From conservation of angular momentum we get

$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega $$

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

$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$

$${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$$

$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$

Solving above we get

$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$

$$R = {l \over \theta }$$

Time $$ = {{4 \times 2\pi R} \over v} = {{4 \times 2\pi } \over v}\left( {{l \over \theta }} \right)$$

put l = 4.4 $$\times$$ 9.46 $$\times$$ 10¹⁵

v = 8 $$\times$$ 1.5 $$\times$$ 10¹¹

$$\theta = {4 \over {3600}} \times {\pi \over {180}}$$ rad.

we get time = 4.5 $$\times$$ 10¹⁰ sec.

Parallel axis theorem

I = I_CM + Md²

$$I = {{M{r^2}} \over 2} + M{\left( {{L \over 2}} \right)^2}$$

$$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$$

$$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$$

M = 15 kg

$$ \Rightarrow \rho = {M \over {\pi {r^2}L}} = {{15} \over {\pi {{(0.2)}^2} \times 0.8}}$$

= 0.1492 $$\times$$ 10³

I_z = I_x + I_y (using perpendicular axis theorem) & I = mk² (K : radius of gyration)

so, mK_z² = mK_x² + mK_y²

K_z² = K_x² + K_y²

so radius of gyration about axes x, y & z won't be same hence assertion A is not correct reason R is correct statement (property of a rigid body)