrt{2}$
$2.$ When disc $B$ is brought in contact with disc $A$,they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is
$(A)$ $\frac{2I\omega}{3t}$ $(B)$ $\frac{9I\omega}{2t}$ $(C)$ $\frac{9I\omega}{4t}$ $(D)$ $\frac{3I\omega}{2t}$
$3.$ The loss of kinetic energy during the above process is
$(A)$ $\frac{I\omega^2}{2}$ $(B)$ $\frac{I\omega^2}{3}$ $(C)$ $\frac{I\omega^2}{4}$ $(D)$ $\frac{I\omega^2}{6}$

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: