JEE Advanced 2019 · Paper 1, Physics question 8

Only the shell charge inside the cylinder counts: $\Phi=Q_{enc}/\epsilon_0$.

(A) With $h>2R$ and $r>R$ the cylinder encloses the whole shell, so $\Phi=Q/\epsilon_0$. True.

(B) With $r=\frac35R$, the cylinder's rim corners are at distance $\sqrt{r^2+(h/2)^2}$ from the centre. That is less than $R$ when $h<\frac85R$, so the cylinder lies entirely inside the shell and $\Phi=0$. True.

(C) With $h>2R$, the cylinder cuts out two polar caps of the shell, with half-angle given by $\sin\theta=r/R$. Each cap holds the fraction $\frac{1-\cos\theta}2$ of the charge. For $r=\frac35R$, $\cos\theta=\frac45$:
$$Q_{enc}=2\times\frac{1-0.8}2Q=\frac Q5,\qquad\Phi=\frac{Q}{5\epsilon_0}.$$
True.

(D) For $r=\frac45R$, $\cos\theta=\frac35$ and $Q_{enc}=2\times0.2Q=\frac25Q$, not $\frac Q5$. False.

(A), (B), (C)