An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

$A$ $2 \ g$ object,located in a region of uniform electric field $E = (300 \ NC^{-1}) \hat{i}$,carries a charge $Q$. The object is released from rest at $x = 0$ and has a kinetic energy of $0.12 \ J$ at $x = 0.5 \ m$. Then,$Q$ is: (in $\mu C$)

$A$ charge $q_2$ of mass $m$ revolves around a stationary charge $q_1$ in a circular orbit of radius $r$. The orbital periodic time of $q_2$ would be . . . . . . .

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

$A$ uniform electric field,$\vec{E} = -400 \sqrt{3} \hat{y} \text{ NC}^{-1}$ is applied in a region. $A$ charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $u = 2 \sqrt{10} \times 10^6 \text{ ms}^{-1}$. This particle is aimed to hit a target $T$,which is $5 \text{ m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m} = 10^{10} \text{ Ckg}^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu\text{s}$ as well as $\sqrt{\frac{5}{2}} \mu\text{s}$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu\text{s}$

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)