(2)

$T_1=27$ days, $R_1=R,$ $R_2=2R$

By conservation of angular momentum

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

$\frac{2}{5}m{R}_1^2\times \frac{2\pi }{T_1}=\frac{2}{5}m{R}_2^2\times \frac{2\pi }{T_2}$

$\frac{R^2}{27}=\frac{4R^2}{T_2}\Rightarrow T_2=27\times 4=108\text{ days}$

(4)

As there is no external torque acting on a sphere, i.e., ${\tau }_{\text{ex}}=0$

So, $\frac{d\mathbf{L}}{dt}={\tau }_{\text{ex}}=0 \text{i.e., }\mathbf{L}=\text{constant}$

So angular momentum remains constant.

(1)

Initial angular momentum $=I{\omega }_1+I{\omega }_2$

Let $\omega$ be angular speed of the combined system.

Final angular momentum = $2I\omega$

$\therefore$      According to conservation of angular momentum

           $I{\omega }_1+I{\omega }_2=2I\omega \text{or} \omega =\frac{{\omega }_1+{\omega }_2}{2}$

Initial rotational kinetic energy,

         $E=\frac{1}{2}I({\omega }_1^2+{\omega }_2^2)$

Final rotational kinetic energy

  $E_f=\frac{1}{2}\left(2I\right){\omega }^2=\frac{1}{2}\left(2I\right){\left(\frac{{\displaystyle {\omega }_1+{\omega }_2}}{{\displaystyle 2}}\right)}^2=\frac{1}{4}I{({\omega }_1+{\omega }_2)}^2$

$\therefore$        Loss of energy $\Delta E=E_i-E_f$

           $=\frac{I}{2}({\omega }_1^2+{\omega }_2^2)-\frac{I}{4}({\omega }_1^2+{\omega }_2^2+2{\omega }_1{\omega }_2)$

            $=\frac{I}{4}[{\omega }_1^2+{\omega }_2^2-2{\omega }_1{\omega }_2]=\frac{I}{4}{({\omega }_1-{\omega }_2)}^2$

(3)

Here, $m_A=m,$ $m_B=2m$

Both bodies A and B have equal kinetic energy of rotation

        $K_A=K_B\Rightarrow \frac{1}{2}{I}_A{\omega }_A^2=\frac{1}{2}{I}_B{\omega }_B^2$

$\Rightarrow \frac{{\omega }_A^2}{{\omega }_B^2}=\frac{I_B}{I_A}$                                                            ...(i)

Ratio of angular momenta, 

        $\frac{L_A}{L_B}=\frac{I_A{\omega }_A}{I_B{\omega }_B}=\frac{I_A}{I_B}\times \sqrt{\frac{I_B}{I_A}} \text{[Using eqn. (i)]}$

                    $=\sqrt{\frac{I_A}{I_B}}<1$                $(\because I_B>I_A)$

$\therefore L_B>L_A$