Correct Answer is Option (A)

We know Rotational Kinetic Energy$=\frac{1}{2}I{\omega }^2,$

Angular Momentum $L=I\omega \Rightarrow I=\frac{L}{\omega }$

$\therefore$ Initial $K.E.=\frac{1}{2}\frac{L}{\omega }\times {\omega }^2=\frac{1}{2}L\omega$

Final $K.E^{\prime }$ = $\frac{K.E}{2}$ = $\frac{1}{2}{L}^{\prime }\times 2\omega$

$\therefore$ $\frac{K.E}{K.E^{\prime }}=\frac{L\times \omega }{L^{\prime }\times {\omega }^{\prime }}$

$\Rightarrow \frac{K.E}{\frac{K.E}{2}}=\frac{L\times \omega }{L^{\prime }\times 2\omega }$

$\therefore$ $L^{\prime }=\frac{L}{4}$

Correct Answer is Option (C)

When two small spheres of mass $m$ are attached gently, the external torque, about the axis of rotation, is zero.

So, $\frac{d\overrightarrow{L}}{dt}=\bar{z}$ = 0

$\overrightarrow{L}$ = conserved

So the angular momentum about the axis of rotation is conserved.

$\therefore$ $I_1{\omega }_1=I_2{\omega }_2$

$\Rightarrow {\omega }_2=\frac{I_1}{I_2}{\omega }_1$

Here Moment of inertia of Disc $I_1=\frac{1}{2}M{R}^2$ and

After adding two sphere Moment of Inertia of disc and two sphere,

$I_2=\frac{1}{2}M{R}^2+$$2\left(\frac{1}{2}m{R}^2+\frac{1}{2}m{R}^2\right)$

$\therefore$ ${\omega }_2=\frac{\frac{1}{2}M{R}^2}{\frac{1}{2}MR+2m{R}^2}\times {\omega }_1=\frac{M}{M+4m}{\omega }_1$

Correct Answer is Option (D)

Angular momentum $(L)$

$=$ ( linear momentum ) $\times$ ( perpendicular distance of the line of action of momentum from the axis of rotation)

$=mv\times r$

$=mv\times 0$

$=0$

[ Here $r=0$ because the particle is moving through the line PQ and r is the perpendicular distance from line PQ of the particle ]