A couple consists of two equal and opposite forces acting at a separation, so that net force becomes zero. When a couple acts on a body it rotates the body but does not produce any translatory motion. Hence, only rotational motion is produced.

The velocity of centre of mass of two particle system is given by

$${v_c} = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} = \frac{{m\left( {2v} \right) + m\left( { - v} \right)}}{{m + m}} = \frac{v}{2}$$

According to right hand thumb rule.

$$\eqalign{ & \rho = \frac{M}{{\frac{4}{3}\pi \left( {R_2^3 - R_1^3} \right)}} \cr & {I_{{\text{shell}}}} = \frac{2}{5}{M_2}R_2^2 - \frac{2}{5}{M_1}R_1^2\,......\left( 1 \right) \cr & {M_2} = \rho \times \frac{4}{3}\pi R_2^3 \cr & = \frac{{MR_2^3}}{{\left( {R_2^3 - R_1^3} \right)}};{M_1} = \frac{{MR_1^3}}{{R_2^3 - R_1^3}} \cr} $$
Putting values of $${M_1}$$ and $${M_2}$$ in eq. (1),
$${I_{{\text{shell}}}} = \frac{{2M}}{5}\frac{{\left( {R_2^5 - R_1^5} \right)}}{{\left( {R_2^3 - R_1^3} \right)}}$$

The velocity of efflux is given $$v = \sqrt {2gh} $$
Where $$h$$  is the height of the free surface of liquid from the hole
$$\therefore \,\,v = \sqrt {2 \times 10 \times 20} = 20\,m/s$$

Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

As the rod is bent into two equal halves, the mass and length of each half is $$\frac{M}{2}$$ and $$\frac{L}{2}$$ respectively.
The moment of inertia about an axis passing through its edge and perpendicular to the rod

$$\eqalign{ & = \frac{{{\text{Mass}} \times {{\left( {{\text{length}}} \right)}^2}}}{3} \cr & \therefore I = 2 \times \frac{1}{3} \times \frac{M}{2}{\left( {\frac{L}{2}} \right)^2} = \frac{{M{L^2}}}{{12}} \cr} $$

As friction is absent at the point of contact,
Acceleration $$ = \frac{{{\text{Force}}}}{{{\text{Mass}}}}$$
It is independent of $$h$$

The mass distribution of this sector is same about the is of rotation as that of the complete disc about the axis. Therefore the formula remains the same as that of the disc.

As we know that radius of gyration $$k = \sqrt {\frac{I}{m}} $$
So, for two different cases $$\frac{{{k_{{\text{ring}}}}}}{{{k_{{\text{disc}}}}}} = \sqrt {\frac{{{I_{{\text{ring}}}}}}{{{I_{{\text{disc}}}}}}} = \sqrt {\frac{{M{R^2}}}{{\frac{1}{2}M{R^2}}}} $$
$$\therefore \frac{{{k_{{\text{ring}}}}}}{{{k_{{\text{disc}}}}}} = \sqrt 2 \Rightarrow \frac{{{k_{{\text{ring}}}}}}{{{k_{{\text{disc}}}}}} = \frac{1}{{\sqrt 2 }}$$