$A$ positively charged particle $q$ of mass $m$ is passed through a velocity selector. It moves horizontally rightward without deviation along the line $y = \frac{2mv}{qB}$ with a speed $v$. The electric field is vertically downwards and the magnetic field is into the plane of the paper. Now,the electric field is switched off at $t = 0$. The angular momentum of the charged particle about the origin $O$ at $t = \frac{\pi m}{qB}$ is:

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

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2} \, T$. If the value of $e/m$ is $1.76 \times 10^{11} \, C/kg$,the frequency of revolution of the electron is

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

$H^+$,$He^+$,and $O^{++}$ are projected in a uniform transverse magnetic field with equal accelerating potential. If their masses are $1 \, a.m.u.$,$4 \, a.m.u.$,and $16 \, a.m.u.$ respectively,find the ratio of their radii.