$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

An electron is accelerated through a potential difference $(p.d.)$ of $45.5 \ V$. The velocity acquired by it is (in $m \ s^{-1}$)

$A$ proton and an $\alpha$-particle start from rest in a uniform electric field. The ratio of times taken by them to travel the same distance in the field is

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. Its final velocity will be

$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be: