Question

A uniform electric field, $\vec{E}=-400\sqrt{3}\hat{y}N{C}^{-1}$ 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 $2\sqrt{10}\times 10^{6}m{s}^{-1}$. This particle is aimed to hit a target $T$, which is $5m$ away from its entry point into the field as shown schematically in the figure.
Take $\frac{q}{m}=10^{10}Ck{g}^{-1}$. 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 $=R=\frac{2u\sin \theta }{g_{\text{eff }}}u\cos \theta =5m$
$\sin 2\theta =\left(\frac{qER}{m{u}^2}\right)=\frac{\sqrt{3}}{2}$
$\theta =30^{\circ }$
So for same range angle of projection is either $30^{\circ }$ or $60^{\circ }$
So Option (B) is correct
$T=\frac{2u\sin \theta }{g_{\text{eff }}}=\left(\sqrt{\frac{10}{3}}\times 10^{-6}\right)\sin \theta$
$T_1\left(\theta =30^{\circ }\right)=\sqrt{\frac{5}{6}}\mu s$
$T_2\left(\theta =60^{\circ }\right)=\sqrt{\frac{5}{2}}\mu s$
So, Option (C) is correct