$A$ 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:

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:

$A$ flat surface of a thin uniform disk $A$ of radius $R$ is glued to a horizontal table. Another thin uniform disk $B$ of mass $M$ and with the same radius $R$ rolls without slipping on the circumference of $A$,as shown in the figure. $A$ flat surface of $B$ also lies on the plane of the table. The center of mass of $B$ has a fixed angular speed $\omega$ about the vertical axis passing through the center of $A$. The angular momentum of $B$ is $n M \omega R^2$ with respect to the center of $A$. Which of the following is the value of $n$?

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 $3$.

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

$A$ uniform rod of length $l$ and density $\rho$ is revolving about a vertical axis passing through its one end. If $\omega$ is the angular velocity of the rod,then the centrifugal force per unit area of the rod is