What is the behaviour of perpendicular electric field ${\rm{\vec E}}$ and magnetic field ${\rm{\vec B}}$ ?

A charge particle moving in magnetic field $B$, has the components of velocity along $B$ as well as perpendicular to $B$. The path of the charge particle will be

Given below are two statements

Statement $I$ : The electric force changes the speed of the charged particle and hence changes its kinetic energy: whereas the magnetic force does not change the kinetic energy of the charged particle

Statement $II$ : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

A particle having a charge of $10.0\,\mu C$ and mass $1\,\mu g$ moves in a circle of radius $10\,cm$ under the influence of a magnetic field of induction $0.1\,T$. When the particle is at a point $P$, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is......$V/m$

Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $q$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then, they will meet again for the first time at a point whose distance from the point of projection is