$A$ body with a moment of inertia of $3\ kg\cdot m^2$ rotating with an angular velocity of $2\ rad/s$ has the same kinetic energy as a mass of $12\ kg$ moving with a velocity of .......... $m/s$.

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 rings are rotating about the same axis, then the loss in kinetic energy is

$A$ $2 \, kg$ steel rod of length $0.6 \, m$ is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end,the speed of the free end of the rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \, ms^{-1}$. (Take $g = 10 \, ms^{-2}$)

$A$ hollow sphere of radius $R$ and mass $m$ is fully filled with water of mass $m$. It is rolled down a horizontal plane such that its centre of mass moves with a velocity $v$. If it purely rolls,which of the following statements is correct?

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_{0}$ in the $x-y$ plane as shown in the figure. For time $t < \frac{v_{0} \sin \theta}{g}$,the angular momentum of the particle is (where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$ axes respectively):

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 $3$.