Two coaxial thin pipes (hollow cylinder) of radius R and 2R are having a common horizontal axis. A thin uniform hollow spherical ball of diameter R is fitted inside them as shown. The inner surfaces are sufficiently rough so that there is no slipping at contact surfaces anywhere. The outer pipe is held fixed but inner pipe is free to rotate about its axis. On slight displacement, the ball starts moving along the gap shown in the downward direction. All the 3 bodies (both pipes & ball) have same mass of 1 kg each. When ball reaches the bottom of the pipe  [Take $g=10\text{ m}/{\text{s}}^2$ ]

Linear velocities (as shown) of points are w.r.t. ground and angular velocities are about respective axis of rotation passing through centre of mass. 
P is a point on outer shell in contact with spherical shell which is the instantaneous point of rest for spherical shell.

$$\begin{gathered} v_1={\omega }_1\frac{R}{2},v=2v_1\&v={\omega }_{2}R \\ \Rightarrow {\omega }_1=\frac{2v_1}{R}=\frac{v}{R}\&{\omega }_2=\frac{v}{R} \end{gathered}$$

Hence option A is correct obviously B is wrong.

$$\begin{gathered} {\text{KE}}_{\text{inner shell }}=\frac{1}{2}\left(m{R}^2\right){\omega }_2^2=\frac{1}{2}m{R}^2{\left(\frac{v}{R}\right)}^2=\frac{1}{2}m{v}^2 K{E}_{\text{ball }}=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left\{\frac{2}{3}m{\left(\frac{R}{2}\right)}^2\right\}\left\{{\omega }_1^2\right\} \\ =\frac{1}{2}m{\left(\frac{v}{2}\right)}^2+\frac{1}{12}m{R}^2{\left(\frac{v}{R}\right)}^2 =\frac{m{v}^2}{8}+\frac{m{v}^2}{12}=\frac{3+2}{24}m{v}^2=\frac{5m{v}^2}{24} \\ \Rightarrow \frac{K_{\text{in,cyl }}}{K_{\text{ball }}}=\frac{\frac{1}{2}m{v}^2}{\frac{5m{v}^2}{24}}=\frac{1}{2}\times \frac{24}{5}=\frac{12}{5} \end{gathered}$$Hence, option C is correct.
For ball + inner shell as a system :
T.M.E. is conserved as there is no dissipative force within system.
$\text{mg}3\text{R}=({\text{KE}}_{\text{ball }}+{\text{KE}}_{\text{shell }})$  at bottom,
$\Rightarrow 3\text{mgR}=5\text{ K}+12\text{ K}=17\text{ K}$
(from above relation)
$\Rightarrow \text{ K}=\frac{3\text{mgR}}{17}$
where KE of inner shell
 $$\begin{gathered} =12K=\frac{12\times 3mgR}{17}=\frac{1}{2}m{v}^2 \\ \Rightarrow v^2=\frac{72gR}{17}\Rightarrow {\left(2v_1\right)}^2=\frac{72gR}{17} \\ {v}_1^2=\frac{18gR}{17} =\text{ma}=\text{m}{(}_{\text{c}}{+}_{\tau })=\text{m}{(}_{\text{c}}+\overrightarrow{0}){=}_{\text{c}} =\frac{m{v}_1^2}{\left(\frac{3}{2}R\right)}\text{ vertically upward} \\ \mathbf{ }=\frac{2\text{ m}}{3\text{R}}\times \frac{18\text{gR}}{17}=\frac{12\text{mg}}{17}=\frac{12\times 1\times 10}{17} =\frac{120}{17}=7.06\text{ N vertically upward.} \end{gathered}$$

Hence option D is correct.