A charged shell of radius R carries a total charge Q. Given $\varphi$ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following options is/are correct?

[${\in }_0$ is the permittivity of free space]

Option-1: If $h=\frac{8R}{5}$  and $r=\frac{3R}{5}$ , the cylindrical surface just fits into the sphere. If $h<\frac{8R}{5}$ , the cylindrical surface will be well inside the sphere.  $\varphi =\frac{q_{in}}{{\in }_0}=\frac{0}{{\in }_0}=0$
  (1) is correct

 
Option -2: If h>2R and r>R the sphere will be well inside the cylinder.$\therefore \varphi =\frac{Q}{{\in }_0}$

 
  (2) is correct 
Option -3:      $$\begin{gathered} q_{in}=\left(Q_{AB}\right)2=\frac{Q}{2}\left(1-\cos \theta \right)\times 2=Q\left[1-\frac{4}{5}\right]=\frac{Q}{5} \\ \therefore \varphi =\frac{q_{in}}{{\in }_0}=\frac{Q}{5{\in }_0} \end{gathered}$$
  (3) is correct

 
Option -4:   $$\begin{gathered} q_{in}=\frac{Q}{2}\left(1-\cos 53º\right)\times 2=\raisebox{1ex}{$2Q$}\!\left/ \!\raisebox{-1ex}{$5 $}\right. \\ \Rightarrow \varphi =2Q/5{ϵ}_0 \end{gathered}$$
   (4) is wrong