$A$ 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?

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third,the new angular momentum becomes:

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule,assuming it to be rigid. The rule to be applied is Bohr's quantization condition.
$1.$ $A$ diatomic molecule has a moment of inertia $I$. By Bohr's quantization condition,its rotational energy in the $n^{\text{th}}$ level $(n=1, 2, 3, \dots)$ is:
$(A) \frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)$ $(B) \frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)$ $(C) n\left(\frac{h^2}{8 \pi^2 I}\right)$ $(D) n^2\left(\frac{h^2}{8 \pi^2 I}\right)$
$2.$ It is found that the excitation frequency from the ground state $(n=1)$ to the first excited state $(n=2)$ of rotation for the $CO$ molecule is close to $\frac{4}{\pi} \times 10^{11} \text{ Hz}$. Then the moment of inertia of the $CO$ molecule about its centre of mass is close to (Take $h=2 \pi \times 10^{-34} \text{ Js}$):
$(A) 2.76 \times 10^{-46} \text{ kg m}^2$ $(B) 1.87 \times 10^{-46} \text{ kg m}^2$ $(C) 4.67 \times 10^{-47} \text{ kg m}^2$ $(D) 1.17 \times 10^{-47} \text{ kg m}^2$
$3.$ In a $CO$ molecule,the distance between $C$ (mass $= 12 \text{ a.m.u.}$) and $O$ (mass $= 16 \text{ a.m.u.}$),where $1 \text{ a.m.u.} = \frac{5}{3} \times 10^{-27} \text{ kg}$,is close to:
$(A) 2.4 \times 10^{-10} \text{ m}$ $(B) 1.9 \times 10^{-10} \text{ m}$ $(C) 1.3 \times 10^{-10} \text{ m}$ $(D) 4.4 \times 10^{-11} \text{ m}$
Give the answer for questions $1, 2,$ and $3$.

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

$A$ horizontal platform is rotating with a uniform angular velocity about a vertical axis passing through its center. At some instant, a viscous liquid of mass $m$ is dropped at its center, which is free to spread and eventually falls off the edge. During this time interval, the angular velocity

$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speed $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after collision. The total rotational kinetic energy about the centre of mass $c$ will be: