Two discs have moments of inertia $I_{1}$ and $I_{2}$ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds $\omega_{1}$ and $\omega_{2}$ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

The portion $AB$ of the wedge shown in the figure is rough and $BC$ is smooth. $A$ solid cylinder rolls without slipping from $A$ to $B$. The ratio of translational kinetic energy to rotational kinetic energy,when the cylinder reaches point $C$,is:

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then, the rise in temperature of the sphere is

$A$ wheel of radius $R$ and mass $M$ is placed at the bottom of a fixed step of height $R$ as shown in the figure. $A$ constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $\tau$ about an axis normal to the plane of the paper passing through the point $Q$. Which of the following options is/are correct?

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 $(2)$:

$A$ rigid uniform bar $AB$ of length $L$ is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time,the angle made by the bar with the vertical is $\theta$. Which of the following statements about its motion is/are correct?
$[A]$ The midpoint of the bar will fall vertically downward
$[B]$ The trajectory of the point $A$ is a parabola
$[C]$ Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$
$[D]$ When the bar makes an angle $\theta$ with the vertical,the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$