Question

A uniform electric field,$\overrightarrow{E}=-400\sqrt{3}$ $\hat{y}$$N{C}^{-1}$is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2\sqrt{10}$ $m$$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. Then
Take $\frac{q}{m}={10}^{10}Ck{g}^{-1}$.

• 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$\sqrt{\frac{5}{6}}\mu s\text{as well as}\sqrt{\frac{5}{2}}\mu s$
• D. time taken by the particle to hit $T$ is$\sqrt{\frac{5}{3}}\mu s$

Answer 1

Answer: (B), (C)

time taken by the particle to hit $T$ could be$\sqrt{\frac{5}{6}}\mu s\text{as well as}\sqrt{\frac{5}{2}}\mu s$

Using $q{E}_y=m{a}_y$

$a_y=-400\sqrt{3}\times {10}^{10}$

Using
$\left[R\right(range)=\frac{u^2\sin 2\theta }{a_y}]$

$R=5=\frac{40\times {10}^{12}\sin 2\theta }{400\sqrt{3}{10}^{10}}$
$\sin 2\theta =\frac{\sqrt{3}}{2}$

$2\theta ={60}^{\circ },120\Rightarrow \theta ={30}^{\circ },{60}^{\circ }$
Time of flight :
$T_1=\frac{2\times 2\sqrt{10}\times {10}^6\times \frac{1}{2}}{400\sqrt{3}{10}^{10}}=\sqrt{\frac{5}{6}}\mu s$
$(for\theta ={30}^{\circ })$

Time of flight :
$T_2=\frac{2\times 2\sqrt{10}\times {10}^6\times \frac{\sqrt{3}}{2}}{400\sqrt{3}{10}^{10}}=\sqrt{\frac{5}{2}}\mu s$

$(for\theta ={60}^{\circ })$

Hence , option (b) and (c) are the correct answers .