Key Idea
$$KE$$  of a rotating rigid body, $$KE = \frac{1}{2}I{\omega ^2}$$
$$\eqalign{ & \therefore KE{\text{ of sphere, }}{K_S} = \frac{1}{2}I\omega _1^2 = \frac{1}{2}\frac{2}{5}m{R^2}\omega _1^2 \cr & = \frac{1}{5}m{R^2}\omega _1^2 \cr & KE\,{\text{of}}\,{\text{cylinder,}}\,\,{K_C} = \frac{1}{2}\frac{1}{2}\;m{R^2}\omega _2^2 = \frac{1}{4}m{R^2}\omega _2^2 \cr & \therefore \frac{{{K_S}}}{{{K_C}}} = \frac{{\frac{{m{R^2}\omega _1^2}}{5}}}{{\frac{{m{R^2}\omega _2^2}}{4}}} = \frac{4}{5}\frac{{\omega _1^2}}{{\omega _2^2}} = \frac{4}{5}\frac{{\omega _1^2}}{{{{\left( {2{\omega _1}} \right)}^2}}} \cr & = \frac{1}{5}\,\,\left( {{\text{given,}}\,\,{\omega _2} = 2{\omega _1}} \right) \cr} $$

$$\eqalign{ & {\text{Here}}\,\alpha = {\text{revolutions}}/{s^2} = 4\pi \,rad/{s^2}\left( {{\text{given}}} \right) \cr & {I_{{\text{cylinder}}}} = \frac{1}{2}M{R^2} = \frac{1}{2}\left( {50} \right){\left( {0.5} \right)^2} \cr & = \frac{{25}}{4}kg - {m^2} \cr & {\text{As}}\,\tau = I\alpha \,{\text{so}}\,TR = \cr & \Rightarrow T = \frac{{I\alpha }}{R} = \frac{{\left( {\frac{{25}}{4}} \right)\left( {4\pi } \right)}}{{\left( {0.5} \right)}}N \cr & = 50\,\pi N \cr & = 157\,N \cr} $$

According to question, torque $$\tau = 0$$
It means that, $$\alpha = 0$$
$$\eqalign{ & \alpha = \frac{{{d^2}\theta }}{{d{t^2}}} \cr & {\text{Given,}}\,\,\theta \left( t \right) = 2{t^3} - 6{t^2} \cr & {\text{So,}}\,\,\frac{{d\theta }}{{dt}} = 6{t^2} - 12t \cr & \alpha = \frac{{{d^2}\theta }}{{d{t^2}}} = 12t - 12 \cr & 12t - 12 = 0 \Rightarrow t = 1\,s \cr} $$

KEY CONCEPT
The disc has two types of motion namely translational and rotational. Therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two.

In this case both the angular momentum have the same direction (perpendicular to the plane of paper and away from the reader).
$$\vec L = {{\vec L}_T} + {{\vec L}_R}$$
$${L_T} = $$   angular momentum due to translational motion.
$${L_R} = $$   angular momentum due to rotational motion about $$C.M.$$
$$L = MV \times R + {I_{cm}}\omega $$
$${I_{cm}} = M.I.$$    about centre of mass C.
$$ = M\left( {R\,\omega } \right) + \frac{1}{2}M{R^2}\,\omega $$
($$v = R\omega $$   in case of rolling motion and surface at rest)
$$ = \frac{3}{2}M{R^2}\,\omega $$

As we know, moment of inertia of a solid cylinder about an axis which is perpendicular bisector

$$\eqalign{ & I = \frac{{m{R^2}}}{4} + \frac{{m{l^2}}}{{12}} \cr & I = \frac{m}{4}\left[ {{R^2} + \frac{{{l^2}}}{3}} \right] \cr & {\text{Let }}V = {\text{ volume of cylinder }} = \pi {R^2}l \cr & = \frac{m}{4}\left[ {\frac{V}{{\pi l}} + \frac{{{l^2}}}{3}} \right] \cr & \Rightarrow \frac{{dl}}{{dl}} = \frac{m}{4}\left[ {\frac{{ - V}}{{\pi {l^2}}} + \frac{{2l}}{3}} \right] = 0 \cr & \frac{V}{{\pi {l^2}}} + \frac{{2l}}{3}\,\,\,\,\,\,\, \Rightarrow V = \frac{{2\pi {l^3}}}{3} \cr & \pi {R^2}l = \frac{{2\pi {l^3}}}{3}\,\,\,\, \Rightarrow \frac{{{l^2}}}{{{R^2}}} = \frac{3}{2}\;{\text{ or, }}\frac{l}{R} = \sqrt {\frac{3}{2}} \cr} $$

Let the length of a small element of tube be $$dx.$$  Mass of this element $$dm = \frac{M}{L}dx$$

where, $$M$$ is mass of filled liquid and $$L$$ is length of tube, Force on this element
$$\eqalign{ & dF = dm \times x{\omega ^2} \cr & \int_0^F {dF = \frac{M}{L}{\omega ^2}} \int_0^L x dx \cr & {\text{or}}\,\,F = \frac{M}{L}{\omega ^2}\left[ {\frac{{{L^2}}}{2}} \right] = \frac{{ML{\omega ^2}}}{2} \cr & {\text{or}}\,\,F = \frac{1}{2}ML{\omega ^2} \cr} $$

Perp axis theorem $${I_x} + {I_y} = {I_z}$$
$$\eqalign{ & \Rightarrow 2I = M{R^2} \cr & {\text{so}}\,\,I = \frac{{M{R^2}}}{2} \cr} $$

For negotiating a circular curve on a levelled road, the maximum velocity of the car is $${v_{{\text{max}}}} = \sqrt {\mu rg} $$
$$\eqalign{ & {\text{Here }}\mu = 0.6,\,\,\,\,r = 150\,m,\,\,\,\,g = 9.8 \cr & \therefore {v_{{\text{max}}}} = \sqrt {0.6 \times 150 \times 9.8} \cr} $$
≃ 30 m/s

When the disc spins, the frictional force between the gramophone record and coin is $$\mu \,mg.$$  The coin will revolve with record, if $${F_{{\text{frictional}}}} \geqslant {F_{{\text{centripetal}}}}$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,\mu \,mg \geqslant m{\omega ^2}r \cr & \frac{{\mu g}}{r} \geqslant {\omega ^2} \cr} $$

Consider an element of length $$dx$$ at a distance $$x$$ from end $$A.$$
Here, mass per unit length $$\lambda $$ of rod

$$\eqalign{ & \lambda \propto x \Rightarrow \lambda = kx \cr & \therefore dm = \lambda dx = kx\,dx \cr} $$
Position of centre of gravity of rod from end $$A.$$
$$\eqalign{ & {x_{CG}} = \frac{{\int_0^L x dm}}{{\int_0^L d m}} \cr & \therefore {x_{CG}} = \frac{{\int_0^3 x \left( {kx\,dx} \right)}}{{\int_0^3 k x\,dx}} = \frac{{\left[ {\frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {\frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{\frac{{{{\left( 3 \right)}^3}}}{3}}}{{\frac{{{{\left( 3 \right)}^2}}}{2}}} = 2\,m \cr} $$