Let $$x$$ = distance between hinge and centre of the door = $$\frac{L}{2} = \frac{1}{2} = 0.5\,m$$
As per conservation of angular momentum $$mvx = I\omega $$
$$ \Rightarrow \omega = \frac{{mvx}}{I} = \frac{{0.01 \times 500 \times 0.5}}{{\frac{1}{3} \times 12 \times {{\left( 1 \right)}^2}}} = 0.625\,rad/s$$

Moment of inertia
$$ = m{\left( {\frac{{3a}}{4}} \right)^2} + {m_1}\frac{{{a^2}}}{3}$$
For the centre of rod
$$\eqalign{ & \left( {\frac{{{m_1}{a^2}}}{{12}} + \frac{{{m_1}{a^2}}}{4}} \right) = \frac{{{m_1}{a^2}}}{3} \cr & \therefore \quad {m_1} = 4m \cr & {\text{Total}}\,I = m{\left( {\frac{{3a}}{4}} \right)^2} + \frac{{4m{a^2}}}{3} \cr & = \frac{{9m{a^2}}}{{16}} + \frac{{4m{a^2}}}{3} \cr & = \frac{{\left( {27 + 64} \right)}}{{48}}m{a^2} \cr & = \frac{{91}}{{48}}m{a^2} \cr} $$

Here $$P$$ rolls down while $$Q$$ rolls up
$$\eqalign{ & \therefore {w_P} > {w_C} > {w_Q} \cr & {\text{Now,}}\,v = wr \to v \propto w \cr & {v_P} > {v_C} > {v_Q}. \cr} $$
Hence, the answer is $${v_P} > {v_C} > {v_Q}.$$

For a ball to move in horizontal circle, the ball should satisfy the condition
Tension in the string = Centripetal force
$$ \Rightarrow {T_{\max }} = \frac{{Mv_{\max }^2}}{R}$$
$$ \Rightarrow {v_{\max }} = \sqrt {\frac{{{T_{\max }} \cdot R}}{R}} \,......\left( {\text{i}} \right)$$
Making substitution, we obtain $${v_{\max }} = \sqrt {\frac{{25 \times 1.96}}{{0.25}}} = \sqrt {196} = 14\,m/s$$
NOTE
In a vertical circle, the tension at the highest point is zero and at lowest point is maximum.

External torque $${\tau _{{\text{ext}}}} = 0$$
So, $$\frac{{dL}}{{dt}} = 0$$
Angular momentum, $$L$$ = constant
or $$I\omega = {\text{constant}}$$
$$\therefore {I_1}{\omega _1} = {I_2}{\omega _2}\,......\left( {\text{i}} \right)$$
So, for two different cases
Here, $${I_1} = M{r^2},{\omega _1}$$
$$ = \omega ,{I_2} = M{r^2} + 4m{r^2}$$
Hence, Eq. (i) can be written as
$$\eqalign{ & M{r^2}\omega = \left( {M{r^2} + 4m{r^2}} \right){\omega _2} \cr & \therefore {\omega _2} = \frac{{M\omega }}{{M + 4m}} \cr} $$

The radial component of linear momentum does not contribute to angular momentum of the particle. It is only the transverse component of linear momentum (perpendicular to position vector $$r$$), which when multiplied by distance from the axis of rotation gives us angular momentum.

Hence, angular momentum is axial vector.

We know that $$\vec \tau = \vec r \times \vec F$$
The angle between $${\vec \tau }$$  and $${\vec r}$$  is $${90^ \circ }$$  and the angle between $${\vec \tau }$$  and $${\vec F}$$  is also $${90^ \circ }.$$  We also know that the dot product of two vectors which have an angle of $${90^ \circ }$$ between them is zero. Therefore (D) is the correct option.

Let friction force $$= f$$
$$\eqalign{ & F + f = ma\,......\left( {\text{i}} \right) \cr & \left( {F - f} \right)R = I\alpha \,......\left( {{\text{ii}}} \right) \cr} $$

From eqns. (i) and (ii),
$$2F = ma + \frac{{I\alpha }}{R}$$
Use $$\alpha = \frac{a}{R}\left( {{\text{for pure rolling}}} \right)$$
$$\eqalign{ & 2F = ma + \frac{{Ia}}{{{R^2}}} \cr & 40 = 4a + \frac{{0.02a}}{{{{\left( {0.1} \right)}^2}}};a = \frac{{40}}{6} = 6.7\,m/{s^2} \cr} $$

$${{\vec v}_{cm}} = \frac{{{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}}}{{{m_1} + {m_2}}} = \frac{{2 \times 2 + 4 \times 10}}{{2 + 4}} = 7.3\,m/s$$

Moment of inertia depends on distribution of mass and about axis of rotation. Density of iron is more than that of aluminium, therefore for moment of inertia to be maximum, the iron should be far away from the axis. Thus, aluminium should be at interior and iron surrounds it.