Given that, the rod is of uniform mass density and $$AB=BC$$

Let mass of one rod is $$m.$$
Balancing torque about hinge point.
$$\eqalign{ & mg\left( {{C_1}P} \right) = mg\left( {{C_2}N} \right) \cr & mg\left( {\frac{L}{2}\sin \,\theta } \right) = mg\left( {\frac{L}{2}cos\,\theta - L\,\sin \,\theta } \right) \cr & \Rightarrow \frac{3}{2}mgL\,\sin \,\theta = \frac{{mgL}}{2}\cos \,\theta \cr & \Rightarrow \frac{{\sin \,\theta }}{{\cos \,\theta }} = \frac{1}{3}\,\,\,\,or,\tan \,\theta = \frac{1}{3} \cr} $$

Moment of inertia of the triangular plate is maximum about the shortest side because effective distance of mass distribution about this side is maximum. Since, distances of centre of mass from the sides are related as $${x_{BC}} < {x_{AB}} < {x_{AC}}$$
Therefore $${I_{BC}} > {I_{AB}} > {I_{AC}}\,\,{\text{or}}\,\,{I_{BC}} > {I_{AC}}$$

This is a standard formula and should be memorized.
$$a = \frac{{g\,\sin \theta }}{{1 + \frac{I}{{M{R^2}}}}}$$

Let $$\sigma $$ be the mass per unit area.

The total mass of the disc $$ = \sigma \times \pi {R^2} = 9M$$
The mass of the circular disc cut
$$ = \sigma \times \pi {\left( {\frac{R}{3}} \right)^2} = \sigma \times \frac{{\pi {R^2}}}{9} = M$$
Let us consider the above system as a complete disc of mass $$9M$$  and a negative mass $$M$$ super imposed on it.
Moment of inertia $$\left( {{I_1}} \right)$$ of the complete disc $$ = \frac{1}{2}9M{R^2}$$   about an axis passing through $$O$$ and perpendicular to the plane of the disc.
$$M.I.$$  of the cut out portion about an axis passing through $$O'$$ and perpendicular to the plane of disc $$ = \frac{1}{2} \times M \times {\left( {\frac{R}{3}} \right)^2}$$
$$\therefore M.I.\,\left( {{I_2}} \right)$$   of the cut out portion about an axis passing through $$O$$ and perpendicular to the plane of disc $$ = \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right]$$         [Using perpendicular axis theorum]
$$\therefore $$ The total $$M.I.$$  of the system about an axis passing through $$O$$ and perpendicular to the plane of the disc is
$$\eqalign{ & I = {I_1} + {I_2} \cr & = \frac{1}{2}9M{R^2} - \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right] \cr & = \frac{{9M{R^2}}}{2} - \frac{{9M{R^2}}}{{18}} \cr & = \frac{{\left( {9 - 1} \right)M{R^2}}}{2} \cr & = 4M{R^2} \cr} $$

When solid sphere rolls on inclined plane, then it has both rotational as well as translational kinetic energy
Total kinetic energy $$K = {K_{{\text{rot}}}} + {K_{{\text{trans}}}} = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}$$
For sphere, moment of inertia about its diameter $$I = \frac{2}{5}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{r^2}{\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{v^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & = \frac{7}{{10}}m{v^2} \cr} $$

On reaching sphere at $$O,$$ it has only kinetic energy
$$\eqalign{ & \therefore PE = {\text{Total}}\,KE \cr & mgh = \frac{7}{{10}}m{v^2} \cr & \Rightarrow v = \sqrt {\frac{{10gh}}{7}} \cr} $$

$$\eqalign{ & {\text{In }}\Delta OAB\,\,\,\,\,\,\,\,\cos \frac{\pi }{n} = \frac{{OA}}{{OB}} \cr & \therefore OB = \frac{h}{{\cos \frac{\pi }{n}}} \cr & \Delta = \frac{h}{{\cos \frac{\pi }{n}}} - h = h\left[ {\frac{1}{{\cos \frac{\pi }{n}}} - 1} \right] \cr} $$

Let $$O\left( {0,0} \right)$$  be the centre of sphere $$A$$ then,
$${x_{cm}} = \frac{{m \times 0 + 2\;m \times 3R}}{{m + 2\;m}} = 2R$$
= at the point of contact

The velocity of top point of the wheel is twice that of centre of mass. And the speed of centre of mass is same for both the wheels. $$\left[ {v = \omega r} \right]$$

As both cars take the same time to complete the circle and as $$\omega = \frac{{2\pi }}{t},$$   therefore ratio of angular speeds of the cars will be $$1:1.$$

$$\eqalign{ & {I_{{y_1}}} = \frac{{M{R^2}}}{4} \cr & \therefore {{I'}_{{y_1}}} = \frac{{M{R^2}}}{4} + M{R^2} = \frac{5}{4}M{R^2} \cr} $$

$$\eqalign{ & {I_{{y_2}}} = \frac{{M{R^2}}}{2} \cr & \therefore {{I'}_{{y_2}}} = \frac{{M{R^2}}}{2} + M{R^2} = \frac{3}{4}M{R^2} \cr & {{I'}_{{y_1}}} = MK_1^2,{{I'}_{{y_2}}} = MK_2^2 \cr & \therefore \frac{{K_1^2}}{{K_2^2}} = \frac{{{{I'}_{{y_1}}}}}{{{{I'}_{{y_2}}}}} \Rightarrow {K_1}:{K_2} = \sqrt 5 :\sqrt 6 \cr} $$