A particle of mass $m$ and charge $q$ is thrown in a region where uniform gravitational field and electric field are present. The path of particle

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 \, \frac{ N }{ C }$ as shown in the figure. A body of mass $1\, kg$ and charge $5\, mC$ is allowed to slide down from rest at a height of $1\, m$. If the coefficient of friction is $0.2,$ find the time (in $s$ )taken by the body to reach the bottom. $\left[ g =9.8 \,m / s ^{2}, \sin 30^{\circ}=\frac{1}{2}\right.$; $\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$

An electric line of force in $X$, $Y-$ plane is given by $x^2+y^2 = 1$. A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $X, Y-$ plane

An electron and a proton are in a uniform electric field, the ratio of their accelerations will be

An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)

A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ 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 $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. 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 $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$

$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$