$A$ uniform disc is rolling on a horizontal surface. At a certain instant,$B$ is the point of contact and $A$ is the topmost point of the disc,where $R$ is the radius of the disc.

$A$ 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:

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 $\frac{1}{6} \ m$

$A$ mass $M = 40 \ kg$ is fixed at the very edge of a long plank of mass $80 \ kg$ and length $1 \ m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a mass of $100 \ kg$ be attached so that the plank starts rotating with an angular acceleration of $1 \ rad/s^2$?

If the earth suddenly stops revolving and all its rotational $KE$ is used up in raising its temperature and if $s$ is taken to be the specific heat of the earth's material,the rise of temperature of the earth will be: ($R =$ radius of the earth and $\omega =$ its angular velocity,$J =$ Joule's constant)

$A$ uniform rod of mass $m$ and length $\ell$ hinged at end $A$ is released from the horizontal position shown in the figure. Just after the rod is released:

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$