This is the free body diagram of pulley and mass
For translation motion of the block,

$$mg - T = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$$

For rotational motion of the pulley,

$$T\times R = I\alpha = I{a \over R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

where $$\alpha = $$ angular acceleration of disc = $$ {a \over R}$$

and $$I = {1 \over 2}m{R^2}$$ (For circular disc)

Solving $$(1)$$ & $$(2),$$

$$a = {{mg} \over {\left( {m + {I \over {{R^2}}}} \right)}} = {{mg} \over {m + {{m{R^2}} \over {2{R^2}}}}}$$

$$ = {{2mg} \over {3m}} = {{2g} \over 3}$$

Here no external force is applied on the disc so Torque ($$\tau $$) = 0.

So angular momentum is conserved.

That means $${I_1}{\omega _1} = {I_2}{\omega _2}$$

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

$$\therefore$$ Angular speed is inversely proportional to Moment of inertia.

For disc $$I = {1 \over 2}M{R^2}$$

$$\therefore$$ Moment of Inertia is proportional to Mass.

As insect moves along a diameter, the effective mass of disc first decreases then increases and hence the moment of inertia first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.

The moment of inertia of the rod about $$O$$ is $${1 \over 2}m{\ell ^2}.$$ The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this conditions is $${1 \over 2}I{\omega ^2}$$ where $$I$$ is the moment of inertia of the rod about $$O.$$ when the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the center of mass is raised through $$h$$, so the increase in potential energy is $$mgh$$. This is equal to kinetic energy $${1 \over 2}I{\omega ^2}$$.

$$\therefore$$ $$mgh = {1 \over 2}I{\omega ^2} = {1 \over 2}\left( {{1 \over 3}m{l^2}} \right){\omega ^2}$$.

$$ \Rightarrow h = {{{\ell ^2}{\omega ^2}} \over {6g}}$$

Given The linear mass density $$\lambda = k{\left( {{x \over L}} \right)^n}$$

$${x_{CM}} = {{\int\limits_0^L {x{\mkern 1mu} dm} } \over {\int\limits_0^L {dm} }}$$

$$ = {{\int\limits_0^L {x\left( {\lambda {\mkern 1mu} dx} \right)} } \over {\int\limits_0^L {\lambda {\mkern 1mu} dx} }}$$

$$ = {{\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}} .xdx} \over {\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}{\mkern 1mu} dx} }}$$

= $${{k\left[ {{{{x^{n + 2}}} \over {\left( {n + 2} \right){L^n}}}} \right]_0^L} \over {\left[ {{{k{\mkern 1mu} {x^{n + 1}}} \over {\left( {n + 1} \right){L^n}}}} \right]_0^L}}$$

$$ = {{L\left( {n + 1} \right)} \over {n + 2}}$$

For $$n=0,$$ $${x_{CM}} = {L \over 2};n = 1,$$

$${x_{CM}} = {{2L} \over 3};n = 2,\,{x_{CM}} = {{3L} \over 4};\,...$$

From here you can see only option (A) can be the right answer.