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 $r$ & with its center same as that of the shell. Here center of cylinder is a point on the axis of the cylinder which is equidistant from its top & bottom surfaces. Which of the following are correct.

For Option (A)

Since $h>2R$ and $r=\frac{3R}{5}$, so we get

$$\begin{gathered} \sin \alpha =\frac{r}{R}=\frac{3}{5} \\ \Rightarrow \cos \alpha =\frac{4}{5} \end{gathered}$$

Since $\varphi =2\left[\frac{Q}{2{\epsilon }_0}\left(1-\cos \alpha \right)\right]$           … (1)

$\Rightarrow =\frac{Q}{5{\epsilon }_0}$

Please note that a factor of 2 is used in equation (1), because we have two surfaces i.e., $AB$ and $CD$ of the shell that are enclosed by the cylinder.

For Option (B)

$h<\frac{8R}{5}$ and $r=\frac{3R}{5}$

$h<2R$ and $r<R$

$$\begin{gathered} \Rightarrow Q_{enclosed}=0 \\ \Rightarrow \varphi =0 \end{gathered}$$

For Option (C)

$h>2R$ and $r>R$

$$\begin{gathered} \Rightarrow Q_{enclosed}=Q \\ \Rightarrow \varphi =\frac{Q}{{\epsilon }_0} \end{gathered}$$

For Option (D)

$h>2R$ and $r=\frac{4R}{5}$ i.e., $r<R$

$Q_{enclosed}=Q\left[1-\cos \alpha \right]$

Since, $\sin \alpha =\frac{r}{R}=\frac{4}{5}$

$$\begin{gathered} \Rightarrow \cos \alpha =\frac{3}{5} \\ \Rightarrow \varphi =\frac{2Q}{2{\epsilon }_0}\left(1-\cos \alpha \right) \\ \Rightarrow \varphi =\frac{Q}{{\epsilon }_0}\left(1-\frac{3}{5}\right) \\ \Rightarrow \varphi =\frac{2Q}{5{\epsilon }_0} \end{gathered}$$