$\alpha=y_{0}+u_{y} t+\frac{1}{2} q_{y} t^{2}$
$0=0-v t+\frac{1}{2} \frac{q E}{m} t^{2}$
$t=\frac{2 m v}{q E}$
In this time charge must complex one oe more revolution in $\mathrm{x}$$-$ $\mathrm{z}$ plane due to magnetic field
$T=\frac{2 \pi m}{q B}=-2 \Rightarrow t=n t \ldots(2)$
$\frac{2 m v}{q E}=n \times \frac{2 \pi m}{q B} \Rightarrow \frac{v B}{\pi E}$




$[A$ For $B>\frac{2}{3} \frac{p}{QR}$, the particle will re-enter region $1$
$[B]$ For $B=\frac{8}{13} \frac{\mathrm{p}}{QR}$, the particle will enter region $3$ through the point $P_2$ on $\mathrm{x}$-axis
$[C]$ When the particle re-enters region 1 through the longest possible path in region $2$ , the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from $y$-axis is $p / \sqrt{2}$
$[D]$ For a fixed $B$, particles of same charge $Q$ and same velocity $v$, the distance between the point $P_1$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle
(Neglect the effect of earth's magnetic field.)