Three solid spheres each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side length $d$. The ratio $I_{0} / I_{A}$ of the moment of inertia $I_{0}$ of the system about an axis passing through the centroid and perpendicular to the plane of the triangle,to the moment of inertia $I_{A}$ about an axis passing through the center of any one of the spheres and perpendicular to the plane of the triangle,is:

$A$ ring starting from rest rotates under a constant angular acceleration of $8 \ rad \ s^{-2}$ due to an applied torque. How many revolutions will the ring complete in $5 \ s$? How many revolutions will it complete in the $6^{th}$ second? If the torque becomes zero after $6 \ s$,how many revolutions will the ring complete in the $7^{th}$ second?

$A$ thin circular coin of mass $5 \text{ g}$ and radius $4/3 \text{ cm}$ is initially in a horizontal $xy$-plane. The coin is tossed vertically up ($+z$ direction) by applying an impulse of $J = \sqrt{\frac{\pi}{2}} \times 10^{-2} \text{ N-s}$ at a distance $r = 2/3 \text{ cm}$ from its center. The coin spins about its diameter and moves along the $+z$ direction. By the time the coin reaches back to its initial position,it completes $n$ rotations. The value of $n$ is. . . . . [Given: The acceleration due to gravity $g = 10 \text{ m/s}^2$]

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is $I$. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis,then the loss in kinetic energy is:

$A$ block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially,the right edge of the block is at $x=0$,in a coordinate system fixed to the table. $A$ point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block,its position is $x$ and the velocity is $v$. At that instant,which of the following options is/are correct?
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$.
$[B]$ The position of the point mass is: $x=-\sqrt{2} \frac{mR}{M+m}$.
$[C]$ The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
$[D]$ The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.

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: