This is a case of sliding (the plane being frictionless) and therefore the acceleration of all the bodies is same $$\left( {g\,\sin \,\theta } \right).$$

$$\eqalign{ & K.E.\, = \frac{1}{2}I{\omega ^2},\,\,{\text{but}}\,L = I\omega \Rightarrow I = \frac{L}{\omega } \cr & \therefore K.E. = \frac{1}{2}\frac{L}{\omega } \times {\omega ^2} = \frac{1}{2}L\omega \cr & \therefore \frac{{K.E.}}{{K.E.}} = \frac{{L \times \omega }}{{L' \times \omega '}} \cr & \Rightarrow \frac{{K.E.}}{{\frac{{K.E.}}{2}}} = \frac{{L \times \omega }}{{L' \times 2\omega }} \cr & \therefore L' = \frac{L}{4} \cr} $$

$$\eqalign{ & {\text{Torque at angle }}\theta \cr & \tau = Mg\,\sin \,\theta .\frac{l}{2} \cr & {\text{Also }}\tau = l\alpha \cr & \therefore l\alpha = Mg\,\sin \,\theta .\frac{l}{2} \cr & \frac{{M{l^2}}}{3}.\alpha = Mg\,\sin \,\theta .\frac{l}{2}\,\,\,\left[ {\because {I_{rod}} = \frac{{M{l^2}}}{3}} \right] \cr & \Rightarrow \frac{{l\alpha }}{3} = g\frac{{\sin \,\theta }}{2} \cr & \therefore \alpha = \frac{{3g\sin \,\theta }}{{2l}} \cr} $$

Centre of mass is at rest
$$\therefore {V_{cm}} = 0$$

By the theorem of perpendicular axes,
$$\eqalign{ & {I_z} = {I_x} + {I_y} \cr & {\text{or}},\,\,{I_z} = 2{I_y} \cr & \left( {\because \,\,{I_x} = {I_y}{\text{ by symmetry of the figure}}} \right) \cr & \therefore {I_{EF}} = \frac{{{I_z}}}{2}\,\,.....(i) \cr} $$
Again, by the same theorem
$$\eqalign{ & {I_z} = {I_{AC}} + {I_{BD}} = 2{I_{AC}} \cr & \left( {\therefore \,{I_{AC}} = {I_{BD}}\,\,{\text{by symmetry of the figure}}} \right) \cr & \therefore {I_{AC}} = \frac{{{I_z}}}{2}\,\,.....(ii) \cr} $$
from (i) and (ii), we get $${I_{EF}} = {I_{AC}}$$

Given,
Moment of inertia, $$I = 1.2\,kg - {m^2}$$
Rotational kinetic energy, $${K_r} = 1500\,J$$
Angular acceleration $$\alpha = 25\,rad/{s^2},\,{\omega _0} = 0,t = ?$$
Kinetic energy of rotation is given by $${K_{{\text{rot}}}} = \frac{1}{2}I{\omega ^2}$$
$$\eqalign{ & \therefore \omega = \sqrt {\frac{{2{K_r}}}{I}} = \sqrt {\frac{{2 \times 1500}}{{1.2}}} \cr & = 50\,rad/s \cr} $$
Now, from equation of rotational motion
$$\eqalign{ & \omega = {\omega _0} + \alpha t \cr & t = \frac{{\omega - {\omega _0}}}{\alpha } \cr & = \frac{{50 - 0}}{{25}} \cr & = 2\,s \cr} $$

The $$M.I.$$  about the axis of rotation is not constant as the perpendicular distance of the bead with the axis of rotation increases.
Also since no external torque is acting.
$$\eqalign{ & \therefore {\tau _{{\text{ext}}}} = \frac{{dL}}{{dt}} \cr & \Rightarrow L = {\text{constant}}\,\,\,\,\, \Rightarrow I\omega = {\text{constant}} \cr} $$
Since, $$I$$  increases, $$\omega $$  decreases.

Let speed of belt be $$v$$ Angular speeds of wheels
$${\omega _B} = \frac{v}{{2\pi {R_B}}},{\omega _A} = \frac{v}{{2\pi {R_A}}};\frac{{{\omega _A}}}{{{\omega _B}}} = \frac{{{R_B}}}{{{R_A}}} = 2$$

$$\eqalign{ & mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}{I_{cm}}{\omega ^2} \cr & = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}{I_{cm}}{\left( {\frac{{{v_{cm}}}}{R}} \right)^2} \cr & = \frac{1}{2}\left( {m + \frac{{{I_{cm}}}}{{{R^2}}}} \right)v_{cm}^2 \cr & = \frac{{1m}}{2}\left[ {1 + \frac{{{K^2}}}{{{R^2}}}} \right] \cr & \therefore h \propto 1 + \frac{{{K^2}}}{{{R^2}}} \cr & {\text{For ring : }}h \propto 2\,\,\,\,\,\left( {\because K = R} \right) \cr & {\text{For solid cylinder, }}h \propto \frac{3}{2}\,\,\,\,\,\left( {\because K = \frac{R}{{\sqrt 2 }}} \right) \cr & {\text{For solid Sphere, }}h \propto \frac{7}{5}\,\,\,\,\,\left( {\because K = \sqrt {\frac{2}{5}} R} \right) \cr & {\text{Ratio of heights }}2:\frac{3}{2}:\frac{7}{5}\,\,\,\, \Rightarrow 20:15:14 \cr} $$

If no external force acts on a system of particles, the centre of mass remains at rest. So, speed of centre of mass is zero.