Consider a body of mass $1.0 \ kg$ at rest at the origin at time $t=0$. $A$ force $\overrightarrow{F}=(\alpha t \hat{i}+\beta \hat{j})$ is applied on the body,where $\alpha=1.0 \ Ns^{-1}$ and $\beta=1.0 \ N$. The torque acting on the body about the origin at time $t=1.0 \ s$ is $\vec{\tau}$. Which of the following statements is (are) true?
$(A)$ $|\vec{\tau}|=\frac{1}{3} \ Nm$
$(B)$ The torque $\vec{\tau}$ is in the direction of the unit vector $+\hat{k}$
$(C)$ The velocity of the body at $t=1 \ s$ is $\overrightarrow{v}=\frac{1}{2}(\hat{i}+2 \hat{j}) \ ms^{-1}$
$(D)$ The magnitude of displacement of the body at $t=1 \ s$ is $\frac{1}{6} \ m$

$Assertion$ : If polar ice melts, days will be shorter.
$Reason$ : Moment of inertia decreases and thus angular velocity increases.

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?

One twirls a circular ring (of mass $M$ and radius $R$) near the tip of one's finger as shown in Figure $1$. In the process,the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone,shown by the dotted line. The radius of the path traced out by the point where the ring and the finger are in contact is $r$. The finger rotates with an angular velocity $\omega_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger are in contact (Figure $2$). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.
$(1)$ The total kinetic energy of the ring is
$[A]$ $M \omega_0^2 R^2$ $[B]$ $\frac{1}{2} M \omega_0^2(R-r)^2$ $[C]$ $M \omega_0^2(R-r)^2$ $[D]$ $\frac{3}{2} M \omega_0^2(R-r)^2$
$(2)$ The minimum value of $\omega_0$ below which the ring will drop down is
$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$ $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$ $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$ $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$
Given the answers to questions $(1)$ and $(2)$:

$A$ hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?