For the pivoted slender rod of length $l$ as shown in the figure,the angular velocity as the bar reaches the vertical position after being released from the horizontal position is:

$A$ particle performs rotational motion with an angular momentum $L$. If the frequency of rotation is doubled and its kinetic energy becomes one-fourth,the new angular momentum becomes:

In the figure shown,the plank is being pulled to the right with a constant speed $v$. If the cylinder does not slip,then:

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 $3$.

$A$ uniform body of mass $M$ and radius $R$ has a small mass $m$ attached at its edge as shown in the figure. The system is placed on a perfectly rough horizontal surface such that mass $m$ is at the same horizontal level as the centre of the body. It is assumed that there is no slipping at point $A$. If $I_A$ is the moment of inertia of the combined system about the point of contact $A$,then the normal reaction at point $A$ just after the system is released from rest is ........ $N$. ($M = 6 \ kg$,$m = 2 \ kg$,$I_A = 4 \ kg \ m^2$,$R = 1 \ m$,$g = 10 \ m/s^2$)

$A$ thin uniform straight rod of mass $2 \, kg$ and length $1 \, m$ is free to rotate about its upper end when at rest. It receives an impulsive blow of $10 \, Ns$ at its lowest point,normal to its length as shown in the figure. The kinetic energy of the rod just after impact is ........ $J$.