Question

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 $\frac{ q }{ m }=10^{10} \,C\,kg ^{-1}$. 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 $\sqrt{\frac{5}{3}} \mu s$

Answer 1

Answer: (B), (C)

Range $= R =\frac{2 u \sin \theta}{ g _{\text {eff }}} u \cos \theta=5 \,m$
$\sin 2 \theta=\left(\frac{q E R}{m u^{2}}\right)=\frac{\sqrt{3}}{2}$
$\theta=30^{\circ}$
So for same range angle of projection is either $30^{\circ}$ or $60^{\circ}$
So Option (B) is correct
$T =\frac{2 u \sin \theta}{ g _{\text {eff }}}=\left(\sqrt{\frac{10}{3}} \times 10^{-6}\right) \sin \theta$
$ T _{1}\left(\theta=30^{\circ}\right)=\sqrt{\frac{5}{6}} \mu s $
$T _{2}\left(\theta=60^{\circ}\right)=\sqrt{\frac{5}{2}} \mu s$
So, Option (C) is correct