$A$ particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform electric and magnetic fields,$E$ and $B$ respectively,which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

$A$ proton (mass $m$ and charge $+e$) and an $\alpha$-particle (mass $4m$ and charge $+2e$) are projected with the same kinetic energy at right angles to a uniform magnetic field. Which one of the following statements will be true?

An electron gun with its collector at a potential of $100 \; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \; mm$ of $Hg)$. $A$ magnetic field of $2.83 \times 10^{-4} \; T$ curves the path of the electrons in a circular orbit of radius $12.0 \; cm$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons,and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e/m$ from the data.

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move

The ratio of periods of $\alpha$-particle and proton moving on circular path in uniform magnetic field is . . . . . . .